3.490 \(\int \frac{1}{(1-a^2 x^2)^{9/2} \tanh ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=80 \[ -\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{35 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac{63 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac{7 \text{Shi}\left (7 \tanh ^{-1}(a x)\right )}{64 a} \]

[Out]

-(1/(a*(1 - a^2*x^2)^(7/2)*ArcTanh[a*x])) + (35*SinhIntegral[ArcTanh[a*x]])/(64*a) + (63*SinhIntegral[3*ArcTan
h[a*x]])/(64*a) + (35*SinhIntegral[5*ArcTanh[a*x]])/(64*a) + (7*SinhIntegral[7*ArcTanh[a*x]])/(64*a)

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Rubi [A]  time = 0.190509, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5966, 6034, 5448, 3298} \[ -\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{35 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac{63 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac{7 \text{Shi}\left (7 \tanh ^{-1}(a x)\right )}{64 a} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^(9/2)*ArcTanh[a*x]^2),x]

[Out]

-(1/(a*(1 - a^2*x^2)^(7/2)*ArcTanh[a*x])) + (35*SinhIntegral[ArcTanh[a*x]])/(64*a) + (63*SinhIntegral[3*ArcTan
h[a*x]])/(64*a) + (35*SinhIntegral[5*ArcTanh[a*x]])/(64*a) + (7*SinhIntegral[7*ArcTanh[a*x]])/(64*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 )^{9/2} \tanh ^{-1}(a x)^2} \, dx &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+(7 a) \int \frac{x}{\left (1-a^2 x^2\right )^{9/2} \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{7 \operatorname{Subst}\left (\int \frac{\cosh ^6(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{7 \operatorname{Subst}\left (\int \left (\frac{5 \sinh (x)}{64 x}+\frac{9 \sinh (3 x)}{64 x}+\frac{5 \sinh (5 x)}{64 x}+\frac{\sinh (7 x)}{64 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{7 \operatorname{Subst}\left (\int \frac{\sinh (7 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac{63 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}\\ &=-\frac{1}{a \left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}+\frac{35 \text{Shi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac{63 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac{35 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac{7 \text{Shi}\left (7 \tanh ^{-1}(a x)\right )}{64 a}\\ \end{align*}

Mathematica [A]  time = 0.209039, size = 65, normalized size = 0.81 \[ \frac{7 \left (5 \text{Shi}\left (\tanh ^{-1}(a x)\right )+9 \text{Shi}\left (3 \tanh ^{-1}(a x)\right )+5 \text{Shi}\left (5 \tanh ^{-1}(a x)\right )+\text{Shi}\left (7 \tanh ^{-1}(a x)\right )\right )-\frac{64}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)}}{64 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^(9/2)*ArcTanh[a*x]^2),x]

[Out]

(-64/((1 - a^2*x^2)^(7/2)*ArcTanh[a*x]) + 7*(5*SinhIntegral[ArcTanh[a*x]] + 9*SinhIntegral[3*ArcTanh[a*x]] + 5
*SinhIntegral[5*ArcTanh[a*x]] + SinhIntegral[7*ArcTanh[a*x]]))/(64*a)

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Maple [B]  time = 0.159, size = 232, normalized size = 2.9 \begin{align*}{\frac{1}{64\,a{\it Artanh} \left ( ax \right ) \left ({a}^{2}{x}^{2}-1 \right ) } \left ( 35\,{\it Artanh} \left ( ax \right ){\it Shi} \left ({\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+63\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+35\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+7\,{\it Artanh} \left ( ax \right ){\it Shi} \left ( 7\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-21\,\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-7\,\cosh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-\cosh \left ( 7\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-35\,{\it Shi} \left ({\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -63\,{\it Shi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -35\,{\it Shi} \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) -7\,{\it Shi} \left ( 7\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) +35\,\sqrt{-{a}^{2}{x}^{2}+1}+21\,\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ) +7\,\cosh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ) +\cosh \left ( 7\,{\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)^(9/2)/arctanh(a*x)^2,x)

[Out]

1/64/a*(35*arctanh(a*x)*Shi(arctanh(a*x))*x^2*a^2+63*arctanh(a*x)*Shi(3*arctanh(a*x))*x^2*a^2+35*arctanh(a*x)*
Shi(5*arctanh(a*x))*x^2*a^2+7*arctanh(a*x)*Shi(7*arctanh(a*x))*x^2*a^2-21*cosh(3*arctanh(a*x))*x^2*a^2-7*cosh(
5*arctanh(a*x))*x^2*a^2-cosh(7*arctanh(a*x))*x^2*a^2-35*Shi(arctanh(a*x))*arctanh(a*x)-63*Shi(3*arctanh(a*x))*
arctanh(a*x)-35*Shi(5*arctanh(a*x))*arctanh(a*x)-7*Shi(7*arctanh(a*x))*arctanh(a*x)+35*(-a^2*x^2+1)^(1/2)+21*c
osh(3*arctanh(a*x))+7*cosh(5*arctanh(a*x))+cosh(7*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{9}{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)^(9/2)/arctanh(a*x)^2,x, algorithm="maxima")

[Out]

integrate(1/((-a^2*x^2 + 1)^(9/2)*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^{10} x^{10} - 5 \, a^{8} x^{8} + 10 \, a^{6} x^{6} - 10 \, a^{4} x^{4} + 5 \, 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)^(9/2)/arctanh(a*x)^2,x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)/((a^10*x^10 - 5*a^8*x^8 + 10*a^6*x^6 - 10*a^4*x^4 + 5*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)**(9/2)/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{9}{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)^(9/2)/arctanh(a*x)^2,x, algorithm="giac")

[Out]

integrate(1/((-a^2*x^2 + 1)^(9/2)*arctanh(a*x)^2), x)