3.5.18 \(\int \frac {x}{(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)^2} \, dx\) [418]

Optimal. Leaf size=36 \[ -\frac {x}{a \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}+\frac {\text {Chi}\left (\tanh ^{-1}(a x)\right )}{a^2} \]

[Out]

Chi(arctanh(a*x))/a^2-x/a/arctanh(a*x)/(-a^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6153, 6115, 3382} \begin {gather*} \frac {\text {Chi}\left (\tanh ^{-1}(a x)\right )}{a^2}-\frac {x}{a \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/((1 - a^2*x^2)^(3/2)*ArcTanh[a*x]^2),x]

[Out]

-(x/(a*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])) + CoshIntegral[ArcTanh[a*x]]/a^2

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 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])

Rule 6153

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[(f*x)^m*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Dist[f*(m/(b*c*(p + 1))), In
t[(f*x)^(m - 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && Eq
Q[c^2*d + e, 0] && EqQ[m + 2*q + 2, 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 34, normalized size = 0.94 \begin {gather*} \frac {-\frac {a x}{\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}+\text {Chi}\left (\tanh ^{-1}(a x)\right )}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/((1 - a^2*x^2)^(3/2)*ArcTanh[a*x]^2),x]

[Out]

(-((a*x)/(Sqrt[1 - a^2*x^2]*ArcTanh[a*x])) + CoshIntegral[ArcTanh[a*x]])/a^2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(34)=68\).
time = 1.14, size = 90, normalized size = 2.50

method	result	size
default	\(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{2} \left (a x -1\right ) \arctanh \left (a x \right )}-\frac {\expIntegral \left (1, -\arctanh \left (a x \right )\right )}{2 a^{2}}+\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 \arctanh \left (a x \right ) \left (a x +1\right ) a^{2}}-\frac {\expIntegral \left (1, \arctanh \left (a x \right )\right )}{2 a^{2}}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(-a^2*x^2+1)^(3/2)/arctanh(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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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(x/(-a^2*x^2+1)^(3/2)/arctanh(a*x)^2,x, algorithm="maxima")

[Out]

integrate(x/((-a^2*x^2 + 1)^(3/2)*arctanh(a*x)^2), 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(x/(-a^2*x^2+1)^(3/2)/arctanh(a*x)^2,x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)*x/((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}^{2}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-a**2*x**2+1)**(3/2)/atanh(a*x)**2,x)

[Out]

Integral(x/((-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x)**2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-a^2*x^2+1)^(3/2)/arctanh(a*x)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(atanh(a*x)^2*(1 - a^2*x^2)^(3/2)),x)

[Out]

int(x/(atanh(a*x)^2*(1 - a^2*x^2)^(3/2)), x)

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