3.1.38 \(\int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx\) [38]
Optimal. Leaf size=145 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\log \left ((1-x) (1+x)^2\right )}{4 \sqrt [3]{2}}+\frac {1}{2} \log \left (x+\sqrt [3]{1-x^3}\right )-\frac {3 \log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \]
[Out]
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)
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Rubi [A]
time = 0.07, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2177, 245,
2174} \begin {gather*} \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 {\sqrt {3} \tan ^{-1}\left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\log \left ((1-x) (x+1)^2\right )}{4 \sqrt [3]{2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x/((1 + x)*(1 - x^3)^(1/3)),x]
[Out]
(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))
Rule 245
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]]/(Sq
rt[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]
Rule 2174
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))))/Sqrt[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]
Rule 2177
Int[((e_.) + (f_.)*(x_))/(((c_.) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Dist[f/d, Int[1/(a +
b*x^3)^(1/3), x], x] + Dist[(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]
Rubi steps
\begin {align*} \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx &=\int \frac {1}{\sqrt [3]{1-x^3}} \, dx-\int \frac {1}{(1+x) \sqrt [3]{1-x^3}} \, dx\\ &=\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\tan ^{-1}\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}}\\ \end {align*}
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Mathematica [F]
time = 3.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{(1+x) \sqrt [3]{1-x^3}} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
Integrate[x/((1 + x)*(1 - x^3)^(1/3)),x]
[Out]
Integrate[x/((1 + x)*(1 - x^3)^(1/3)), x]
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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded in comparison} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
mathics('Integrate[x/((1 + x)*(1 - x^3)^(1/3)),x]')
[Out]
cought exception: maximum recursion depth exceeded in comparison
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 12.29, size = 1790, normalized size = 12.34
| | |
| method |
result |
size |
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| trager |
\(\text {Expression too large to display}\) |
\(1790\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x/(1+x)/(-x^3+1)^(1/3),x,method=_RETURNVERBOSE)
[Out]
-1/4*ln(-(10*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x+12*RootOf(RootOf(_Z^3+4)^2
+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)^2*x-8*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf
(_Z^3+4)^2*(-x^3+1)^(2/3)-13*(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x-8*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*R
ootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)*x+13*(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2+8*(-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*x^2*RootOf(_Z^3+4)-42*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_
Z^3+4)+4*_Z^2)*x^2-30*x*RootOf(_Z^3+4)-36*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+52*(-x^3+1)^(2
/3)-35*RootOf(_Z^3+4)-42*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2))/(1+x)^2)*RootOf(_Z^3+4)-1/2*ln(-
(10*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x+12*RootOf(RootOf(_Z^3+4)^2+2*_Z*Roo
tOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)^2*x-8*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^
2*(-x^3+1)^(2/3)-13*(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x-8*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^
3+4)+4*_Z^2)*RootOf(_Z^3+4)*x+13*(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2+8*(-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*x^2*RootOf(_Z^3+4)-42*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*
_Z^2)*x^2-30*x*RootOf(_Z^3+4)-36*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+52*(-x^3+1)^(2/3)-35*Ro
otOf(_Z^3+4)-42*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2))/(1+x)^2)*RootOf(RootOf(_Z^3+4)^2+2*_Z*Roo
tOf(_Z^3+4)+4*_Z^2)+1/2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*ln((4*RootOf(RootOf(_Z^3+4)^2+2*_Z
*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x-12*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^
3+4)^2*x-8*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^2*(-x^3+1)^(2/3)+9*(-x^3+1)^(1/3
)*RootOf(_Z^3+4)^2*x-8*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)*x-9*(
-x^3+1)^(1/3)*RootOf(_Z^3+4)^2+8*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^
3+4)+14*x^2*RootOf(_Z^3+4)-42*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2+4*x*RootOf(_Z^3+4)-12*Ro
otOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x-36*(-x^3+1)^(2/3)+14*RootOf(_Z^3+4)-42*RootOf(RootOf(_Z^3+
4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2))/(1+x)^2)-1/6*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*
_Z^2)*ln(-RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)^4*x^3+2*RootOf(_Z^3+4)^2*RootOf
(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^3-2*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+
4)+4*_Z^2)+12*x*(-x^3+1)^(2/3)-12*x^2*(-x^3+1)^(1/3)+8*x^3-4)+1/6*ln(RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+
4)+4*_Z^2)^2*RootOf(_Z^3+4)^4*x^3+6*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^2*(-x^3
+1)^(2/3)*x-6*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*(-x^3+1)^(1/3)*x^2+4*RootOf
(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^3-2*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+
2*_Z*RootOf(_Z^3+4)+4*_Z^2)+4*x^3-4)*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)-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+6*RootOf(RootOf(_Z^3+4)^2+2*_Z*R
ootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^2*(-x^3+1)^(2/3)*x-6*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf
(_Z^3+4)+4*_Z^2)*(-x^3+1)^(1/3)*x^2+4*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^3
-2*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)+4*x^3-4)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(1+x)/(-x^3+1)^(1/3),x, algorithm="maxima")
[Out]
integrate(x/((-x^3 + 1)^(1/3)*(x + 1)), x)
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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(1+x)/(-x^3+1)^(1/3),x, algorithm="fricas")
[Out]
Exception raised: TypeError >> Error detected within library code: integrate: implementation incomplete (re
sidue poly has multiple non-linear factors)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(1+x)/(-x**3+1)**(1/3),x)
[Out]
Integral(x/((-(x - 1)*(x**2 + x + 1))**(1/3)*(x + 1)), x)
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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(1+x)/(-x^3+1)^(1/3),x)
[Out]
Could not integrate
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (1-x^3\right )}^{1/3}\,\left (x+1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x/((1 - x^3)^(1/3)*(x + 1)),x)
[Out]
int(x/((1 - x^3)^(1/3)*(x + 1)), x)
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