\(\int \frac {1}{(a+b x^2)^{3/2} (c+d x^2)^3} \, dx\) [102]

Optimal result

Integrand size = 21, antiderivative size = 214 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx=\frac {b (4 b c-a d) (2 b c+3 a d) x}{8 a c^2 (b c-a d)^3 \sqrt {a+b x^2}}-\frac {d x}{4 c (b c-a d) \sqrt {a+b x^2} \left (c+d x^2\right )^2}-\frac {d (8 b c-3 a d) x}{8 c^2 (b c-a d)^2 \sqrt {a+b x^2} \left (c+d x^2\right )}-\frac {3 d \left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{8 c^{5/2} (b c-a d)^{7/2}} \] Output:

1/8*b*(-a*d+4*b*c)*(3*a*d+2*b*c)*x/a/c^2/(-a*d+b*c)^3/(b*x^2+a)^(1/2)-1/4* 
d*x/c/(-a*d+b*c)/(b*x^2+a)^(1/2)/(d*x^2+c)^2-1/8*d*(-3*a*d+8*b*c)*x/c^2/(- 
a*d+b*c)^2/(b*x^2+a)^(1/2)/(d*x^2+c)-3/8*d*(a^2*d^2-4*a*b*c*d+8*b^2*c^2)*a 
rctanh((-a*d+b*c)^(1/2)*x/c^(1/2)/(b*x^2+a)^(1/2))/c^(5/2)/(-a*d+b*c)^(7/2 
)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 13.04 (sec) , antiderivative size = 1392, normalized size of antiderivative = 6.50 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[1/((a + b*x^2)^(3/2)*(c + d*x^2)^3),x]
 

Output:

(x*(-108045*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] - (324135*d*x^2*Sqrt[( 
(b*c - a*d)*x^2)/(c*(a + b*x^2))])/c - (324135*d^2*x^4*Sqrt[((b*c - a*d)*x 
^2)/(c*(a + b*x^2))])/c^2 - (103320*d^3*x^6*Sqrt[((b*c - a*d)*x^2)/(c*(a + 
 b*x^2))])/c^3 + 42735*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2) + (128205 
*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2))/c + (139545*d^2*x^4*(((b 
*c - a*d)*x^2)/(c*(a + b*x^2)))^(3/2))/c^2 + (46200*d^3*x^6*(((b*c - a*d)* 
x^2)/(c*(a + b*x^2)))^(3/2))/c^3 - 3864*(((b*c - a*d)*x^2)/(c*(a + b*x^2)) 
)^(5/2) - (4032*d*x^2*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2))/c - (4032 
*d^2*x^4*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2))/c^2 - (1344*d^3*x^6*(( 
(b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2))/c^3 + 108045*ArcTanh[Sqrt[((b*c - 
 a*d)*x^2)/(c*(a + b*x^2))]] + (324135*d*x^2*ArcTanh[Sqrt[((b*c - a*d)*x^2 
)/(c*(a + b*x^2))]])/c + (324135*d^2*x^4*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c 
*(a + b*x^2))]])/c^2 + (103320*d^3*x^6*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*( 
a + b*x^2))]])/c^3 + (8505*(b*c - a*d)^2*x^4*ArcTanh[Sqrt[((b*c - a*d)*x^2 
)/(c*(a + b*x^2))]])/(c^2*(a + b*x^2)^2) + (17955*d*(b*c - a*d)^2*x^6*ArcT 
anh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c^3*(a + b*x^2)^2) + (21735 
*d^2*(b*c - a*d)^2*x^8*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/( 
c^4*(a + b*x^2)^2) + (7560*d^3*(b*c - a*d)^2*x^10*ArcTanh[Sqrt[((b*c - a*d 
)*x^2)/(c*(a + b*x^2))]])/(c^5*(a + b*x^2)^2) - (78750*(b*c - a*d)*x^2*Arc 
Tanh[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]])/(c*(a + b*x^2)) + (23625...
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {316, 402, 27, 402, 27, 291, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {\int \frac {-4 b d x^2+4 b c-3 a d}{\left (b x^2+a\right )^{3/2} \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}-\frac {d x}{4 c \sqrt {a+b x^2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {b x (a d+4 b c)}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}-\frac {\int \frac {d \left (a (8 b c-3 a d)-2 b (4 b c+a d) x^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )^2}dx}{a (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \sqrt {a+b x^2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {b x (a d+4 b c)}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}-\frac {d \int \frac {a (8 b c-3 a d)-2 b (4 b c+a d) x^2}{\sqrt {b x^2+a} \left (d x^2+c\right )^2}dx}{a (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \sqrt {a+b x^2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {b x (a d+4 b c)}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}-\frac {d \left (\frac {\int \frac {3 a \left (8 b^2 c^2-4 a b d c+a^2 d^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {x \sqrt {a+b x^2} (4 b c-a d) (3 a d+2 b c)}{2 c \left (c+d x^2\right ) (b c-a d)}\right )}{a (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \sqrt {a+b x^2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {b x (a d+4 b c)}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}-\frac {d \left (\frac {3 a \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {x \sqrt {a+b x^2} (4 b c-a d) (3 a d+2 b c)}{2 c \left (c+d x^2\right ) (b c-a d)}\right )}{a (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \sqrt {a+b x^2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {b x (a d+4 b c)}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}-\frac {d \left (\frac {3 a \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \int \frac {1}{c-\frac {(b c-a d) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 c (b c-a d)}-\frac {x \sqrt {a+b x^2} (4 b c-a d) (3 a d+2 b c)}{2 c \left (c+d x^2\right ) (b c-a d)}\right )}{a (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \sqrt {a+b x^2} \left (c+d x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {b x (a d+4 b c)}{a \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}-\frac {d \left (\frac {3 a \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{3/2}}-\frac {x \sqrt {a+b x^2} (4 b c-a d) (3 a d+2 b c)}{2 c \left (c+d x^2\right ) (b c-a d)}\right )}{a (b c-a d)}}{4 c (b c-a d)}-\frac {d x}{4 c \sqrt {a+b x^2} \left (c+d x^2\right )^2 (b c-a d)}\)

Input:

Int[1/((a + b*x^2)^(3/2)*(c + d*x^2)^3),x]
 

Output:

-1/4*(d*x)/(c*(b*c - a*d)*Sqrt[a + b*x^2]*(c + d*x^2)^2) + ((b*(4*b*c + a* 
d)*x)/(a*(b*c - a*d)*Sqrt[a + b*x^2]*(c + d*x^2)) - (d*(-1/2*((4*b*c - a*d 
)*(2*b*c + 3*a*d)*x*Sqrt[a + b*x^2])/(c*(b*c - a*d)*(c + d*x^2)) + (3*a*(8 
*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[ 
a + b*x^2])])/(2*c^(3/2)*(b*c - a*d)^(3/2))))/(a*(b*c - a*d)))/(4*c*(b*c - 
 a*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

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

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 316

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

rule 402

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

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(\frac {-\frac {3 \sqrt {b \,x^{2}+a}\, a d \left (x^{2} d +c \right )^{2} \left (a^{2} d^{2}-4 a b c d +8 b^{2} c^{2}\right ) \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )}{8}+\frac {5 \sqrt {\left (a d -b c \right ) c}\, \left (-\frac {8 c^{4} b^{3}}{5}-\frac {16 b^{3} c^{3} d \,x^{2}}{5}-\frac {12 d^{2} b \left (\frac {2}{3} b^{2} x^{4}+a b \,x^{2}+a^{2}\right ) c^{2}}{5}+a \,d^{3} \left (b \,x^{2}+a \right ) \left (-2 b \,x^{2}+a \right ) c +\frac {3 a^{2} d^{4} x^{2} \left (b \,x^{2}+a \right )}{5}\right ) x}{8}}{\left (x^{2} d +c \right )^{2} c^{2} \left (a d -b c \right )^{3} \sqrt {\left (a d -b c \right ) c}\, \sqrt {b \,x^{2}+a}\, a}\)	\(222\)
default	\(\text {Expression too large to display}\)	\(4035\)

Input:

int(1/(b*x^2+a)^(3/2)/(d*x^2+c)^3,x,method=_RETURNVERBOSE)
 

Output:

5/8*(-3/5*(b*x^2+a)^(1/2)*a*d*(d*x^2+c)^2*(a^2*d^2-4*a*b*c*d+8*b^2*c^2)*ar 
ctan(c*(b*x^2+a)^(1/2)/x/((a*d-b*c)*c)^(1/2))+((a*d-b*c)*c)^(1/2)*(-8/5*c^ 
4*b^3-16/5*b^3*c^3*d*x^2-12/5*d^2*b*(2/3*b^2*x^4+a*b*x^2+a^2)*c^2+a*d^3*(b 
*x^2+a)*(-2*b*x^2+a)*c+3/5*a^2*d^4*x^2*(b*x^2+a))*x)/((a*d-b*c)*c)^(1/2)/( 
b*x^2+a)^(1/2)/(d*x^2+c)^2/c^2/(a*d-b*c)^3/a
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 721 vs. \(2 (190) = 380\).

Time = 0.76 (sec) , antiderivative size = 1482, normalized size of antiderivative = 6.93 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c)^3,x, algorithm="fricas")
 

Output:

[-1/32*(3*(8*a^2*b^2*c^4*d - 4*a^3*b*c^3*d^2 + a^4*c^2*d^3 + (8*a*b^3*c^2* 
d^3 - 4*a^2*b^2*c*d^4 + a^3*b*d^5)*x^6 + (16*a*b^3*c^3*d^2 - 2*a^3*b*c*d^4 
 + a^4*d^5)*x^4 + (8*a*b^3*c^4*d + 12*a^2*b^2*c^3*d^2 - 7*a^3*b*c^2*d^3 + 
2*a^4*c*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*((8*b^4*c^4*d^2 + 2*a*b^3*c^3*d^3 - 13*a^2*b^2*c^2*d^4 + 3*a^3*b*c*d^5) 
*x^5 + (16*b^4*c^5*d - 4*a*b^3*c^4*d^2 - 7*a^2*b^2*c^3*d^3 - 8*a^3*b*c^2*d 
^4 + 3*a^4*c*d^5)*x^3 + (8*b^4*c^6 - 8*a*b^3*c^5*d + 12*a^2*b^2*c^4*d^2 - 
17*a^3*b*c^3*d^3 + 5*a^4*c^2*d^4)*x)*sqrt(b*x^2 + a))/(a^2*b^4*c^9 - 4*a^3 
*b^3*c^8*d + 6*a^4*b^2*c^7*d^2 - 4*a^5*b*c^6*d^3 + a^6*c^5*d^4 + (a*b^5*c^ 
7*d^2 - 4*a^2*b^4*c^6*d^3 + 6*a^3*b^3*c^5*d^4 - 4*a^4*b^2*c^4*d^5 + a^5*b* 
c^3*d^6)*x^6 + (2*a*b^5*c^8*d - 7*a^2*b^4*c^7*d^2 + 8*a^3*b^3*c^6*d^3 - 2* 
a^4*b^2*c^5*d^4 - 2*a^5*b*c^4*d^5 + a^6*c^3*d^6)*x^4 + (a*b^5*c^9 - 2*a^2* 
b^4*c^8*d - 2*a^3*b^3*c^7*d^2 + 8*a^4*b^2*c^6*d^3 - 7*a^5*b*c^5*d^4 + 2*a^ 
6*c^4*d^5)*x^2), 1/16*(3*(8*a^2*b^2*c^4*d - 4*a^3*b*c^3*d^2 + a^4*c^2*d^3 
+ (8*a*b^3*c^2*d^3 - 4*a^2*b^2*c*d^4 + a^3*b*d^5)*x^6 + (16*a*b^3*c^3*d^2 
- 2*a^3*b*c*d^4 + a^4*d^5)*x^4 + (8*a*b^3*c^4*d + 12*a^2*b^2*c^3*d^2 - 7*a 
^3*b*c^2*d^3 + 2*a^4*c*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...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x**2+a)**(3/2)/(d*x**2+c)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (d x^{2} + c\right )}^{3}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c)^3,x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(3/2)*(d*x^2 + c)^3), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 643 vs. \(2 (190) = 380\).

Time = 0.68 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.00 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx=\frac {b^{3} x}{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {b x^{2} + a}} + \frac {3 \, {\left (8 \, b^{\frac {5}{2}} c^{2} d - 4 \, a b^{\frac {3}{2}} c d^{2} + a^{2} \sqrt {b} d^{3}\right )} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt {-b^{2} c^{2} + a b c d}}\right )}{8 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} \sqrt {-b^{2} c^{2} + a b c d}} + \frac {16 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} b^{\frac {5}{2}} c^{2} d^{2} - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a b^{\frac {3}{2}} c d^{3} + 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{2} \sqrt {b} d^{4} + 80 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} b^{\frac {7}{2}} c^{3} d - 104 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a b^{\frac {5}{2}} c^{2} d^{2} + 54 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{2} b^{\frac {3}{2}} c d^{3} - 9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{3} \sqrt {b} d^{4} + 64 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{2} b^{\frac {5}{2}} c^{2} d^{2} - 52 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{3} b^{\frac {3}{2}} c d^{3} + 9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{4} \sqrt {b} d^{4} + 10 \, a^{4} b^{\frac {3}{2}} c d^{3} - 3 \, a^{5} \sqrt {b} d^{4}}{4 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )} {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} d + 4 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b c - 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a d + a^{2} d\right )}^{2}} \] Input:

integrate(1/(b*x^2+a)^(3/2)/(d*x^2+c)^3,x, algorithm="giac")
 

Output:

b^3*x/((a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3)*sqrt(b*x^2 
+ a)) + 3/8*(8*b^(5/2)*c^2*d - 4*a*b^(3/2)*c*d^2 + a^2*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^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*sqrt(-b^ 
2*c^2 + a*b*c*d)) + 1/4*(16*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(5/2)*c^2*d^ 
2 - 12*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*b^(3/2)*c*d^3 + 3*(sqrt(b)*x - sq 
rt(b*x^2 + a))^6*a^2*sqrt(b)*d^4 + 80*(sqrt(b)*x - sqrt(b*x^2 + a))^4*b^(7 
/2)*c^3*d - 104*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a*b^(5/2)*c^2*d^2 + 54*(sq 
rt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(3/2)*c*d^3 - 9*(sqrt(b)*x - sqrt(b*x^2 
 + a))^4*a^3*sqrt(b)*d^4 + 64*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^2*b^(5/2)* 
c^2*d^2 - 52*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*b^(3/2)*c*d^3 + 9*(sqrt(b 
)*x - sqrt(b*x^2 + a))^2*a^4*sqrt(b)*d^4 + 10*a^4*b^(3/2)*c*d^3 - 3*a^5*sq 
rt(b)*d^4)/((b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 - a^3*c^2*d^3)*((sq 
rt(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)^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{3/2}\,{\left (d\,x^2+c\right )}^3} \,d x \] Input:

int(1/((a + b*x^2)^(3/2)*(c + d*x^2)^3),x)
 

Output:

int(1/((a + b*x^2)^(3/2)*(c + d*x^2)^3), x)
 

Reduce [B] (verification not implemented)

Time = 5.18 (sec) , antiderivative size = 3380, normalized size of antiderivative = 15.79 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int(1/(b*x^2+a)^(3/2)/(d*x^2+c)^3,x)
 

Output:

( - 9*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x 
**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**5*c**2*d**4 - 18*sqrt(c)*s 
qrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)* 
sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**5*c*d**5*x**2 - 9*sqrt(c)*sqrt(a*d - b*c) 
*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sq 
rt(c)*sqrt(b)))*a**5*d**6*x**4 + 60*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d 
 - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b))) 
*a**4*b*c**3*d**3 + 111*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sq 
rt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**4*b*c**2 
*d**4*x**2 + 42*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sq 
rt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**4*b*c*d**5*x**4 
- 9*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x** 
2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**4*b*d**6*x**6 - 168*sqrt(c)* 
sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d) 
*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**3*b**2*c**4*d**2 - 276*sqrt(c)*sqrt(a*d 
- b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)* 
x)/(sqrt(c)*sqrt(b)))*a**3*b**2*c**3*d**3*x**2 - 48*sqrt(c)*sqrt(a*d - b*c 
)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(s 
qrt(c)*sqrt(b)))*a**3*b**2*c**2*d**4*x**4 + 60*sqrt(c)*sqrt(a*d - b*c)*ata 
n((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqr...