3.726 \(\int \frac{1}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=122 \[ \frac{b^2 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} (b c-a d)^{5/2}}-\frac{d x (5 b c-2 a d)}{3 c^2 \sqrt{c+d x^2} (b c-a d)^2}-\frac{d x}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)} \]

[Out]

-(d*x)/(3*c*(b*c - a*d)*(c + d*x^2)^(3/2)) - (d*(5*b*c - 2*a*d)*x)/(3*c^2*(b*c - a*d)^2*Sqrt[c + d*x^2]) + (b^
2*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(Sqrt[a]*(b*c - a*d)^(5/2))

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Rubi [A]  time = 0.100544, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {414, 527, 12, 377, 205} \[ \frac{b^2 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} (b c-a d)^{5/2}}-\frac{d x (5 b c-2 a d)}{3 c^2 \sqrt{c+d x^2} (b c-a d)^2}-\frac{d x}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

-(d*x)/(3*c*(b*c - a*d)*(c + d*x^2)^(3/2)) - (d*(5*b*c - 2*a*d)*x)/(3*c^2*(b*c - a*d)^2*Sqrt[c + d*x^2]) + (b^
2*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(Sqrt[a]*(b*c - a*d)^(5/2))

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx &=-\frac{d x}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac{\int \frac{3 b c-2 a d-2 b d x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{3 c (b c-a d)}\\ &=-\frac{d x}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{d (5 b c-2 a d) x}{3 c^2 (b c-a d)^2 \sqrt{c+d x^2}}+\frac{\int \frac{3 b^2 c^2}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{3 c^2 (b c-a d)^2}\\ &=-\frac{d x}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{d (5 b c-2 a d) x}{3 c^2 (b c-a d)^2 \sqrt{c+d x^2}}+\frac{b^2 \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{(b c-a d)^2}\\ &=-\frac{d x}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{d (5 b c-2 a d) x}{3 c^2 (b c-a d)^2 \sqrt{c+d x^2}}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{(b c-a d)^2}\\ &=-\frac{d x}{3 c (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{d (5 b c-2 a d) x}{3 c^2 (b c-a d)^2 \sqrt{c+d x^2}}+\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 5.46133, size = 1385, normalized size = 11.35 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

(x*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*(-1575*Sqrt[(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)] - (2
100*d*x^2*Sqrt[(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)])/c - (840*d^2*x^4*Sqrt[(a*(b*c - a*d)*x^2*
(c + d*x^2))/(c^2*(a + b*x^2)^2)])/c^2 + 2100*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2)*Sqrt[(a*(c + d*x^2))/(
c*(a + b*x^2))] + (2800*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))])
/c + (1120*d^2*x^4*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))])/c^2 + 1575
*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]] + (2100*d*x^2*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]]
)/c + (840*d^2*x^4*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/c^2 + (1575*(b*c - a*d)^2*x^4*ArcSin[Sqrt[
((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c^2*(a + b*x^2)^2) + (2100*d*(b*c - a*d)^2*x^6*ArcSin[Sqrt[((b*c - a*d)*
x^2)/(c*(a + b*x^2))]])/(c^3*(a + b*x^2)^2) + (840*d^2*(b*c - a*d)^2*x^8*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a +
b*x^2))]])/(c^4*(a + b*x^2)^2) - (3150*(b*c - a*d)*x^2*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c*(a
+ b*x^2)) + (4200*d*(-(b*c) + a*d)*x^4*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c^2*(a + b*x^2)) + (
1680*d^2*(-(b*c) + a*d)*x^6*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c^3*(a + b*x^2)) + 96*(((b*c - a
*d)*x^2)/(c*(a + b*x^2)))^(7/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*Hypergeometric2F1[2, 2, 9/2, ((b*c - a*d
)*x^2)/(c*(a + b*x^2))] + (168*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(7/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*
x^2))]*Hypergeometric2F1[2, 2, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/c + (72*d^2*x^4*(((b*c - a*d)*x^2)/(c*
(a + b*x^2)))^(7/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*Hypergeometric2F1[2, 2, 9/2, ((b*c - a*d)*x^2)/(c*(a
+ b*x^2))])/c^2 + 24*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(7/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*Hypergeo
metricPFQ[{2, 2, 2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + (48*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2
)))^(7/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a
+ b*x^2))])/c + (24*d^2*x^4*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(7/2)*Sqrt[(a*(c + d*x^2))/(c*(a + b*x^2))]*H
ypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/c^2))/(315*a*(((b*c - a*d)*x^2)/(c*(
a + b*x^2)))^(5/2)*(c + d*x^2)^(5/2))

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Maple [B]  time = 0.01, size = 1086, normalized size = 8.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

1/6/(-a*b)^(1/2)/(a*d-b*c)*b/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3
/2)+1/6*d/(a*d-b*c)/c/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)*x+1
/3*d/(a*d-b*c)/c^2/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x-1/2/
(-a*b)^(1/2)*b^2/(a*d-b*c)^2/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1
/2)-1/2*b/(a*d-b*c)^2/c/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x
*d+1/2/(-a*b)^(1/2)*b^2/(a*d-b*c)^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(
1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1
/2))/(x+1/b*(-a*b)^(1/2)))-1/6/(-a*b)^(1/2)/(a*d-b*c)*b/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-
a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+1/6*d/(a*d-b*c)/c/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1
/2))-(a*d-b*c)/b)^(3/2)*x+1/3*d/(a*d-b*c)/c^2/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2)
)-(a*d-b*c)/b)^(1/2)*x+1/2/(-a*b)^(1/2)*b^2/(a*d-b*c)^2/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-
a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-1/2*b/(a*d-b*c)^2/c/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^
(1/2))-(a*d-b*c)/b)^(1/2)*x*d-1/2/(-a*b)^(1/2)*b^2/(a*d-b*c)^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a
*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-
a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.22251, size = 1530, normalized size = 12.54 \begin{align*} \left [-\frac{3 \,{\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c^{3} d x^{2} + b^{2} c^{4}\right )} \sqrt{-a b c + a^{2} d} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \,{\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt{-a b c + a^{2} d} \sqrt{d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \,{\left ({\left (5 \, a b^{2} c^{2} d^{2} - 7 \, a^{2} b c d^{3} + 2 \, a^{3} d^{4}\right )} x^{3} + 3 \,{\left (2 \, a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{12 \,{\left (a b^{3} c^{7} - 3 \, a^{2} b^{2} c^{6} d + 3 \, a^{3} b c^{5} d^{2} - a^{4} c^{4} d^{3} +{\left (a b^{3} c^{5} d^{2} - 3 \, a^{2} b^{2} c^{4} d^{3} + 3 \, a^{3} b c^{3} d^{4} - a^{4} c^{2} d^{5}\right )} x^{4} + 2 \,{\left (a b^{3} c^{6} d - 3 \, a^{2} b^{2} c^{5} d^{2} + 3 \, a^{3} b c^{4} d^{3} - a^{4} c^{3} d^{4}\right )} x^{2}\right )}}, \frac{3 \,{\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c^{3} d x^{2} + b^{2} c^{4}\right )} \sqrt{a b c - a^{2} d} \arctan \left (\frac{\sqrt{a b c - a^{2} d}{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - 2 \,{\left ({\left (5 \, a b^{2} c^{2} d^{2} - 7 \, a^{2} b c d^{3} + 2 \, a^{3} d^{4}\right )} x^{3} + 3 \,{\left (2 \, a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{6 \,{\left (a b^{3} c^{7} - 3 \, a^{2} b^{2} c^{6} d + 3 \, a^{3} b c^{5} d^{2} - a^{4} c^{4} d^{3} +{\left (a b^{3} c^{5} d^{2} - 3 \, a^{2} b^{2} c^{4} d^{3} + 3 \, a^{3} b c^{3} d^{4} - a^{4} c^{2} d^{5}\right )} x^{4} + 2 \,{\left (a b^{3} c^{6} d - 3 \, a^{2} b^{2} c^{5} d^{2} + 3 \, a^{3} b c^{4} d^{3} - a^{4} c^{3} d^{4}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(b^2*c^2*d^2*x^4 + 2*b^2*c^3*d*x^2 + b^2*c^4)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2
*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(
d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*((5*a*b^2*c^2*d^2 - 7*a^2*b*c*d^3 + 2*a^3*d^4)*x^3 + 3*(2*a*b^2*c
^3*d - 3*a^2*b*c^2*d^2 + a^3*c*d^3)*x)*sqrt(d*x^2 + c))/(a*b^3*c^7 - 3*a^2*b^2*c^6*d + 3*a^3*b*c^5*d^2 - a^4*c
^4*d^3 + (a*b^3*c^5*d^2 - 3*a^2*b^2*c^4*d^3 + 3*a^3*b*c^3*d^4 - a^4*c^2*d^5)*x^4 + 2*(a*b^3*c^6*d - 3*a^2*b^2*
c^5*d^2 + 3*a^3*b*c^4*d^3 - a^4*c^3*d^4)*x^2), 1/6*(3*(b^2*c^2*d^2*x^4 + 2*b^2*c^3*d*x^2 + b^2*c^4)*sqrt(a*b*c
- a^2*d)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 +
(a*b*c^2 - a^2*c*d)*x)) - 2*((5*a*b^2*c^2*d^2 - 7*a^2*b*c*d^3 + 2*a^3*d^4)*x^3 + 3*(2*a*b^2*c^3*d - 3*a^2*b*c^
2*d^2 + a^3*c*d^3)*x)*sqrt(d*x^2 + c))/(a*b^3*c^7 - 3*a^2*b^2*c^6*d + 3*a^3*b*c^5*d^2 - a^4*c^4*d^3 + (a*b^3*c
^5*d^2 - 3*a^2*b^2*c^4*d^3 + 3*a^3*b*c^3*d^4 - a^4*c^2*d^5)*x^4 + 2*(a*b^3*c^6*d - 3*a^2*b^2*c^5*d^2 + 3*a^3*b
*c^4*d^3 - a^4*c^3*d^4)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)/(d*x**2+c)**(5/2),x)

[Out]

Integral(1/((a + b*x**2)*(c + d*x**2)**(5/2)), x)

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Giac [B]  time = 1.1623, size = 433, normalized size = 3.55 \begin{align*} -\frac{b^{2} \sqrt{d} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{a b c d - a^{2} d^{2}}} - \frac{{\left (\frac{{\left (5 \, b^{3} c^{3} d^{3} - 12 \, a b^{2} c^{2} d^{4} + 9 \, a^{2} b c d^{5} - 2 \, a^{3} d^{6}\right )} x^{2}}{b^{4} c^{6} d - 4 \, a b^{3} c^{5} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{3} - 4 \, a^{3} b c^{3} d^{4} + a^{4} c^{2} d^{5}} + \frac{3 \,{\left (2 \, b^{3} c^{4} d^{2} - 5 \, a b^{2} c^{3} d^{3} + 4 \, a^{2} b c^{2} d^{4} - a^{3} c d^{5}\right )}}{b^{4} c^{6} d - 4 \, a b^{3} c^{5} d^{2} + 6 \, a^{2} b^{2} c^{4} d^{3} - 4 \, a^{3} b c^{3} d^{4} + a^{4} c^{2} d^{5}}\right )} x}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

-b^2*sqrt(d)*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/((b^2*c^2 -
2*a*b*c*d + a^2*d^2)*sqrt(a*b*c*d - a^2*d^2)) - 1/3*((5*b^3*c^3*d^3 - 12*a*b^2*c^2*d^4 + 9*a^2*b*c*d^5 - 2*a^
3*d^6)*x^2/(b^4*c^6*d - 4*a*b^3*c^5*d^2 + 6*a^2*b^2*c^4*d^3 - 4*a^3*b*c^3*d^4 + a^4*c^2*d^5) + 3*(2*b^3*c^4*d^
2 - 5*a*b^2*c^3*d^3 + 4*a^2*b*c^2*d^4 - a^3*c*d^5)/(b^4*c^6*d - 4*a*b^3*c^5*d^2 + 6*a^2*b^2*c^4*d^3 - 4*a^3*b*
c^3*d^4 + a^4*c^2*d^5))*x/(d*x^2 + c)^(3/2)