Integrand size = 9, antiderivative size = 134 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {x}{3 a \left (a+b x^3\right )}-\frac {2 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} \sqrt [3]{b}}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} \sqrt [3]{b}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{9 a^{5/3} \sqrt [3]{b}} \] Output:
1/3*x/a/(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^(5/3)/b^(1/3)+2/9*ln(a^(1/3)+b^(1/3)*x)/a^(5/3)/b^(1/3)-1/9*ln(a^( 2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(5/3)/b^(1/3)
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {\frac {3 a^{2/3} x}{a+b x^3}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}}{9 a^{5/3}} \] Input:
Integrate[(a + b*x^3)^(-2),x]
Output:
((3*a^(2/3)*x)/(a + b*x^3) - (2*Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3)) /Sqrt[3]])/b^(1/3) + (2*Log[a^(1/3) + b^(1/3)*x])/b^(1/3) - Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/b^(1/3))/(9*a^(5/3))
Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {749, 750, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle \frac {2 \int \frac {1}{b x^3+a}dx}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {2 \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 a^{2/3}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {2 \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {2 \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {2 \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{3 a}+\frac {x}{3 a \left (a+b x^3\right )}\) |
Input:
Int[(a + b*x^3)^(-2),x]
Output:
x/(3*a*(a + b*x^3)) + (2*(Log[a^(1/3) + b^(1/3)*x]/(3*a^(2/3)*b^(1/3)) + ( -((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^(2/3))))/(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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*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 = 0.43 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.34
| method | result | size |
| risch | \(\frac {x}{3 a \left (b \,x^{3}+a \right )}+\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{9 a b}\) | \(46\) |
| default | \(\frac {x}{3 a \left (b \,x^{3}+a \right )}+\frac {\frac {2 \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}}{a}\) | \(112\) |
Input:
int(1/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
1/3*x/a/(b*x^3+a)+2/9/a/b*sum(1/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))
Time = 0.08 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.90 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\left [\frac {3 \, a^{2} b x + 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) - {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{9 \, {\left (a^{3} b^{2} x^{3} + a^{4} b\right )}}, \frac {3 \, a^{2} b x + 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} x^{3} + a^{2} b\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) - {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, {\left (b x^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{9 \, {\left (a^{3} b^{2} x^{3} + a^{4} b\right )}}\right ] \] Input:
integrate(1/(b*x^3+a)^2,x, algorithm="fricas")
Output:
[1/9*(3*a^2*b*x + 3*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt(-(a^2*b)^(1/3)/b)*l og((2*a*b*x^3 - 3*(a^2*b)^(1/3)*a*x - a^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a^2* b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) - (b*x^ 3 + a)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) + 2* (b*x^3 + a)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)))/(a^3*b^2*x^3 + a^4*b ), 1/9*(3*a^2*b*x + 6*sqrt(1/3)*(a*b^2*x^3 + a^2*b)*sqrt((a^2*b)^(1/3)/b)* arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)/ b)/a^2) - (b*x^3 + a)*(a^2*b)^(2/3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b )^(1/3)*a) + 2*(b*x^3 + a)*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)))/(a^3* b^2*x^3 + a^4*b)]
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {x}{3 a^{2} + 3 a b x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a^{5} b - 8, \left ( t \mapsto t \log {\left (\frac {9 t a^{2}}{2} + x \right )} \right )\right )} \] Input:
integrate(1/(b*x**3+a)**2,x)
Output:
x/(3*a**2 + 3*a*b*x**3) + RootSum(729*_t**3*a**5*b - 8, Lambda(_t, _t*log( 9*_t*a**2/2 + x)))
Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {x}{3 \, {\left (a b x^{3} + a^{2}\right )}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate(1/(b*x^3+a)^2,x, algorithm="maxima")
Output:
1/3*x/(a*b*x^3 + a^2) + 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3)) /(a/b)^(1/3))/(a*b*(a/b)^(2/3)) - 1/9*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3 ))/(a*b*(a/b)^(2/3)) + 2/9*log(x + (a/b)^(1/3))/(a*b*(a/b)^(2/3))
Time = 0.12 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=-\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a^{2}} + \frac {x}{3 \, {\left (b x^{3} + a\right )} a} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2} b} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \, a^{2} b} \] Input:
integrate(1/(b*x^3+a)^2,x, algorithm="giac")
Output:
-2/9*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^2 + 1/3*x/((b*x^3 + a)*a) + 2/9*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b) ^(1/3))/(a^2*b) + 1/9*(-a*b^2)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/ 3))/(a^2*b)
Time = 0.15 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {x}{3\,a\,\left (b\,x^3+a\right )}+\frac {2\,\ln \left (\frac {2\,b^{5/3}}{a^{2/3}}+\frac {2\,b^2\,x}{a}\right )}{9\,a^{5/3}\,b^{1/3}}+\frac {\ln \left (\frac {2\,b^2\,x}{a}+\frac {b^{5/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{a^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,a^{5/3}\,b^{1/3}}-\frac {\ln \left (\frac {2\,b^2\,x}{a}-\frac {b^{5/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{a^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,a^{5/3}\,b^{1/3}} \] Input:
int(1/(a + b*x^3)^2,x)
Output:
x/(3*a*(a + b*x^3)) + (2*log((2*b^(5/3))/a^(2/3) + (2*b^2*x)/a))/(9*a^(5/3 )*b^(1/3)) + (log((2*b^2*x)/a + (b^(5/3)*(3^(1/2)*1i - 1))/a^(2/3))*(3^(1/ 2)*1i - 1))/(9*a^(5/3)*b^(1/3)) - (log((2*b^2*x)/a - (b^(5/3)*(3^(1/2)*1i + 1))/a^(2/3))*(3^(1/2)*1i + 1))/(9*a^(5/3)*b^(1/3))
Time = 0.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (a+b x^3\right )^2} \, dx=\frac {-2 a^{\frac {4}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right )-2 a^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b \,x^{3}-a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right )-a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b \,x^{3}+2 a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )+2 a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b \,x^{3}+3 b^{\frac {1}{3}} a x}{9 b^{\frac {1}{3}} a^{2} \left (b \,x^{3}+a \right )} \] Input:
int(1/(b*x^3+a)^2,x)
Output:
( - 2*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))* a - 2*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))* b*x**3 - a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a - a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b*x**3 + 2*a* *(1/3)*log(a**(1/3) + b**(1/3)*x)*a + 2*a**(1/3)*log(a**(1/3) + b**(1/3)*x )*b*x**3 + 3*b**(1/3)*a*x)/(9*b**(1/3)*a**2*(a + b*x**3))