3.489 \(\int \frac{1}{(1-a^2 x^2)^{7/2} \tanh ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=66 \[ -\frac{1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac{15 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac{5 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a} \]

[Out]

-(1/(a*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x])) + (5*SinhIntegral[ArcTanh[a*x]])/(8*a) + (15*SinhIntegral[3*ArcTanh[
a*x]])/(16*a) + (5*SinhIntegral[5*ArcTanh[a*x]])/(16*a)

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Rubi [A]  time = 0.176986, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5966, 6034, 5448, 3298} \[ -\frac{1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac{15 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac{5 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^(7/2)*ArcTanh[a*x]^2),x]

[Out]

-(1/(a*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x])) + (5*SinhIntegral[ArcTanh[a*x]])/(8*a) + (15*SinhIntegral[3*ArcTanh[
a*x]])/(16*a) + (5*SinhIntegral[5*ArcTanh[a*x]])/(16*a)

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{1}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)^2} \, dx &=-\frac{1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+(5 a) \int \frac{x}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh ^4(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{8 x}+\frac{3 \sinh (3 x)}{16 x}+\frac{\sinh (5 x)}{16 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac{5 \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a}+\frac{15 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{8 a}+\frac{15 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{16 a}+\frac{5 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{16 a}\\ \end{align*}

Mathematica [A]  time = 0.166656, size = 56, normalized size = 0.85 \[ \frac{5 \left (2 \text{Shi}\left (\tanh ^{-1}(a x)\right )+3 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )+\text{Shi}\left (5 \tanh ^{-1}(a x)\right )\right )-\frac{16}{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}}{16 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^(7/2)*ArcTanh[a*x]^2),x]

[Out]

(-16/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]) + 5*(2*SinhIntegral[ArcTanh[a*x]] + 3*SinhIntegral[3*ArcTanh[a*x]] + S
inhIntegral[5*ArcTanh[a*x]]))/(16*a)

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Maple [B]  time = 0.171, size = 176, normalized size = 2.7 \begin{align*}{\frac{1}{16\,a{\it Artanh} \left ( ax \right ) \left ({a}^{2}{x}^{2}-1 \right ) } \left ( 5\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+10\,{\it Artanh} \left ( ax \right ){\it Shi} \left ({\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+15\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-5\,\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-\cosh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-5\,{\it Shi} \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -10\,{\it Shi} \left ({\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -15\,{\it Shi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) +10\,\sqrt{-{a}^{2}{x}^{2}+1}+5\,\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ) +\cosh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^2,x)

[Out]

1/16/a*(5*arctanh(a*x)*Shi(5*arctanh(a*x))*x^2*a^2+10*arctanh(a*x)*Shi(arctanh(a*x))*x^2*a^2+15*arctanh(a*x)*S
hi(3*arctanh(a*x))*x^2*a^2-5*cosh(3*arctanh(a*x))*x^2*a^2-cosh(5*arctanh(a*x))*x^2*a^2-5*Shi(5*arctanh(a*x))*a
rctanh(a*x)-10*Shi(arctanh(a*x))*arctanh(a*x)-15*Shi(3*arctanh(a*x))*arctanh(a*x)+10*(-a^2*x^2+1)^(1/2)+5*cosh
(3*arctanh(a*x))+cosh(5*arctanh(a*x)))/arctanh(a*x)/(a^2*x^2-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{7}{2}} \operatorname{artanh}\left (a x\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^2,x, algorithm="maxima")

[Out]

integrate(1/((-a^2*x^2 + 1)^(7/2)*arctanh(a*x)^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} x^{2} + 1}}{{\left (a^{8} x^{8} - 4 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 1\right )} \operatorname{artanh}\left (a x\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^2,x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)/((a^8*x^8 - 4*a^6*x^6 + 6*a^4*x^4 - 4*a^2*x^2 + 1)*arctanh(a*x)^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a**2*x**2+1)**(7/2)/atanh(a*x)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{7}{2}} \operatorname{artanh}\left (a x\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^2,x, algorithm="giac")

[Out]

integrate(1/((-a^2*x^2 + 1)^(7/2)*arctanh(a*x)^2), x)