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

Optimal. Leaf size=130 \[ \frac {1}{2} \left (\frac {x-1}{x}\right )^{2/3} \left (\frac {1}{x}+1\right )^{4/3}+\frac {1}{3} \left (\frac {x-1}{x}\right )^{2/3} \sqrt [3]{\frac {1}{x}+1}-\frac {1}{3} \log \left (\frac {\sqrt [3]{\frac {x-1}{x}}}{\sqrt [3]{\frac {1}{x}+1}}+1\right )-\frac {1}{9} \log \left (\frac {1}{x}+1\right )-\frac {2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}\right )}{3 \sqrt {3}} \]

[Out]

1/3*(1/x+1)^(1/3)*((-1+x)/x)^(2/3)+1/2*(1/x+1)^(4/3)*((-1+x)/x)^(2/3)-1/3*ln(1+((-1+x)/x)^(1/3)/(1/x+1)^(1/3))
-1/9*ln(1/x+1)+2/9*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 = 0.05, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6171, 80, 50, 60} \[ \frac {1}{2} \left (\frac {x-1}{x}\right )^{2/3} \left (\frac {1}{x}+1\right )^{4/3}+\frac {1}{3} \left (\frac {x-1}{x}\right )^{2/3} \sqrt [3]{\frac {1}{x}+1}-\frac {1}{3} \log \left (\frac {\sqrt [3]{\frac {x-1}{x}}}{\sqrt [3]{\frac {1}{x}+1}}+1\right )-\frac {1}{9} \log \left (\frac {1}{x}+1\right )-\frac {2 \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{\frac {x-1}{x}}}{\sqrt {3} \sqrt [3]{\frac {1}{x}+1}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + x^(-1))^(1/3)*((-1 + x)/x)^(2/3))/3 + ((1 + x^(-1))^(4/3)*((-1 + x)/x)^(2/3))/2 - (2*ArcTan[1/Sqrt[3] -
(2*((-1 + x)/x)^(1/3))/(Sqrt[3]*(1 + x^(-1))^(1/3))])/(3*Sqrt[3]) - Log[1 + ((-1 + x)/x)^(1/3)/(1 + x^(-1))^(1
/3)]/3 - Log[1 + x^(-1)]/9

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 80

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

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^3} \, dx &=-\operatorname {Subst}\left (\int \frac {x \sqrt [3]{1+x}}{\sqrt [3]{1-x}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{2} \left (1+\frac {1}{x}\right )^{4/3} \left (\frac {-1+x}{x}\right )^{2/3}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt [3]{1+x}}{\sqrt [3]{1-x}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3}+\frac {1}{2} \left (1+\frac {1}{x}\right )^{4/3} \left (\frac {-1+x}{x}\right )^{2/3}-\frac {2}{9} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (1+x)^{2/3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} \sqrt [3]{1+\frac {1}{x}} \left (-\frac {1-x}{x}\right )^{2/3}+\frac {1}{2} \left (1+\frac {1}{x}\right )^{4/3} \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 )}{3 \sqrt {3}}-\frac {1}{3} \log \left (1+\frac {\sqrt [3]{-\frac {1-x}{x}}}{\sqrt [3]{1+\frac {1}{x}}}\right )-\frac {1}{9} \log \left (1+\frac {1}{x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.32, size = 134, normalized size = 1.03 \[ -\frac {2}{27} \left (-\text {RootSum}\left [\text {$\#$1}^4-\text {$\#$1}^2+1\& ,\frac {\text {$\#$1}^2 \coth ^{-1}(x)-3 \text {$\#$1}^2 \log \left (e^{\frac {1}{3} \coth ^{-1}(x)}-\text {$\#$1}\right )-3 \log \left (e^{\frac {1}{3} \coth ^{-1}(x)}-\text {$\#$1}\right )+\coth ^{-1}(x)}{\text {$\#$1}^2-2}\& \right ]-2 \coth ^{-1}(x)-\frac {36 e^{\frac {2}{3} \coth ^{-1}(x)}}{e^{2 \coth ^{-1}(x)}+1}+\frac {27 e^{\frac {2}{3} \coth ^{-1}(x)}}{\left (e^{2 \coth ^{-1}(x)}+1\right )^2}+3 \log \left (e^{\frac {2}{3} \coth ^{-1}(x)}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*((27*E^((2*ArcCoth[x])/3))/(1 + E^(2*ArcCoth[x]))^2 - (36*E^((2*ArcCoth[x])/3))/(1 + E^(2*ArcCoth[x])) - 2
*ArcCoth[x] + 3*Log[1 + E^((2*ArcCoth[x])/3)] - RootSum[1 - #1^2 + #1^4 & , (ArcCoth[x] - 3*Log[E^(ArcCoth[x]/
3) - #1] + ArcCoth[x]*#1^2 - 3*Log[E^(ArcCoth[x]/3) - #1]*#1^2)/(-2 + #1^2) & ]))/27

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fricas [A]  time = 0.58, size = 111, normalized size = 0.85 \[ \frac {4 \, \sqrt {3} x^{2} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + 2 \, x^{2} \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - 4 \, x^{2} \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) + 3 \, {\left (5 \, x^{2} + 8 \, x + 3\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{18 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(4*sqrt(3)*x^2*arctan(2/3*sqrt(3)*((x - 1)/(x + 1))^(1/3) - 1/3*sqrt(3)) + 2*x^2*log(((x - 1)/(x + 1))^(2
/3) - ((x - 1)/(x + 1))^(1/3) + 1) - 4*x^2*log(((x - 1)/(x + 1))^(1/3) + 1) + 3*(5*x^2 + 8*x + 3)*((x - 1)/(x
+ 1))^(2/3))/x^2

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giac [A]  time = 0.14, size = 122, normalized size = 0.94 \[ \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right )}\right ) + \frac {2 \, {\left (\frac {{\left (x - 1\right )} \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}}{x + 1} + 4 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}\right )}}{3 \, {\left (\frac {x - 1}{x + 1} + 1\right )}^{2}} + \frac {1}{9} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {2}{9} \, \log \left ({\left | \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(1/3) - 1)) + 2/3*((x - 1)*((x - 1)/(x + 1))^(2/3)/(x + 1)
+ 4*((x - 1)/(x + 1))^(2/3))/((x - 1)/(x + 1) + 1)^2 + 1/9*log(((x - 1)/(x + 1))^(2/3) - ((x - 1)/(x + 1))^(1
/3) + 1) - 2/9*log(abs(((x - 1)/(x + 1))^(1/3) + 1))

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maple [C]  time = 0.65, size = 738, normalized size = 5.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-1+x)*(5*x+3)/x^2/((-1+x)/(1+x))^(1/3)+(2/27*RootOf(_Z^2-3*_Z+9)*ln((-10*RootOf(_Z^2-3*_Z+9)^2*x^2+27*Roo
tOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(2/3)-27*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(1/3)*x+10*RootOf(_Z^2-3*_Z+9)^2*x+6
9*RootOf(_Z^2-3*_Z+9)*x^2-216*(x^3+x^2-x-1)^(2/3)-27*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(1/3)+216*(x^3+x^2-x-1)
^(1/3)*x+20*RootOf(_Z^2-3*_Z+9)^2+36*RootOf(_Z^2-3*_Z+9)*x-108*x^2+216*(x^3+x^2-x-1)^(1/3)-33*RootOf(_Z^2-3*_Z
+9)-144*x-36)/x/(1+x))-2/27*ln(-(10*RootOf(_Z^2-3*_Z+9)^2*x^2+27*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(2/3)-27*Ro
otOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(1/3)*x-10*RootOf(_Z^2-3*_Z+9)^2*x+9*RootOf(_Z^2-3*_Z+9)*x^2+135*(x^3+x^2-x-1)
^(2/3)-27*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(1/3)-135*(x^3+x^2-x-1)^(1/3)*x-20*RootOf(_Z^2-3*_Z+9)^2+96*RootOf
(_Z^2-3*_Z+9)*x-9*x^2-135*(x^3+x^2-x-1)^(1/3)+87*RootOf(_Z^2-3*_Z+9)-54*x-45)/x/(1+x))*RootOf(_Z^2-3*_Z+9)+2/9
*ln(-(10*RootOf(_Z^2-3*_Z+9)^2*x^2+27*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-x-1)^(2/3)-27*RootOf(_Z^2-3*_Z+9)*(x^3+x^2-
x-1)^(1/3)*x-10*RootOf(_Z^2-3*_Z+9)^2*x+9*RootOf(_Z^2-3*_Z+9)*x^2+135*(x^3+x^2-x-1)^(2/3)-27*RootOf(_Z^2-3*_Z+
9)*(x^3+x^2-x-1)^(1/3)-135*(x^3+x^2-x-1)^(1/3)*x-20*RootOf(_Z^2-3*_Z+9)^2+96*RootOf(_Z^2-3*_Z+9)*x-9*x^2-135*(
x^3+x^2-x-1)^(1/3)+87*RootOf(_Z^2-3*_Z+9)-54*x-45)/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 = 124, normalized size = 0.95 \[ \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} - 1\right )}\right ) + \frac {2 \, {\left (\left (\frac {x - 1}{x + 1}\right )^{\frac {5}{3}} + 4 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}}\right )}}{3 \, {\left (\frac {2 \, {\left (x - 1\right )}}{x + 1} + \frac {{\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + 1\right )}} + \frac {1}{9} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {2}{3}} - \left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) - \frac {2}{9} \, \log \left (\left (\frac {x - 1}{x + 1}\right )^{\frac {1}{3}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*((x - 1)/(x + 1))^(1/3) - 1)) + 2/3*(((x - 1)/(x + 1))^(5/3) + 4*((x - 1)/(x
+ 1))^(2/3))/(2*(x - 1)/(x + 1) + (x - 1)^2/(x + 1)^2 + 1) + 1/9*log(((x - 1)/(x + 1))^(2/3) - ((x - 1)/(x +
1))^(1/3) + 1) - 2/9*log(((x - 1)/(x + 1))^(1/3) + 1)

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mupad [B]  time = 0.03, size = 145, normalized size = 1.12 \[ \frac {\frac {8\,{\left (\frac {x-1}{x+1}\right )}^{2/3}}{3}+\frac {2\,{\left (\frac {x-1}{x+1}\right )}^{5/3}}{3}}{\frac {2\,\left (x-1\right )}{x+1}+\frac {{\left (x-1\right )}^2}{{\left (x+1\right )}^2}+1}-\frac {2\,\ln \left (\frac {4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}}{9}+\frac {4}{9}\right )}{9}-\ln \left (9\,{\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}^2+\frac {4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}}{9}\right )\,\left (-\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )+\ln \left (9\,{\left (\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right )}^2+\frac {4\,{\left (\frac {x-1}{x+1}\right )}^{1/3}}{9}\right )\,\left (\frac {1}{9}+\frac {\sqrt {3}\,1{}\mathrm {i}}{9}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((8*((x - 1)/(x + 1))^(2/3))/3 + (2*((x - 1)/(x + 1))^(5/3))/3)/((2*(x - 1))/(x + 1) + (x - 1)^2/(x + 1)^2 + 1
) - (2*log((4*((x - 1)/(x + 1))^(1/3))/9 + 4/9))/9 - log(9*((3^(1/2)*1i)/9 - 1/9)^2 + (4*((x - 1)/(x + 1))^(1/
3))/9)*((3^(1/2)*1i)/9 - 1/9) + log(9*((3^(1/2)*1i)/9 + 1/9)^2 + (4*((x - 1)/(x + 1))^(1/3))/9)*((3^(1/2)*1i)/
9 + 1/9)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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