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

Optimal. Leaf size=55 \[ -\frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^5}-\frac{\text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{16 a^5}+\frac{\text{Chi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^5}+\frac{\log \left (\tanh ^{-1}(a x)\right )}{16 a^5} \]

[Out]

-CoshIntegral[2*ArcTanh[a*x]]/(32*a^5) - CoshIntegral[4*ArcTanh[a*x]]/(16*a^5) + CoshIntegral[6*ArcTanh[a*x]]/
(32*a^5) + Log[ArcTanh[a*x]]/(16*a^5)

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Rubi [A]  time = 0.154642, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6034, 5448, 3301} \[ -\frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^5}-\frac{\text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{16 a^5}+\frac{\text{Chi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^5}+\frac{\log \left (\tanh ^{-1}(a x)\right )}{16 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-CoshIntegral[2*ArcTanh[a*x]]/(32*a^5) - CoshIntegral[4*ArcTanh[a*x]]/(16*a^5) + CoshIntegral[6*ArcTanh[a*x]]/
(32*a^5) + Log[ArcTanh[a*x]]/(16*a^5)

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 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{x^4}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh ^2(x) \sinh ^4(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{16 x}-\frac{\cosh (2 x)}{32 x}-\frac{\cosh (4 x)}{16 x}+\frac{\cosh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}\\ &=\frac{\log \left (\tanh ^{-1}(a x)\right )}{16 a^5}-\frac{\operatorname{Subst}\left (\int \frac{\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^5}+\frac{\operatorname{Subst}\left (\int \frac{\cosh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^5}-\frac{\operatorname{Subst}\left (\int \frac{\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a^5}\\ &=-\frac{\text{Chi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^5}-\frac{\text{Chi}\left (4 \tanh ^{-1}(a x)\right )}{16 a^5}+\frac{\text{Chi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^5}+\frac{\log \left (\tanh ^{-1}(a x)\right )}{16 a^5}\\ \end{align*}

Mathematica [A]  time = 0.134723, size = 40, normalized size = 0.73 \[ \frac{-\text{Chi}\left (2 \tanh ^{-1}(a x)\right )-2 \text{Chi}\left (4 \tanh ^{-1}(a x)\right )+\text{Chi}\left (6 \tanh ^{-1}(a x)\right )+2 \log \left (\tanh ^{-1}(a x)\right )}{32 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-CoshIntegral[2*ArcTanh[a*x]] - 2*CoshIntegral[4*ArcTanh[a*x]] + CoshIntegral[6*ArcTanh[a*x]] + 2*Log[ArcTanh
[a*x]])/(32*a^5)

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Maple [A]  time = 0.076, size = 48, normalized size = 0.9 \begin{align*} -{\frac{{\it Chi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{a}^{5}}}-{\frac{{\it Chi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{16\,{a}^{5}}}+{\frac{{\it Chi} \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{a}^{5}}}+{\frac{\ln \left ({\it Artanh} \left ( ax \right ) \right ) }{16\,{a}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*Chi(2*arctanh(a*x))/a^5-1/16*Chi(4*arctanh(a*x))/a^5+1/32*Chi(6*arctanh(a*x))/a^5+1/16*ln(arctanh(a*x))/
a^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{{\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^4/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.95745, size = 558, normalized size = 10.15 \begin{align*} \frac{4 \, \log \left (\log \left (-\frac{a x + 1}{a x - 1}\right )\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 ) + \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 ) - 2 \, \logintegral \left (\frac{a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 2 \, \logintegral \left (\frac{a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) - \logintegral \left (-\frac{a x - 1}{a x + 1}\right )}{64 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(4*log(log(-(a*x + 1)/(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)) + 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)) - 2*log_int
egral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) - 2*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1
)) - log_integral(-(a*x + 1)/(a*x - 1)) - log_integral(-(a*x - 1)/(a*x + 1)))/a^5

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\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**4/(-a**2*x**2+1)**4/atanh(a*x),x)

[Out]

Integral(x**4/((a*x - 1)**4*(a*x + 1)**4*atanh(a*x)), 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 )}^{4} \operatorname{artanh}\left (a x\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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