3.5.49 \(\int x^2 (1-a^2 x^2)^{3/2} \tanh ^{-1}(a x) \, dx\) [449]

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

[Out]

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

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Rubi [A]
time = 0.41, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6161, 6157, 6163, 267, 6097, 272, 45} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{8 a^3}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{72 a^3}+\frac {\sqrt {1-a^2 x^2}}{16 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/(16*a^3) + (1 - a^2*x^2)^(3/2)/(72*a^3) - (1 - a^2*x^2)^(5/2)/(30*a^3) - (x*Sqrt[1 - a^2*x^2
]*ArcTanh[a*x])/(16*a^2) + (7*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^3) - ((I/16)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqr
t[1 + a*x]])/a^3 + ((I/16)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^3

Rule 45

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 267

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 272

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 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 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 6163

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(-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^2 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x) \, dx &=-\left (a^2 \int x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\right )+\int x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{4} \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{4} a \int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{6} a^2 \int \frac {x^4 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx+\frac {1}{6} a^3 \int \frac {x^5}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2}+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx+\frac {\int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{8 a}-\frac {1}{24} a \int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{8} a \text {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )+\frac {1}{12} a^3 \text {Subst}\left (\int \frac {x^2}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-a^2 x^2}}{8 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 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)}{4 a^3}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^3}-\frac {\int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{16 a^2}-\frac {\int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{16 a}-\frac {1}{48} a \text {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )-\frac {1}{8} a \text {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 )+\frac {1}{12} a^3 \text {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 {\sqrt {1-a^2 x^2}}{48 a^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{36 a^3}-\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 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^3}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^3}-\frac {1}{48} a \text {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 )\\ &=\frac {\sqrt {1-a^2 x^2}}{16 a^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{72 a^3}-\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 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^3}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^3}\\ \end {align*}

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Mathematica [A]
time = 0.67, size = 224, normalized size = 0.92 \begin {gather*} \frac {31 \sqrt {1-a^2 x^2}+38 a^2 x^2 \sqrt {1-a^2 x^2}-24 a^4 x^4 \sqrt {1-a^2 x^2}-45 a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+210 a^3 x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-120 a^5 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-45 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+45 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-45 i \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )+45 i \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )}{720 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(31*Sqrt[1 - a^2*x^2] + 38*a^2*x^2*Sqrt[1 - a^2*x^2] - 24*a^4*x^4*Sqrt[1 - a^2*x^2] - 45*a*x*Sqrt[1 - a^2*x^2]
*ArcTanh[a*x] + 210*a^3*x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 120*a^5*x^5*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - (45*
I)*ArcTanh[a*x]*Log[1 - I/E^ArcTanh[a*x]] + (45*I)*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] - (45*I)*PolyLog[2,
(-I)/E^ArcTanh[a*x]] + (45*I)*PolyLog[2, I/E^ArcTanh[a*x]])/(720*a^3)

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Maple [A]
time = 1.38, size = 195, normalized size = 0.80

method	result	size
default	\(-\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (120 \arctanh \left (a x \right ) a^{5} x^{5}+24 a^{4} x^{4}-210 a^{3} x^{3} \arctanh \left (a x \right )-38 a^{2} x^{2}+45 a x \arctanh \left (a x \right )-31\right )}{720 a^{3}}-\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{16 a^{3}}+\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{16 a^{3}}-\frac {i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a^{3}}+\frac {i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a^{3}}\)	\(195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/720/a^3*(-(a*x-1)*(a*x+1))^(1/2)*(120*arctanh(a*x)*a^5*x^5+24*a^4*x^4-210*a^3*x^3*arctanh(a*x)-38*a^2*x^2+4
5*a*x*arctanh(a*x)-31)-1/16*I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^3+1/16*I*ln(1-I*(a*x+1)/(-a^2*
x^2+1)^(1/2))*arctanh(a*x)/a^3-1/16*I*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^3+1/16*I*dilog(1-I*(a*x+1)/(-a^2
*x^2+1)^(1/2))/a^3

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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

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