\(\int \frac {x^4}{(a+b x^3)^2} \, dx\) [131]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 136 \[ \int \frac {x^4}{\left (a+b x^3\right )^2} \, dx=-\frac {x^2}{3 b \left (a+b x^3\right )}-\frac {2 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{5/3}}-\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 \sqrt [3]{a} b^{5/3}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{9 \sqrt [3]{a} b^{5/3}} \] Output:

-1/3*x^2/b/(b*x^3+a)-2/9*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3)) 
*3^(1/2)/a^(1/3)/b^(5/3)-2/9*ln(a^(1/3)+b^(1/3)*x)/a^(1/3)/b^(5/3)+1/9*ln( 
a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(1/3)/b^(5/3)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88 \[ \int \frac {x^4}{\left (a+b x^3\right )^2} \, dx=\frac {-\frac {3 b^{2/3} x^2}{a+b x^3}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}-\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{a}}}{9 b^{5/3}} \] Input:

Integrate[x^4/(a + b*x^3)^2,x]
 

Output:

((-3*b^(2/3)*x^2)/(a + b*x^3) - (2*Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/ 
3))/Sqrt[3]])/a^(1/3) - (2*Log[a^(1/3) + b^(1/3)*x])/a^(1/3) + Log[a^(2/3) 
 - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/a^(1/3))/(9*b^(5/3))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {817, 821, 16, 1142, 25, 27, 1082, 217, 1103}

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.

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

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 821

\(\displaystyle \frac {2 \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{3 b}-\frac {x^2}{3 b \left (a+b x^3\right )}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {2 \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{3 b}-\frac {x^2}{3 b \left (a+b x^3\right )}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{3 b}-\frac {x^2}{3 b \left (a+b x^3\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{3 b}-\frac {x^2}{3 b \left (a+b x^3\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{3 b}-\frac {x^2}{3 b \left (a+b x^3\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2 \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{3 b}-\frac {x^2}{3 b \left (a+b x^3\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{3 b}-\frac {x^2}{3 b \left (a+b x^3\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{3 b}-\frac {x^2}{3 b \left (a+b x^3\right )}\)

Input:

Int[x^4/(a + b*x^3)^2,x]
 

Output:

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

Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 817
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 821
Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 
*x^2), x], x] /; FreeQ[{a, b}, x]
 

rule 1082
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 
Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.44 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.33

method result size
risch \(-\frac {x^{2}}{3 b \left (b \,x^{3}+a \right )}+\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{9 b^{2}}\) \(45\)
default \(-\frac {x^{2}}{3 b \left (b \,x^{3}+a \right )}+\frac {-\frac {2 \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}}{b}\) \(114\)

Input:

int(x^4/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/3*x^2/b/(b*x^3+a)+2/9/b^2*sum(1/_R*ln(x-_R),_R=RootOf(_Z^3*b+a))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.94 \[ \int \frac {x^4}{\left (a+b x^3\right )^2} \, dx=\left [-\frac {3 \, a b^{2} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} x}{b x^{3} + a}\right ) - {\left (b x^{3} + a\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) + 2 \, {\left (b x^{3} + a\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{9 \, {\left (a b^{4} x^{3} + a^{2} b^{3}\right )}}, -\frac {3 \, a b^{2} x^{2} - 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) - {\left (b x^{3} + a\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) + 2 \, {\left (b x^{3} + a\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{9 \, {\left (a b^{4} x^{3} + a^{2} b^{3}\right )}}\right ] \] Input:

integrate(x^4/(b*x^3+a)^2,x, algorithm="fricas")
 

Output:

[-1/9*(3*a*b^2*x^2 - 3*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt((-a*b^2)^(1/3)/a 
)*log((2*b^2*x^3 - a*b + 3*sqrt(1/3)*(a*b*x + 2*(-a*b^2)^(2/3)*x^2 + (-a*b 
^2)^(1/3)*a)*sqrt((-a*b^2)^(1/3)/a) - 3*(-a*b^2)^(2/3)*x)/(b*x^3 + a)) - ( 
b*x^3 + a)*(-a*b^2)^(2/3)*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3 
)) + 2*(b*x^3 + a)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a*b^4*x^3 + 
a^2*b^3), -1/9*(3*a*b^2*x^2 - 6*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt(-(-a*b^ 
2)^(1/3)/a)*arctan(sqrt(1/3)*(2*b*x + (-a*b^2)^(1/3))*sqrt(-(-a*b^2)^(1/3) 
/a)/b) - (b*x^3 + a)*(-a*b^2)^(2/3)*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a 
*b^2)^(2/3)) + 2*(b*x^3 + a)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a* 
b^4*x^3 + a^2*b^3)]
 

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.32 \[ \int \frac {x^4}{\left (a+b x^3\right )^2} \, dx=- \frac {x^{2}}{3 a b + 3 b^{2} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a b^{5} + 8, \left ( t \mapsto t \log {\left (\frac {81 t^{2} a b^{3}}{4} + x \right )} \right )\right )} \] Input:

integrate(x**4/(b*x**3+a)**2,x)
 

Output:

-x**2/(3*a*b + 3*b**2*x**3) + RootSum(729*_t**3*a*b**5 + 8, Lambda(_t, _t* 
log(81*_t**2*a*b**3/4 + x)))
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {x^4}{\left (a+b x^3\right )^2} \, dx=-\frac {x^{2}}{3 \, {\left (b^{2} x^{3} + a b\right )}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {2 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \] Input:

integrate(x^4/(b*x^3+a)^2,x, algorithm="maxima")
 

Output:

-1/3*x^2/(b^2*x^3 + a*b) + 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/ 
3))/(a/b)^(1/3))/(b^2*(a/b)^(1/3)) + 1/9*log(x^2 - x*(a/b)^(1/3) + (a/b)^( 
2/3))/(b^2*(a/b)^(1/3)) - 2/9*log(x + (a/b)^(1/3))/(b^2*(a/b)^(1/3))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97 \[ \int \frac {x^4}{\left (a+b x^3\right )^2} \, dx=-\frac {x^{2}}{3 \, {\left (b x^{3} + a\right )} b} - \frac {2 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b} - \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b^{3}} + \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a b^{3}} \] Input:

integrate(x^4/(b*x^3+a)^2,x, algorithm="giac")
 

Output:

-1/3*x^2/((b*x^3 + a)*b) - 2/9*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/(a* 
b) - 2/9*sqrt(3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(- 
a/b)^(1/3))/(a*b^3) + 1/9*(-a*b^2)^(2/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b) 
^(2/3))/(a*b^3)
 

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int \frac {x^4}{\left (a+b x^3\right )^2} \, dx=\frac {2\,\ln \left (\frac {4\,x}{9\,b}-\frac {4\,{\left (-a\right )}^{1/3}}{9\,b^{4/3}}\right )}{9\,{\left (-a\right )}^{1/3}\,b^{5/3}}-\frac {x^2}{3\,b\,\left (b\,x^3+a\right )}+\frac {\ln \left (\frac {4\,x}{9\,b}-\frac {{\left (-a\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{9\,b^{4/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,{\left (-a\right )}^{1/3}\,b^{5/3}}-\frac {\ln \left (\frac {4\,x}{9\,b}-\frac {{\left (-a\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{9\,b^{4/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,{\left (-a\right )}^{1/3}\,b^{5/3}} \] Input:

int(x^4/(a + b*x^3)^2,x)
 

Output:

(2*log((4*x)/(9*b) - (4*(-a)^(1/3))/(9*b^(4/3))))/(9*(-a)^(1/3)*b^(5/3)) - 
 x^2/(3*b*(a + b*x^3)) + (log((4*x)/(9*b) - ((-a)^(1/3)*(3^(1/2)*1i - 1)^2 
)/(9*b^(4/3)))*(3^(1/2)*1i - 1))/(9*(-a)^(1/3)*b^(5/3)) - (log((4*x)/(9*b) 
 - ((-a)^(1/3)*(3^(1/2)*1i + 1)^2)/(9*b^(4/3)))*(3^(1/2)*1i + 1))/(9*(-a)^ 
(1/3)*b^(5/3))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.16 \[ \int \frac {x^4}{\left (a+b x^3\right )^2} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) a -2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b \,x^{3}-3 b^{\frac {2}{3}} a^{\frac {1}{3}} x^{2}+\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a +\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b \,x^{3}-2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a -2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b \,x^{3}}{9 b^{\frac {5}{3}} a^{\frac {1}{3}} \left (b \,x^{3}+a \right )} \] Input:

int(x^4/(b*x^3+a)^2,x)
 

Output:

( - 2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a - 2*sqr 
t(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*b*x**3 - 3*b**(2/3 
)*a**(1/3)*x**2 + log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a + 
log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b*x**3 - 2*log(a**(1/3 
) + b**(1/3)*x)*a - 2*log(a**(1/3) + b**(1/3)*x)*b*x**3)/(9*b**(2/3)*a**(1 
/3)*b*(a + b*x**3))