3.135 \(\int e^{\coth ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=74 \[ \frac{x^m \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{m}{2},1-\frac{m}{2},\frac{1}{a^2 x^2}\right )}{a m}+\frac{x^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (-m-1),\frac{1-m}{2},\frac{1}{a^2 x^2}\right )}{m+1} \]

[Out]

(x^(1 + m)*Hypergeometric2F1[1/2, (-1 - m)/2, (1 - m)/2, 1/(a^2*x^2)])/(1 + m) + (x^m*Hypergeometric2F1[1/2, -
m/2, 1 - m/2, 1/(a^2*x^2)])/(a*m)

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Rubi [A]  time = 0.0615149, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6172, 808, 364} \[ \frac{x^m \, _2F_1\left (\frac{1}{2},-\frac{m}{2};1-\frac{m}{2};\frac{1}{a^2 x^2}\right )}{a m}+\frac{x^{m+1} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-m-1);\frac{1-m}{2};\frac{1}{a^2 x^2}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]*x^m,x]

[Out]

(x^(1 + m)*Hypergeometric2F1[1/2, (-1 - m)/2, (1 - m)/2, 1/(a^2*x^2)])/(1 + m) + (x^m*Hypergeometric2F1[1/2, -
m/2, 1 - m/2, 1/(a^2*x^2)])/(a*m)

Rule 6172

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

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int e^{\coth ^{-1}(a x)} x^m \, dx &=-\left (\left (\left (\frac{1}{x}\right )^m x^m\right ) \operatorname{Subst}\left (\int \frac{x^{-2-m} \left (1+\frac{x}{a}\right )}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\right )\\ &=-\left (\left (\left (\frac{1}{x}\right )^m x^m\right ) \operatorname{Subst}\left (\int \frac{x^{-2-m}}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\right )-\frac{\left (\left (\frac{1}{x}\right )^m x^m\right ) \operatorname{Subst}\left (\int \frac{x^{-1-m}}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\frac{x^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1}{2} (-1-m);\frac{1-m}{2};\frac{1}{a^2 x^2}\right )}{1+m}+\frac{x^m \, _2F_1\left (\frac{1}{2},-\frac{m}{2};1-\frac{m}{2};\frac{1}{a^2 x^2}\right )}{a m}\\ \end{align*}

Mathematica [C]  time = 0.38729, size = 128, normalized size = 1.73 \[ x^{m+1} \left (\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},-\frac{m}{2}-\frac{1}{2},\frac{1}{2}-\frac{m}{2},\frac{1}{a^2 x^2}\right )}{m+1}-\frac{\sqrt{1-\frac{1}{a^2 x^2}} \sqrt{x^2-\frac{1}{a^2}} F_1\left (m;-\frac{1}{2},\frac{1}{2};m+1;-a x,a x\right )}{m \sqrt{a x-1} \sqrt{\frac{a x+1}{a^2}} \sqrt{1-a^2 x^2}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[a*x]*x^m,x]

[Out]

x^(1 + m)*(-((Sqrt[1 - 1/(a^2*x^2)]*Sqrt[-a^(-2) + x^2]*AppellF1[m, -1/2, 1/2, 1 + m, -(a*x), a*x])/(m*Sqrt[-1
+ a*x]*Sqrt[(1 + a*x)/a^2]*Sqrt[1 - a^2*x^2])) + Hypergeometric2F1[-1/2, -1/2 - m/2, 1/2 - m/2, 1/(a^2*x^2)]/
(1 + m))

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Maple [F]  time = 0.184, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m}{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{\sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a x + 1\right )} x^{m} \sqrt{\frac{a x - 1}{a x + 1}}}{a x - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{\sqrt{\frac{a x - 1}{a x + 1}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**m/sqrt((a*x - 1)/(a*x + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{\sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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