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

Optimal. Leaf size=114 \[ \frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac{9 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \]

[Out]

-x/(2*a*(1 - a^2*x^2)^3*ArcTanh[a*x]^2) - 3/(a^2*(1 - a^2*x^2)^3*ArcTanh[a*x]) + 5/(2*a^2*(1 - a^2*x^2)^2*ArcT
anh[a*x]) + (5*SinhIntegral[2*ArcTanh[a*x]])/(16*a^2) + SinhIntegral[4*ArcTanh[a*x]]/a^2 + (9*SinhIntegral[6*A
rcTanh[a*x]])/(16*a^2)

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Rubi [A]  time = 0.585044, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6032, 6028, 5966, 6034, 5448, 3298} \[ \frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac{9 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(2*a*(1 - a^2*x^2)^3*ArcTanh[a*x]^2) - 3/(a^2*(1 - a^2*x^2)^3*ArcTanh[a*x]) + 5/(2*a^2*(1 - a^2*x^2)^2*ArcT
anh[a*x]) + (5*SinhIntegral[2*ArcTanh[a*x]])/(16*a^2) + SinhIntegral[4*ArcTanh[a*x]]/a^2 + (9*SinhIntegral[6*A
rcTanh[a*x]])/(16*a^2)

Rule 6032

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

Rule 6028

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int
[x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*A
rcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegersQ[p, 2*q] && LtQ[q, -1] &&
IGtQ[m, 1] && NeQ[p, -1]

Rule 5966

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((d + e*x^2)^(q + 1
)*(a + b*ArcTanh[c*x])^(p + 1))/(b*c*d*(p + 1)), x] + Dist[(2*c*(q + 1))/(b*(p + 1)), Int[x*(d + e*x^2)^q*(a +
b*ArcTanh[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && 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}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^3} \, dx &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac{\int \frac{1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx}{2 a}+\frac{1}{2} (5 a) \int \frac{x^2}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{1}{2 a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+3 \int \frac{x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\frac{5 \int \frac{1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx}{2 a}-\frac{5 \int \frac{1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-10 \int \frac{x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+15 \int \frac{x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \left (\frac{5 \sinh (2 x)}{32 x}+\frac{\sinh (4 x)}{8 x}+\frac{\sinh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}-\frac{10 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac{10 \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^2}+\frac{15 \operatorname{Subst}\left (\int \left (\frac{5 \sinh (2 x)}{32 x}+\frac{\sinh (4 x)}{8 x}+\frac{\sinh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{15 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac{3 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2}+\frac{3 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^2}+\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac{75 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^2}\\ &=-\frac{x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac{3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac{5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac{5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac{\text{Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac{9 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}\\ \end{align*}

Mathematica [A]  time = 0.358251, size = 73, normalized size = 0.64 \[ \frac{\frac{8 \left (\left (5 a^2 x^2+1\right ) \tanh ^{-1}(a x)+a x\right )}{\left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2}+5 \text{Shi}\left (2 \tanh ^{-1}(a x)\right )+16 \text{Shi}\left (4 \tanh ^{-1}(a x)\right )+9 \text{Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*(a*x + (1 + 5*a^2*x^2)*ArcTanh[a*x]))/((-1 + a^2*x^2)^3*ArcTanh[a*x]^2) + 5*SinhIntegral[2*ArcTanh[a*x]] +
16*SinhIntegral[4*ArcTanh[a*x]] + 9*SinhIntegral[6*ArcTanh[a*x]])/(16*a^2)

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Maple [A]  time = 0.076, size = 121, normalized size = 1.1 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{5\,\sinh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{64\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{5\,\cosh \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{\it Artanh} \left ( ax \right ) }}+{\frac{5\,{\it Shi} \left ( 2\,{\it Artanh} \left ( ax \right ) \right ) }{16}}-{\frac{\sinh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{16\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{\cosh \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) }{4\,{\it Artanh} \left ( ax \right ) }}+{\it Shi} \left ( 4\,{\it Artanh} \left ( ax \right ) \right ) -{\frac{\sinh \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{64\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}}-{\frac{3\,\cosh \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{32\,{\it Artanh} \left ( ax \right ) }}+{\frac{9\,{\it Shi} \left ( 6\,{\it Artanh} \left ( ax \right ) \right ) }{16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(-5/64/arctanh(a*x)^2*sinh(2*arctanh(a*x))-5/32/arctanh(a*x)*cosh(2*arctanh(a*x))+5/16*Shi(2*arctanh(a*x
))-1/16/arctanh(a*x)^2*sinh(4*arctanh(a*x))-1/4/arctanh(a*x)*cosh(4*arctanh(a*x))+Shi(4*arctanh(a*x))-1/64/arc
tanh(a*x)^2*sinh(6*arctanh(a*x))-3/32/arctanh(a*x)*cosh(6*arctanh(a*x))+9/16*Shi(6*arctanh(a*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.84023, size = 1037, normalized size = 9.1 \begin{align*} \frac{{\left (9 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 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 ) - 9 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 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 ) + 16 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, 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 ) - 16 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, 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 ) + 5 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \logintegral \left (-\frac{a x + 1}{a x - 1}\right ) - 5 \,{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, 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 )^{2} + 64 \, a x + 32 \,{\left (5 \, a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )}{32 \,{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*((9*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 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)) - 9*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 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)) + 16*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral((a^2*x^2 + 2*a*x +
1)/(a^2*x^2 - 2*a*x + 1)) - 16*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x
^2 + 2*a*x + 1)) + 5*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a*x + 1)/(a*x - 1)) - 5*(a^6*x^6 - 3
*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x - 1))^2 + 64*a*x + 32*(5*a^2
*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))/((a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)*log(-(a*x + 1)/(a*x - 1))^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname{atanh}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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