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

Optimal. Leaf size=99 \[ \sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3}-\log \left (\frac {\sqrt [3]{\frac {x-1}{x}}}{\sqrt [3]{\frac {1}{x}+1}}+1\right )-\frac {1}{3} \log \left (\frac {1}{x}+1\right )-\frac {2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}\right )}{\sqrt {3}} \]

[Out]

(1/x+1)^(1/3)*((-1+x)/x)^(2/3)-ln(1+((-1+x)/x)^(1/3)/(1/x+1)^(1/3))-1/3*ln(1/x+1)+2/3*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 = 0.04, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6171, 50, 60} \[ \sqrt [3]{\frac {1}{x}+1} \left (\frac {x-1}{x}\right )^{2/3}-\log \left (\frac {\sqrt [3]{\frac {x-1}{x}}}{\sqrt [3]{\frac {1}{x}+1}}+1\right )-\frac {1}{3} \log \left (\frac {1}{x}+1\right )-\frac {2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + x^(-1))^(1/3)*((-1 + x)/x)^(2/3) - (2*ArcTan[1/Sqrt[3] - (2*((-1 + x)/x)^(1/3))/(Sqrt[3]*(1 + x^(-1))^(1/
3))])/Sqrt[3] - Log[1 + ((-1 + x)/x)^(1/3)/(1 + x^(-1))^(1/3)] - Log[1 + x^(-1)]/3

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 60

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(d/b), 3]}, Simp[(Sq
rt[3]*q*ArcTan[1/Sqrt[3] - (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3))])/d, x] + (Simp[(3*q*Log[(q*(a + b*
x)^(1/3))/(c + d*x)^(1/3) + 1])/(2*d), x] + Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0] && NegQ[d/b]

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 \frac {e^{\frac {2}{3} \coth ^{-1}(x)}}{x^2} \, dx &=-\operatorname {Subst}\left (\int \frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x}} \, dx,x,\frac {1}{x}\right )\\ &=\sqrt [3]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{2/3}-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (1+x)^{2/3}} \, dx,x,\frac {1}{x}\right )\\ &=\sqrt [3]{1+\frac {1}{x}} \left (\frac {-1+x}{x}\right )^{2/3}-\frac {2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-\frac {1-x}{x}}}{\sqrt {3} \sqrt [3]{1+\frac {1}{x}}}\right )}{\sqrt {3}}-\log \left (1+\frac {\sqrt [3]{-\frac {1-x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )-\frac {1}{3} \log \left (1+\frac {1}{x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 87, normalized size = 0.88 \[ \frac {2 e^{\frac {2}{3} \coth ^{-1}(x)}}{e^{2 \coth ^{-1}(x)}+1}-\frac {2}{3} \log \left (e^{\frac {2}{3} \coth ^{-1}(x)}+1\right )+\frac {1}{3} \log \left (-e^{\frac {2}{3} \coth ^{-1}(x)}+e^{\frac {4}{3} \coth ^{-1}(x)}+1\right )-\frac {2 \tan ^{-1}\left (\frac {2 e^{\frac {2}{3} \coth ^{-1}(x)}-1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*E^((2*ArcCoth[x])/3))/(1 + E^(2*ArcCoth[x])) - (2*ArcTan[(-1 + 2*E^((2*ArcCoth[x])/3))/Sqrt[3]])/Sqrt[3] -
(2*Log[1 + E^((2*ArcCoth[x])/3)])/3 + Log[1 - E^((2*ArcCoth[x])/3) + E^((4*ArcCoth[x])/3)]/3

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fricas [A]  time = 0.68, size = 97, normalized size = 0.98 \[ \frac {2 \, \sqrt {3} x \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + x \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - 2 \, x \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + 3 \, {\left (x + 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{3 \, x} \]

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]

1/3*(2*sqrt(3)*x*arctan(2/3*sqrt(3)*((x - 1)/(x + 1))^(1/3) - 1/3*sqrt(3)) + x*log(((x - 1)/(x + 1))^(2/3) - (
(x - 1)/(x + 1))^(1/3) + 1) - 2*x*log(((x - 1)/(x + 1))^(1/3) + 1) + 3*(x + 1)*((x - 1)/(x + 1))^(2/3))/x

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giac [A]  time = 0.14, size = 99, normalized size = 1.00 \[ \frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right )}\right ) + \frac {2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{\frac {x - 1}{x + 1} + 1} + \frac {1}{3} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {2}{3} \, \log \left ({\left | \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1 \right |}\right ) \]

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]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(1/3) - 1)) + 2*((x - 1)/(x + 1))^(2/3)/((x - 1)/(x + 1) +
1) + 1/3*log(((x - 1)/(x + 1))^(2/3) - ((x - 1)/(x + 1))^(1/3) + 1) - 2/3*log(abs(((x - 1)/(x + 1))^(1/3) + 1
))

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maple [C]  time = 0.79, size = 501, normalized size = 5.06 \[ \frac {-1+x}{x \left (\frac {-1+x}{1+x}\right )^{\frac {1}{3}}}+\frac {\left (-\frac {2 \ln \left (\frac {8 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x^{2}+27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}-45 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x -8 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x -30 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x^{2}-216 \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}-45 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-81 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x -16 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2}-54 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x -27 x^{2}-81 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-24 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )-36 x -9}{x \left (1+x \right )}\right )}{3}+\frac {2 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x^{2}+27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}+72 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x +2 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2} x -27 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x^{2}+135 \left (x^{3}+x^{2}-x -1\right )^{\frac {2}{3}}+72 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}-81 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}} x +4 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )^{2}+6 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right ) x -36 x^{2}-81 \left (x^{3}+x^{2}-x -1\right )^{\frac {1}{3}}+33 \RootOf \left (\textit {\_Z}^{2}-3 \textit {\_Z} +9\right )-216 x -180}{x \left (1+x \right )}\right )}{9}\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+x)/x/((-1+x)/(1+x))^(1/3)+(-2/3*ln((8*RootOf(_Z^2-3*_Z+9)^2*x^2+27*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(2/3)
-45*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(1/3)*x-8*RootOf(_Z^2-3*_Z+9)^2*x-30*RootOf(_Z^2-3*_Z+9)*x^2-216*(x^3+x^
2-x-1)^(2/3)-45*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(1/3)-81*(x^3+x^2-x-1)^(1/3)*x-16*RootOf(_Z^2-3*_Z+9)^2-54*R
ootOf(_Z^2-3*_Z+9)*x-27*x^2-81*(x^3+x^2-x-1)^(1/3)-24*RootOf(_Z^2-3*_Z+9)-36*x-9)/x/(1+x))+2/9*RootOf(_Z^2-3*_
Z+9)*ln(-(-2*RootOf(_Z^2-3*_Z+9)^2*x^2+27*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(2/3)+72*RootOf(_Z^2-3*_Z+9)*(x^3+
x^2-x-1)^(1/3)*x+2*RootOf(_Z^2-3*_Z+9)^2*x-27*RootOf(_Z^2-3*_Z+9)*x^2+135*(x^3+x^2-x-1)^(2/3)+72*RootOf(_Z^2-3
*_Z+9)*(x^3+x^2-x-1)^(1/3)-81*(x^3+x^2-x-1)^(1/3)*x+4*RootOf(_Z^2-3*_Z+9)^2+6*RootOf(_Z^2-3*_Z+9)*x-36*x^2-81*
(x^3+x^2-x-1)^(1/3)+33*RootOf(_Z^2-3*_Z+9)-216*x-180)/x/(1+x)))/((-1+x)/(1+x))^(1/3)*((-1+x)*(1+x)^2)^(1/3)/(1
+x)

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maxima [A]  time = 0.41, size = 98, normalized size = 0.99 \[ \frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right )}\right ) + \frac {2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{\frac {x - 1}{x + 1} + 1} + \frac {1}{3} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {2}{3} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) \]

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]

2/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(1/3) - 1)) + 2*((x - 1)/(x + 1))^(2/3)/((x - 1)/(x + 1) +
1) + 1/3*log(((x - 1)/(x + 1))^(2/3) - ((x - 1)/(x + 1))^(1/3) + 1) - 2/3*log(((x - 1)/(x + 1))^(1/3) + 1)

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mupad [B]  time = 0.02, size = 118, normalized size = 1.19 \[ \frac {2\,{\left (\frac {x-1}{x+1}\right )}^{2/3}}{\frac {x-1}{x+1}+1}-\ln \left (9\,{\left (-\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )}^2+4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )+\ln \left (9\,{\left (\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )}^2+4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )-\frac {2\,\ln \left (4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}+4\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(9*((3^(1/2)*1i)/3 + 1/3)^2 + 4*((x - 1)/(x + 1))^(1/3))*((3^(1/2)*1i)/3 + 1/3) - log(9*((3^(1/2)*1i)/3 - 1
/3)^2 + 4*((x - 1)/(x + 1))^(1/3))*((3^(1/2)*1i)/3 - 1/3) - (2*log(4*((x - 1)/(x + 1))^(1/3) + 4))/3 + (2*((x
- 1)/(x + 1))^(2/3))/((x - 1)/(x + 1) + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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