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]

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

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Rubi [A]  time = 0.13, antiderivative size = 29, 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, 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 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]

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 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^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*}

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Mathematica [A]  time = 0.15, 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)

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fricas [B]  time = 0.47, size = 136, normalized size = 4.69 \[ \frac {\operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) - 3 \, \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + 3 \, \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )}{64 \, a^{4}} \]

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

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

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)

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maple [A]  time = 0.19, size = 24, normalized size = 0.83 \[ \frac {\frac {\Shi \left (6 \arctanh \left (a x \right )\right )}{32}-\frac {3 \Shi \left (2 \arctanh \left (a x \right )\right )}{32}}{a^{4}} \]

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

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

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^3}{\mathrm {atanh}\left (a\,x\right )\,{\left (a^2\,x^2-1\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname {atanh}{\left (a x \right )}}\, dx \]

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)

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