\(\int \frac {x^2}{(c+d x)^2 (a+b x^2)^2} \, dx\) [202]

Optimal result

Integrand size = 20, antiderivative size = 225 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^2} \, dx=-\frac {c^2 d}{\left (b c^2+a d^2\right )^2 (c+d x)}-\frac {2 a c d \left (b c^2+a d^2\right )+\left (b^2 c^4-a^2 d^4\right ) x}{2 \left (b c^2+a d^2\right )^3 \left (a+b x^2\right )}+\frac {\left (b^2 c^4-6 a b c^2 d^2+a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} \sqrt {b} \left (b c^2+a d^2\right )^3}+\frac {2 c d \left (b c^2-a d^2\right ) \log (c+d x)}{\left (b c^2+a d^2\right )^3}-\frac {c d \left (b c^2-a d^2\right ) \log \left (a+b x^2\right )}{\left (b c^2+a d^2\right )^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^2} \, dx=\frac {-\frac {2 c^2 d \left (b c^2+a d^2\right )}{c+d x}+\frac {\left (b c^2+a d^2\right ) \left (-b c^2 x+a d (-2 c+d x)\right )}{a+b x^2}+\frac {\left (b^2 c^4-6 a b c^2 d^2+a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+4 \left (b c^3 d-a c d^3\right ) \log (c+d x)-2 \left (b c^3 d-a c d^3\right ) \log \left (a+b x^2\right )}{2 \left (b c^2+a d^2\right )^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {601, 25, 2160, 2009}

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.

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

\(\Big \downarrow \) 601

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2160

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 601

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2160

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] 
:> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, 
 d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {\frac {\left (\frac {a^{2} d^{4}}{2}-\frac {b^{2} c^{4}}{2}\right ) x -a c d \left (a \,d^{2}+b \,c^{2}\right )}{b \,x^{2}+a}+\frac {\left (4 a b c \,d^{3}-4 b^{2} c^{3} d \right ) \ln \left (b \,x^{2}+a \right )}{4 b}+\frac {\left (a^{2} d^{4}-6 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}}{\left (a \,d^{2}+b \,c^{2}\right )^{3}}-\frac {c^{2} d}{\left (a \,d^{2}+b \,c^{2}\right )^{2} \left (d x +c \right )}-\frac {2 c d \left (a \,d^{2}-b \,c^{2}\right ) \ln \left (d x +c \right )}{\left (a \,d^{2}+b \,c^{2}\right )^{3}}\)	\(196\)
risch	\(\text {Expression too large to display}\)	\(3192\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 677 vs. \(2 (213) = 426\).

Time = 1.02 (sec) , antiderivative size = 1375, normalized size of antiderivative = 6.11 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.69 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (b c^{3} d - a c d^{3}\right )} \log \left (b x^{2} + a\right )}{b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}} + \frac {2 \, {\left (b c^{3} d - a c d^{3}\right )} \log \left (d x + c\right )}{b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}} + \frac {{\left (b^{2} c^{4} - 6 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}\right )} \sqrt {a b}} - \frac {4 \, a c^{2} d + {\left (3 \, b c^{2} d - a d^{3}\right )} x^{2} + {\left (b c^{3} + a c d^{2}\right )} x}{2 \, {\left (a b^{2} c^{5} + 2 \, a^{2} b c^{3} d^{2} + a^{3} c d^{4} + {\left (b^{3} c^{4} d + 2 \, a b^{2} c^{2} d^{3} + a^{2} b d^{5}\right )} x^{3} + {\left (b^{3} c^{5} + 2 \, a b^{2} c^{3} d^{2} + a^{2} b c d^{4}\right )} x^{2} + {\left (a b^{2} c^{4} d + 2 \, a^{2} b c^{2} d^{3} + a^{3} d^{5}\right )} x\right )}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 399, normalized size of antiderivative = 1.77 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^2} \, dx=-\frac {c^{2} d^{5}}{{\left (b^{2} c^{4} d^{4} + 2 \, a b c^{2} d^{6} + a^{2} d^{8}\right )} {\left (d x + c\right )}} - \frac {{\left (b c^{3} d - a c d^{3}\right )} \log \left (b - \frac {2 \, b c}{d x + c} + \frac {b c^{2}}{{\left (d x + c\right )}^{2}} + \frac {a d^{2}}{{\left (d x + c\right )}^{2}}\right )}{b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}} + \frac {{\left (b^{2} c^{4} d^{2} - 6 \, a b c^{2} d^{4} + a^{2} d^{6}\right )} \arctan \left (\frac {b c - \frac {b c^{2}}{d x + c} - \frac {a d^{2}}{d x + c}}{\sqrt {a b} d}\right )}{2 \, {\left (b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}\right )} \sqrt {a b} d^{2}} - \frac {\frac {b^{2} c^{3} d - 3 \, a b c d^{3}}{b c^{2} + a d^{2}} - \frac {b^{2} c^{4} d^{2} - 6 \, a b c^{2} d^{4} + a^{2} d^{6}}{{\left (b c^{2} + a d^{2}\right )} {\left (d x + c\right )} d}}{2 \, {\left (b c^{2} + a d^{2}\right )}^{2} {\left (b - \frac {2 \, b c}{d x + c} + \frac {b c^{2}}{{\left (d x + c\right )}^{2}} + \frac {a d^{2}}{{\left (d x + c\right )}^{2}}\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 7.64 (sec) , antiderivative size = 806, normalized size of antiderivative = 3.58 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^2} \, dx=\frac {\ln \left (a^7\,d^{12}\,\sqrt {-a\,b}-b^5\,c^{12}\,{\left (-a\,b\right )}^{3/2}+15\,a^3\,c^4\,d^8\,{\left (-a\,b\right )}^{5/2}-134\,a^5\,c^2\,d^{10}\,{\left (-a\,b\right )}^{3/2}+134\,b^3\,c^{10}\,d^2\,{\left (-a\,b\right )}^{5/2}+a\,b^7\,c^{12}\,x+a^7\,b\,d^{12}\,x+236\,a\,c^6\,d^6\,{\left (-a\,b\right )}^{7/2}-15\,b\,c^8\,d^4\,{\left (-a\,b\right )}^{7/2}+134\,a^2\,b^6\,c^{10}\,d^2\,x+15\,a^3\,b^5\,c^8\,d^4\,x-236\,a^4\,b^4\,c^6\,d^6\,x+15\,a^5\,b^3\,c^4\,d^8\,x+134\,a^6\,b^2\,c^2\,d^{10}\,x\right )\,\left (a^2\,\left (\frac {d^4\,\sqrt {-a\,b}}{4}+b\,c\,d^3\right )-a\,\left (b^2\,c^3\,d+\frac {3\,b\,c^2\,d^2\,\sqrt {-a\,b}}{2}\right )+\frac {b^2\,c^4\,\sqrt {-a\,b}}{4}\right )}{a^4\,b\,d^6+3\,a^3\,b^2\,c^2\,d^4+3\,a^2\,b^3\,c^4\,d^2+a\,b^4\,c^6}-\frac {\ln \left (c+d\,x\right )\,\left (2\,a\,c\,d^3-2\,b\,c^3\,d\right )}{a^3\,d^6+3\,a^2\,b\,c^2\,d^4+3\,a\,b^2\,c^4\,d^2+b^3\,c^6}-\frac {\ln \left (b^5\,c^{12}\,{\left (-a\,b\right )}^{3/2}-a^7\,d^{12}\,\sqrt {-a\,b}-15\,a^3\,c^4\,d^8\,{\left (-a\,b\right )}^{5/2}+134\,a^5\,c^2\,d^{10}\,{\left (-a\,b\right )}^{3/2}-134\,b^3\,c^{10}\,d^2\,{\left (-a\,b\right )}^{5/2}+a\,b^7\,c^{12}\,x+a^7\,b\,d^{12}\,x-236\,a\,c^6\,d^6\,{\left (-a\,b\right )}^{7/2}+15\,b\,c^8\,d^4\,{\left (-a\,b\right )}^{7/2}+134\,a^2\,b^6\,c^{10}\,d^2\,x+15\,a^3\,b^5\,c^8\,d^4\,x-236\,a^4\,b^4\,c^6\,d^6\,x+15\,a^5\,b^3\,c^4\,d^8\,x+134\,a^6\,b^2\,c^2\,d^{10}\,x\right )\,\left (b^2\,\left (\frac {c^4\,\sqrt {-a\,b}}{4}+a\,c^3\,d\right )-b\,\left (a^2\,c\,d^3+\frac {3\,a\,c^2\,d^2\,\sqrt {-a\,b}}{2}\right )+\frac {a^2\,d^4\,\sqrt {-a\,b}}{4}\right )}{a^4\,b\,d^6+3\,a^3\,b^2\,c^2\,d^4+3\,a^2\,b^3\,c^4\,d^2+a\,b^4\,c^6}-\frac {\frac {c\,x}{2\,\left (b\,c^2+a\,d^2\right )}-\frac {x^2\,\left (a\,d^3-3\,b\,c^2\,d\right )}{2\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}+\frac {2\,a\,c^2\,d}{{\left (b\,c^2+a\,d^2\right )}^2}}{b\,d\,x^3+b\,c\,x^2+a\,d\,x+a\,c} \] Input:

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

Output:

(log(a^7*d^12*(-a*b)^(1/2) - b^5*c^12*(-a*b)^(3/2) + 15*a^3*c^4*d^8*(-a*b) 
^(5/2) - 134*a^5*c^2*d^10*(-a*b)^(3/2) + 134*b^3*c^10*d^2*(-a*b)^(5/2) + a 
*b^7*c^12*x + a^7*b*d^12*x + 236*a*c^6*d^6*(-a*b)^(7/2) - 15*b*c^8*d^4*(-a 
*b)^(7/2) + 134*a^2*b^6*c^10*d^2*x + 15*a^3*b^5*c^8*d^4*x - 236*a^4*b^4*c^ 
6*d^6*x + 15*a^5*b^3*c^4*d^8*x + 134*a^6*b^2*c^2*d^10*x)*(a^2*((d^4*(-a*b) 
^(1/2))/4 + b*c*d^3) - a*(b^2*c^3*d + (3*b*c^2*d^2*(-a*b)^(1/2))/2) + (b^2 
*c^4*(-a*b)^(1/2))/4))/(a*b^4*c^6 + a^4*b*d^6 + 3*a^2*b^3*c^4*d^2 + 3*a^3* 
b^2*c^2*d^4) - (log(c + d*x)*(2*a*c*d^3 - 2*b*c^3*d))/(a^3*d^6 + b^3*c^6 + 
 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4) - (log(b^5*c^12*(-a*b)^(3/2) - a^7*d^1 
2*(-a*b)^(1/2) - 15*a^3*c^4*d^8*(-a*b)^(5/2) + 134*a^5*c^2*d^10*(-a*b)^(3/ 
2) - 134*b^3*c^10*d^2*(-a*b)^(5/2) + a*b^7*c^12*x + a^7*b*d^12*x - 236*a*c 
^6*d^6*(-a*b)^(7/2) + 15*b*c^8*d^4*(-a*b)^(7/2) + 134*a^2*b^6*c^10*d^2*x + 
 15*a^3*b^5*c^8*d^4*x - 236*a^4*b^4*c^6*d^6*x + 15*a^5*b^3*c^4*d^8*x + 134 
*a^6*b^2*c^2*d^10*x)*(b^2*((c^4*(-a*b)^(1/2))/4 + a*c^3*d) - b*(a^2*c*d^3 
+ (3*a*c^2*d^2*(-a*b)^(1/2))/2) + (a^2*d^4*(-a*b)^(1/2))/4))/(a*b^4*c^6 + 
a^4*b*d^6 + 3*a^2*b^3*c^4*d^2 + 3*a^3*b^2*c^2*d^4) - ((c*x)/(2*(a*d^2 + b* 
c^2)) - (x^2*(a*d^3 - 3*b*c^2*d))/(2*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2)) 
+ (2*a*c^2*d)/(a*d^2 + b*c^2)^2)/(a*c + a*d*x + b*c*x^2 + b*d*x^3)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1004, normalized size of antiderivative = 4.46 \[ \int \frac {x^2}{(c+d x)^2 \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*c**2*d**4 + sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*c*d**5*x - 6*sqrt(b)*sqrt(a)*atan 
((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*c**4*d**2 - 6*sqrt(b)*sqrt(a)*atan((b*x)/ 
(sqrt(b)*sqrt(a)))*a**2*b*c**3*d**3*x + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b 
)*sqrt(a)))*a**2*b*c**2*d**4*x**2 + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sq 
rt(a)))*a**2*b*c*d**5*x**3 + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a))) 
*a*b**2*c**6 + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*c**5*d 
*x - 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*c**4*d**2*x**2 
 - 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*c**3*d**3*x**3 + 
 sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**3*c**6*x**2 + sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**3*c**5*d*x**3 + 2*log(a + b*x**2)*a 
**3*b*c**3*d**3 + 2*log(a + b*x**2)*a**3*b*c**2*d**4*x - 2*log(a + b*x**2) 
*a**2*b**2*c**5*d - 2*log(a + b*x**2)*a**2*b**2*c**4*d**2*x + 2*log(a + b* 
x**2)*a**2*b**2*c**3*d**3*x**2 + 2*log(a + b*x**2)*a**2*b**2*c**2*d**4*x** 
3 - 2*log(a + b*x**2)*a*b**3*c**5*d*x**2 - 2*log(a + b*x**2)*a*b**3*c**4*d 
**2*x**3 - 4*log(c + d*x)*a**3*b*c**3*d**3 - 4*log(c + d*x)*a**3*b*c**2*d* 
*4*x + 4*log(c + d*x)*a**2*b**2*c**5*d + 4*log(c + d*x)*a**2*b**2*c**4*d** 
2*x - 4*log(c + d*x)*a**2*b**2*c**3*d**3*x**2 - 4*log(c + d*x)*a**2*b**2*c 
**2*d**4*x**3 + 4*log(c + d*x)*a*b**3*c**5*d*x**2 + 4*log(c + d*x)*a*b**3* 
c**4*d**2*x**3 - a**4*c*d**5 - a**4*d**6*x - 2*a**3*b*c**3*d**3 + a**3*...