Integrand size = 18, antiderivative size = 145 \[ \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\log \left ((1-x) (1+x)^2\right )}{4 \sqrt [3]{2}}+\frac {1}{2} \log \left (x+\sqrt [3]{1-x^3}\right )-\frac {3 \log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \] Output:
1/8*ln((1-x)*(1+x)^2)*2^(2/3)+1/2*ln(x+(-x^3+1)^(1/3))-3/8*ln(-1+x+2^(2/3) *(-x^3+1)^(1/3))*2^(2/3)-1/3*arctan(1/3*(1-2*x/(-x^3+1)^(1/3))*3^(1/2))*3^ (1/2)+1/4*arctan(1/3*(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)*2^( 2/3)
\[ \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx=\int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx \] Input:
Integrate[x/((1 + x)*(1 - x^3)^(1/3)),x]
Output:
Integrate[x/((1 + x)*(1 - x^3)^(1/3)), x]
Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2577, 769, 2574}
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.
\(\displaystyle \int \frac {x}{(x+1) \sqrt [3]{1-x^3}} \, dx\) |
\(\Big \downarrow \) 2577 |
\(\displaystyle \int \frac {1}{\sqrt [3]{1-x^3}}dx-\int \frac {1}{(x+1) \sqrt [3]{1-x^3}}dx\) |
\(\Big \downarrow \) 769 |
\(\displaystyle -\int \frac {1}{(x+1) \sqrt [3]{1-x^3}}dx-\frac {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (\sqrt [3]{1-x^3}+x\right )\) |
\(\Big \downarrow \) 2574 |
\(\displaystyle \frac {\sqrt {3} \arctan \left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\arctan \left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (\sqrt [3]{1-x^3}+x\right )-\frac {3 \log \left (2^{2/3} \sqrt [3]{1-x^3}+x-1\right )}{4 \sqrt [3]{2}}+\frac {\log \left ((1-x) (x+1)^2\right )}{4 \sqrt [3]{2}}\) |
Input:
Int[x/((1 + x)*(1 - x^3)^(1/3)),x]
Output:
(Sqrt[3]*ArcTan[(1 + (2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]])/(2*2^(1/ 3)) - ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]]/Sqrt[3] + Log[(1 - x)*(1 + x)^2]/(4*2^(1/3)) + Log[x + (1 - x^3)^(1/3)]/2 - (3*Log[-1 + x + 2^(2/3 )*(1 - x^3)^(1/3)])/(4*2^(1/3))
Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* (x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[ Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b, 3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sq rt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2^(7/3 )*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^ (1/3)])/(2^(7/3)*Rt[b, 3]*c), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 0]
Int[((e_.) + (f_.)*(x_))/(((c_.) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)) , x_Symbol] :> Simp[f/d Int[1/(a + b*x^3)^(1/3), x], x] + Simp[(d*e - c*f )/d Int[1/((c + d*x)*(a + b*x^3)^(1/3)), x], x] /; FreeQ[{a, b, c, d, e, f}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 16.86 (sec) , antiderivative size = 1421, normalized size of antiderivative = 9.80
Input:
int(x/(1+x)/(-x^3+1)^(1/3),x,method=_RETURNVERBOSE)
Output:
1/3*ln(-RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+ 4)^4*x^3+3*(-x^3+1)^(2/3)*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*Ro otOf(_Z^3+4)+4*_Z^2)*x-RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2) *RootOf(_Z^3+4)^2*x^3+2*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*Root Of(_Z^3+4)+4*_Z^2)+6*x^2*(-x^3+1)^(1/3)+2*x^3-2)+1/2*RootOf(RootOf(_Z^3+4) ^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*ln((10*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf( _Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)^2*x+2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf (_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x+4*(-x^3+1)^(2/3)*RootOf(_Z^3+4)^2*Root Of(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)+9*(-x^3+1)^(1/3)*RootOf(Ro otOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)*x-2*(-x^3+1)^(1/ 3)*RootOf(_Z^3+4)^2*x-9*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf (_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)-35*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^ 3+4)+4*_Z^2)*x^2+2*(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2-7*x^2*RootOf(_Z^3+4)-30 *RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x-6*x*RootOf(_Z^3+4)- 26*(-x^3+1)^(2/3)-35*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)-7 *RootOf(_Z^3+4))/(1+x)^2)+1/4*RootOf(_Z^3+4)*ln(-(8*RootOf(RootOf(_Z^3+4)^ 2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)^2*x+10*RootOf(RootOf(_Z^3+4 )^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x+8*(-x^3+1)^(2/3)*RootOf (_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)-26*(-x^3+1) ^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+...
Exception generated. \[ \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x/(1+x)/(-x^3+1)^(1/3),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (residue poly has multiple non-linear fac tors)
\[ \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx=\int \frac {x}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right )}\, dx \] Input:
integrate(x/(1+x)/(-x**3+1)**(1/3),x)
Output:
Integral(x/((-(x - 1)*(x**2 + x + 1))**(1/3)*(x + 1)), x)
\[ \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx=\int { \frac {x}{{\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )}} \,d x } \] Input:
integrate(x/(1+x)/(-x^3+1)^(1/3),x, algorithm="maxima")
Output:
integrate(x/((-x^3 + 1)^(1/3)*(x + 1)), x)
\[ \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx=\int { \frac {x}{{\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )}} \,d x } \] Input:
integrate(x/(1+x)/(-x^3+1)^(1/3),x, algorithm="giac")
Output:
integrate(x/((-x^3 + 1)^(1/3)*(x + 1)), x)
Timed out. \[ \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx=\int \frac {x}{{\left (1-x^3\right )}^{1/3}\,\left (x+1\right )} \,d x \] Input:
int(x/((1 - x^3)^(1/3)*(x + 1)),x)
Output:
int(x/((1 - x^3)^(1/3)*(x + 1)), x)
\[ \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx=\int \frac {x}{\left (-x^{3}+1\right )^{\frac {1}{3}} x +\left (-x^{3}+1\right )^{\frac {1}{3}}}d x \] Input:
int(x/(1+x)/(-x^3+1)^(1/3),x)
Output:
int(x/(( - x**3 + 1)**(1/3)*x + ( - x**3 + 1)**(1/3)),x)