Integrand size = 11, antiderivative size = 136 \[ \int \frac {x}{\left (a+b x^3\right )^2} \, dx=\frac {x^2}{3 a \left (a+b x^3\right )}-\frac {\arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{4/3} b^{2/3}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{4/3} b^{2/3}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{4/3} b^{2/3}} \] Output:
1/3*x^2/a/(b*x^3+a)-1/9*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))* 3^(1/2)/a^(4/3)/b^(2/3)-1/9*ln(a^(1/3)+b^(1/3)*x)/a^(4/3)/b^(2/3)+1/18*ln( a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(4/3)/b^(2/3)
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88 \[ \int \frac {x}{\left (a+b x^3\right )^2} \, dx=\frac {\frac {6 \sqrt [3]{a} x^2}{a+b x^3}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{2/3}}-\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}+\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{2/3}}}{18 a^{4/3}} \] Input:
Integrate[x/(a + b*x^3)^2,x]
Output:
((6*a^(1/3)*x^2)/(a + b*x^3) - (2*Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3 ))/Sqrt[3]])/b^(2/3) - (2*Log[a^(1/3) + b^(1/3)*x])/b^(2/3) + Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/b^(2/3))/(18*a^(4/3))
Time = 0.45 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, Rules used = {819, 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.
| \(\displaystyle \int \frac {x}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 819 |
| \(\displaystyle \frac {\int \frac {x}{b x^3+a}dx}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle \frac {\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}}{3 a}+\frac {x^2}{3 a \left (a+b x^3\right )}\) |
Input:
Int[x/(a + b*x^3)^2,x]
Output:
x^2/(3*a*(a + b*x^3)) + (-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*a)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
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]
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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.44 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.35
| method | result | size |
| risch | \(\frac {x^{2}}{3 a \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{9 a b}\) | \(48\) |
| default | \(\frac {x^{2}}{3 a \left (b \,x^{3}+a \right )}+\frac {-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}}{3 a}\) | \(114\) |
Input:
int(x/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
1/3*x^2/a/(b*x^3+a)+1/9/a/b*sum(1/_R*ln(x-_R),_R=RootOf(_Z^3*b+a))
Time = 0.09 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.84 \[ \int \frac {x}{\left (a+b x^3\right )^2} \, dx=\left [\frac {6 \, a b^{2} x^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {-\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (a b^{2}\right )^{\frac {2}{3}} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (a b^{2}\right )^{\frac {2}{3}} x}{b x^{3} + a}\right ) + {\left (b x^{3} + a\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} b x + \left (a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left (b x^{3} + a\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b x + \left (a b^{2}\right )^{\frac {1}{3}}\right )}{18 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}, \frac {6 \, a b^{2} x^{2} - 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x - \left (a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {\frac {\left (a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) + {\left (b x^{3} + a\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} - \left (a b^{2}\right )^{\frac {1}{3}} b x + \left (a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left (b x^{3} + a\right )} \left (a b^{2}\right )^{\frac {2}{3}} \log \left (b x + \left (a b^{2}\right )^{\frac {1}{3}}\right )}{18 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}\right ] \] Input:
integrate(x/(b*x^3+a)^2,x, algorithm="fricas")
Output:
[1/18*(6*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^2*b^3*x^3 + a^3*b^ 2), 1/18*(6*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^2*b^3*x^3 + a^3 *b^2)]
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.32 \[ \int \frac {x}{\left (a+b x^3\right )^2} \, dx=\frac {x^{2}}{3 a^{2} + 3 a b x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a^{4} b^{2} + 1, \left ( t \mapsto t \log {\left (81 t^{2} a^{3} b + x \right )} \right )\right )} \] Input:
integrate(x/(b*x**3+a)**2,x)
Output:
x**2/(3*a**2 + 3*a*b*x**3) + RootSum(729*_t**3*a**4*b**2 + 1, Lambda(_t, _ t*log(81*_t**2*a**3*b + x)))
Time = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \frac {x}{\left (a+b x^3\right )^2} \, dx=\frac {x^{2}}{3 \, {\left (a b x^{3} + a^{2}\right )}} + \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a b \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {\log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {1}{3}}} \] Input:
integrate(x/(b*x^3+a)^2,x, algorithm="maxima")
Output:
1/3*x^2/(a*b*x^3 + a^2) + 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3 ))/(a/b)^(1/3))/(a*b*(a/b)^(1/3)) + 1/18*log(x^2 - x*(a/b)^(1/3) + (a/b)^( 2/3))/(a*b*(a/b)^(1/3)) - 1/9*log(x + (a/b)^(1/3))/(a*b*(a/b)^(1/3))
Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\left (a+b x^3\right )^2} \, dx=\frac {x^{2}}{3 \, {\left (b x^{3} + a\right )} a} - \frac {\left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{2}} - \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b^{2}} + \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, a^{2} b^{2}} \] Input:
integrate(x/(b*x^3+a)^2,x, algorithm="giac")
Output:
1/3*x^2/((b*x^3 + a)*a) - 1/9*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/a^2 - 1/9*sqrt(3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b )^(1/3))/(a^2*b^2) + 1/18*(-a*b^2)^(2/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b) ^(2/3))/(a^2*b^2)
Time = 0.40 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int \frac {x}{\left (a+b x^3\right )^2} \, dx=\frac {x^2}{3\,a\,\left (b\,x^3+a\right )}+\frac {{\left (-1\right )}^{1/3}\,\ln \left (\frac {{\left (-1\right )}^{2/3}\,b^{2/3}}{9\,a^{5/3}}+\frac {b\,x}{9\,a^2}\right )}{9\,a^{4/3}\,b^{2/3}}-\frac {{\left (-1\right )}^{1/3}\,\ln \left ({\left (-1\right )}^{2/3}\,a^{1/3}-2\,b^{1/3}\,x+{\left (-1\right )}^{1/6}\,\sqrt {3}\,a^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{4/3}\,b^{2/3}}+\frac {{\left (-1\right )}^{1/3}\,\ln \left (2\,b^{1/3}\,x-{\left (-1\right )}^{2/3}\,a^{1/3}+{\left (-1\right )}^{1/6}\,\sqrt {3}\,a^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{4/3}\,b^{2/3}} \] Input:
int(x/(a + b*x^3)^2,x)
Output:
x^2/(3*a*(a + b*x^3)) + ((-1)^(1/3)*log(((-1)^(2/3)*b^(2/3))/(9*a^(5/3)) + (b*x)/(9*a^2)))/(9*a^(4/3)*b^(2/3)) - ((-1)^(1/3)*log((-1)^(2/3)*a^(1/3) - 2*b^(1/3)*x + (-1)^(1/6)*3^(1/2)*a^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(9*a^( 4/3)*b^(2/3)) + ((-1)^(1/3)*log(2*b^(1/3)*x - (-1)^(2/3)*a^(1/3) + (-1)^(1 /6)*3^(1/2)*a^(1/3))*((3^(1/2)*1i)/2 - 1/2))/(9*a^(4/3)*b^(2/3))
Time = 0.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.16 \[ \int \frac {x}{\left (a+b x^3\right )^2} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) a -2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b \,x^{3}+6 b^{\frac {2}{3}} a^{\frac {1}{3}} x^{2}+\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a +\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b \,x^{3}-2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a -2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b \,x^{3}}{18 b^{\frac {2}{3}} a^{\frac {4}{3}} \left (b \,x^{3}+a \right )} \] Input:
int(x/(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 + 6*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)/(18*b**(2/3)*a**( 1/3)*a*(a + b*x**3))