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

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

[Out]

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

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Rubi [A]  time = 0.102706, 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, 208} \[ \frac{b x (2 b c-5 a d)}{3 a^2 \sqrt{a+b x^2} (b c-a d)^2}+\frac{d^2 \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} (b c-a d)^{5/2}}+\frac{b x}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x)/(3*a*(b*c - a*d)*(a + b*x^2)^(3/2)) + (b*(2*b*c - 5*a*d)*x)/(3*a^2*(b*c - a*d)^2*Sqrt[a + b*x^2]) + (d^2
*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(Sqrt[c]*(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 208

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

Rubi steps

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

Mathematica [C]  time = 2.62801, size = 775, normalized size = 6.35 \[ \frac{x \left (12 c^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \text{HypergeometricPFQ}\left (\left \{2,2,\frac{7}{2}\right \},\left \{1,\frac{9}{2}\right \},\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )+12 d^2 x^4 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \text{HypergeometricPFQ}\left (\left \{2,2,\frac{7}{2}\right \},\left \{1,\frac{9}{2}\right \},\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )+24 c d x^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \text{HypergeometricPFQ}\left (\left \{2,2,\frac{7}{2}\right \},\left \{1,\frac{9}{2}\right \},\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )+48 c^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )-105 c^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{3/2}-315 c^2 \sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}+315 c^2 \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )+36 d^2 x^4 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )-56 d^2 x^4 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{3/2}-168 d^2 x^4 \sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}+168 d^2 x^4 \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )+84 c d x^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )-140 c d x^2 \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{3/2}-420 c d x^2 \sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}+420 c d x^2 \tanh ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )\right )}{63 c^3 \left (a+b x^2\right )^{5/2} \left (\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-315*c^2*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 420*c*d*x^2*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 1
68*d^2*x^4*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 105*c^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2) - 140*c
*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2) - 56*d^2*x^4*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2) + 315*
c^2*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]] + 420*c*d*x^2*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^
2))]] + 168*d^2*x^4*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]] + 48*c^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2
)))^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 84*c*d*x^2*(((b*c - a*d)*x^2)/(c
*(a + b*x^2)))^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 36*d^2*x^4*(((b*c - a
*d)*x^2)/(c*(a + b*x^2)))^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 12*c^2*(((
b*c - a*d)*x^2)/(c*(a + b*x^2)))^(7/2)*HypergeometricPFQ[{2, 2, 7/2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^
2))] + 24*c*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(7/2)*HypergeometricPFQ[{2, 2, 7/2}, {1, 9/2}, ((b*c - a
*d)*x^2)/(c*(a + b*x^2))] + 12*d^2*x^4*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(7/2)*HypergeometricPFQ[{2, 2, 7/2}
, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(63*c^3*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2)*(a + b*x^2)
^(5/2))

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Maple [B]  time = 0.014, size = 1070, normalized size = 8.8 \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)^(5/2)/(d*x^2+c),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.24145, size = 1530, normalized size = 12.54 \begin{align*} \left [\frac{3 \,{\left (a^{2} b^{2} d^{2} x^{4} + 2 \, a^{3} b d^{2} x^{2} + a^{4} d^{2}\right )} \sqrt{b c^{2} - a c d} \log \left (\frac{{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \,{\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} + 4 \,{\left ({\left (2 \, b c - a d\right )} x^{3} + a c x\right )} \sqrt{b c^{2} - a c d} \sqrt{b x^{2} + a}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right ) + 4 \,{\left ({\left (2 \, b^{4} c^{3} - 7 \, a b^{3} c^{2} d + 5 \, a^{2} b^{2} c d^{2}\right )} x^{3} + 3 \,{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 2 \, a^{3} b c d^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{12 \,{\left (a^{4} b^{3} c^{4} - 3 \, a^{5} b^{2} c^{3} d + 3 \, a^{6} b c^{2} d^{2} - a^{7} c d^{3} +{\left (a^{2} b^{5} c^{4} - 3 \, a^{3} b^{4} c^{3} d + 3 \, a^{4} b^{3} c^{2} d^{2} - a^{5} b^{2} c d^{3}\right )} x^{4} + 2 \,{\left (a^{3} b^{4} c^{4} - 3 \, a^{4} b^{3} c^{3} d + 3 \, a^{5} b^{2} c^{2} d^{2} - a^{6} b c d^{3}\right )} x^{2}\right )}}, -\frac{3 \,{\left (a^{2} b^{2} d^{2} x^{4} + 2 \, a^{3} b d^{2} x^{2} + a^{4} d^{2}\right )} \sqrt{-b c^{2} + a c d} \arctan \left (\frac{\sqrt{-b c^{2} + a c d}{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt{b x^{2} + a}}{2 \,{\left ({\left (b^{2} c^{2} - a b c d\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - 2 \,{\left ({\left (2 \, b^{4} c^{3} - 7 \, a b^{3} c^{2} d + 5 \, a^{2} b^{2} c d^{2}\right )} x^{3} + 3 \,{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 2 \, a^{3} b c d^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{6 \,{\left (a^{4} b^{3} c^{4} - 3 \, a^{5} b^{2} c^{3} d + 3 \, a^{6} b c^{2} d^{2} - a^{7} c d^{3} +{\left (a^{2} b^{5} c^{4} - 3 \, a^{3} b^{4} c^{3} d + 3 \, a^{4} b^{3} c^{2} d^{2} - a^{5} b^{2} c d^{3}\right )} x^{4} + 2 \,{\left (a^{3} b^{4} c^{4} - 3 \, a^{4} b^{3} c^{3} d + 3 \, a^{5} b^{2} c^{2} d^{2} - a^{6} b c d^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^2)*sqrt(b*c^2 - a*c*d)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d
^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 + 4*((2*b*c - a*d)*x^3 + a*c*x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x
^2 + a))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + 4*((2*b^4*c^3 - 7*a*b^3*c^2*d + 5*a^2*b^2*c*d^2)*x^3 + 3*(a*b^3*c^3 -
3*a^2*b^2*c^2*d + 2*a^3*b*c*d^2)*x)*sqrt(b*x^2 + a))/(a^4*b^3*c^4 - 3*a^5*b^2*c^3*d + 3*a^6*b*c^2*d^2 - a^7*c*
d^3 + (a^2*b^5*c^4 - 3*a^3*b^4*c^3*d + 3*a^4*b^3*c^2*d^2 - a^5*b^2*c*d^3)*x^4 + 2*(a^3*b^4*c^4 - 3*a^4*b^3*c^3
*d + 3*a^5*b^2*c^2*d^2 - a^6*b*c*d^3)*x^2), -1/6*(3*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^2)*sqrt(-b*c^2
+ a*c*d)*arctan(1/2*sqrt(-b*c^2 + a*c*d)*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)/((b^2*c^2 - a*b*c*d)*x^3 +
(a*b*c^2 - a^2*c*d)*x)) - 2*((2*b^4*c^3 - 7*a*b^3*c^2*d + 5*a^2*b^2*c*d^2)*x^3 + 3*(a*b^3*c^3 - 3*a^2*b^2*c^2*
d + 2*a^3*b*c*d^2)*x)*sqrt(b*x^2 + a))/(a^4*b^3*c^4 - 3*a^5*b^2*c^3*d + 3*a^6*b*c^2*d^2 - a^7*c*d^3 + (a^2*b^5
*c^4 - 3*a^3*b^4*c^3*d + 3*a^4*b^3*c^2*d^2 - a^5*b^2*c*d^3)*x^4 + 2*(a^3*b^4*c^4 - 3*a^4*b^3*c^3*d + 3*a^5*b^2
*c^2*d^2 - a^6*b*c*d^3)*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 )^{\frac{5}{2}} \left (c + d x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.17883, size = 432, normalized size = 3.54 \begin{align*} -\frac{\sqrt{b} d^{2} \arctan \left (\frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt{-b^{2} c^{2} + a b c d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-b^{2} c^{2} + a b c d}} + \frac{{\left (\frac{{\left (2 \, b^{6} c^{3} - 9 \, a b^{5} c^{2} d + 12 \, a^{2} b^{4} c d^{2} - 5 \, a^{3} b^{3} d^{3}\right )} x^{2}}{a^{2} b^{5} c^{4} - 4 \, a^{3} b^{4} c^{3} d + 6 \, a^{4} b^{3} c^{2} d^{2} - 4 \, a^{5} b^{2} c d^{3} + a^{6} b d^{4}} + \frac{3 \,{\left (a b^{5} c^{3} - 4 \, a^{2} b^{4} c^{2} d + 5 \, a^{3} b^{3} c d^{2} - 2 \, a^{4} b^{2} d^{3}\right )}}{a^{2} b^{5} c^{4} - 4 \, a^{3} b^{4} c^{3} d + 6 \, a^{4} b^{3} c^{2} d^{2} - 4 \, a^{5} b^{2} c d^{3} + a^{6} b d^{4}}\right )} x}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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