∫(dx/((αx^2 +px+β)(√(αx^2 +qx+β))))=?


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

$\int \frac{d x}{(α x^{2} + p x + β) \sqrt{α x^{2} + q x + β}} = ?$
Commented by bemath last updated on 07/Sep/20

$waw . . . . .$
	Answered by ajfour last updated on 07/Sep/20
	$say I=∫(dx/((ax^2 +px+b)(√(ax^2 +qx+b)))) =∫((dx/(a(√a)))/({(x+(p/(2a)))^2 +(b/a)−(p^2 /(4a^2 ))}(√((x+(q/(2a)))^2 +(b/a)−(q^2 /(4a^2 )))))) let x+(q/(2a))=(√(((b/a)−(q^2 /(4a^2 ))))) tan θ=λtan θ I=(1/(a(√a)))∫((sec θdθ)/((λtan θ+((p−q)/(2a)))^2 +λ^2 −(((p^2 −q^2 ))/(4a^2 )))) =(1/(a(√a)))∫((cos θdθ)/((λsin θ+(((p−q))/(2a))cos θ)^2 +(λ^2 −(((p^2 −q^2 ))/(4a^2 )))cos^2 θ)) =(1/(a(√a)))∫((cos θdθ)/(λ^2 −((q(p−q))/(2a^2 ))cos^2 θ+((λ(p−q))/(2a))sin 2θ)) let θ=(π/4)−φ I=(1/(a(√a)))∫((−cos ((π/4)−φ)dφ)/(λ^2 −((q(p−q))/(4a^2 ))−(((p−q))/(4a^2 )){qsin 2φ−2λacos 2φ})) now let (√(q^2 +4λ^2 a^2 )) = h=(√(4ab)) tan δ=(q/(2λa)) ; λ^2 −((q(p−q))/(4a^2 ))=k=(b/a)−((pq)/(4a^2 )) I=(1/(a(√a)))∫((−cos ((π/4)−φ)dφ)/(k+(((p−q)h)/(4a^2 ))cos (2φ+δ))) =(1/(a(√a)))∫((−cos (φ+(δ/2)−(π/4)−(δ/2))d(φ+(δ/2)))/(k+(((p−q)h)/(4a^2 )){cos^2 (φ+(δ/2))−sin^2 (φ+(δ/2))})) now say φ+(δ/2)=z , (π/4)+(δ/2)=β , then ((h(q−p))/(4(√a)))I=cos β∫((cos zdz)/(((4a^2 k)/(p−q))+1−2sin^2 z)) +sin β∫((sin zdz)/(((4a^2 k)/(p−q))−1+2cos^2 z)) .......$
	$s a y I = \int \frac{d x}{({a x}^{2} + p x + b) \sqrt{{a x}^{2} + q x + b}}$ $= \int \frac{\frac{d x}{a \sqrt{a}}}{{{(x + \frac{p}{2 a})}^{2} + \frac{b}{a} - \frac{p^{2}}{4 a^{2}}} \sqrt{{(x + \frac{q}{2 a})}^{2} + \frac{b}{a} - \frac{q^{2}}{4 a^{2}}}}$ $l e t x + \frac{q}{2 a} = \sqrt{(\frac{b}{a} - \frac{q^{2}}{4 a^{2}})} \tan θ = λ \tan θ$ $I = \frac{1}{a \sqrt{a}} \int \frac{\sec θ d θ}{{(λ \tan θ + \frac{p - q}{2 a})}^{2} + λ^{2} - \frac{(p^{2} - q^{2})}{4 a^{2}}}$ $= \frac{1}{a \sqrt{a}} \int \frac{\cos θ d θ}{{(λ \sin θ + \frac{(p - q)}{2 a} \cos θ)}^{2} + (λ^{2} - \frac{(p^{2} - q^{2})}{4 a^{2}}) \cos^{2} θ}$ $= \frac{1}{a \sqrt{a}} \int \frac{\cos θ d θ}{λ^{2} - \frac{q (p - q)}{2 a^{2}} \cos^{2} θ + \frac{λ (p - q)}{2 a} \sin 2 θ}$ $l e t θ = \frac{π}{4} - ϕ$ $I = \frac{1}{a \sqrt{a}} \int \frac{- \cos (\frac{π}{4} - ϕ) d ϕ}{λ^{2} - \frac{q (p - q)}{4 a^{2}} - \frac{(p - q)}{4 a^{2}} {q \sin 2 ϕ - 2 λ a \cos 2 ϕ}}$ $n o w l e t \sqrt{q^{2} + 4 λ^{2} a^{2}} = h = \sqrt{4 a b}$ $\tan δ = \frac{q}{2 λ a}; λ^{2} - \frac{q (p - q)}{4 a^{2}} = k = \frac{b}{a} - \frac{p q}{4 a^{2}}$ $I = \frac{1}{a \sqrt{a}} \int \frac{- \cos (\frac{π}{4} - ϕ) d ϕ}{k + \frac{(p - q) h}{4 a^{2}} \cos (2 ϕ + δ)}$ $= \frac{1}{a \sqrt{a}} \int \frac{- \cos (ϕ + \frac{δ}{2} - \frac{π}{4} - \frac{δ}{2}) d (ϕ + \frac{δ}{2})}{k + \frac{(p - q) h}{4 a^{2}} {\cos^{2} (ϕ + \frac{δ}{2}) - \sin^{2} (ϕ + \frac{δ}{2})}}$ $n o w s a y ϕ + \frac{δ}{2} = z, \frac{π}{4} + \frac{δ}{2} = β, t h e n$ $\frac{h (q - p)}{4 \sqrt{a}} I = \cos β \int \frac{\cos z d z}{\frac{4 a^{2} k}{p - q} + 1 - 2 \sin^{2} z}$ $+ \sin β \int \frac{\sin z d z}{\frac{4 a^{2} k}{p - q} - 1 + 2 \cos^{2} z}$ $. . . . . . .$
	Commented by MJS_new last updated on 07/Sep/20

	$thank you!$
	Answered by MJS_new last updated on 07/Sep/20

	$Mr . Mathdave showed us another path . It$ $works with \int \frac{d x}{({a x}^{2} + b x + c) \sqrt{{d x}^{2} + e x + f}} only$ $if a = d \land c = f$ $\int \frac{d x}{(α x^{2} + p x + β) \sqrt{α x^{2} + q x + β}} =$ $= α^{- 3 / 2} \int \frac{d x}{(x^{2} + \frac{p}{α} x + \frac{β}{α}) \sqrt{x^{2} + \frac{q}{α} x + \frac{β}{α}}} =$ $[t = \frac{\sqrt{β} - x \sqrt{α}}{\sqrt{β} + x \sqrt{α}} \to d x = - \frac{2 \sqrt{β}}{{(t + 1)}^{2} \sqrt{α}} d t]$ $= 2 α^{7 / 4} \int \frac{t + 1}{((p - 2 \sqrt{α β}) t^{2} - (p + 2 \sqrt{α β})) \sqrt{(2 \sqrt{α β} - q) \sqrt{β} t^{2} + (2 \sqrt{α β} + q) \sqrt{β}}} d t =$ $[a = p - 2 \sqrt{α β} b = - (p + 2 \sqrt{α β})$ $c = (2 \sqrt{α β} - q) \sqrt{β} d = (2 \sqrt{α β} + q) \sqrt{β}]$ $= 2 α^{7 / 4} \int \frac{t + 1}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} d t =$ $= 2 α^{7 / 4} (\int \frac{t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} d t + \int \frac{d t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}})$ $both are easy to solve :$ $\int \frac{t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} d t =$ $[u = \frac{\sqrt{d}}{\sqrt{{c t}^{2} + d}} \to d t = - \frac{{({c t}^{2} + d)}^{3 / 2}}{c t \sqrt{d}} d u]$ $= \sqrt{d} \int \frac{d u}{(a d - b c) u^{2} - a d} = . . .$ $\int \frac{d t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} =$ $[u = \frac{t \sqrt{a d - b c}}{\sqrt{{c t}^{2} + d}} \to d t = \frac{{({c t}^{2} + d)}^{3 / 2}}{d \sqrt{a d - b c}} d u]$ $= \frac{1}{\sqrt{a d - b c}} \int \frac{d u}{u^{2} + b} = . . .$
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