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/(-a^2*x^2+1)^(5/2)/arctanh(a*x)+5/8*Shi(arctanh(a*x))/a+15/16*Shi(3*arctanh(a*x))/a+5/16*Shi(5*arctanh(a*
x))/a

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Rubi [A]  time = 0.18, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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 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])

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*}

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Mathematica [A]  time = 0.18, 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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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a^{8} x^{8} - 4 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 1\right )} \operatorname {artanh}\left (a x\right )^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}} \operatorname {artanh}\left (a x\right )^{2}}\,{d x} \]

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)

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maple [B]  time = 0.53, size = 176, normalized size = 2.67 \[ \frac {10 \arctanh \left (a x \right ) \Shi \left (\arctanh \left (a x \right )\right ) x^{2} a^{2}+15 \arctanh \left (a x \right ) \Shi \left (3 \arctanh \left (a x \right )\right ) x^{2} a^{2}+5 \arctanh \left (a x \right ) \Shi \left (5 \arctanh \left (a x \right )\right ) x^{2} a^{2}-5 \cosh \left (3 \arctanh \left (a x \right )\right ) x^{2} a^{2}-\cosh \left (5 \arctanh \left (a x \right )\right ) x^{2} a^{2}-10 \Shi \left (\arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-15 \Shi \left (3 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-5 \Shi \left (5 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )+10 \sqrt {-a^{2} x^{2}+1}+5 \cosh \left (3 \arctanh \left (a x \right )\right )+\cosh \left (5 \arctanh \left (a x \right )\right )}{16 a \arctanh \left (a x \right ) \left (a^{2} x^{2}-1\right )} \]

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*(10*arctanh(a*x)*Shi(arctanh(a*x))*x^2*a^2+15*arctanh(a*x)*Shi(3*arctanh(a*x))*x^2*a^2+5*arctanh(a*x)*S
hi(5*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^2-10*Shi(arctanh(a*x))*ar
ctanh(a*x)-15*Shi(3*arctanh(a*x))*arctanh(a*x)-5*Shi(5*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}} \operatorname {artanh}\left (a x\right )^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \]

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]

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

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