3.426 \(\int x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=243 \[ -\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^5}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^5}-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{8 a^5}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{24 a^2}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^5}-\frac {7 \left (1-a^2 x^2\right )^{3/2}}{72 a^5}+\frac {\sqrt {1-a^2 x^2}}{16 a^5}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^4} \]

[Out]

-7/72*(-a^2*x^2+1)^(3/2)/a^5+1/30*(-a^2*x^2+1)^(5/2)/a^5-1/8*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)
/a^5-1/16*I*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^5+1/16*I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^
5+1/16*(-a^2*x^2+1)^(1/2)/a^5-1/16*x*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^4-1/24*x^3*arctanh(a*x)*(-a^2*x^2+1)^(1
/2)/a^2+1/6*x^5*arctanh(a*x)*(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6010, 6016, 266, 43, 261, 5950} \[ -\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^5}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^5}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^5}-\frac {7 \left (1-a^2 x^2\right )^{3/2}}{72 a^5}+\frac {\sqrt {1-a^2 x^2}}{16 a^5}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{24 a^2}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^4}-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{8 a^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[1 - a^2*x^2]*ArcTanh[a*x],x]

[Out]

Sqrt[1 - a^2*x^2]/(16*a^5) - (7*(1 - a^2*x^2)^(3/2))/(72*a^5) + (1 - a^2*x^2)^(5/2)/(30*a^5) - (x*Sqrt[1 - a^2
*x^2]*ArcTanh[a*x])/(16*a^4) - (x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(24*a^2) + (x^5*Sqrt[1 - a^2*x^2]*ArcTanh[
a*x])/6 - (ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(8*a^5) - ((I/16)*PolyLog[2, ((-I)*Sqrt[1 - a*x])
/Sqrt[1 + a*x]])/a^5 + ((I/16)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^5

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 266

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

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 6010

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 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 x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx &=\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{6} \int \frac {x^4 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{6} a \int \frac {x^5}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {\int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}+\frac {\int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx}{24 a}-\frac {1}{12} a \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {\int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{16 a^4}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{16 a^3}+\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )}{48 a}-\frac {1}{12} a \operatorname {Subst}\left (\int \left (\frac {1}{a^4 \sqrt {1-a^2 x}}-\frac {2 \sqrt {1-a^2 x}}{a^4}+\frac {\left (1-a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )\\ &=\frac {5 \sqrt {1-a^2 x^2}}{48 a^5}-\frac {\left (1-a^2 x^2\right )^{3/2}}{9 a^5}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^5}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{8 a^5}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^5}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^5}+\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a^2 \sqrt {1-a^2 x}}-\frac {\sqrt {1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )}{48 a}\\ &=\frac {\sqrt {1-a^2 x^2}}{16 a^5}-\frac {7 \left (1-a^2 x^2\right )^{3/2}}{72 a^5}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^5}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{24 a^2}+\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{8 a^5}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^5}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 178, normalized size = 0.73 \[ \frac {\sqrt {1-a^2 x^2} \left (-\frac {45 i \left (\text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-\text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )\right )}{\sqrt {1-a^2 x^2}}+24 \left (a^2 x^2-1\right )^2+70 \left (a^2 x^2-1\right )+120 a x \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)+210 a x \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)+45 a x \tanh ^{-1}(a x)+45\right )}{720 a^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4*Sqrt[1 - a^2*x^2]*ArcTanh[a*x],x]

[Out]

(Sqrt[1 - a^2*x^2]*(45 + 70*(-1 + a^2*x^2) + 24*(-1 + a^2*x^2)^2 + 45*a*x*ArcTanh[a*x] + 210*a*x*(-1 + a^2*x^2
)*ArcTanh[a*x] + 120*a*x*(-1 + a^2*x^2)^2*ArcTanh[a*x] - ((45*I)*(ArcTanh[a*x]*(Log[1 - I/E^ArcTanh[a*x]] - Lo
g[1 + I/E^ArcTanh[a*x]]) + PolyLog[2, (-I)/E^ArcTanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2])
)/(720*a^5)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a^{2} x^{2} + 1} x^{4} \operatorname {artanh}\left (a x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(a*x)*(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} x^{2} + 1} x^{4} \operatorname {artanh}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(a*x)*(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.46, size = 195, normalized size = 0.80 \[ \frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (120 \arctanh \left (a x \right ) x^{5} a^{5}+24 x^{4} a^{4}-30 a^{3} x^{3} \arctanh \left (a x \right )+22 a^{2} x^{2}-45 a x \arctanh \left (a x \right )-1\right )}{720 a^{5}}-\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{16 a^{5}}+\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{16 a^{5}}-\frac {i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a^{5}}+\frac {i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/720/a^5*(-(a*x-1)*(a*x+1))^(1/2)*(120*arctanh(a*x)*x^5*a^5+24*x^4*a^4-30*a^3*x^3*arctanh(a*x)+22*a^2*x^2-45*
a*x*arctanh(a*x)-1)-1/16*I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^5+1/16*I*ln(1-I*(a*x+1)/(-a^2*x^2
+1)^(1/2))*arctanh(a*x)/a^5-1/16*I*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^5+1/16*I*dilog(1-I*(a*x+1)/(-a^2*x^
2+1)^(1/2))/a^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} x^{2} + 1} x^{4} \operatorname {artanh}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(a*x)*(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,\mathrm {atanh}\left (a\,x\right )\,\sqrt {1-a^2\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*atanh(a*x)*(1 - a^2*x^2)^(1/2),x)

[Out]

int(x^4*atanh(a*x)*(1 - a^2*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*atanh(a*x)*(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**4*sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x), x)

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