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

Optimal. Leaf size=41 \[ -\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^5}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^5}+\frac {3 \log \left (\tanh ^{-1}(a x)\right )}{8 a^5} \]

[Out]

-1/2*Chi(2*arctanh(a*x))/a^5+1/8*Chi(4*arctanh(a*x))/a^5+3/8*ln(arctanh(a*x))/a^5

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Rubi [A]  time = 0.12, antiderivative size = 41, normalized size of antiderivative = 1.00, 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, 3312, 3301} \[ -\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^5}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^5}+\frac {3 \log \left (\tanh ^{-1}(a x)\right )}{8 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-CoshIntegral[2*ArcTanh[a*x]]/(2*a^5) + CoshIntegral[4*ArcTanh[a*x]]/(8*a^5) + (3*Log[ArcTanh[a*x]])/(8*a^5)

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]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 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])

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 31, normalized size = 0.76 \[ \frac {-4 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )+\text {Chi}\left (4 \tanh ^{-1}(a x)\right )+3 \log \left (\tanh ^{-1}(a x)\right )}{8 a^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*CoshIntegral[2*ArcTanh[a*x]] + CoshIntegral[4*ArcTanh[a*x]] + 3*Log[ArcTanh[a*x]])/(8*a^5)

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fricas [B]  time = 0.59, size = 118, normalized size = 2.88 \[ \frac {6 \, \log \left (\log \left (-\frac {a x + 1}{a x - 1}\right )\right ) + \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - 4 \, \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - 4 \, \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )}{16 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(6*log(log(-(a*x + 1)/(a*x - 1))) + log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) + log_integ
ral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) - 4*log_integral(-(a*x + 1)/(a*x - 1)) - 4*log_integral(-(a*x
- 1)/(a*x + 1)))/a^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{4}}{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 36, normalized size = 0.88 \[ -\frac {\Chi \left (2 \arctanh \left (a x \right )\right )}{2 a^{5}}+\frac {\Chi \left (4 \arctanh \left (a x \right )\right )}{8 a^{5}}+\frac {3 \ln \left (\arctanh \left (a x \right )\right )}{8 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*Chi(2*arctanh(a*x))/a^5+1/8*Chi(4*arctanh(a*x))/a^5+3/8*ln(arctanh(a*x))/a^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {x^{4}}{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int \frac {x^4}{\mathrm {atanh}\left (a\,x\right )\,{\left (a^2\,x^2-1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{4}}{a^{6} x^{6} \operatorname {atanh}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}{\left (a x \right )} - \operatorname {atanh}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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