3.141 \(\int e^{4 \tanh ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=45 \[ -4 x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,a x)+\frac{4 x^{m+1}}{1-a x}+\frac{x^{m+1}}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0431906, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6126, 89, 80, 64} \[ -4 x^{m+1} \, _2F_1(1,m+1;m+2;a x)+\frac{4 x^{m+1}}{1-a x}+\frac{x^{m+1}}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])*x^m,x]

[Out]

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

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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

Mathematica [A]  time = 0.0227577, size = 47, normalized size = 1.04 \[ \frac{x^{m+1} (-4 (m+1) (a x-1) \text{Hypergeometric2F1}(1,m+1,m+2,a x)+a x-4 m-5)}{(m+1) (a x-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])*x^m,x]

[Out]

(x^(1 + m)*(-5 - 4*m + a*x - 4*(1 + m)*(-1 + a*x)*Hypergeometric2F1[1, 1 + m, 2 + m, a*x]))/((1 + m)*(-1 + a*x
))

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Maple [C]  time = 0.444, size = 461, normalized size = 10.2 \begin{align*}{\frac{1}{2} \left ( -{a}^{2} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( -{\frac{{x}^{1+m} \left ( 2\,{a}^{2}{x}^{2}-m-3 \right ) }{ \left ( -{a}^{2}{x}^{2}+1 \right ){a}^{4} \left ( 1+m \right ) } \left ( -{a}^{2} \right ) ^{{\frac{5}{2}}+{\frac{m}{2}}}}-{\frac{{x}^{1+m} \left ( 3+m \right ) }{2\,{a}^{4}} \left ( -{a}^{2} \right ) ^{{\frac{5}{2}}+{\frac{m}{2}}}{\it LerchPhi} \left ({a}^{2}{x}^{2},1,{\frac{1}{2}}+{\frac{m}{2}} \right ) } \right ) }+2\,{\frac{ \left ( -{a}^{2} \right ) ^{-m/2}}{a} \left ( -{\frac{{x}^{m} \left ( -{a}^{2} \right ) ^{m/2} \left ( 2\,{a}^{2}{x}^{2}-m-2 \right ) }{ \left ( -{a}^{2}{x}^{2}+1 \right ) m}}-1/2\,{x}^{m} \left ( -{a}^{2} \right ) ^{m/2} \left ( 2+m \right ){\it LerchPhi} \left ({a}^{2}{x}^{2},1,m/2 \right ) \right ) }-3\, \left ( -{a}^{2} \right ) ^{-1/2-m/2} \left ( -{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{3/2+m/2} \left ( -3-m \right ) }{{a}^{2} \left ( 3+m \right ) \left ( -{a}^{2}{x}^{2}+1 \right ) }}-1/2\,{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{3/2+m/2} \left ( 1+m \right ){\it LerchPhi} \left ({a}^{2}{x}^{2},1,1/2+m/2 \right ) }{{a}^{2}}} \right ) -2\,{\frac{ \left ( -{a}^{2} \right ) ^{-m/2}}{a} \left ({\frac{{x}^{m} \left ( -{a}^{2} \right ) ^{m/2} \left ( -m-2 \right ) }{ \left ( 2+m \right ) \left ( -{a}^{2}{x}^{2}+1 \right ) }}+1/2\,{x}^{m} \left ( -{a}^{2} \right ) ^{m/2}m{\it LerchPhi} \left ({a}^{2}{x}^{2},1,m/2 \right ) \right ) }+{\frac{1}{2} \left ( -{a}^{2} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( -2\,{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{1/2+m/2} \left ( -1-m \right ) }{ \left ( 1+m \right ) \left ( -2\,{a}^{2}{x}^{2}+2 \right ) }}+2\,{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{1/2+m/2} \left ( -1/4\,{m}^{2}+1/4 \right ){\it LerchPhi} \left ({a}^{2}{x}^{2},1,1/2+m/2 \right ) }{1+m}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-a^2)^(-1/2-1/2*m)*(-x^(1+m)*(-a^2)^(5/2+1/2*m)*(2*a^2*x^2-m-3)/(-a^2*x^2+1)/a^4/(1+m)-1/2*x^(1+m)*(-a^2)
^(5/2+1/2*m)*(3+m)/a^4*LerchPhi(a^2*x^2,1,1/2+1/2*m))+2/a*(-a^2)^(-1/2*m)*(-x^m*(-a^2)^(1/2*m)*(2*a^2*x^2-m-2)
/(-a^2*x^2+1)/m-1/2*x^m*(-a^2)^(1/2*m)*(2+m)*LerchPhi(a^2*x^2,1,1/2*m))-3*(-a^2)^(-1/2-1/2*m)*(-1/(3+m)*x^(1+m
)*(-a^2)^(3/2+1/2*m)*(-3-m)/a^2/(-a^2*x^2+1)-1/2*x^(1+m)*(-a^2)^(3/2+1/2*m)*(1+m)/a^2*LerchPhi(a^2*x^2,1,1/2+1
/2*m))-2/a*(-a^2)^(-1/2*m)*(1/(2+m)*x^m*(-a^2)^(1/2*m)*(-m-2)/(-a^2*x^2+1)+1/2*x^m*(-a^2)^(1/2*m)*m*LerchPhi(a
^2*x^2,1,1/2*m))+1/2*(-a^2)^(-1/2-1/2*m)*(-2/(1+m)*x^(1+m)*(-a^2)^(1/2+1/2*m)*(-1-m)/(-2*a^2*x^2+2)+2/(1+m)*x^
(1+m)*(-a^2)^(1/2+1/2*m)*(-1/4*m^2+1/4)*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 )}^{4} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x^m,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x^m,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2*x**m,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}^{4} x^{m}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*x^m,x, algorithm="giac")

[Out]

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