\(\int \frac {x^2}{(a+b x^2) (c+d x^2)^3} \, dx\) [648]

Optimal result

Integrand size = 22, antiderivative size = 155 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {x}{4 (b c-a d) \left (c+d x^2\right )^2}+\frac {(3 b c+a d) x}{8 c (b c-a d)^2 \left (c+d x^2\right )}-\frac {\sqrt {a} b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{(b c-a d)^3}+\frac {\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{3/2} \sqrt {d} (b c-a d)^3} \] Output:

1/4*x/(-a*d+b*c)/(d*x^2+c)^2+1/8*(a*d+3*b*c)*x/c/(-a*d+b*c)^2/(d*x^2+c)-a^ 
(1/2)*b^(3/2)*arctan(b^(1/2)*x/a^(1/2))/(-a*d+b*c)^3+1/8*(-a^2*d^2+6*a*b*c 
*d+3*b^2*c^2)*arctan(d^(1/2)*x/c^(1/2))/c^(3/2)/d^(1/2)/(-a*d+b*c)^3
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.95 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {\sqrt {a} b^{3/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{(-b c+a d)^3}+\frac {\frac {\sqrt {c} (b c-a d) x \left (a d \left (-c+d x^2\right )+b c \left (5 c+3 d x^2\right )\right )}{\left (c+d x^2\right )^2}+\frac {\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {d}}}{8 c^{3/2} (b c-a d)^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {373, 402, 397, 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.

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 397

\(\displaystyle \frac {x}{4 \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\frac {8 a b^2 c \int \frac {1}{b x^2+a}dx}{b c-a d}-\frac {\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \int \frac {1}{d x^2+c}dx}{b c-a d}}{2 c (b c-a d)}-\frac {x (a d+3 b c)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x}{4 \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\frac {8 \sqrt {a} b^{3/2} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b c-a d}-\frac {\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d} (b c-a d)}}{2 c (b c-a d)}-\frac {x (a d+3 b c)}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 (b c-a d)}\)

Input:

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

Output:

x/(4*(b*c - a*d)*(c + d*x^2)^2) - (-1/2*((3*b*c + a*d)*x)/(c*(b*c - a*d)*( 
c + d*x^2)) + ((8*Sqrt[a]*b^(3/2)*c*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(b*c - a* 
d) - ((3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt 
[c]*Sqrt[d]*(b*c - a*d)))/(2*c*(b*c - a*d)))/(4*(b*c - a*d))
 

Defintions of rubi rules used

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 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, 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 = 0.92 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {a \,b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\left (a d -b c \right )^{3} \sqrt {a b}}+\frac {\frac {\frac {d \left (a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}\right ) x^{3}}{8 c}+\left (\frac {3}{4} a b c d -\frac {5}{8} b^{2} c^{2}-\frac {1}{8} a^{2} d^{2}\right ) x}{\left (x^{2} d +c \right )^{2}}+\frac {\left (a^{2} d^{2}-6 a b c d -3 b^{2} c^{2}\right ) \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{8 c \sqrt {c d}}}{\left (a d -b c \right )^{3}}\)	\(151\)
risch	\(\frac {\frac {d \left (a d +3 b c \right ) x^{3}}{8 c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {\left (a d -5 b c \right ) x}{8 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}}{\left (x^{2} d +c \right )^{2}}-\frac {\ln \left (c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) a^{2} d^{2}}{16 \sqrt {-c d}\, \left (a d -b c \right )^{3} c}+\frac {3 \ln \left (c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) a b d}{8 \sqrt {-c d}\, \left (a d -b c \right )^{3}}+\frac {3 c \ln \left (c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) b^{2}}{16 \sqrt {-c d}\, \left (a d -b c \right )^{3}}+\frac {\ln \left (-c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) a^{2} d^{2}}{16 \sqrt {-c d}\, \left (a d -b c \right )^{3} c}-\frac {3 \ln \left (-c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) a b d}{8 \sqrt {-c d}\, \left (a d -b c \right )^{3}}-\frac {3 c \ln \left (-c \,d^{2} x -\left (-c d \right )^{\frac {3}{2}}\right ) b^{2}}{16 \sqrt {-c d}\, \left (a d -b c \right )^{3}}+\frac {\sqrt {-a b}\, b \ln \left (\left (-64 \left (-a b \right )^{\frac {3}{2}} a b \,c^{2} d^{2}-64 \left (-a b \right )^{\frac {3}{2}} b^{2} c^{3} d -\sqrt {-a b}\, a^{4} d^{4}+12 \sqrt {-a b}\, a^{3} b c \,d^{3}-94 \sqrt {-a b}\, a^{2} b^{2} c^{2} d^{2}-36 \sqrt {-a b}\, a \,b^{3} c^{3} d -9 \sqrt {-a b}\, b^{4} c^{4}\right ) x +a^{5} d^{4}-12 a^{4} b c \,d^{3}+30 a^{3} b^{2} c^{2} d^{2}-28 a^{2} b^{3} c^{3} d +9 a \,b^{4} c^{4}\right )}{2 \left (a d -b c \right )^{3}}-\frac {\sqrt {-a b}\, b \ln \left (\left (64 \left (-a b \right )^{\frac {3}{2}} a b \,c^{2} d^{2}+64 \left (-a b \right )^{\frac {3}{2}} b^{2} c^{3} d +\sqrt {-a b}\, a^{4} d^{4}-12 \sqrt {-a b}\, a^{3} b c \,d^{3}+94 \sqrt {-a b}\, a^{2} b^{2} c^{2} d^{2}+36 \sqrt {-a b}\, a \,b^{3} c^{3} d +9 \sqrt {-a b}\, b^{4} c^{4}\right ) x +a^{5} d^{4}-12 a^{4} b c \,d^{3}+30 a^{3} b^{2} c^{2} d^{2}-28 a^{2} b^{3} c^{3} d +9 a \,b^{4} c^{4}\right )}{2 \left (a d -b c \right )^{3}}\)	\(699\)

Input:

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

Output:

a*b^2/(a*d-b*c)^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))+1/(a*d-b*c)^3*((1/8* 
d*(a^2*d^2+2*a*b*c*d-3*b^2*c^2)/c*x^3+(3/4*a*b*c*d-5/8*b^2*c^2-1/8*a^2*d^2 
)*x)/(d*x^2+c)^2+1/8*(a^2*d^2-6*a*b*c*d-3*b^2*c^2)/c/(c*d)^(1/2)*arctan(x* 
d/(c*d)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (133) = 266\).

Time = 0.31 (sec) , antiderivative size = 1587, normalized size of antiderivative = 10.24 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/16*(2*(3*b^2*c^3*d^2 - 2*a*b*c^2*d^3 - a^2*c*d^4)*x^3 - 8*(b*c^2*d^3*x^ 
4 + 2*b*c^3*d^2*x^2 + b*c^4*d)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a) 
/(b*x^2 + a)) - (3*b^2*c^4 + 6*a*b*c^3*d - a^2*c^2*d^2 + (3*b^2*c^2*d^2 + 
6*a*b*c*d^3 - a^2*d^4)*x^4 + 2*(3*b^2*c^3*d + 6*a*b*c^2*d^2 - a^2*c*d^3)*x 
^2)*sqrt(-c*d)*log((d*x^2 - 2*sqrt(-c*d)*x - c)/(d*x^2 + c)) + 2*(5*b^2*c^ 
4*d - 6*a*b*c^3*d^2 + a^2*c^2*d^3)*x)/(b^3*c^7*d - 3*a*b^2*c^6*d^2 + 3*a^2 
*b*c^5*d^3 - a^3*c^4*d^4 + (b^3*c^5*d^3 - 3*a*b^2*c^4*d^4 + 3*a^2*b*c^3*d^ 
5 - a^3*c^2*d^6)*x^4 + 2*(b^3*c^6*d^2 - 3*a*b^2*c^5*d^3 + 3*a^2*b*c^4*d^4 
- a^3*c^3*d^5)*x^2), 1/8*((3*b^2*c^3*d^2 - 2*a*b*c^2*d^3 - a^2*c*d^4)*x^3 
+ (3*b^2*c^4 + 6*a*b*c^3*d - a^2*c^2*d^2 + (3*b^2*c^2*d^2 + 6*a*b*c*d^3 - 
a^2*d^4)*x^4 + 2*(3*b^2*c^3*d + 6*a*b*c^2*d^2 - a^2*c*d^3)*x^2)*sqrt(c*d)* 
arctan(sqrt(c*d)*x/c) - 4*(b*c^2*d^3*x^4 + 2*b*c^3*d^2*x^2 + b*c^4*d)*sqrt 
(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + (5*b^2*c^4*d - 6*a* 
b*c^3*d^2 + a^2*c^2*d^3)*x)/(b^3*c^7*d - 3*a*b^2*c^6*d^2 + 3*a^2*b*c^5*d^3 
 - a^3*c^4*d^4 + (b^3*c^5*d^3 - 3*a*b^2*c^4*d^4 + 3*a^2*b*c^3*d^5 - a^3*c^ 
2*d^6)*x^4 + 2*(b^3*c^6*d^2 - 3*a*b^2*c^5*d^3 + 3*a^2*b*c^4*d^4 - a^3*c^3* 
d^5)*x^2), 1/16*(2*(3*b^2*c^3*d^2 - 2*a*b*c^2*d^3 - a^2*c*d^4)*x^3 - 16*(b 
*c^2*d^3*x^4 + 2*b*c^3*d^2*x^2 + b*c^4*d)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) 
- (3*b^2*c^4 + 6*a*b*c^3*d - a^2*c^2*d^2 + (3*b^2*c^2*d^2 + 6*a*b*c*d^3 - 
a^2*d^4)*x^4 + 2*(3*b^2*c^3*d + 6*a*b*c^2*d^2 - a^2*c*d^3)*x^2)*sqrt(-c...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.72 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=-\frac {a b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b}} + \frac {{\left (3 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} \sqrt {c d}} + \frac {{\left (3 \, b c d + a d^{2}\right )} x^{3} + {\left (5 \, b c^{2} - a c d\right )} x}{8 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2} + {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{4} + 2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} x^{2}\right )}} \] Input:

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

Output:

-a*b^2*arctan(b*x/sqrt(a*b))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a 
^3*d^3)*sqrt(a*b)) + 1/8*(3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*arctan(d*x/sqrt 
(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(c*d)) 
 + 1/8*((3*b*c*d + a*d^2)*x^3 + (5*b*c^2 - a*c*d)*x)/(b^2*c^5 - 2*a*b*c^4* 
d + a^2*c^3*d^2 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^4 + 2*(b^2*c 
^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*x^2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.33 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=-\frac {a b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b}} + \frac {{\left (3 \, b^{2} c^{2} + 6 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} \sqrt {c d}} + \frac {3 \, b c d x^{3} + a d^{2} x^{3} + 5 \, b c^{2} x - a c d x}{8 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} {\left (d x^{2} + c\right )}^{2}} \] Input:

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

Output:

-a*b^2*arctan(b*x/sqrt(a*b))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a 
^3*d^3)*sqrt(a*b)) + 1/8*(3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*arctan(d*x/sqrt 
(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(c*d)) 
 + 1/8*(3*b*c*d*x^3 + a*d^2*x^3 + 5*b*c^2*x - a*c*d*x)/((b^2*c^3 - 2*a*b*c 
^2*d + a^2*c*d^2)*(d*x^2 + c)^2)
 

Mupad [B] (verification not implemented)

Time = 2.33 (sec) , antiderivative size = 5898, normalized size of antiderivative = 38.05 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((-a*b^3)^(1/2)*((((160*a*b^9*c^8*d^2 - 32*a^8*b^2*c*d^9 - 992*a^2* 
b^8*c^7*d^3 + 2592*a^3*b^7*c^6*d^4 - 3680*a^4*b^6*c^5*d^5 + 3040*a^5*b^5*c 
^4*d^6 - 1440*a^6*b^4*c^3*d^7 + 352*a^7*b^3*c^2*d^8)/(64*(b^6*c^8 + a^6*c^ 
2*d^6 - 6*a^5*b*c^3*d^5 + 15*a^2*b^4*c^6*d^2 - 20*a^3*b^3*c^5*d^3 + 15*a^4 
*b^2*c^4*d^4 - 6*a*b^5*c^7*d)) - (x*(-a*b^3)^(1/2)*(256*b^9*c^9*d^2 - 1280 
*a*b^8*c^8*d^3 + 2304*a^2*b^7*c^7*d^4 - 1280*a^3*b^6*c^6*d^5 - 1280*a^4*b^ 
5*c^5*d^6 + 2304*a^5*b^4*c^4*d^7 - 1280*a^6*b^3*c^3*d^8 + 256*a^7*b^2*c^2* 
d^9))/(64*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2)*(b^4*c^6 + a 
^4*c^2*d^4 - 4*a^3*b*c^3*d^3 + 6*a^2*b^2*c^4*d^2 - 4*a*b^3*c^5*d)))*(-a*b^ 
3)^(1/2))/(2*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2)) - (x*(9* 
b^7*c^4*d + a^4*b^3*d^5 + 36*a*b^6*c^3*d^2 - 12*a^3*b^4*c*d^4 + 94*a^2*b^5 
*c^2*d^3))/(32*(b^4*c^6 + a^4*c^2*d^4 - 4*a^3*b*c^3*d^3 + 6*a^2*b^2*c^4*d^ 
2 - 4*a*b^3*c^5*d)))*1i)/(2*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c 
*d^2)) - ((-a*b^3)^(1/2)*((((160*a*b^9*c^8*d^2 - 32*a^8*b^2*c*d^9 - 992*a^ 
2*b^8*c^7*d^3 + 2592*a^3*b^7*c^6*d^4 - 3680*a^4*b^6*c^5*d^5 + 3040*a^5*b^5 
*c^4*d^6 - 1440*a^6*b^4*c^3*d^7 + 352*a^7*b^3*c^2*d^8)/(64*(b^6*c^8 + a^6* 
c^2*d^6 - 6*a^5*b*c^3*d^5 + 15*a^2*b^4*c^6*d^2 - 20*a^3*b^3*c^5*d^3 + 15*a 
^4*b^2*c^4*d^4 - 6*a*b^5*c^7*d)) + (x*(-a*b^3)^(1/2)*(256*b^9*c^9*d^2 - 12 
80*a*b^8*c^8*d^3 + 2304*a^2*b^7*c^7*d^4 - 1280*a^3*b^6*c^6*d^5 - 1280*a^4* 
b^5*c^5*d^6 + 2304*a^5*b^4*c^4*d^7 - 1280*a^6*b^3*c^3*d^8 + 256*a^7*b^2...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 550, normalized size of antiderivative = 3.55 \[ \int \frac {x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx=\frac {8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,c^{4} d +16 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,c^{3} d^{2} x^{2}+8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b \,c^{2} d^{3} x^{4}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} c^{2} d^{2}+2 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} c \,d^{3} x^{2}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} d^{4} x^{4}-6 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a b \,c^{3} d -12 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a b \,c^{2} d^{2} x^{2}-6 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a b c \,d^{3} x^{4}-3 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{4}-6 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{3} d \,x^{2}-3 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) b^{2} c^{2} d^{2} x^{4}-a^{2} c^{2} d^{3} x +a^{2} c \,d^{4} x^{3}+6 a b \,c^{3} d^{2} x +2 a b \,c^{2} d^{3} x^{3}-5 b^{2} c^{4} d x -3 b^{2} c^{3} d^{2} x^{3}}{8 c^{2} d \left (a^{3} d^{5} x^{4}-3 a^{2} b c \,d^{4} x^{4}+3 a \,b^{2} c^{2} d^{3} x^{4}-b^{3} c^{3} d^{2} x^{4}+2 a^{3} c \,d^{4} x^{2}-6 a^{2} b \,c^{2} d^{3} x^{2}+6 a \,b^{2} c^{3} d^{2} x^{2}-2 b^{3} c^{4} d \,x^{2}+a^{3} c^{2} d^{3}-3 a^{2} b \,c^{3} d^{2}+3 a \,b^{2} c^{4} d -b^{3} c^{5}\right )} \] Input:

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

Output:

(8*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b*c**4*d + 16*sqrt(b)*sqr 
t(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b*c**3*d**2*x**2 + 8*sqrt(b)*sqrt(a)*at 
an((b*x)/(sqrt(b)*sqrt(a)))*b*c**2*d**3*x**4 + sqrt(d)*sqrt(c)*atan((d*x)/ 
(sqrt(d)*sqrt(c)))*a**2*c**2*d**2 + 2*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)* 
sqrt(c)))*a**2*c*d**3*x**2 + sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c))) 
*a**2*d**4*x**4 - 6*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b*c**3 
*d - 12*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b*c**2*d**2*x**2 - 
 6*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b*c*d**3*x**4 - 3*sqrt( 
d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**2*c**4 - 6*sqrt(d)*sqrt(c)*ata 
n((d*x)/(sqrt(d)*sqrt(c)))*b**2*c**3*d*x**2 - 3*sqrt(d)*sqrt(c)*atan((d*x) 
/(sqrt(d)*sqrt(c)))*b**2*c**2*d**2*x**4 - a**2*c**2*d**3*x + a**2*c*d**4*x 
**3 + 6*a*b*c**3*d**2*x + 2*a*b*c**2*d**3*x**3 - 5*b**2*c**4*d*x - 3*b**2* 
c**3*d**2*x**3)/(8*c**2*d*(a**3*c**2*d**3 + 2*a**3*c*d**4*x**2 + a**3*d**5 
*x**4 - 3*a**2*b*c**3*d**2 - 6*a**2*b*c**2*d**3*x**2 - 3*a**2*b*c*d**4*x** 
4 + 3*a*b**2*c**4*d + 6*a*b**2*c**3*d**2*x**2 + 3*a*b**2*c**2*d**3*x**4 - 
b**3*c**5 - 2*b**3*c**4*d*x**2 - b**3*c**3*d**2*x**4))