\(\int \frac {x^3}{(c+d x) (a+b x^2)^2} \, dx\) [190]

Optimal result

Integrand size = 20, antiderivative size = 142 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {a (c-d x)}{2 b \left (b c^2+a d^2\right ) \left (a+b x^2\right )}+\frac {\sqrt {a} d \left (3 b c^2+a d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{3/2} \left (b c^2+a d^2\right )^2}-\frac {c^3 \log (c+d x)}{\left (b c^2+a d^2\right )^2}+\frac {c^3 \log \left (a+b x^2\right )}{2 \left (b c^2+a d^2\right )^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {\sqrt {a} d \left (3 b c^2+a d^2\right ) \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+\sqrt {b} \left (a \left (b c^2+a d^2\right ) (c-d x)-2 b c^3 \left (a+b x^2\right ) \log (c+d x)+b c^3 \left (a+b x^2\right ) \log \left (a+b x^2\right )\right )}{2 b^{3/2} \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {601, 25, 27, 657, 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.

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

\(\Big \downarrow \) 601

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 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 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {\frac {-\frac {a d \left (a \,d^{2}+b \,c^{2}\right ) x}{2 b}+\frac {a c \left (a \,d^{2}+b \,c^{2}\right )}{2 b}}{b \,x^{2}+a}+\frac {c^{3} b \ln \left (b \,x^{2}+a \right )+\frac {\left (a^{2} d^{3}+3 a b \,c^{2} d \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{2 b}}{\left (a \,d^{2}+b \,c^{2}\right )^{2}}-\frac {c^{3} \ln \left (d x +c \right )}{\left (a \,d^{2}+b \,c^{2}\right )^{2}}\)	\(139\)
risch	\(\frac {-\frac {a d x}{2 b \left (a \,d^{2}+b \,c^{2}\right )}+\frac {a c}{2 b \left (a \,d^{2}+b \,c^{2}\right )}}{b \,x^{2}+a}-\frac {c^{3} \ln \left (d x +c \right )}{a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{2} b^{3} d^{4}+2 a \,b^{4} c^{2} d^{2}+b^{5} c^{4}\right ) \textit {\_Z}^{2}-4 b^{3} c^{3} \textit {\_Z} +a \,d^{2}+4 b \,c^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 a^{3} b^{3} d^{6}+5 a^{2} b^{4} c^{2} d^{4}+a \,b^{5} c^{4} d^{2}-b^{6} c^{6}\right ) \textit {\_R}^{2}+\left (2 a^{2} b^{2} c \,d^{4}-2 b^{4} c^{5}\right ) \textit {\_R} +2 a^{2} d^{4}+8 b \,c^{2} d^{2} a +8 b^{2} c^{4}\right ) x +\left (4 a^{3} b^{3} c \,d^{5}+8 a^{2} b^{4} c^{3} d^{3}+4 a \,b^{5} c^{5} d \right ) \textit {\_R}^{2}+\left (-a^{3} b \,d^{5}-2 a^{2} b^{2} c^{2} d^{3}-a \,b^{3} c^{4} d \right ) \textit {\_R} +2 d^{3} c \,a^{2}+4 a b \,c^{3} d \right )\right )}{4}\)	\(352\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 445, normalized size of antiderivative = 3.13 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^2} \, dx=\left [\frac {2 \, a b c^{3} + 2 \, a^{2} c d^{2} + {\left (3 \, a b c^{2} d + a^{2} d^{3} + {\left (3 \, b^{2} c^{2} d + a b d^{3}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 2 \, {\left (a b c^{2} d + a^{2} d^{3}\right )} x + 2 \, {\left (b^{2} c^{3} x^{2} + a b c^{3}\right )} \log \left (b x^{2} + a\right ) - 4 \, {\left (b^{2} c^{3} x^{2} + a b c^{3}\right )} \log \left (d x + c\right )}{4 \, {\left (a b^{3} c^{4} + 2 \, a^{2} b^{2} c^{2} d^{2} + a^{3} b d^{4} + {\left (b^{4} c^{4} + 2 \, a b^{3} c^{2} d^{2} + a^{2} b^{2} d^{4}\right )} x^{2}\right )}}, \frac {a b c^{3} + a^{2} c d^{2} + {\left (3 \, a b c^{2} d + a^{2} d^{3} + {\left (3 \, b^{2} c^{2} d + a b d^{3}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - {\left (a b c^{2} d + a^{2} d^{3}\right )} x + {\left (b^{2} c^{3} x^{2} + a b c^{3}\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b^{2} c^{3} x^{2} + a b c^{3}\right )} \log \left (d x + c\right )}{2 \, {\left (a b^{3} c^{4} + 2 \, a^{2} b^{2} c^{2} d^{2} + a^{3} b d^{4} + {\left (b^{4} c^{4} + 2 \, a b^{3} c^{2} d^{2} + a^{2} b^{2} d^{4}\right )} x^{2}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.35 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {c^{3} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )}} - \frac {c^{3} \log \left (d x + c\right )}{b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}} + \frac {{\left (3 \, a b c^{2} d + a^{2} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (b^{3} c^{4} + 2 \, a b^{2} c^{2} d^{2} + a^{2} b d^{4}\right )} \sqrt {a b}} - \frac {a d x - a c}{2 \, {\left (a b^{2} c^{2} + a^{2} b d^{2} + {\left (b^{3} c^{2} + a b^{2} d^{2}\right )} x^{2}\right )}} \] Input:

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

Output:

1/2*c^3*log(b*x^2 + a)/(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4) - c^3*log(d*x + 
 c)/(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4) + 1/2*(3*a*b*c^2*d + a^2*d^3)*arct 
an(b*x/sqrt(a*b))/((b^3*c^4 + 2*a*b^2*c^2*d^2 + a^2*b*d^4)*sqrt(a*b)) - 1/ 
2*(a*d*x - a*c)/(a*b^2*c^2 + a^2*b*d^2 + (b^3*c^2 + a*b^2*d^2)*x^2)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.44 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^2} \, dx=-\frac {c^{3} d \log \left ({\left | d x + c \right |}\right )}{b^{2} c^{4} d + 2 \, a b c^{2} d^{3} + a^{2} d^{5}} + \frac {c^{3} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )}} + \frac {{\left (3 \, a b c^{2} d + a^{2} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (b^{3} c^{4} + 2 \, a b^{2} c^{2} d^{2} + a^{2} b d^{4}\right )} \sqrt {a b}} + \frac {a b c^{3} + a^{2} c d^{2} - {\left (a b c^{2} d + a^{2} d^{3}\right )} x}{2 \, {\left (b c^{2} + a d^{2}\right )}^{2} {\left (b x^{2} + a\right )} b} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.41 (sec) , antiderivative size = 603, normalized size of antiderivative = 4.25 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {\frac {a\,c}{2\,b\,\left (b\,c^2+a\,d^2\right )}-\frac {a\,d\,x}{2\,b\,\left (b\,c^2+a\,d^2\right )}}{b\,x^2+a}+\frac {\ln \left (36\,b^8\,c^{10}\,x+36\,b^6\,c^{10}\,\sqrt {-a\,b^3}+a^5\,b^3\,d^{10}\,x+a^5\,b\,d^{10}\,\sqrt {-a\,b^3}-22\,a^2\,c^4\,d^6\,{\left (-a\,b^3\right )}^{3/2}-81\,b^2\,c^8\,d^2\,{\left (-a\,b^3\right )}^{3/2}+8\,a^4\,b^2\,c^2\,d^8\,\sqrt {-a\,b^3}+81\,a\,b^7\,c^8\,d^2\,x-60\,a\,b\,c^6\,d^4\,{\left (-a\,b^3\right )}^{3/2}+60\,a^2\,b^6\,c^6\,d^4\,x+22\,a^3\,b^5\,c^4\,d^6\,x+8\,a^4\,b^4\,c^2\,d^8\,x\right )\,\left (2\,b^3\,c^3+a\,d^3\,\sqrt {-a\,b^3}+3\,b\,c^2\,d\,\sqrt {-a\,b^3}\right )}{4\,\left (a^2\,b^3\,d^4+2\,a\,b^4\,c^2\,d^2+b^5\,c^4\right )}-\frac {\ln \left (36\,b^8\,c^{10}\,x-36\,b^6\,c^{10}\,\sqrt {-a\,b^3}+a^5\,b^3\,d^{10}\,x-a^5\,b\,d^{10}\,\sqrt {-a\,b^3}+22\,a^2\,c^4\,d^6\,{\left (-a\,b^3\right )}^{3/2}+81\,b^2\,c^8\,d^2\,{\left (-a\,b^3\right )}^{3/2}-8\,a^4\,b^2\,c^2\,d^8\,\sqrt {-a\,b^3}+81\,a\,b^7\,c^8\,d^2\,x+60\,a\,b\,c^6\,d^4\,{\left (-a\,b^3\right )}^{3/2}+60\,a^2\,b^6\,c^6\,d^4\,x+22\,a^3\,b^5\,c^4\,d^6\,x+8\,a^4\,b^4\,c^2\,d^8\,x\right )\,\left (a\,d^3\,\sqrt {-a\,b^3}-2\,b^3\,c^3+3\,b\,c^2\,d\,\sqrt {-a\,b^3}\right )}{4\,\left (a^2\,b^3\,d^4+2\,a\,b^4\,c^2\,d^2+b^5\,c^4\right )}-\frac {c^3\,\ln \left (c+d\,x\right )}{{\left (b\,c^2+a\,d^2\right )}^2} \] Input:

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

Output:

((a*c)/(2*b*(a*d^2 + b*c^2)) - (a*d*x)/(2*b*(a*d^2 + b*c^2)))/(a + b*x^2) 
+ (log(36*b^8*c^10*x + 36*b^6*c^10*(-a*b^3)^(1/2) + a^5*b^3*d^10*x + a^5*b 
*d^10*(-a*b^3)^(1/2) - 22*a^2*c^4*d^6*(-a*b^3)^(3/2) - 81*b^2*c^8*d^2*(-a* 
b^3)^(3/2) + 8*a^4*b^2*c^2*d^8*(-a*b^3)^(1/2) + 81*a*b^7*c^8*d^2*x - 60*a* 
b*c^6*d^4*(-a*b^3)^(3/2) + 60*a^2*b^6*c^6*d^4*x + 22*a^3*b^5*c^4*d^6*x + 8 
*a^4*b^4*c^2*d^8*x)*(2*b^3*c^3 + a*d^3*(-a*b^3)^(1/2) + 3*b*c^2*d*(-a*b^3) 
^(1/2)))/(4*(b^5*c^4 + a^2*b^3*d^4 + 2*a*b^4*c^2*d^2)) - (log(36*b^8*c^10* 
x - 36*b^6*c^10*(-a*b^3)^(1/2) + a^5*b^3*d^10*x - a^5*b*d^10*(-a*b^3)^(1/2 
) + 22*a^2*c^4*d^6*(-a*b^3)^(3/2) + 81*b^2*c^8*d^2*(-a*b^3)^(3/2) - 8*a^4* 
b^2*c^2*d^8*(-a*b^3)^(1/2) + 81*a*b^7*c^8*d^2*x + 60*a*b*c^6*d^4*(-a*b^3)^ 
(3/2) + 60*a^2*b^6*c^6*d^4*x + 22*a^3*b^5*c^4*d^6*x + 8*a^4*b^4*c^2*d^8*x) 
*(a*d^3*(-a*b^3)^(1/2) - 2*b^3*c^3 + 3*b*c^2*d*(-a*b^3)^(1/2)))/(4*(b^5*c^ 
4 + a^2*b^3*d^4 + 2*a*b^4*c^2*d^2)) - (c^3*log(c + d*x))/(a*d^2 + b*c^2)^2
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.99 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^2} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} d^{3}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,c^{2} d +\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,d^{3} x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c^{2} d \,x^{2}+\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} c^{3}+\mathrm {log}\left (b \,x^{2}+a \right ) b^{3} c^{3} x^{2}-2 \,\mathrm {log}\left (d x +c \right ) a \,b^{2} c^{3}-2 \,\mathrm {log}\left (d x +c \right ) b^{3} c^{3} x^{2}-a^{2} b \,d^{3} x -a \,b^{2} c^{2} d x -a \,b^{2} c \,d^{2} x^{2}-b^{3} c^{3} x^{2}}{2 b^{2} \left (a^{2} b \,d^{4} x^{2}+2 a \,b^{2} c^{2} d^{2} x^{2}+b^{3} c^{4} x^{2}+a^{3} d^{4}+2 a^{2} b \,c^{2} d^{2}+a \,b^{2} c^{4}\right )} \] Input:

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

Output:

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