\(\int \frac {1}{(a+b x^2)^2 (c+d x^2)^{5/2}} \, dx\) [1036]

Optimal result

Integrand size = 21, antiderivative size = 201 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {d (3 b c+2 a d) x}{6 a c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac {d \left (3 b^2 c^2+16 a b c d-4 a^2 d^2\right ) x}{6 a c^2 (b c-a d)^3 \sqrt {c+d x^2}}+\frac {b^2 (b c-6 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} (b c-a d)^{7/2}} \] Output:

1/6*d*(2*a*d+3*b*c)*x/a/c/(-a*d+b*c)^2/(d*x^2+c)^(3/2)+1/2*b*x/a/(-a*d+b*c 
)/(b*x^2+a)/(d*x^2+c)^(3/2)+1/6*d*(-4*a^2*d^2+16*a*b*c*d+3*b^2*c^2)*x/a/c^ 
2/(-a*d+b*c)^3/(d*x^2+c)^(1/2)+1/2*b^2*(-6*a*d+b*c)*arctan((-a*d+b*c)^(1/2 
)*x/a^(1/2)/(d*x^2+c)^(1/2))/a^(3/2)/(-a*d+b*c)^(7/2)
 

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {x \left (3 b^3 c^2 \left (c+d x^2\right )^2-2 a^3 d^3 \left (3 c+2 d x^2\right )+2 a b^2 c d^2 x^2 \left (9 c+8 d x^2\right )+2 a^2 b d^2 \left (9 c^2+5 c d x^2-2 d^2 x^4\right )\right )}{6 a c^2 (b c-a d)^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac {b^2 (b c-6 a d) \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 a^{3/2} (b c-a d)^{7/2}} \] Input:

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

Output:

(x*(3*b^3*c^2*(c + d*x^2)^2 - 2*a^3*d^3*(3*c + 2*d*x^2) + 2*a*b^2*c*d^2*x^ 
2*(9*c + 8*d*x^2) + 2*a^2*b*d^2*(9*c^2 + 5*c*d*x^2 - 2*d^2*x^4)))/(6*a*c^2 
*(b*c - a*d)^3*(a + b*x^2)*(c + d*x^2)^(3/2)) - (b^2*(b*c - 6*a*d)*ArcTan[ 
(a*Sqrt[d] + b*x*(Sqrt[d]*x - Sqrt[c + d*x^2]))/(Sqrt[a]*Sqrt[b*c - a*d])] 
)/(2*a^(3/2)*(b*c - a*d)^(7/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {316, 25, 402, 402, 27, 291, 218}

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.

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

\(\Big \downarrow \) 316

\(\displaystyle \frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {\int -\frac {4 b d x^2+b c-2 a d}{\left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx}{2 a (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {4 b d x^2+b c-2 a d}{\left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\int \frac {3 b^2 c^2-12 a b d c+4 a^2 d^2+2 b d (3 b c+2 a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )^{3/2}}dx}{3 c (b c-a d)}+\frac {d x (2 a d+3 b c)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int \frac {3 b^2 c^2 (b c-6 a d)}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{c (b c-a d)}+\frac {d x \left (-4 a^2 d^2+16 a b c d+3 b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}}{3 c (b c-a d)}+\frac {d x (2 a d+3 b c)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 b^2 c (b c-6 a d) \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b c-a d}+\frac {d x \left (-4 a^2 d^2+16 a b c d+3 b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}}{3 c (b c-a d)}+\frac {d x (2 a d+3 b c)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {\frac {3 b^2 c (b c-6 a d) \int \frac {1}{a-\frac {(a d-b c) x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{b c-a d}+\frac {d x \left (-4 a^2 d^2+16 a b c d+3 b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}}{3 c (b c-a d)}+\frac {d x (2 a d+3 b c)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {d x \left (-4 a^2 d^2+16 a b c d+3 b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}+\frac {3 b^2 c (b c-6 a d) \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} (b c-a d)^{3/2}}}{3 c (b c-a d)}+\frac {d x (2 a d+3 b c)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a (b c-a d)}+\frac {b x}{2 a \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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 = 0.84 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {-\frac {b^{2} c^{2} \left (\frac {\sqrt {x^{2} d +c}\, b x}{b \,x^{2}+a}-\frac {\left (6 a d -b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{2 a \left (a d -b c \right )^{3}}+\frac {\left (a d -3 b c \right ) d^{2} x}{\left (a d -b c \right )^{3} \sqrt {x^{2} d +c}}-\frac {d^{3} x^{3}}{3 \left (a d -b c \right )^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}}{c^{2}}\)	\(158\)
default	\(\text {Expression too large to display}\)	\(3489\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 697 vs. \(2 (177) = 354\).

Time = 0.76 (sec) , antiderivative size = 1434, normalized size of antiderivative = 7.13 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (177) = 354\).

Time = 0.40 (sec) , antiderivative size = 619, normalized size of antiderivative = 3.08 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {{\left (\frac {2 \, {\left (4 \, b^{4} c^{4} d^{4} - 13 \, a b^{3} c^{3} d^{5} + 15 \, a^{2} b^{2} c^{2} d^{6} - 7 \, a^{3} b c d^{7} + a^{4} d^{8}\right )} x^{2}}{b^{6} c^{8} d - 6 \, a b^{5} c^{7} d^{2} + 15 \, a^{2} b^{4} c^{6} d^{3} - 20 \, a^{3} b^{3} c^{5} d^{4} + 15 \, a^{4} b^{2} c^{4} d^{5} - 6 \, a^{5} b c^{3} d^{6} + a^{6} c^{2} d^{7}} + \frac {3 \, {\left (3 \, b^{4} c^{5} d^{3} - 10 \, a b^{3} c^{4} d^{4} + 12 \, a^{2} b^{2} c^{3} d^{5} - 6 \, a^{3} b c^{2} d^{6} + a^{4} c d^{7}\right )}}{b^{6} c^{8} d - 6 \, a b^{5} c^{7} d^{2} + 15 \, a^{2} b^{4} c^{6} d^{3} - 20 \, a^{3} b^{3} c^{5} d^{4} + 15 \, a^{4} b^{2} c^{4} d^{5} - 6 \, a^{5} b c^{3} d^{6} + a^{6} c^{2} d^{7}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {{\left (b^{3} c \sqrt {d} - 6 \, a b^{2} d^{\frac {3}{2}}\right )} \arctan \left (-\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{3} c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{2} d^{\frac {3}{2}} - b^{3} c^{2} \sqrt {d}}{{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}} \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 3742, normalized size of antiderivative = 18.62 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(72*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) 
- 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a**3*b**2*c 
**4*d**2 + 144*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt( 
a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)* 
a**3*b**2*c**3*d**3*x**2 + 72*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d 
)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt 
(d)*sqrt(b)*x)*a**3*b**2*c**2*d**4*x**4 + 6*sqrt(a)*sqrt(a*d - b*c)*log( - 
 sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + 
d*x**2) + sqrt(d)*sqrt(b)*x)*a**2*b**3*c**5*d + 84*sqrt(a)*sqrt(a*d - b*c) 
*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sq 
rt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a**2*b**3*c**4*d**2*x**2 + 150*sqrt(a) 
*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b 
*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a**2*b**3*c**3*d**3*x* 
*4 + 72*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b 
*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a**2*b* 
*3*c**2*d**4*x**6 - 3*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a 
)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt 
(b)*x)*a*b**4*c**6 + 9*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt( 
a)*sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqr 
t(b)*x)*a*b**4*c**4*d**2*x**4 + 6*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2...