3.144 \(\int e^{\tanh ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=74 \[ \frac{x^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{m+2} \]

[Out]

(x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F1[1/2
, (2 + m)/2, (4 + m)/2, a^2*x^2])/(2 + m)

________________________________________________________________________________________

Rubi [A]  time = 0.0419184, 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 = {6124, 808, 364} \[ \frac{x^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F1[1/2
, (2 + m)/2, (4 + m)/2, a^2*x^2])/(2 + m)

Rule 6124

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

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^{\tanh ^{-1}(a x)} x^m \, dx &=\int \frac{x^m (1+a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=a \int \frac{x^{1+m}}{\sqrt{1-a^2 x^2}} \, dx+\int \frac{x^m}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{x^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};a^2 x^2\right )}{1+m}+\frac{a x^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{2+m}\\ \end{align*}

Mathematica [C]  time = 0.0299778, size = 70, normalized size = 0.95 \[ -\frac{\sqrt{-a x-1} \sqrt{1-a x} x^{m+1} F_1\left (m+1;-\frac{1}{2},\frac{1}{2};m+2;-a x,a x\right )}{(m+1) \sqrt{a x-1} \sqrt{a x+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.234, size = 67, normalized size = 0.9 \begin{align*}{\frac{{x}^{1+m}}{1+m}{\mbox{$_2$F$_1$}({\frac{1}{2}},{\frac{1}{2}}+{\frac{m}{2}};\,{\frac{3}{2}}+{\frac{m}{2}};\,{a}^{2}{x}^{2})}}+{\frac{a{x}^{2+m}}{2+m}{\mbox{$_2$F$_1$}({\frac{1}{2}},1+{\frac{m}{2}};\,2+{\frac{m}{2}};\,{a}^{2}{x}^{2})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1+m)*hypergeom([1/2,1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)+a*x^(2+m)*hypergeom([1/2,1+1/2*m],[2+1/2*m],a^2*x
^2)/(2+m)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )} x^{m}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1} x^{m}}{a x - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 3.7963, size = 97, normalized size = 1.31 \begin{align*} \frac{a x^{2} x^{m} \Gamma \left (\frac{m}{2} + 1\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac{m}{2} + 2\right )} + \frac{x x^{m} \Gamma \left (\frac{m}{2} + \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**2*x**m*gamma(m/2 + 1)*hyper((1/2, m/2 + 1), (m/2 + 2,), a**2*x**2*exp_polar(2*I*pi))/(2*gamma(m/2 + 2)) +
x*x**m*gamma(m/2 + 1/2)*hyper((1/2, m/2 + 1/2), (m/2 + 3/2,), a**2*x**2*exp_polar(2*I*pi))/(2*gamma(m/2 + 3/2
))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )} x^{m}}{\sqrt{-a^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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