3.120 \(\int e^{\frac {2}{3} \coth ^{-1}(x)} x^2 \, dx\)

Optimal. Leaf size=157 \[ \frac {1}{3} \sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3} x^3+\frac {4}{9} \sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3} x^2+\frac {14}{27} \sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3} x-\frac {11}{27} \log \left (\sqrt [3]{\frac {1}{x}+1}-\sqrt [3]{\frac {x-1}{x}}\right )-\frac {11 \log (x)}{81}-\frac {22 \tan ^{-1}\left (\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3}} \]

[Out]

14/27*(1/x+1)^(1/3)*((-1+x)/x)^(2/3)*x+4/9*(1/x+1)^(1/3)*((-1+x)/x)^(2/3)*x^2+1/3*(1/x+1)^(1/3)*((-1+x)/x)^(2/
3)*x^3-11/27*ln((1/x+1)^(1/3)-((-1+x)/x)^(1/3))-11/81*ln(x)-22/81*arctan(1/3*3^(1/2)+2/3*((-1+x)/x)^(1/3)/(1/x
+1)^(1/3)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6171, 99, 151, 12, 91} \[ \frac {1}{3} \sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3} x^3+\frac {4}{9} \sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3} x^2+\frac {14}{27} \sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3} x-\frac {11}{27} \log \left (\sqrt [3]{\frac {1}{x}+1}-\sqrt [3]{\frac {x-1}{x}}\right )-\frac {11 \log (x)}{81}-\frac {22 \tan ^{-1}\left (\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[E^((2*ArcCoth[x])/3)*x^2,x]

[Out]

(14*(1 + x^(-1))^(1/3)*((-1 + x)/x)^(2/3)*x)/27 + (4*(1 + x^(-1))^(1/3)*((-1 + x)/x)^(2/3)*x^2)/9 + ((1 + x^(-
1))^(1/3)*((-1 + x)/x)^(2/3)*x^3)/3 - (22*ArcTan[1/Sqrt[3] + (2*((-1 + x)/x)^(1/3))/(Sqrt[3]*(1 + x^(-1))^(1/3
))])/(27*Sqrt[3]) - (11*Log[(1 + x^(-1))^(1/3) - ((-1 + x)/x)^(1/3)])/27 - (11*Log[x])/81

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 91

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.)*(x_))), x_Symbol] :> With[{q = Rt[
(d*e - c*f)/(b*e - a*f), 3]}, -Simp[(Sqrt[3]*q*ArcTan[1/Sqrt[3] + (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/
3))])/(d*e - c*f), x] + (Simp[(q*Log[e + f*x])/(2*(d*e - c*f)), x] - Simp[(3*q*Log[q*(a + b*x)^(1/3) - (c + d*
x)^(1/3)])/(2*(d*e - c*f)), x])] /; FreeQ[{a, b, c, d, e, f}, x]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 6171

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x/a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2
)), x], x, 1/x] /; FreeQ[{a, n}, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int e^{\frac {2}{3} \coth ^{-1}(x)} x^2 \, dx &=-\operatorname {Subst}\left (\int \frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x} x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} \sqrt [3]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{2/3} x^3-\frac {1}{3} \operatorname {Subst}\left (\int \frac {\frac {8}{3}+2 x}{\sqrt [3]{1-x} x^3 (1+x)^{2/3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {4}{9} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3} x^2+\frac {1}{3} \sqrt [3]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{2/3} x^3+\frac {1}{6} \operatorname {Subst}\left (\int \frac {-\frac {28}{9}-\frac {8 x}{3}}{\sqrt [3]{1-x} x^2 (1+x)^{2/3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {14}{27} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3} x+\frac {4}{9} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3} x^2+\frac {1}{3} \sqrt [3]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{2/3} x^3-\frac {1}{6} \operatorname {Subst}\left (\int \frac {44}{27 \sqrt [3]{1-x} x (1+x)^{2/3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {14}{27} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3} x+\frac {4}{9} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3} x^2+\frac {1}{3} \sqrt [3]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{2/3} x^3-\frac {22}{81} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} x (1+x)^{2/3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {14}{27} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3} x+\frac {4}{9} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3} x^2+\frac {1}{3} \sqrt [3]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{2/3} x^3-\frac {22 \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{-\frac {1-x}{x}}}{\sqrt {3} \sqrt [3]{1+\frac {1}{x}}}\right )}{27 \sqrt {3}}-\frac {11}{27} \log \left (\sqrt [3]{1+\frac {1}{x}}-\sqrt [3]{-\frac {1-x}{x}}\right )-\frac {11 \log (x)}{81}\\ \end {align*}

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Mathematica [C]  time = 7.77, size = 340, normalized size = 2.17 \[ -\frac {e^{-\frac {10}{3} \coth ^{-1}(x)} \left (54 e^{8 \coth ^{-1}(x)} \left (782 e^{2 \coth ^{-1}(x)}+325 e^{4 \coth ^{-1}(x)}+475\right ) \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {16}{3};e^{2 \coth ^{-1}(x)}\right )+162 e^{8 \coth ^{-1}(x)} \left (64 e^{2 \coth ^{-1}(x)}+29 e^{4 \coth ^{-1}(x)}+35\right ) \, _5F_4\left (2,2,2,2,\frac {7}{3};1,1,1,\frac {16}{3};e^{2 \coth ^{-1}(x)}\right )+486 e^{8 \coth ^{-1}(x)} \, _6F_5\left (2,2,2,2,2,\frac {7}{3};1,1,1,1,\frac {16}{3};e^{2 \coth ^{-1}(x)}\right )+972 e^{10 \coth ^{-1}(x)} \, _6F_5\left (2,2,2,2,2,\frac {7}{3};1,1,1,1,\frac {16}{3};e^{2 \coth ^{-1}(x)}\right )+486 e^{12 \coth ^{-1}(x)} \, _6F_5\left (2,2,2,2,2,\frac {7}{3};1,1,1,1,\frac {16}{3};e^{2 \coth ^{-1}(x)}\right )[phone] e^{2 \coth ^{-1}(x)} \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};e^{2 \coth ^{-1}(x)}\right )-14083160 e^{4 \coth ^{-1}(x)} \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};e^{2 \coth ^{-1}(x)}\right )-8250060 e^{6 \coth ^{-1}(x)} \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};e^{2 \coth ^{-1}(x)}\right )+1456000 e^{8 \coth ^{-1}(x)} \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};e^{2 \coth ^{-1}(x)}\right )[phone] \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};e^{2 \coth ^{-1}(x)}\right )-20915440 e^{2 \coth ^{-1}(x)}+7026175 e^{4 \coth ^{-1}(x)}+7394140 e^{6 \coth ^{-1}(x)}-433485 e^{8 \coth ^{-1}(x)}-22750000\right )}{49140} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((2*ArcCoth[x])/3)*x^2,x]

[Out]

-1/49140*(-22750000 - 20915440*E^(2*ArcCoth[x]) + 7026175*E^(4*ArcCoth[x]) + 7394140*E^(6*ArcCoth[x]) - 433485
*E^(8*ArcCoth[x]) + 22750000*Hypergeometric2F1[1/3, 1, 4/3, E^(2*ArcCoth[x])] + 15227940*E^(2*ArcCoth[x])*Hype
rgeometric2F1[1/3, 1, 4/3, E^(2*ArcCoth[x])] - 14083160*E^(4*ArcCoth[x])*Hypergeometric2F1[1/3, 1, 4/3, E^(2*A
rcCoth[x])] - 8250060*E^(6*ArcCoth[x])*Hypergeometric2F1[1/3, 1, 4/3, E^(2*ArcCoth[x])] + 1456000*E^(8*ArcCoth
[x])*Hypergeometric2F1[1/3, 1, 4/3, E^(2*ArcCoth[x])] + 54*E^(8*ArcCoth[x])*(475 + 782*E^(2*ArcCoth[x]) + 325*
E^(4*ArcCoth[x]))*HypergeometricPFQ[{2, 2, 2, 7/3}, {1, 1, 16/3}, E^(2*ArcCoth[x])] + 162*E^(8*ArcCoth[x])*(35
+ 64*E^(2*ArcCoth[x]) + 29*E^(4*ArcCoth[x]))*HypergeometricPFQ[{2, 2, 2, 2, 7/3}, {1, 1, 1, 16/3}, E^(2*ArcCo
th[x])] + 486*E^(8*ArcCoth[x])*HypergeometricPFQ[{2, 2, 2, 2, 2, 7/3}, {1, 1, 1, 1, 16/3}, E^(2*ArcCoth[x])] +
972*E^(10*ArcCoth[x])*HypergeometricPFQ[{2, 2, 2, 2, 2, 7/3}, {1, 1, 1, 1, 16/3}, E^(2*ArcCoth[x])] + 486*E^(
12*ArcCoth[x])*HypergeometricPFQ[{2, 2, 2, 2, 2, 7/3}, {1, 1, 1, 1, 16/3}, E^(2*ArcCoth[x])])/E^((10*ArcCoth[x
])/3)

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fricas [A]  time = 0.40, size = 100, normalized size = 0.64 \[ \frac {1}{27} \, {\left (9 \, x^{3} + 21 \, x^{2} + 26 \, x + 14\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \frac {22}{81} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + \frac {11}{81} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {22}{81} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/3)*x^2,x, algorithm="fricas")

[Out]

1/27*(9*x^3 + 21*x^2 + 26*x + 14)*((x - 1)/(x + 1))^(2/3) - 22/81*sqrt(3)*arctan(2/3*sqrt(3)*((x - 1)/(x + 1))
^(1/3) + 1/3*sqrt(3)) + 11/81*log(((x - 1)/(x + 1))^(2/3) + ((x - 1)/(x + 1))^(1/3) + 1) - 22/81*log(((x - 1)/
(x + 1))^(1/3) - 1)

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giac [A]  time = 0.18, size = 144, normalized size = 0.92 \[ -\frac {22}{81} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right )}\right ) + \frac {2 \, {\left (\frac {10 \, {\left (x - 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{x + 1} - \frac {11 \, {\left (x - 1\right )}^{2} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{{\left (x + 1\right )}^{2}} - 35 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}\right )}}{27 \, {\left (\frac {x - 1}{x + 1} - 1\right )}^{3}} + \frac {11}{81} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {22}{81} \, \log \left ({\left | \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/3)*x^2,x, algorithm="giac")

[Out]

-22/81*sqrt(3)*arctan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(1/3) + 1)) + 2/27*(10*(x - 1)*((x - 1)/(x + 1))^(2/3)/
(x + 1) - 11*(x - 1)^2*((x - 1)/(x + 1))^(2/3)/(x + 1)^2 - 35*((x - 1)/(x + 1))^(2/3))/((x - 1)/(x + 1) - 1)^3
+ 11/81*log(((x - 1)/(x + 1))^(2/3) + ((x - 1)/(x + 1))^(1/3) + 1) - 22/81*log(abs(((x - 1)/(x + 1))^(1/3) -
1))

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maple [C]  time = 0.70, size = 408, normalized size = 2.60 \[ \frac {\left (9 x^{2}+12 x +14\right ) \left (-1+x \right )}{27 \left (\frac {-1+x}{1+x}\right )^{\frac {1}{3}}}+\frac {\left (-\frac {22 \ln \left (-\frac {4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x +4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}+3 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x -2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +x^{2}+3 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-1}{1+x}\right )}{81}+\frac {22 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -5 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-3 \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}+3 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x -6 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +2 x^{2}+3 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+4 x +2}{1+x}\right )}{81}\right ) \left (\left (-1+x \right ) \left (1+x \right )^{2}\right )^{\frac {1}{3}}}{\left (\frac {-1+x}{1+x}\right )^{\frac {1}{3}} \left (1+x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((-1+x)/(1+x))^(1/3)*x^2,x)

[Out]

1/27*(9*x^2+12*x+14)*(-1+x)/((-1+x)/(1+x))^(1/3)+(-22/81*ln(-(4*RootOf(_Z^2-_Z+1)^2*x^2+3*RootOf(_Z^2-_Z+1)*(x
^3+x^2-x-1)^(2/3)-3*RootOf(_Z^2-_Z+1)*(x^3+x^2-x-1)^(1/3)*x+4*RootOf(_Z^2-_Z+1)^2*x-4*RootOf(_Z^2-_Z+1)*x^2-3*
RootOf(_Z^2-_Z+1)*(x^3+x^2-x-1)^(1/3)+3*(x^3+x^2-x-1)^(1/3)*x-2*RootOf(_Z^2-_Z+1)*x+x^2+3*(x^3+x^2-x-1)^(1/3)+
2*RootOf(_Z^2-_Z+1)-1)/(1+x))+22/81*RootOf(_Z^2-_Z+1)*ln((2*RootOf(_Z^2-_Z+1)^2*x^2+3*RootOf(_Z^2-_Z+1)*(x^3+x
^2-x-1)^(2/3)+2*RootOf(_Z^2-_Z+1)^2*x-5*RootOf(_Z^2-_Z+1)*x^2-3*(x^3+x^2-x-1)^(2/3)+3*(x^3+x^2-x-1)^(1/3)*x-6*
RootOf(_Z^2-_Z+1)*x+2*x^2+3*(x^3+x^2-x-1)^(1/3)-RootOf(_Z^2-_Z+1)+4*x+2)/(1+x)))/((-1+x)/(1+x))^(1/3)*((-1+x)*
(1+x)^2)^(1/3)/(1+x)

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maxima [A]  time = 0.42, size = 149, normalized size = 0.95 \[ -\frac {22}{81} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right )}\right ) - \frac {2 \, {\left (11 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {8}{3}} - 10 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {5}{3}} + 35 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}\right )}}{27 \, {\left (\frac {3 \, {\left (x - 1\right )}}{x + 1} - \frac {3 \, {\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac {{\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - 1\right )}} + \frac {11}{81} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} + \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {22}{81} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/3)*x^2,x, algorithm="maxima")

[Out]

-22/81*sqrt(3)*arctan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(1/3) + 1)) - 2/27*(11*((x - 1)/(x + 1))^(8/3) - 10*((x
- 1)/(x + 1))^(5/3) + 35*((x - 1)/(x + 1))^(2/3))/(3*(x - 1)/(x + 1) - 3*(x - 1)^2/(x + 1)^2 + (x - 1)^3/(x +
1)^3 - 1) + 11/81*log(((x - 1)/(x + 1))^(2/3) + ((x - 1)/(x + 1))^(1/3) + 1) - 22/81*log(((x - 1)/(x + 1))^(1
/3) - 1)

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mupad [B]  time = 1.19, size = 171, normalized size = 1.09 \[ -\frac {22\,\ln \left (\frac {484\,{\left (\frac {x-1}{x+1}\right )}^{1/3}}{729}-\frac {484}{729}\right )}{81}-\frac {\frac {70\,{\left (\frac {x-1}{x+1}\right )}^{2/3}}{27}-\frac {20\,{\left (\frac {x-1}{x+1}\right )}^{5/3}}{27}+\frac {22\,{\left (\frac {x-1}{x+1}\right )}^{8/3}}{27}}{\frac {3\,\left (x-1\right )}{x+1}-\frac {3\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}+\frac {{\left (x-1\right )}^3}{{\left (x+1\right )}^3}-1}-\ln \left (\frac {484\,{\left (\frac {x-1}{x+1}\right )}^{1/3}}{729}-9\,{\left (-\frac {11}{81}+\frac {\sqrt {3}\,11{}\mathrm {i}}{81}\right )}^2\right )\,\left (-\frac {11}{81}+\frac {\sqrt {3}\,11{}\mathrm {i}}{81}\right )+\ln \left (\frac {484\,{\left (\frac {x-1}{x+1}\right )}^{1/3}}{729}-9\,{\left (\frac {11}{81}+\frac {\sqrt {3}\,11{}\mathrm {i}}{81}\right )}^2\right )\,\left (\frac {11}{81}+\frac {\sqrt {3}\,11{}\mathrm {i}}{81}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((x - 1)/(x + 1))^(1/3),x)

[Out]

log((484*((x - 1)/(x + 1))^(1/3))/729 - 9*((3^(1/2)*11i)/81 + 11/81)^2)*((3^(1/2)*11i)/81 + 11/81) - ((70*((x
- 1)/(x + 1))^(2/3))/27 - (20*((x - 1)/(x + 1))^(5/3))/27 + (22*((x - 1)/(x + 1))^(8/3))/27)/((3*(x - 1))/(x +
1) - (3*(x - 1)^2)/(x + 1)^2 + (x - 1)^3/(x + 1)^3 - 1) - log((484*((x - 1)/(x + 1))^(1/3))/729 - 9*((3^(1/2)
*11i)/81 - 11/81)^2)*((3^(1/2)*11i)/81 - 11/81) - (22*log((484*((x - 1)/(x + 1))^(1/3))/729 - 484/729))/81

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt [3]{\frac {x - 1}{x + 1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))**(1/3)*x**2,x)

[Out]

Integral(x**2/((x - 1)/(x + 1))**(1/3), x)

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