\(\int \frac {x^5}{(c+d x) (a+b x^2)^2} \, dx\) [188]

Optimal result

Integrand size = 20, antiderivative size = 170 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {x}{b^2 d}-\frac {a^2 (c-d x)}{2 b^2 \left (b c^2+a d^2\right ) \left (a+b x^2\right )}-\frac {a^{3/2} d \left (5 b c^2+3 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{5/2} \left (b c^2+a d^2\right )^2}-\frac {c^5 \log (c+d x)}{d^2 \left (b c^2+a d^2\right )^2}-\frac {a c \left (2 b c^2+a d^2\right ) \log \left (a+b x^2\right )}{2 b^2 \left (b c^2+a d^2\right )^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.98 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {1}{2} \left (\frac {2 x}{b^2 d}+\frac {a^2 (-c+d x)}{b^2 \left (b c^2+a d^2\right ) \left (a+b x^2\right )}-\frac {a^{3/2} d \left (5 b c^2+3 a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2} \left (b c^2+a d^2\right )^2}-\frac {2 c^5 \log (c+d x)}{\left (b c^2 d+a d^3\right )^2}-\frac {a c \left (2 b c^2+a d^2\right ) \log \left (a+b x^2\right )}{b^2 \left (b c^2+a d^2\right )^2}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {601, 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.

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

\(\Big \downarrow \) 601

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

\(\Big \downarrow \) 2160

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^2 (c-d x)}{2 b^2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )}-\frac {\frac {a^{5/2} d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (3 a d^2+5 b c^2\right )}{b^{5/2} \left (a d^2+b c^2\right )^2}+\frac {a^2 c \left (a d^2+2 b c^2\right ) \log \left (a+b x^2\right )}{b^2 \left (a d^2+b c^2\right )^2}-\frac {2 a x}{b^2 d}+\frac {2 a c^5 \log (c+d x)}{d^2 \left (a d^2+b c^2\right )^2}}{2 a}\)

Input:

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

Output:

-1/2*(a^2*(c - d*x))/(b^2*(b*c^2 + a*d^2)*(a + b*x^2)) - ((-2*a*x)/(b^2*d) 
 + (a^(5/2)*d*(5*b*c^2 + 3*a*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(b^(5/2)*(b 
*c^2 + a*d^2)^2) + (2*a*c^5*Log[c + d*x])/(d^2*(b*c^2 + a*d^2)^2) + (a^2*c 
*(2*b*c^2 + a*d^2)*Log[a + b*x^2])/(b^2*(b*c^2 + a*d^2)^2))/(2*a)
 

Defintions of rubi rules used

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 = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {x}{b^{2} d}-\frac {a \left (\frac {\left (-\frac {1}{2} a^{2} d^{3}-\frac {1}{2} a b \,c^{2} d \right ) x +\frac {a c \left (a \,d^{2}+b \,c^{2}\right )}{2}}{b \,x^{2}+a}+\frac {\left (2 a b c \,d^{2}+4 c^{3} b^{2}\right ) \ln \left (b \,x^{2}+a \right )}{4 b}+\frac {\left (3 a^{2} d^{3}+5 a b \,c^{2} d \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a \,d^{2}+b \,c^{2}\right )^{2} b^{2}}-\frac {c^{5} \ln \left (d x +c \right )}{d^{2} \left (a \,d^{2}+b \,c^{2}\right )^{2}}\)	\(165\)
risch	\(\frac {x}{b^{2} d}+\frac {\frac {a^{2} d^{2} x}{2 a \,d^{2}+2 b \,c^{2}}-\frac {c \,a^{2} d}{2 \left (a \,d^{2}+b \,c^{2}\right )}}{b^{2} d \left (b \,x^{2}+a \right )}-\frac {c^{5} \ln \left (d x +c \right )}{d^{2} \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{2} b \,d^{4}+2 a \,b^{2} c^{2} d^{2}+b^{3} c^{4}\right ) \textit {\_Z}^{2}+\left (4 a^{2} b c \,d^{3}+8 a \,b^{2} c^{3} d \right ) \textit {\_Z} +9 a^{3} d^{4}+16 a^{2} b \,c^{2} d^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 a^{3} b \,d^{7}+5 a^{2} b^{2} c^{2} d^{5}+a \,b^{3} c^{4} d^{3}-b^{4} c^{6} d \right ) \textit {\_R}^{2}+\left (4 a^{3} b c \,d^{6}+8 a^{2} b^{2} c^{3} d^{4}-4 b^{4} c^{7}\right ) \textit {\_R} +18 a^{4} d^{7}+32 a^{3} b \,c^{2} d^{5}-16 a \,b^{3} c^{6} d \right ) x +\left (4 a^{3} b c \,d^{6}+8 a^{2} b^{2} c^{3} d^{4}+4 a \,b^{3} c^{5} d^{2}\right ) \textit {\_R}^{2}+\left (3 a^{4} d^{7}+14 a^{3} b \,c^{2} d^{5}+19 a^{2} b^{2} c^{4} d^{3}+8 a \,b^{3} c^{6} d \right ) \textit {\_R} +18 a^{4} c \,d^{6}+32 a^{3} b \,c^{3} d^{4}+12 a^{2} b^{2} c^{5} d^{2}\right )}{4 b^{2} d}\)	\(441\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (157) = 314\).

Time = 0.78 (sec) , antiderivative size = 658, normalized size of antiderivative = 3.87 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^2} \, dx=\left [-\frac {2 \, a^{2} b c^{3} d^{2} + 2 \, a^{3} c d^{4} - 4 \, {\left (b^{3} c^{4} d + 2 \, a b^{2} c^{2} d^{3} + a^{2} b d^{5}\right )} x^{3} - {\left (5 \, a^{2} b c^{2} d^{3} + 3 \, a^{3} d^{5} + {\left (5 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b d^{5}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 2 \, {\left (2 \, a b^{2} c^{4} d + 5 \, a^{2} b c^{2} d^{3} + 3 \, a^{3} d^{5}\right )} x + 2 \, {\left (2 \, a^{2} b c^{3} d^{2} + a^{3} c d^{4} + {\left (2 \, a b^{2} c^{3} d^{2} + a^{2} b c d^{4}\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left (b^{3} c^{5} x^{2} + a b^{2} c^{5}\right )} \log \left (d x + c\right )}{4 \, {\left (a b^{4} c^{4} d^{2} + 2 \, a^{2} b^{3} c^{2} d^{4} + a^{3} b^{2} d^{6} + {\left (b^{5} c^{4} d^{2} + 2 \, a b^{4} c^{2} d^{4} + a^{2} b^{3} d^{6}\right )} x^{2}\right )}}, -\frac {a^{2} b c^{3} d^{2} + a^{3} c d^{4} - 2 \, {\left (b^{3} c^{4} d + 2 \, a b^{2} c^{2} d^{3} + a^{2} b d^{5}\right )} x^{3} + {\left (5 \, a^{2} b c^{2} d^{3} + 3 \, a^{3} d^{5} + {\left (5 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b d^{5}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - {\left (2 \, a b^{2} c^{4} d + 5 \, a^{2} b c^{2} d^{3} + 3 \, a^{3} d^{5}\right )} x + {\left (2 \, a^{2} b c^{3} d^{2} + a^{3} c d^{4} + {\left (2 \, a b^{2} c^{3} d^{2} + a^{2} b c d^{4}\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) + 2 \, {\left (b^{3} c^{5} x^{2} + a b^{2} c^{5}\right )} \log \left (d x + c\right )}{2 \, {\left (a b^{4} c^{4} d^{2} + 2 \, a^{2} b^{3} c^{2} d^{4} + a^{3} b^{2} d^{6} + {\left (b^{5} c^{4} d^{2} + 2 \, a b^{4} c^{2} d^{4} + a^{2} b^{3} d^{6}\right )} x^{2}\right )}}\right ] \] Input:

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

Output:

[-1/4*(2*a^2*b*c^3*d^2 + 2*a^3*c*d^4 - 4*(b^3*c^4*d + 2*a*b^2*c^2*d^3 + a^ 
2*b*d^5)*x^3 - (5*a^2*b*c^2*d^3 + 3*a^3*d^5 + (5*a*b^2*c^2*d^3 + 3*a^2*b*d 
^5)*x^2)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) - 2*(2 
*a*b^2*c^4*d + 5*a^2*b*c^2*d^3 + 3*a^3*d^5)*x + 2*(2*a^2*b*c^3*d^2 + a^3*c 
*d^4 + (2*a*b^2*c^3*d^2 + a^2*b*c*d^4)*x^2)*log(b*x^2 + a) + 4*(b^3*c^5*x^ 
2 + a*b^2*c^5)*log(d*x + c))/(a*b^4*c^4*d^2 + 2*a^2*b^3*c^2*d^4 + a^3*b^2* 
d^6 + (b^5*c^4*d^2 + 2*a*b^4*c^2*d^4 + a^2*b^3*d^6)*x^2), -1/2*(a^2*b*c^3* 
d^2 + a^3*c*d^4 - 2*(b^3*c^4*d + 2*a*b^2*c^2*d^3 + a^2*b*d^5)*x^3 + (5*a^2 
*b*c^2*d^3 + 3*a^3*d^5 + (5*a*b^2*c^2*d^3 + 3*a^2*b*d^5)*x^2)*sqrt(a/b)*ar 
ctan(b*x*sqrt(a/b)/a) - (2*a*b^2*c^4*d + 5*a^2*b*c^2*d^3 + 3*a^3*d^5)*x + 
(2*a^2*b*c^3*d^2 + a^3*c*d^4 + (2*a*b^2*c^3*d^2 + a^2*b*c*d^4)*x^2)*log(b* 
x^2 + a) + 2*(b^3*c^5*x^2 + a*b^2*c^5)*log(d*x + c))/(a*b^4*c^4*d^2 + 2*a^ 
2*b^3*c^2*d^4 + a^3*b^2*d^6 + (b^5*c^4*d^2 + 2*a*b^4*c^2*d^4 + a^2*b^3*d^6 
)*x^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.36 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^2} \, dx=-\frac {c^{5} \log \left (d x + c\right )}{b^{2} c^{4} d^{2} + 2 \, a b c^{2} d^{4} + a^{2} d^{6}} - \frac {{\left (2 \, a b c^{3} + a^{2} c d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{4} c^{4} + 2 \, a b^{3} c^{2} d^{2} + a^{2} b^{2} d^{4}\right )}} - \frac {{\left (5 \, a^{2} b c^{2} d + 3 \, a^{3} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (b^{4} c^{4} + 2 \, a b^{3} c^{2} d^{2} + a^{2} b^{2} d^{4}\right )} \sqrt {a b}} + \frac {a^{2} d x - a^{2} c}{2 \, {\left (a b^{3} c^{2} + a^{2} b^{2} d^{2} + {\left (b^{4} c^{2} + a b^{3} d^{2}\right )} x^{2}\right )}} + \frac {x}{b^{2} d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.41 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^2} \, dx=-\frac {c^{5} \log \left ({\left | d x + c \right |}\right )}{b^{2} c^{4} d^{2} + 2 \, a b c^{2} d^{4} + a^{2} d^{6}} - \frac {{\left (2 \, a b c^{3} + a^{2} c d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{4} c^{4} + 2 \, a b^{3} c^{2} d^{2} + a^{2} b^{2} d^{4}\right )}} - \frac {{\left (5 \, a^{2} b c^{2} d + 3 \, a^{3} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (b^{4} c^{4} + 2 \, a b^{3} c^{2} d^{2} + a^{2} b^{2} d^{4}\right )} \sqrt {a b}} + \frac {x}{b^{2} d} - \frac {a^{2} b c^{3} + a^{3} c d^{2} - {\left (a^{2} b c^{2} d + a^{3} d^{3}\right )} x}{2 \, {\left (b c^{2} + a d^{2}\right )}^{2} {\left (b x^{2} + a\right )} b^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.24 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.96 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {\frac {a^2\,d^2\,x}{2\,\left (b\,c^2+a\,d^2\right )}-\frac {a^2\,c\,d}{2\,\left (b\,c^2+a\,d^2\right )}}{d\,b^3\,x^2+a\,d\,b^2}-\frac {\ln \left (\sqrt {-a^3\,b^5}+a\,b^3\,x\right )\,\left (4\,a\,b^4\,c^3+3\,a\,d^3\,\sqrt {-a^3\,b^5}+2\,a^2\,b^3\,c\,d^2+5\,b\,c^2\,d\,\sqrt {-a^3\,b^5}\right )}{4\,\left (a^2\,b^5\,d^4+2\,a\,b^6\,c^2\,d^2+b^7\,c^4\right )}-\frac {\ln \left (\sqrt {-a^3\,b^5}-a\,b^3\,x\right )\,\left (4\,a\,b^4\,c^3-3\,a\,d^3\,\sqrt {-a^3\,b^5}+2\,a^2\,b^3\,c\,d^2-5\,b\,c^2\,d\,\sqrt {-a^3\,b^5}\right )}{4\,\left (a^2\,b^5\,d^4+2\,a\,b^6\,c^2\,d^2+b^7\,c^4\right )}+\frac {x}{b^2\,d}-\frac {c^5\,\ln \left (c+d\,x\right )}{a^2\,d^6+2\,a\,b\,c^2\,d^4+b^2\,c^4\,d^2} \] Input:

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

Output:

((a^2*d^2*x)/(2*(a*d^2 + b*c^2)) - (a^2*c*d)/(2*(a*d^2 + b*c^2)))/(b^3*d*x 
^2 + a*b^2*d) - (log((-a^3*b^5)^(1/2) + a*b^3*x)*(4*a*b^4*c^3 + 3*a*d^3*(- 
a^3*b^5)^(1/2) + 2*a^2*b^3*c*d^2 + 5*b*c^2*d*(-a^3*b^5)^(1/2)))/(4*(b^7*c^ 
4 + a^2*b^5*d^4 + 2*a*b^6*c^2*d^2)) - (log((-a^3*b^5)^(1/2) - a*b^3*x)*(4* 
a*b^4*c^3 - 3*a*d^3*(-a^3*b^5)^(1/2) + 2*a^2*b^3*c*d^2 - 5*b*c^2*d*(-a^3*b 
^5)^(1/2)))/(4*(b^7*c^4 + a^2*b^5*d^4 + 2*a*b^6*c^2*d^2)) + x/(b^2*d) - (c 
^5*log(c + d*x))/(a^2*d^6 + b^2*c^4*d^2 + 2*a*b*c^2*d^4)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.41 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} d^{5}-5 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,c^{2} d^{3}-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,d^{5} x^{2}-5 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} c^{2} d^{3} x^{2}-\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b c \,d^{4}-2 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{2} c^{3} d^{2}-\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{2} c \,d^{4} x^{2}-2 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{3} c^{3} d^{2} x^{2}-2 \,\mathrm {log}\left (d x +c \right ) a \,b^{3} c^{5}-2 \,\mathrm {log}\left (d x +c \right ) b^{4} c^{5} x^{2}+3 a^{3} b \,d^{5} x +5 a^{2} b^{2} c^{2} d^{3} x +a^{2} b^{2} c \,d^{4} x^{2}+2 a^{2} b^{2} d^{5} x^{3}+2 a \,b^{3} c^{4} d x +a \,b^{3} c^{3} d^{2} x^{2}+4 a \,b^{3} c^{2} d^{3} x^{3}+2 b^{4} c^{4} d \,x^{3}}{2 b^{3} d^{2} \left (a^{2} b \,d^{4} x^{2}+2 a \,b^{2} c^{2} d^{2} x^{2}+b^{3} c^{4} x^{2}+a^{3} d^{4}+2 a^{2} b \,c^{2} d^{2}+a \,b^{2} c^{4}\right )} \] Input:

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

Output:

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