3.373 \(\int \frac {x^3 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=205 \[ -\frac {5 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{3 a^4}+\frac {5 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{3 a^4}-\frac {10 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^2}-\frac {\sqrt {1-a^2 x^2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^3} \]

[Out]

-10/3*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^4-5/3*I*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a
^4+5/3*I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^4-1/3*(-a^2*x^2+1)^(1/2)/a^4-1/3*x*arctanh(a*x)*(-a^2*x^2
+1)^(1/2)/a^3-2/3*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/a^4-1/3*x^2*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.31, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6016, 261, 5950, 5994} \[ -\frac {5 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{3 a^4}+\frac {5 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{3 a^4}-\frac {\sqrt {1-a^2 x^2}}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^2}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^3}-\frac {10 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{3 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - a^2*x^2]/(3*a^4) - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(3*a^3) - (10*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*
x]]*ArcTanh[a*x])/(3*a^4) - (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(3*a^4) - (x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]
^2)/(3*a^2) - (((5*I)/3)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4 + (((5*I)/3)*PolyLog[2, (I*Sqrt[1
- a*x])/Sqrt[1 + a*x]])/a^4

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 5950

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 5994

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 6016

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

Rubi steps

\begin {align*} \int \frac {x^3 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^2}+\frac {2 \int \frac {x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{3 a^2}+\frac {2 \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{3 a}\\ &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^3}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^2}+\frac {\int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{3 a^3}+\frac {4 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{3 a^3}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{3 a^2}\\ &=-\frac {\sqrt {1-a^2 x^2}}{3 a^4}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^3}-\frac {10 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{3 a^2}-\frac {5 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{3 a^4}+\frac {5 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{3 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 160, normalized size = 0.78 \[ \frac {\sqrt {1-a^2 x^2} \left (-\frac {5 i \left (\text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-\text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}+\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2-\frac {5 i \tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}-3 \tanh ^{-1}(a x)^2-a x \tanh ^{-1}(a x)-1\right )}{3 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(-1 - a*x*ArcTanh[a*x] - 3*ArcTanh[a*x]^2 + (1 - a^2*x^2)*ArcTanh[a*x]^2 - ((5*I)*ArcTanh[a
*x]*(Log[1 - I/E^ArcTanh[a*x]] - Log[1 + I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2] - ((5*I)*(PolyLog[2, (-I)/E^Arc
Tanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/(3*a^4)

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x^{3} \operatorname {artanh}\left (a x\right )^{2}}{a^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.39, size = 175, normalized size = 0.85 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (a^{2} x^{2} \arctanh \left (a x \right )^{2}+a x \arctanh \left (a x \right )+2 \arctanh \left (a x \right )^{2}+1\right )}{3 a^{4}}-\frac {5 i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{3 a^{4}}+\frac {5 i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{3 a^{4}}-\frac {5 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{3 a^{4}}+\frac {5 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{3 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/a^4*(-(a*x-1)*(a*x+1))^(1/2)*(a^2*x^2*arctanh(a*x)^2+a*x*arctanh(a*x)+2*arctanh(a*x)^2+1)-5/3*I*ln(1+I*(a
*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^4+5/3*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^4-5/3*I*dil
og(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4+5/3*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 \[ \int \frac {x^{3} \operatorname {artanh}\left (a x\right )^{2}}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,{\mathrm {atanh}\left (a\,x\right )}^2}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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