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Question Number 112249 by MJS_new last updated on 07/Sep/20

∫(dx/((αx^2 +px+β)(√(αx^2 +qx+β))))=?

$$\int\frac{{dx}}{\left(\alpha{x}^{\mathrm{2}} +{px}+\beta\right)\sqrt{\alpha{x}^{\mathrm{2}} +{qx}+\beta}}=? \\ $$

Commented by bemath last updated on 07/Sep/20

waw.....

$$\mathrm{waw}..... \\ $$

Answered by ajfour last updated on 07/Sep/20

say I=∫(dx/((ax^2 +px+b)(√(ax^2 +qx+b)))) =∫((dx/(a(√a)))/({(x+(p/(2a)))^2 +(b/a)−(p^2 /(4a^2 ))}(√((x+(q/(2a)))^2 +(b/a)−(q^2 /(4a^2 )))))) let x+(q/(2a))=(√(((b/a)−(q^2 /(4a^2 ))))) tan θ=λtan θ I=(1/(a(√a)))∫((sec θdθ)/((λtan θ+((p−q)/(2a)))^2 +λ^2 −(((p^2 −q^2 ))/(4a^2 )))) =(1/(a(√a)))∫((cos θdθ)/((λsin θ+(((p−q))/(2a))cos θ)^2 +(λ^2 −(((p^2 −q^2 ))/(4a^2 )))cos^2 θ)) =(1/(a(√a)))∫((cos θdθ)/(λ^2 −((q(p−q))/(2a^2 ))cos^2 θ+((λ(p−q))/(2a))sin 2θ)) let θ=(π/4)−φ I=(1/(a(√a)))∫((−cos ((π/4)−φ)dφ)/(λ^2 −((q(p−q))/(4a^2 ))−(((p−q))/(4a^2 )){qsin 2φ−2λacos 2φ})) now let (√(q^2 +4λ^2 a^2 )) = h=(√(4ab)) tan δ=(q/(2λa)) ; λ^2 −((q(p−q))/(4a^2 ))=k=(b/a)−((pq)/(4a^2 )) I=(1/(a(√a)))∫((−cos ((π/4)−φ)dφ)/(k+(((p−q)h)/(4a^2 ))cos (2φ+δ))) =(1/(a(√a)))∫((−cos (φ+(δ/2)−(π/4)−(δ/2))d(φ+(δ/2)))/(k+(((p−q)h)/(4a^2 )){cos^2 (φ+(δ/2))−sin^2 (φ+(δ/2))})) now say φ+(δ/2)=z , (π/4)+(δ/2)=β , then ((h(q−p))/(4(√a)))I=cos β∫((cos zdz)/(((4a^2 k)/(p−q))+1−2sin^2 z)) +sin β∫((sin zdz)/(((4a^2 k)/(p−q))−1+2cos^2 z)) .......

$${say}\:\:{I}=\int\frac{{dx}}{\left({ax}^{\mathrm{2}} +{px}+{b}\right)\sqrt{{ax}^{\mathrm{2}} +{qx}+{b}}} \\ $$$$=\int\frac{\frac{{dx}}{{a}\sqrt{{a}}}}{\left\{\left({x}+\frac{{p}}{\mathrm{2}{a}}\right)^{\mathrm{2}} +\frac{{b}}{{a}}−\frac{{p}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }\right\}\sqrt{\left({x}+\frac{{q}}{\mathrm{2}{a}}\right)^{\mathrm{2}} +\frac{{b}}{{a}}−\frac{{q}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }}} \\ $$$${let}\:\:\:{x}+\frac{{q}}{\mathrm{2}{a}}=\sqrt{\left(\frac{{b}}{{a}}−\frac{{q}^{\mathrm{2}} }{\mathrm{4}{a}^{\mathrm{2}} }\right)}\:\mathrm{tan}\:\theta=\lambda\mathrm{tan}\:\theta \\ $$$${I}=\frac{\mathrm{1}}{{a}\sqrt{{a}}}\int\frac{\mathrm{sec}\:\theta{d}\theta}{\left(\lambda\mathrm{tan}\:\theta+\frac{{p}−{q}}{\mathrm{2}{a}}\right)^{\mathrm{2}} +\lambda^{\mathrm{2}} −\frac{\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)}{\mathrm{4}{a}^{\mathrm{2}} }} \\ $$$$\:\:=\frac{\mathrm{1}}{{a}\sqrt{{a}}}\int\frac{\mathrm{cos}\:\theta{d}\theta}{\left(\lambda\mathrm{sin}\:\theta+\frac{\left({p}−{q}\right)}{\mathrm{2}{a}}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left(\lambda^{\mathrm{2}} −\frac{\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)}{\mathrm{4}{a}^{\mathrm{2}} }\right)\mathrm{cos}\:^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{1}}{{a}\sqrt{{a}}}\int\frac{\mathrm{cos}\:\theta{d}\theta}{\lambda^{\mathrm{2}} −\frac{{q}\left({p}−{q}\right)}{\mathrm{2}{a}^{\mathrm{2}} }\mathrm{cos}\:^{\mathrm{2}} \theta+\frac{\lambda\left({p}−{q}\right)}{\mathrm{2}{a}}\mathrm{sin}\:\mathrm{2}\theta} \\ $$$${let}\:\:\:\:\theta=\frac{\pi}{\mathrm{4}}−\phi \\ $$$${I}=\frac{\mathrm{1}}{{a}\sqrt{{a}}}\int\frac{−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\phi\right){d}\phi}{\lambda^{\mathrm{2}} −\frac{{q}\left({p}−{q}\right)}{\mathrm{4}{a}^{\mathrm{2}} }−\frac{\left({p}−{q}\right)}{\mathrm{4}{a}^{\mathrm{2}} }\left\{{q}\mathrm{sin}\:\mathrm{2}\phi−\mathrm{2}\lambda{a}\mathrm{cos}\:\mathrm{2}\phi\right\}} \\ $$$$\:{now}\:{let}\:\:\sqrt{{q}^{\mathrm{2}} +\mathrm{4}\lambda^{\mathrm{2}} {a}^{\mathrm{2}} }\:=\:{h}=\sqrt{\mathrm{4}{ab}} \\ $$$$\:\:\mathrm{tan}\:\delta=\frac{{q}}{\mathrm{2}\lambda{a}}\:;\:\:\lambda^{\mathrm{2}} −\frac{{q}\left({p}−{q}\right)}{\mathrm{4}{a}^{\mathrm{2}} }={k}=\frac{{b}}{{a}}−\frac{{pq}}{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${I}=\frac{\mathrm{1}}{{a}\sqrt{{a}}}\int\frac{−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\phi\right){d}\phi}{{k}+\frac{\left({p}−{q}\right){h}}{\mathrm{4}{a}^{\mathrm{2}} }\mathrm{cos}\:\left(\mathrm{2}\phi+\delta\right)} \\ $$$$\:=\frac{\mathrm{1}}{{a}\sqrt{{a}}}\int\frac{−\mathrm{cos}\:\left(\phi+\frac{\delta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}−\frac{\delta}{\mathrm{2}}\right){d}\left(\phi+\frac{\delta}{\mathrm{2}}\right)}{{k}+\frac{\left({p}−{q}\right){h}}{\mathrm{4}{a}^{\mathrm{2}} }\left\{\mathrm{cos}\:^{\mathrm{2}} \left(\phi+\frac{\delta}{\mathrm{2}}\right)−\mathrm{sin}\:^{\mathrm{2}} \left(\phi+\frac{\delta}{\mathrm{2}}\right)\right\}} \\ $$$${now}\:\:{say}\:\:\:\phi+\frac{\delta}{\mathrm{2}}={z}\:,\:\:\frac{\pi}{\mathrm{4}}+\frac{\delta}{\mathrm{2}}=\beta\:,\:{then} \\ $$$$\frac{{h}\left({q}−{p}\right)}{\mathrm{4}\sqrt{{a}}}{I}=\mathrm{cos}\:\beta\int\frac{\mathrm{cos}\:{zdz}}{\frac{\mathrm{4}{a}^{\mathrm{2}} {k}}{{p}−{q}}+\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {z}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{sin}\:\beta\int\frac{\mathrm{sin}\:{zdz}}{\frac{\mathrm{4}{a}^{\mathrm{2}} {k}}{{p}−{q}}−\mathrm{1}+\mathrm{2cos}\:^{\mathrm{2}} {z}}\: \\ $$$$....... \\ $$

Commented by MJS_new last updated on 07/Sep/20

thank you!

$$\mathrm{thank}\:\mathrm{you}! \\ $$

Answered by MJS_new last updated on 07/Sep/20

Mr. Mathdave showed us another path. It works with ∫(dx/((ax^2 +bx+c)(√(dx^2 +ex+f)))) only if a=d∧c=f ∫(dx/((αx^2 +px+β)(√(αx^2 +qx+β))))= =α^(−3/2) ∫(dx/( (x^2 +(p/α)x+(β/α))(√(x^2 +(q/α)x+(β/α)))))= [t=(((√β)−x(√α))/( (√β)+x(√α))) → dx=−((2(√β))/((t+1)^2 (√α)))dt] =2α^(7/4) ∫((t+1)/(((p−2(√(αβ)))t^2 −(p+2(√(αβ))))(√((2(√(αβ))−q)(√β)t^2 +(2(√(αβ))+q)(√β)))))dt= [a=p−2(√(αβ)) b=−(p+2(√(αβ))) c=(2(√(αβ))−q)(√β) d=(2(√(αβ))+q)(√β)] =2α^(7/4) ∫((t+1)/((at^2 +b)(√(ct^2 +d))))dt= =2α^(7/4) (∫(t/((at^2 +b)(√(ct^2 +d))))dt+∫(dt/((at^2 +b)(√(ct^2 +d))))) both are easy to solve: ∫(t/((at^2 +b)(√(ct^2 +d))))dt= [u=((√d)/( (√(ct^2 +d)))) → dt=−(((ct^2 +d)^(3/2) )/(ct(√d)))du] =(√d)∫(du/((ad−bc)u^2 −ad))=... ∫(dt/((at^2 +b)(√(ct^2 +d))))= [u=((t(√(ad−bc)))/( (√(ct^2 +d)))) → dt=(((ct^2 +d)^(3/2) )/( d(√(ad−bc))))du] =(1/( (√(ad−bc))))∫(du/(u^2 +b))=...

$$\mathrm{Mr}.\:\mathrm{Mathdave}\:\mathrm{showed}\:\mathrm{us}\:\mathrm{another}\:\mathrm{path}.\:\mathrm{It} \\ $$$$\mathrm{works}\:\mathrm{with}\:\int\frac{{dx}}{\left({ax}^{\mathrm{2}} +{bx}+{c}\right)\sqrt{{dx}^{\mathrm{2}} +{ex}+{f}}}\:\mathrm{only} \\ $$$$\mathrm{if}\:{a}={d}\wedge{c}={f} \\ $$$$\int\frac{{dx}}{\left(\alpha{x}^{\mathrm{2}} +{px}+\beta\right)\sqrt{\alpha{x}^{\mathrm{2}} +{qx}+\beta}}= \\ $$$$=\alpha^{−\mathrm{3}/\mathrm{2}} \int\frac{{dx}}{\:\left({x}^{\mathrm{2}} +\frac{{p}}{\alpha}{x}+\frac{\beta}{\alpha}\right)\sqrt{{x}^{\mathrm{2}} +\frac{{q}}{\alpha}{x}+\frac{\beta}{\alpha}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\beta}−{x}\sqrt{\alpha}}{\:\sqrt{\beta}+{x}\sqrt{\alpha}}\:\rightarrow\:{dx}=−\frac{\mathrm{2}\sqrt{\beta}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{\alpha}}{dt}\right] \\ $$$$=\mathrm{2}\alpha^{\mathrm{7}/\mathrm{4}} \int\frac{{t}+\mathrm{1}}{\left(\left({p}−\mathrm{2}\sqrt{\alpha\beta}\right){t}^{\mathrm{2}} −\left({p}+\mathrm{2}\sqrt{\alpha\beta}\right)\right)\sqrt{\left(\mathrm{2}\sqrt{\alpha\beta}−{q}\right)\sqrt{\beta}{t}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\alpha\beta}+{q}\right)\sqrt{\beta}}}{dt}= \\ $$$$\:\:\:\:\:\left[{a}={p}−\mathrm{2}\sqrt{\alpha\beta}\:\:{b}=−\left({p}+\mathrm{2}\sqrt{\alpha\beta}\right)\right. \\ $$$$\left.\:\:\:\:\:\:{c}=\left(\mathrm{2}\sqrt{\alpha\beta}−{q}\right)\sqrt{\beta}\:\:{d}=\left(\mathrm{2}\sqrt{\alpha\beta}+{q}\right)\sqrt{\beta}\right] \\ $$$$=\mathrm{2}\alpha^{\mathrm{7}/\mathrm{4}} \int\frac{{t}+\mathrm{1}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}{dt}= \\ $$$$=\mathrm{2}\alpha^{\mathrm{7}/\mathrm{4}} \left(\int\frac{{t}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}{dt}+\int\frac{{dt}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}\right) \\ $$$$ \\ $$$$\mathrm{both}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}: \\ $$$$\int\frac{{t}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\sqrt{{d}}}{\:\sqrt{{ct}^{\mathrm{2}} +{d}}}\:\rightarrow\:{dt}=−\frac{\left({ct}^{\mathrm{2}} +{d}\right)^{\mathrm{3}/\mathrm{2}} }{{ct}\sqrt{{d}}}{du}\right] \\ $$$$=\sqrt{{d}}\int\frac{{du}}{\left({ad}−{bc}\right){u}^{\mathrm{2}} −{ad}}=... \\ $$$$\int\frac{{dt}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{{t}\sqrt{{ad}−{bc}}}{\:\sqrt{{ct}^{\mathrm{2}} +{d}}}\:\rightarrow\:{dt}=\frac{\left({ct}^{\mathrm{2}} +{d}\right)^{\mathrm{3}/\mathrm{2}} }{\:{d}\sqrt{{ad}−{bc}}}{du}\right] \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{ad}−{bc}}}\int\frac{{du}}{{u}^{\mathrm{2}} +{b}}=... \\ $$

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