3.5.82 \(\int \frac {1}{(1-a^2 x^2)^{5/2} \tanh ^{-1}(a x)} \, dx\) [482]

Optimal. Leaf size=27 \[ \frac {3 \text {Chi}\left (\tanh ^{-1}(a x)\right )}{4 a}+\frac {\text {Chi}\left (3 \tanh ^{-1}(a x)\right )}{4 a} \]

[Out]

3/4*Chi(arctanh(a*x))/a+1/4*Chi(3*arctanh(a*x))/a

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Rubi [A]
time = 0.07, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6115, 3393, 3382} \begin {gather*} \frac {3 \text {Chi}\left (\tanh ^{-1}(a x)\right )}{4 a}+\frac {\text {Chi}\left (3 \tanh ^{-1}(a x)\right )}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*CoshIntegral[ArcTanh[a*x]])/(4*a) + CoshIntegral[3*ArcTanh[a*x]]/(4*a)

Rule 3382

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 3393

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 6115

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 )^{5/2} \tanh ^{-1}(a x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\cosh ^3(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {\text {Subst}\left (\int \left (\frac {3 \cosh (x)}{4 x}+\frac {\cosh (3 x)}{4 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {\text {Subst}\left (\int \frac {\cosh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a}+\frac {3 \text {Subst}\left (\int \frac {\cosh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a}\\ &=\frac {3 \text {Chi}\left (\tanh ^{-1}(a x)\right )}{4 a}+\frac {\text {Chi}\left (3 \tanh ^{-1}(a x)\right )}{4 a}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 22, normalized size = 0.81 \begin {gather*} \frac {3 \text {Chi}\left (\tanh ^{-1}(a x)\right )+\text {Chi}\left (3 \tanh ^{-1}(a x)\right )}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*CoshIntegral[ArcTanh[a*x]] + CoshIntegral[3*ArcTanh[a*x]])/(4*a)

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Maple [A]
time = 0.88, size = 21, normalized size = 0.78

method	result	size
default	\(\frac {3 \hyperbolicCosineIntegral \left (\arctanh \left (a x \right )\right )+\hyperbolicCosineIntegral \left (3 \arctanh \left (a x \right )\right )}{4 a}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*Chi(arctanh(a*x))+Chi(3*arctanh(a*x)))/a

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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