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/(-a^2*x^2+1)^2/arctanh(a*x)-Shi(2*arctanh(a*x))/a^5+1/2*Shi(4*arctanh(a*x))/a^5

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Rubi [A]  time = 0.19, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 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 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])

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*}

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Mathematica [A]  time = 0.20, 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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fricas [B]  time = 0.69, size = 232, normalized size = 4.38 \[ -\frac {8 \, a^{4} x^{4} - {\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{4 \, {\left (a^{9} x^{4} - 2 \, a^{7} x^{2} + a^{5}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 62, normalized size = 1.17 \[ \frac {-\frac {3}{8 \arctanh \left (a x \right )}+\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{2 \arctanh \left (a x \right )}-\Shi \left (2 \arctanh \left (a x \right )\right )-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )}+\frac {\Shi \left (4 \arctanh \left (a x \right )\right )}{2}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(-a^2*x^2+1)^3/arctanh(a*x)^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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, x^{4}}{{\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (a x + 1\right ) - {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-a x + 1\right )} + 8 \, \int -\frac {x^{3}}{{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) - {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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