3.996 \(\int \frac{e^{\tanh ^{-1}(a x)} x^m}{(1-a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=70 \[ \frac{x^{m+1} \text{Hypergeometric2F1}\left (2,\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \text{Hypergeometric2F1}\left (2,\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{m+2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10401, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6150, 82, 73, 364} \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^m)/(1 - a^2*x^2)^(3/2),x]

[Out]

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

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 82

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[a, Int[(a + b*
x)^n*(c + d*x)^n*(f*x)^p, x], x] + Dist[b/f, Int[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b,
c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] &&  !RationalQ[p] &&  !IGtQ[m, 0] && NeQ[m +
n + p + 2, 0]

Rule 73

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m]

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

Mathematica [A]  time = 0.0347338, size = 67, normalized size = 0.96 \[ x^{m+1} \left (\frac{a x \text{Hypergeometric2F1}\left (2,\frac{m}{2}+1,\frac{m}{2}+2,a^2 x^2\right )}{m+2}+\frac{\text{Hypergeometric2F1}\left (2,\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^m)/(1 - a^2*x^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.207, size = 177, normalized size = 2.5 \begin{align*} -{\frac{1}{2\,a} \left ( -{a}^{2} \right ) ^{-{\frac{m}{2}}} \left ({\frac{{x}^{m} \left ( -m-2 \right ) }{ \left ( 2+m \right ) \left ( -{a}^{2}{x}^{2}+1 \right ) } \left ( -{a}^{2} \right ) ^{{\frac{m}{2}}}}+{\frac{{x}^{m}m}{2} \left ( -{a}^{2} \right ) ^{{\frac{m}{2}}}{\it LerchPhi} \left ({a}^{2}{x}^{2},1,{\frac{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)/(-a^2*x^2+1)^2*x^m,x)

[Out]

-1/2/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))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m}}{a^{3} x^{3} - a^{2} x^{2} - 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)^2*x^m,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 4.73726, size = 673, normalized size = 9.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*m**2*x**3*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m
/2 + 3/2) - 8*gamma(m/2 + 3/2)) + a**2*x**3*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2
+ 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) + a*(-a**2*m**2*x**4*x**m*lerchphi(a**2*x**2*exp_p
olar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) - 2*a**2*m*x**4*x**m*
lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2
)) + m**2*x**2*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 +
2) - 8*gamma(m/2 + 2)) + 2*m*x**2*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**
2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) - 2*m*x**2*x**m*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma
(m/2 + 2)) - 4*x**2*x**m*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2))) + m**2*x*x**m*lerchph
i(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/
2)) - 2*m*x*x**m*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - x*x**m*lerchphi(a**2*x
**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - 2*
x*x**m*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*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}}{{\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)/(-a^2*x^2+1)^2*x^m,x, algorithm="giac")

[Out]

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