3.354 \(\int \frac{x^3}{(1-a^2 x^2)^4 \tanh ^{-1}(a x)} \, dx\)

Optimal. Leaf size=29 \[ \frac{\text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^4}-\frac{3 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^4} \]

[Out]

(-3*SinhIntegral[2*ArcTanh[a*x]])/(32*a^4) + SinhIntegral[6*ArcTanh[a*x]]/(32*a^4)

________________________________________________________________________________________

Rubi [A]  time = 0.125032, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6034, 5448, 3298} \[ \frac{\text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^4}-\frac{3 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((1 - a^2*x^2)^4*ArcTanh[a*x]),x]

[Out]

(-3*SinhIntegral[2*ArcTanh[a*x]])/(32*a^4) + SinhIntegral[6*ArcTanh[a*x]]/(32*a^4)

Rule 6034

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(
m + 1), Subst[Int[((a + b*x)^p*Sinh[x]^m)/Cosh[x]^(m + 2*(q + 1)), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c,
d, e, p}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{x^3}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh ^3(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{3 \sinh (2 x)}{32 x}+\frac{\sinh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^4}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^4}\\ &=-\frac{3 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^4}+\frac{\text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^4}\\ \end{align*}

Mathematica [A]  time = 0.140588, size = 24, normalized size = 0.83 \[ \frac{\text{Shi}\left (6 \tanh ^{-1}(a x)\right )-3 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((1 - a^2*x^2)^4*ArcTanh[a*x]),x]

[Out]

(-3*SinhIntegral[2*ArcTanh[a*x]] + SinhIntegral[6*ArcTanh[a*x]])/(32*a^4)

________________________________________________________________________________________

Maple [A]  time = 0.066, size = 24, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{4}} \left ( -{\frac{3\,{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{32}}+{\frac{{\it Shi} \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{32}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(-a^2*x^2+1)^4/arctanh(a*x),x)

[Out]

1/a^4*(-3/32*Shi(2*arctanh(a*x))+1/32*Shi(6*arctanh(a*x)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname{artanh}\left (a x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="maxima")

[Out]

integrate(x^3/((a^2*x^2 - 1)^4*arctanh(a*x)), x)

________________________________________________________________________________________

Fricas [B]  time = 1.94995, size = 348, normalized size = 12. \begin{align*} \frac{\logintegral \left (-\frac{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - \logintegral \left (-\frac{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) - 3 \, \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) + 3 \, \logintegral \left (-\frac{a x - 1}{a x + 1}\right )}{64 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="fricas")

[Out]

1/64*(log_integral(-(a^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)/(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)) - log_integral(-(a^3*
x^3 - 3*a^2*x^2 + 3*a*x - 1)/(a^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)) - 3*log_integral(-(a*x + 1)/(a*x - 1)) + 3*log
_integral(-(a*x - 1)/(a*x + 1)))/a^4

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname{atanh}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(-a**2*x**2+1)**4/atanh(a*x),x)

[Out]

Integral(x**3/((a*x - 1)**4*(a*x + 1)**4*atanh(a*x)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname{artanh}\left (a x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="giac")

[Out]

integrate(x^3/((a^2*x^2 - 1)^4*arctanh(a*x)), x)