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

Optimal. Leaf size=202 \[ \frac{b x \left (-3 a^2 d^2-16 a b c d+4 b^2 c^2\right )}{6 a^2 c \sqrt{a+b x^2} (b c-a d)^3}+\frac{d^2 (6 b c-a d) \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{7/2}}-\frac{d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}+\frac{b x (3 a d+2 b c)}{6 a c \left (a+b x^2\right )^{3/2} (b c-a d)^2} \]

[Out]

(b*(2*b*c + 3*a*d)*x)/(6*a*c*(b*c - a*d)^2*(a + b*x^2)^(3/2)) + (b*(4*b^2*c^2 - 16*a*b*c*d - 3*a^2*d^2)*x)/(6*
a^2*c*(b*c - a*d)^3*Sqrt[a + b*x^2]) - (d*x)/(2*c*(b*c - a*d)*(a + b*x^2)^(3/2)*(c + d*x^2)) + (d^2*(6*b*c - a
*d)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(2*c^(3/2)*(b*c - a*d)^(7/2))

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Rubi [A]  time = 0.228392, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{b x \left (-3 a^2 d^2-16 a b c d+4 b^2 c^2\right )}{6 a^2 c \sqrt{a+b x^2} (b c-a d)^3}+\frac{d^2 (6 b c-a d) \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{7/2}}-\frac{d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}+\frac{b x (3 a d+2 b c)}{6 a c \left (a+b x^2\right )^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(2*b*c + 3*a*d)*x)/(6*a*c*(b*c - a*d)^2*(a + b*x^2)^(3/2)) + (b*(4*b^2*c^2 - 16*a*b*c*d - 3*a^2*d^2)*x)/(6*
a^2*c*(b*c - a*d)^3*Sqrt[a + b*x^2]) - (d*x)/(2*c*(b*c - a*d)*(a + b*x^2)^(3/2)*(c + d*x^2)) + (d^2*(6*b*c - a
*d)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(2*c^(3/2)*(b*c - a*d)^(7/2))

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 )^2} \, dx &=-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}+\frac{\int \frac{2 b c-a d-4 b d x^2}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx}{2 c (b c-a d)}\\ &=\frac{b (2 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/2}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}-\frac{\int \frac{-4 b^2 c^2+12 a b c d-3 a^2 d^2-2 b d (2 b c+3 a d) x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )} \, dx}{6 a c (b c-a d)^2}\\ &=\frac{b (2 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac{b \left (4 b^2 c^2-16 a b c d-3 a^2 d^2\right ) x}{6 a^2 c (b c-a d)^3 \sqrt{a+b x^2}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}+\frac{\int \frac{3 a^2 d^2 (6 b c-a d)}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{6 a^2 c (b c-a d)^3}\\ &=\frac{b (2 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac{b \left (4 b^2 c^2-16 a b c d-3 a^2 d^2\right ) x}{6 a^2 c (b c-a d)^3 \sqrt{a+b x^2}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}+\frac{\left (d^2 (6 b c-a d)\right ) \int \frac{1}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{2 c (b c-a d)^3}\\ &=\frac{b (2 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac{b \left (4 b^2 c^2-16 a b c d-3 a^2 d^2\right ) x}{6 a^2 c (b c-a d)^3 \sqrt{a+b x^2}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}+\frac{\left (d^2 (6 b c-a d)\right ) \operatorname{Subst}\left (\int \frac{1}{c-(b c-a d) x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{2 c (b c-a d)^3}\\ &=\frac{b (2 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac{b \left (4 b^2 c^2-16 a b c d-3 a^2 d^2\right ) x}{6 a^2 c (b c-a d)^3 \sqrt{a+b x^2}}-\frac{d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}+\frac{d^2 (6 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{c} \sqrt{a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{7/2}}\\ \end{align*}

Mathematica [A]  time = 5.48197, size = 170, normalized size = 0.84 \[ \frac{1}{6} \left (x \sqrt{a+b x^2} \left (\frac{4 b^2 (4 a d-b c)}{a^2 \left (a+b x^2\right ) (a d-b c)^3}+\frac{2 b^2}{a \left (a+b x^2\right )^2 (b c-a d)^2}-\frac{3 d^3}{c \left (c+d x^2\right ) (b c-a d)^3}\right )+\frac{3 d^2 (a d-6 b c) \tan ^{-1}\left (\frac{x \sqrt{a d-b c}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{c^{3/2} (a d-b c)^{7/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*((2*b^2)/(a*(b*c - a*d)^2*(a + b*x^2)^2) + (4*b^2*(-(b*c) + 4*a*d))/(a^2*(-(b*c) + a*d)^3*(
a + b*x^2)) - (3*d^3)/(c*(b*c - a*d)^3*(c + d*x^2))) + (3*d^2*(-6*b*c + a*d)*ArcTan[(Sqrt[-(b*c) + a*d]*x)/(Sq
rt[c]*Sqrt[a + b*x^2])])/(c^(3/2)*(-(b*c) + a*d)^(7/2)))/6

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Maple [B]  time = 0.02, size = 2371, normalized size = 11.7 \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)^2,x)

[Out]

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

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

[Out]

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

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Fricas [B]  time = 9.3097, size = 2849, normalized size = 14.1 \begin{align*} \text{result too large to display} \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)^2,x, algorithm="fricas")

[Out]

[1/24*(3*(6*a^4*b*c^2*d^2 - a^5*c*d^3 + (6*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^6 + (6*a^2*b^3*c^2*d^2 + 11*a^3*b^2*
c*d^3 - 2*a^4*b*d^4)*x^4 + (12*a^3*b^2*c^2*d^2 + 4*a^4*b*c*d^3 - a^5*d^4)*x^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*((4*b^5*c^4*d - 20*a*b^4*c^3*d^2 + 13*a^2*b^
3*c^2*d^3 + 3*a^3*b^2*c*d^4)*x^5 + 2*(2*b^5*c^5 - 7*a*b^4*c^4*d - 4*a^2*b^3*c^3*d^2 + 6*a^3*b^2*c^2*d^3 + 3*a^
4*b*c*d^4)*x^3 + 3*(2*a*b^4*c^5 - 8*a^2*b^3*c^4*d + 6*a^3*b^2*c^3*d^2 - a^4*b*c^2*d^3 + a^5*c*d^4)*x)*sqrt(b*x
^2 + a))/(a^4*b^4*c^7 - 4*a^5*b^3*c^6*d + 6*a^6*b^2*c^5*d^2 - 4*a^7*b*c^4*d^3 + a^8*c^3*d^4 + (a^2*b^6*c^6*d -
4*a^3*b^5*c^5*d^2 + 6*a^4*b^4*c^4*d^3 - 4*a^5*b^3*c^3*d^4 + a^6*b^2*c^2*d^5)*x^6 + (a^2*b^6*c^7 - 2*a^3*b^5*c
^6*d - 2*a^4*b^4*c^5*d^2 + 8*a^5*b^3*c^4*d^3 - 7*a^6*b^2*c^3*d^4 + 2*a^7*b*c^2*d^5)*x^4 + (2*a^3*b^5*c^7 - 7*a
^4*b^4*c^6*d + 8*a^5*b^3*c^5*d^2 - 2*a^6*b^2*c^4*d^3 - 2*a^7*b*c^3*d^4 + a^8*c^2*d^5)*x^2), -1/12*(3*(6*a^4*b*
c^2*d^2 - a^5*c*d^3 + (6*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^6 + (6*a^2*b^3*c^2*d^2 + 11*a^3*b^2*c*d^3 - 2*a^4*b*d^
4)*x^4 + (12*a^3*b^2*c^2*d^2 + 4*a^4*b*c*d^3 - a^5*d^4)*x^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*((4*b^5*c
^4*d - 20*a*b^4*c^3*d^2 + 13*a^2*b^3*c^2*d^3 + 3*a^3*b^2*c*d^4)*x^5 + 2*(2*b^5*c^5 - 7*a*b^4*c^4*d - 4*a^2*b^3
*c^3*d^2 + 6*a^3*b^2*c^2*d^3 + 3*a^4*b*c*d^4)*x^3 + 3*(2*a*b^4*c^5 - 8*a^2*b^3*c^4*d + 6*a^3*b^2*c^3*d^2 - a^4
*b*c^2*d^3 + a^5*c*d^4)*x)*sqrt(b*x^2 + a))/(a^4*b^4*c^7 - 4*a^5*b^3*c^6*d + 6*a^6*b^2*c^5*d^2 - 4*a^7*b*c^4*d
^3 + a^8*c^3*d^4 + (a^2*b^6*c^6*d - 4*a^3*b^5*c^5*d^2 + 6*a^4*b^4*c^4*d^3 - 4*a^5*b^3*c^3*d^4 + a^6*b^2*c^2*d^
5)*x^6 + (a^2*b^6*c^7 - 2*a^3*b^5*c^6*d - 2*a^4*b^4*c^5*d^2 + 8*a^5*b^3*c^4*d^3 - 7*a^6*b^2*c^3*d^4 + 2*a^7*b*
c^2*d^5)*x^4 + (2*a^3*b^5*c^7 - 7*a^4*b^4*c^6*d + 8*a^5*b^3*c^5*d^2 - 2*a^6*b^2*c^4*d^3 - 2*a^7*b*c^3*d^4 + a^
8*c^2*d^5)*x^2)]

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Sympy [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)**2,x)

[Out]

Exception raised: ValueError

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Giac [B]  time = 7.72122, size = 837, normalized size = 4.14 \begin{align*} \frac{{\left (\frac{2 \,{\left (b^{8} c^{4} - 7 \, a b^{7} c^{3} d + 15 \, a^{2} b^{6} c^{2} d^{2} - 13 \, a^{3} b^{5} c d^{3} + 4 \, a^{4} b^{4} d^{4}\right )} x^{2}}{a^{2} b^{7} c^{6} - 6 \, a^{3} b^{6} c^{5} d + 15 \, a^{4} b^{5} c^{4} d^{2} - 20 \, a^{5} b^{4} c^{3} d^{3} + 15 \, a^{6} b^{3} c^{2} d^{4} - 6 \, a^{7} b^{2} c d^{5} + a^{8} b d^{6}} + \frac{3 \,{\left (a b^{7} c^{4} - 6 \, a^{2} b^{6} c^{3} d + 12 \, a^{3} b^{5} c^{2} d^{2} - 10 \, a^{4} b^{4} c d^{3} + 3 \, a^{5} b^{3} d^{4}\right )}}{a^{2} b^{7} c^{6} - 6 \, a^{3} b^{6} c^{5} d + 15 \, a^{4} b^{5} c^{4} d^{2} - 20 \, a^{5} b^{4} c^{3} d^{3} + 15 \, a^{6} b^{3} c^{2} d^{4} - 6 \, a^{7} b^{2} c d^{5} + a^{8} b d^{6}}\right )} x}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} + \frac{{\left (6 \, b^{\frac{3}{2}} c d^{2} - a \sqrt{b} d^{3}\right )} \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 )}{2 \,{\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} \sqrt{-b^{2} c^{2} + a b c d}} - \frac{2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} b^{\frac{3}{2}} c d^{2} -{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a \sqrt{b} d^{3} + a^{2} \sqrt{b} d^{3}}{{\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )}{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} d + 4 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} b c - 2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a d + a^{2} d\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)^2,x, algorithm="giac")

[Out]

1/3*(2*(b^8*c^4 - 7*a*b^7*c^3*d + 15*a^2*b^6*c^2*d^2 - 13*a^3*b^5*c*d^3 + 4*a^4*b^4*d^4)*x^2/(a^2*b^7*c^6 - 6*
a^3*b^6*c^5*d + 15*a^4*b^5*c^4*d^2 - 20*a^5*b^4*c^3*d^3 + 15*a^6*b^3*c^2*d^4 - 6*a^7*b^2*c*d^5 + a^8*b*d^6) +
3*(a*b^7*c^4 - 6*a^2*b^6*c^3*d + 12*a^3*b^5*c^2*d^2 - 10*a^4*b^4*c*d^3 + 3*a^5*b^3*d^4)/(a^2*b^7*c^6 - 6*a^3*b
^6*c^5*d + 15*a^4*b^5*c^4*d^2 - 20*a^5*b^4*c^3*d^3 + 15*a^6*b^3*c^2*d^4 - 6*a^7*b^2*c*d^5 + a^8*b*d^6))*x/(b*x
^2 + a)^(3/2) + 1/2*(6*b^(3/2)*c*d^2 - a*sqrt(b)*d^3)*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^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - a^3*c*d^3)*sqrt(-b^2*c^2 + a*b*c
*d)) - (2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*b^(3/2)*c*d^2 - (sqrt(b)*x - sqrt(b*x^2 + a))^2*a*sqrt(b)*d^3 + a^2*
sqrt(b)*d^3)/((b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - a^3*c*d^3)*((sqrt(b)*x - sqrt(b*x^2 + a))^4*d + 4*(
sqrt(b)*x - sqrt(b*x^2 + a))^2*b*c - 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a*d + a^2*d))