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

Optimal. Leaf size=55 \[ \frac {35 \text {Chi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac {21 \text {Chi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac {7 \text {Chi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\text {Chi}\left (7 \tanh ^{-1}(a x)\right )}{64 a} \]

[Out]

35/64*Chi(arctanh(a*x))/a+21/64*Chi(3*arctanh(a*x))/a+7/64*Chi(5*arctanh(a*x))/a+1/64*Chi(7*arctanh(a*x))/a

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Rubi [A]  time = 0.12, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5968, 3312, 3301} \[ \frac {35 \text {Chi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac {21 \text {Chi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac {7 \text {Chi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\text {Chi}\left (7 \tanh ^{-1}(a x)\right )}{64 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*CoshIntegral[ArcTanh[a*x]])/(64*a) + (21*CoshIntegral[3*ArcTanh[a*x]])/(64*a) + (7*CoshIntegral[5*ArcTanh[
a*x]])/(64*a) + CoshIntegral[7*ArcTanh[a*x]]/(64*a)

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5968

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c, Subst[Int[(
a + b*x)^p/Cosh[x]^(2*(q + 1)), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0]
&& ILtQ[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^{9/2} \tanh ^{-1}(a x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\cosh ^7(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {35 \cosh (x)}{64 x}+\frac {21 \cosh (3 x)}{64 x}+\frac {7 \cosh (5 x)}{64 x}+\frac {\cosh (7 x)}{64 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\cosh (7 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac {7 \operatorname {Subst}\left (\int \frac {\cosh (5 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac {21 \operatorname {Subst}\left (\int \frac {\cosh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}+\frac {35 \operatorname {Subst}\left (\int \frac {\cosh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}\\ &=\frac {35 \text {Chi}\left (\tanh ^{-1}(a x)\right )}{64 a}+\frac {21 \text {Chi}\left (3 \tanh ^{-1}(a x)\right )}{64 a}+\frac {7 \text {Chi}\left (5 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\text {Chi}\left (7 \tanh ^{-1}(a x)\right )}{64 a}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 40, normalized size = 0.73 \[ \frac {35 \text {Chi}\left (\tanh ^{-1}(a x)\right )+21 \text {Chi}\left (3 \tanh ^{-1}(a x)\right )+7 \text {Chi}\left (5 \tanh ^{-1}(a x)\right )+\text {Chi}\left (7 \tanh ^{-1}(a x)\right )}{64 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*CoshIntegral[ArcTanh[a*x]] + 21*CoshIntegral[3*ArcTanh[a*x]] + 7*CoshIntegral[5*ArcTanh[a*x]] + CoshIntegr
al[7*ArcTanh[a*x]])/(64*a)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a^{10} x^{10} - 5 \, a^{8} x^{8} + 10 \, a^{6} x^{6} - 10 \, a^{4} x^{4} + 5 \, a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {9}{2}} \operatorname {artanh}\left (a x\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(9/2)/arctanh(a*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.43, size = 39, normalized size = 0.71 \[ \frac {35 \Chi \left (\arctanh \left (a x \right )\right )+21 \Chi \left (3 \arctanh \left (a x \right )\right )+7 \Chi \left (5 \arctanh \left (a x \right )\right )+\Chi \left (7 \arctanh \left (a x \right )\right )}{64 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(35*Chi(arctanh(a*x))+21*Chi(3*arctanh(a*x))+7*Chi(5*arctanh(a*x))+Chi(7*arctanh(a*x)))/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {9}{2}} \operatorname {artanh}\left (a x\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(9/2)/arctanh(a*x),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(atanh(a*x)*(1 - a^2*x^2)^(9/2)),x)

[Out]

int(1/(atanh(a*x)*(1 - a^2*x^2)^(9/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {9}{2}} \operatorname {atanh}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((-(a*x - 1)*(a*x + 1))**(9/2)*atanh(a*x)), x)

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