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

Optimal. Leaf size=53 \[ -\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^5}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^5}-\frac{x^4}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]

[Out]

-(x^4/(a*(1 - a^2*x^2)^2*ArcTanh[a*x])) - SinhIntegral[2*ArcTanh[a*x]]/a^5 + SinhIntegral[4*ArcTanh[a*x]]/(2*a
^5)

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Rubi [A]  time = 0.186329, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6006, 6034, 5448, 3298} \[ -\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^5}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^5}-\frac{x^4}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^4/(a*(1 - a^2*x^2)^2*ArcTanh[a*x])) - SinhIntegral[2*ArcTanh[a*x]]/a^5 + SinhIntegral[4*ArcTanh[a*x]]/(2*a
^5)

Rule 6006

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^m*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^(p + 1))/(b*c*d*(p + 1)), x] - Dist[(f*m)/(b*c*(p + 1)), In
t[(f*x)^(m - 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && Eq
Q[c^2*d + e, 0] && EqQ[m + 2*q + 2, 0] && LtQ[p, -1]

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^4}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx &=-\frac{x^4}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{4 \int \frac{x^3}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx}{a}\\ &=-\frac{x^4}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^3(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}\\ &=-\frac{x^4}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{4 \operatorname{Subst}\left (\int \left (-\frac{\sinh (2 x)}{4 x}+\frac{\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}\\ &=-\frac{x^4}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^5}-\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}\\ &=-\frac{x^4}{a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac{\text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^5}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^5}\\ \end{align*}

Mathematica [A]  time = 0.185935, size = 49, normalized size = 0.92 \[ \frac{-\frac{2 a^4 x^4}{\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)}-2 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )+\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a^4*x^4)/((-1 + a^2*x^2)^2*ArcTanh[a*x]) - 2*SinhIntegral[2*ArcTanh[a*x]] + SinhIntegral[4*ArcTanh[a*x]])
/(2*a^5)

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Maple [A]  time = 0.069, size = 62, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{5}} \left ( -{\frac{3}{8\,{\it Artanh} \left ( ax \right ) }}+{\frac{\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{2\,{\it Artanh} \left ( ax \right ) }}-{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) -{\frac{\cosh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{8\,{\it Artanh} \left ( ax \right ) }}+{\frac{{\it Shi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^5*(-3/8/arctanh(a*x)+1/2/arctanh(a*x)*cosh(2*arctanh(a*x))-Shi(2*arctanh(a*x))-1/8/arctanh(a*x)*cosh(4*arc
tanh(a*x))+1/2*Shi(4*arctanh(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, x^{4}}{{\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right ) -{\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-a x + 1\right )} + 8 \, \int -\frac{x^{3}}{{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) -{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^4/((a^5*x^4 - 2*a^3*x^2 + a)*log(a*x + 1) - (a^5*x^4 - 2*a^3*x^2 + a)*log(-a*x + 1)) + 8*integrate(-x^3/(
(a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)*log(a*x + 1) - (a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)*log(-a*x + 1)), x)

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Fricas [B]  time = 2.32265, size = 545, normalized size = 10.28 \begin{align*} -\frac{8 \, a^{4} x^{4} -{\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (\frac{a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) -{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (\frac{a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) + 2 \,{\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \logintegral \left (-\frac{a x - 1}{a x + 1}\right )\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{4 \,{\left (a^{9} x^{4} - 2 \, a^{7} x^{2} + a^{5}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*a^4*x^4 - ((a^4*x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) - (a^4*
x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) - 2*(a^4*x^4 - 2*a^2*x^2 + 1)*l
og_integral(-(a*x + 1)/(a*x - 1)) + 2*(a^4*x^4 - 2*a^2*x^2 + 1)*log_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x
+ 1)/(a*x - 1)))/((a^9*x^4 - 2*a^7*x^2 + a^5)*log(-(a*x + 1)/(a*x - 1)))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{4}}{a^{6} x^{6} \operatorname{atanh}^{2}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname{atanh}^{2}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname{atanh}^{2}{\left (a x \right )} - \operatorname{atanh}^{2}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**4/(a**6*x**6*atanh(a*x)**2 - 3*a**4*x**4*atanh(a*x)**2 + 3*a**2*x**2*atanh(a*x)**2 - atanh(a*x)**
2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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