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

Optimal. Leaf size=28 \[ \frac {x^{m+1} F_1\left (m+1;\frac {1}{6},-\frac {1}{6};m+2;x,-x\right )}{m+1} \]

[Out]

x^(1+m)*AppellF1(1+m,1/6,-1/6,2+m,x,-x)/(1+m)

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6126, 133} \[ \frac {x^{m+1} F_1\left (m+1;\frac {1}{6},-\frac {1}{6};m+2;x,-x\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^(ArcTanh[x]/3)*x^m,x]

[Out]

(x^(1 + m)*AppellF1[1 + m, 1/6, -1/6, 2 + m, x, -x])/(1 + m)

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 6126

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

Rubi steps

\begin {align*} \int e^{\frac {1}{3} \tanh ^{-1}(x)} x^m \, dx &=\int \frac {x^m \sqrt [6]{1+x}}{\sqrt [6]{1-x}} \, dx\\ &=\frac {x^{1+m} F_1\left (1+m;\frac {1}{6},-\frac {1}{6};2+m;x,-x\right )}{1+m}\\ \end {align*}

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Mathematica [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int e^{\frac {1}{3} \tanh ^{-1}(x)} x^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[E^(ArcTanh[x]/3)*x^m,x]

[Out]

Integrate[E^(ArcTanh[x]/3)*x^m, x]

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{m} \left (-\frac {\sqrt {-x^{2} + 1}}{x - 1}\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m*(-sqrt(-x^2 + 1)/(x - 1))^(1/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\frac {x + 1}{\sqrt {-x^{2} + 1}}\right )^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \left (\frac {1+x}{\sqrt {-x^{2}+1}}\right )^{\frac {1}{3}} x^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \left (\frac {x + 1}{\sqrt {-x^{2} + 1}}\right )^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int x^m\,{\left (\frac {x+1}{\sqrt {1-x^2}}\right )}^{1/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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