3.488 \(\int \frac {1}{(1-a^2 x^2)^{5/2} \tanh ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=52 \[ -\frac {1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+\frac {3 \text {Shi}\left (\tanh ^{-1}(a x)\right )}{4 a}+\frac {3 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )}{4 a} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5966, 6034, 5448, 3298} \[ -\frac {1}{a \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}+\frac {3 \text {Shi}\left (\tanh ^{-1}(a x)\right )}{4 a}+\frac {3 \text {Shi}\left (3 \tanh ^{-1}(a x)\right )}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]^2),x]

[Out]

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

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Mathematica [A]  time = 0.14, size = 45, normalized size = 0.87 \[ \frac {3 \left (\text {Shi}\left (\tanh ^{-1}(a x)\right )+\text {Shi}\left (3 \tanh ^{-1}(a x)\right )\right )-\frac {4}{\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}}{4 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]^2),x]

[Out]

(-4/((1 - a^2*x^2)^(3/2)*ArcTanh[a*x]) + 3*(SinhIntegral[ArcTanh[a*x]] + SinhIntegral[3*ArcTanh[a*x]]))/(4*a)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x)^2,x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)/((a^6*x^6 - 3*a^4*x^4 + 3*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 \[ \int \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x)^2,x, algorithm="giac")

[Out]

integrate(1/((-a^2*x^2 + 1)^(5/2)*arctanh(a*x)^2), x)

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maple [B]  time = 0.43, size = 120, normalized size = 2.31 \[ \frac {3 \arctanh \left (a x \right ) \Shi \left (\arctanh \left (a x \right )\right ) x^{2} a^{2}+3 \arctanh \left (a x \right ) \Shi \left (3 \arctanh \left (a x \right )\right ) x^{2} a^{2}-\cosh \left (3 \arctanh \left (a x \right )\right ) x^{2} a^{2}-3 \Shi \left (\arctanh \left (a x \right )\right ) \arctanh \left (a x \right )-3 \Shi \left (3 \arctanh \left (a x \right )\right ) \arctanh \left (a x \right )+3 \sqrt {-a^{2} x^{2}+1}+\cosh \left (3 \arctanh \left (a x \right )\right )}{4 a \arctanh \left (a x \right ) \left (a^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x)^2,x)

[Out]

1/4/a*(3*arctanh(a*x)*Shi(arctanh(a*x))*x^2*a^2+3*arctanh(a*x)*Shi(3*arctanh(a*x))*x^2*a^2-cosh(3*arctanh(a*x)
)*x^2*a^2-3*Shi(arctanh(a*x))*arctanh(a*x)-3*Shi(3*arctanh(a*x))*arctanh(a*x)+3*(-a^2*x^2+1)^(1/2)+cosh(3*arct
anh(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 \[ \int \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x)^2,x, algorithm="maxima")

[Out]

integrate(1/((-a^2*x^2 + 1)^(5/2)*arctanh(a*x)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(atanh(a*x)^2*(1 - a^2*x^2)^(5/2)),x)

[Out]

int(1/(atanh(a*x)^2*(1 - a^2*x^2)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a**2*x**2+1)**(5/2)/atanh(a*x)**2,x)

[Out]

Integral(1/((-(a*x - 1)*(a*x + 1))**(5/2)*atanh(a*x)**2), x)

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