3.4.96 \(\int \frac {x^3 \tanh ^{-1}(a x)^2}{(1-a^2 x^2)^{3/2}} \, dx\) [396]

Optimal. Leaf size=186 \[ \frac {2}{a^4 \sqrt {1-a^2 x^2}}-\frac {2 x \tanh ^{-1}(a x)}{a^3 \sqrt {1-a^2 x^2}}+\frac {4 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^4}+\frac {\tanh ^{-1}(a x)^2}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a^4}+\frac {2 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^4}-\frac {2 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^4} \]

[Out]

4*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^4+2*I*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^4-2*I
*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^4+2/a^4/(-a^2*x^2+1)^(1/2)-2*x*arctanh(a*x)/a^3/(-a^2*x^2+1)^(1/2
)+arctanh(a*x)^2/a^4/(-a^2*x^2+1)^(1/2)+arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/a^4

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Rubi [A]
time = 0.22, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6175, 6141, 6097, 6105} \begin {gather*} \frac {4 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a^4}+\frac {2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^4}-\frac {2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a^4}+\frac {2}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a^4}+\frac {\tanh ^{-1}(a x)^2}{a^4 \sqrt {1-a^2 x^2}}-\frac {2 x \tanh ^{-1}(a x)}{a^3 \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2/(a^4*Sqrt[1 - a^2*x^2]) - (2*x*ArcTanh[a*x])/(a^3*Sqrt[1 - a^2*x^2]) + (4*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]
]*ArcTanh[a*x])/a^4 + ArcTanh[a*x]^2/(a^4*Sqrt[1 - a^2*x^2]) + (Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/a^4 + ((2*I)
*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4 - ((2*I)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4

Rule 6097

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

Rule 6105

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-b/(c*d*Sqrt[d + e*x^2]
), x] + Simp[x*((a + b*ArcTanh[c*x])/(d*Sqrt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0
]

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 6175

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

Rubi steps

\begin {align*} \int \frac {x^3 \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=\frac {\int \frac {x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^2}-\frac {\int \frac {x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=\frac {\tanh ^{-1}(a x)^2}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a^4}-\frac {2 \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{a^3}-\frac {2 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{a^3}\\ &=\frac {2}{a^4 \sqrt {1-a^2 x^2}}-\frac {2 x \tanh ^{-1}(a x)}{a^3 \sqrt {1-a^2 x^2}}+\frac {4 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a^4}+\frac {\tanh ^{-1}(a x)^2}{a^4 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{a^4}+\frac {2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^4}-\frac {2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{a^4}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 165, normalized size = 0.89 \begin {gather*} \frac {2 i \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )+\frac {2+\left (2-a^2 x^2\right ) \tanh ^{-1}(a x)^2-2 \tanh ^{-1}(a x) \left (a x-i \sqrt {1-a^2 x^2} \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+i \sqrt {1-a^2 x^2} \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )-2 i \sqrt {1-a^2 x^2} \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )}{\sqrt {1-a^2 x^2}}}{a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*PolyLog[2, (-I)/E^ArcTanh[a*x]] + (2 + (2 - a^2*x^2)*ArcTanh[a*x]^2 - 2*ArcTanh[a*x]*(a*x - I*Sqrt[1 -
a^2*x^2]*Log[1 - I/E^ArcTanh[a*x]] + I*Sqrt[1 - a^2*x^2]*Log[1 + I/E^ArcTanh[a*x]]) - (2*I)*Sqrt[1 - a^2*x^2]*
PolyLog[2, I/E^ArcTanh[a*x]])/Sqrt[1 - a^2*x^2])/a^4

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Maple [A]
time = 0.69, size = 230, normalized size = 1.24

method	result	size
default	\(-\frac {\left (\arctanh \left (a x \right )^{2}-2 \arctanh \left (a x \right )+2\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{4} \left (a x -1\right )}+\frac {\left (\arctanh \left (a x \right )^{2}+2 \arctanh \left (a x \right )+2\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right )}}{2 a^{4} \left (a x +1\right )}+\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \arctanh \left (a x \right )^{2}}{a^{4}}+\frac {2 i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{a^{4}}-\frac {2 i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{a^{4}}+\frac {2 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{4}}-\frac {2 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{4}}\)	\(230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(arctanh(a*x)^2-2*arctanh(a*x)+2)*(-(a*x-1)*(a*x+1))^(1/2)/a^4/(a*x-1)+1/2*(arctanh(a*x)^2+2*arctanh(a*x)
+2)*(-(a*x-1)*(a*x+1))^(1/2)/a^4/(a*x+1)+(-(a*x-1)*(a*x+1))^(1/2)*arctanh(a*x)^2/a^4+2*I*ln(1+I*(a*x+1)/(-a^2*
x^2+1)^(1/2))*arctanh(a*x)/a^4-2*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^4+2*I*dilog(1+I*(a*x+1)/(
-a^2*x^2+1)^(1/2))/a^4-2*I*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4

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

[Out]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*x^3*arctanh(a*x)^2/(a^4*x^4 - 2*a^2*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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