3.5.52 \(\int \frac {(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)}{x} \, dx\) [452]

Optimal. Leaf size=144 \[ -\frac {1}{6} a x \sqrt {1-a^2 x^2}-\frac {7}{6} \text {ArcSin}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{3} \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \]

[Out]

-7/6*arcsin(a*x)+1/3*(-a^2*x^2+1)^(3/2)*arctanh(a*x)-2*arctanh(a*x)*arctanh((-a*x+1)^(1/2)/(a*x+1)^(1/2))+poly
log(2,-(-a*x+1)^(1/2)/(a*x+1)^(1/2))-polylog(2,(-a*x+1)^(1/2)/(a*x+1)^(1/2))-1/6*a*x*(-a^2*x^2+1)^(1/2)+(-a^2*
x^2+1)^(1/2)*arctanh(a*x)

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Rubi [A]
time = 0.17, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6161, 6157, 6165, 222, 6141, 201} \begin {gather*} -\frac {1}{6} a x \sqrt {1-a^2 x^2}+\frac {1}{3} \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {7}{6} \text {ArcSin}(a x)+\text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a*x*Sqrt[1 - a^2*x^2]) - (7*ArcSin[a*x])/6 + Sqrt[1 - a^2*x^2]*ArcTanh[a*x] + ((1 - a^2*x^2)^(3/2)*ArcTa
nh[a*x])/3 - 2*ArcTanh[a*x]*ArcTanh[Sqrt[1 - a*x]/Sqrt[1 + a*x]] + PolyLog[2, -(Sqrt[1 - a*x]/Sqrt[1 + a*x])]
- PolyLog[2, Sqrt[1 - a*x]/Sqrt[1 + a*x]]

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 6141

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

Rule 6157

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(
m + 1)*Sqrt[d + e*x^2]*((a + b*ArcTanh[c*x])/(f*(m + 2))), x] + (Dist[d/(m + 2), Int[(f*x)^m*((a + b*ArcTanh[c
*x])/Sqrt[d + e*x^2]), x], x] - Dist[b*c*(d/(f*(m + 2))), Int[(f*x)^(m + 1)/Sqrt[d + e*x^2], x], x]) /; FreeQ[
{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && NeQ[m, -2]

Rule 6161

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

Rule 6165

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Simp[(-2/Sqrt[d])*(a +
b*ArcTanh[c*x])*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]], x] + (Simp[(b/Sqrt[d])*PolyLog[2, -Sqrt[1 - c*x]/Sqrt[1
+ c*x]], x] - Simp[(b/Sqrt[d])*PolyLog[2, Sqrt[1 - c*x]/Sqrt[1 + c*x]], x]) /; FreeQ[{a, b, c, d, e}, x] && Eq
Q[c^2*d + e, 0] && GtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{x} \, dx &=-\left (a^2 \int x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\right )+\int \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x} \, dx\\ &=\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{3} \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)-\frac {1}{3} a \int \sqrt {1-a^2 x^2} \, dx-a \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx+\int \frac {\tanh ^{-1}(a x)}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {1}{6} a x \sqrt {1-a^2 x^2}-\sin ^{-1}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{3} \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\frac {1}{6} a \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {1}{6} a x \sqrt {1-a^2 x^2}-\frac {7}{6} \sin ^{-1}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{3} \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)-2 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 143, normalized size = 0.99 \begin {gather*} \frac {1}{6} \left (-a x \sqrt {1-a^2 x^2}-14 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )+8 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-2 a^2 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+6 \tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-6 \tanh ^{-1}(a x) \log \left (1+e^{-\tanh ^{-1}(a x)}\right )+6 \text {PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )-6 \text {PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*x*Sqrt[1 - a^2*x^2]) - 14*ArcTan[Tanh[ArcTanh[a*x]/2]] + 8*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 2*a^2*x^2*Sqr
t[1 - a^2*x^2]*ArcTanh[a*x] + 6*ArcTanh[a*x]*Log[1 - E^(-ArcTanh[a*x])] - 6*ArcTanh[a*x]*Log[1 + E^(-ArcTanh[a
*x])] + 6*PolyLog[2, -E^(-ArcTanh[a*x])] - 6*PolyLog[2, E^(-ArcTanh[a*x])])/6

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Maple [A]
time = 1.51, size = 132, normalized size = 0.92

method	result	size
default	\(-\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (2 a^{2} x^{2} \arctanh \left (a x \right )+a x -8 \arctanh \left (a x \right )\right )}{6}-\frac {7 \arctan \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}-\dilog \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )-\dilog \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(-(a*x-1)*(a*x+1))^(1/2)*(2*a^2*x^2*arctanh(a*x)+a*x-8*arctanh(a*x))-7/3*arctan((a*x+1)/(-a^2*x^2+1)^(1/2
))-dilog(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-arctanh(a*x)*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-dilog((a*x+1)/(-a^2*x^2+1
)^(1/2))

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x)/x, 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((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x,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 = -0.01 \begin {gather*} \int \frac {\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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