3.2.39 \(\int e^{\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx\) [139]

Optimal. Leaf size=41 \[ \frac {x^{1+m} F_1\left (-1-m;\frac {5}{4},-\frac {5}{4};-m;\frac {1}{a x},-\frac {1}{a x}\right )}{1+m} \]

[Out]

x^(1+m)*AppellF1(-1-m,5/4,-5/4,-m,1/a/x,-1/a/x)/(1+m)

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Rubi [A]
time = 0.03, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6308, 138} \begin {gather*} \frac {x^{m+1} F_1\left (-m-1;\frac {5}{4},-\frac {5}{4};-m;\frac {1}{a x},-\frac {1}{a x}\right )}{m+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((5*ArcCoth[a*x])/2)*x^m,x]

[Out]

(x^(1 + m)*AppellF1[-1 - m, 5/4, -5/4, -m, 1/(a*x), -(1/(a*x))])/(1 + m)

Rule 138

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

Rule 6308

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

Rubi steps

\begin {align*} \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx &=-\left (\left (\left (\frac {1}{x}\right )^m x^m\right ) \text {Subst}\left (\int \frac {x^{-2-m} \left (1+\frac {x}{a}\right )^{5/4}}{\left (1-\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )\right )\\ &=\frac {x^{1+m} F_1\left (-1-m;\frac {5}{4},-\frac {5}{4};-m;\frac {1}{a x},-\frac {1}{a x}\right )}{1+m}\\ \end {align*}

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Mathematica [F]
time = 0.42, size = 0, normalized size = 0.00 \begin {gather*} \int e^{\frac {5}{2} \coth ^{-1}(a x)} x^m \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[E^((5*ArcCoth[a*x])/2)*x^m,x]

[Out]

Integrate[E^((5*ArcCoth[a*x])/2)*x^m, x]

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(5/4)*x^m,x)

[Out]

int(1/((a*x-1)/(a*x+1))^(5/4)*x^m,x)

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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(1/((a*x-1)/(a*x+1))^(5/4)*x^m,x, algorithm="maxima")

[Out]

integrate(x^m/((a*x - 1)/(a*x + 1))^(5/4), x)

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Fricas [F]
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(1/((a*x-1)/(a*x+1))^(5/4)*x^m,x, algorithm="fricas")

[Out]

integral((a^2*x^2 + 2*a*x + 1)*x^m*((a*x - 1)/(a*x + 1))^(3/4)/(a^2*x^2 - 2*a*x + 1), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))**(5/4)*x**m,x)

[Out]

Timed out

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Giac [F]
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(1/((a*x-1)/(a*x+1))^(5/4)*x^m,x, algorithm="giac")

[Out]

integrate(x^m/((a*x - 1)/(a*x + 1))^(5/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^m}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/((a*x - 1)/(a*x + 1))^(5/4),x)

[Out]

int(x^m/((a*x - 1)/(a*x + 1))^(5/4), x)

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