3.143 \(\int e^{2 \tanh ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=36 \[ \frac{2 x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,a x)}{m+1}-\frac{x^{m+1}}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0245886, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6126, 80, 64} \[ \frac{2 x^{m+1} \, _2F_1(1,m+1;m+2;a x)}{m+1}-\frac{x^{m+1}}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])*x^m,x]

[Out]

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

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]

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 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int e^{2 \tanh ^{-1}(a x)} x^m \, dx &=\int \frac{x^m (1+a x)}{1-a x} \, dx\\ &=-\frac{x^{1+m}}{1+m}+2 \int \frac{x^m}{1-a x} \, dx\\ &=-\frac{x^{1+m}}{1+m}+\frac{2 x^{1+m} \, _2F_1(1,1+m;2+m;a x)}{1+m}\\ \end{align*}

Mathematica [A]  time = 0.0068547, size = 26, normalized size = 0.72 \[ \frac{x^{m+1} (2 \text{Hypergeometric2F1}(1,m+1,m+2,a x)-1)}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])*x^m,x]

[Out]

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

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Maple [C]  time = 0.359, size = 184, normalized size = 5.1 \begin{align*} -{\frac{1}{2} \left ( -{a}^{2} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( 2\,{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{3/2+m/2} \left ( -3-m \right ) }{ \left ( 1+m \right ) \left ( 3+m \right ){a}^{2}}}+{\frac{{x}^{1+m}}{{a}^{2}} \left ( -{a}^{2} \right ) ^{{\frac{3}{2}}+{\frac{m}{2}}}{\it LerchPhi} \left ({a}^{2}{x}^{2},1,{\frac{1}{2}}+{\frac{m}{2}} \right ) } \right ) }-{\frac{1}{a} \left ( -{a}^{2} \right ) ^{-{\frac{m}{2}}} \left ( -2\,{\frac{{x}^{m} \left ( -{a}^{2} \right ) ^{m/2} \left ( -m-2 \right ) }{m \left ( 2+m \right ) }}-{x}^{m} \left ( -{a}^{2} \right ) ^{{\frac{m}{2}}}{\it LerchPhi} \left ({a}^{2}{x}^{2},1,{\frac{m}{2}} \right ) \right ) }+{\frac{{x}^{1+m}}{1+m} \left ({\frac{1}{2}}+{\frac{m}{2}} \right ){\it LerchPhi} \left ({a}^{2}{x}^{2},1,{\frac{1}{2}}+{\frac{m}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-a^2)^(-1/2-1/2*m)*(2/(3+m)*x^(1+m)*(-a^2)^(3/2+1/2*m)*(-3-m)/(1+m)/a^2+x^(1+m)*(-a^2)^(3/2+1/2*m)/a^2*L
erchPhi(a^2*x^2,1,1/2+1/2*m))-1/a*(-a^2)^(-1/2*m)*(-2/(2+m)*x^m*(-a^2)^(1/2*m)*(-m-2)/m-x^m*(-a^2)^(1/2*m)*Ler
chPhi(a^2*x^2,1,1/2*m))+1/(1+m)*x^(1+m)*(1/2+1/2*m)*LerchPhi(a^2*x^2,1,1/2+1/2*m)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((a*x + 1)^2*x^m/(a^2*x^2 - 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}}{a x - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a*x + 1)*x^m/(a*x - 1), x)

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Sympy [B]  time = 3.26814, size = 99, normalized size = 2.75 \begin{align*} \frac{a m x^{2} x^{m} \Phi \left (a x, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} + \frac{2 a x^{2} x^{m} \Phi \left (a x, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} + \frac{m x x^{m} \Phi \left (a x, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} + \frac{x x^{m} \Phi \left (a x, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*m*x**2*x**m*lerchphi(a*x, 1, m + 2)*gamma(m + 2)/gamma(m + 3) + 2*a*x**2*x**m*lerchphi(a*x, 1, m + 2)*gamma(
m + 2)/gamma(m + 3) + m*x*x**m*lerchphi(a*x, 1, m + 1)*gamma(m + 1)/gamma(m + 2) + x*x**m*lerchphi(a*x, 1, m +
1)*gamma(m + 1)/gamma(m + 2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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