3.5.39 \(\int x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx\) [439]

Optimal. Leaf size=281 \[ \frac {11 \sqrt {1-a^2 x^2}}{60 a^4}-\frac {\left (1-a^2 x^2\right )^{3/2}}{30 a^4}+\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac {11 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{30 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {11 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{60 a^4}+\frac {11 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{60 a^4} \]

[Out]

-1/30*(-a^2*x^2+1)^(3/2)/a^4-11/30*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^4-11/60*I*polylog(2,-I*
(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^4+11/60*I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^4+11/60*(-a^2*x^2+1)^(1/
2)/a^4+1/12*x*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^3+1/10*x^3*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a-2/15*arctanh(a*x)
^2*(-a^2*x^2+1)^(1/2)/a^4-1/15*x^2*arctanh(a*x)^2*(-a^2*x^2+1)^(1/2)/a^2+1/5*x^4*arctanh(a*x)^2*(-a^2*x^2+1)^(
1/2)

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Rubi [A]
time = 0.71, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {6161, 6163, 267, 6097, 6141, 272, 45} \begin {gather*} -\frac {11 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{30 a^4}-\frac {11 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{60 a^4}+\frac {11 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{60 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac {\left (1-a^2 x^2\right )^{3/2}}{30 a^4}+\frac {11 \sqrt {1-a^2 x^2}}{60 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}+\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11*Sqrt[1 - a^2*x^2])/(60*a^4) - (1 - a^2*x^2)^(3/2)/(30*a^4) + (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(12*a^3) +
(x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(10*a) - (11*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(30*a^4)
- (2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(15*a^4) - (x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/(15*a^2) + (x^4*Sqrt[
1 - a^2*x^2]*ArcTanh[a*x]^2)/5 - (((11*I)/60)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4 + (((11*I)/6
0)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a^4

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 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 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^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2 \, dx &=-\left (a^2 \int \frac {x^5 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\right )+\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 {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {4}{5} \int \frac {x^3 \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx+\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 {1}{5} (2 a) \int \frac {x^4 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 a^3}+\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\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}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {1}{10} \int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx+\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 {8 \int \frac {x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{15 a^2}-\frac {3 \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{10 a}-\frac {8 \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{15 a}\\ &=-\frac {\sqrt {1-a^2 x^2}}{3 a^4}+\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\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}{15 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^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}-\frac {1}{20} \text {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )-\frac {3 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{20 a^3}-\frac {4 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{15 a^3}-\frac {16 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{15 a^3}-\frac {3 \int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{20 a^2}-\frac {4 \int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{15 a^2}\\ &=\frac {\sqrt {1-a^2 x^2}}{12 a^4}+\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac {11 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{30 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {11 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{60 a^4}+\frac {11 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{60 a^4}-\frac {1}{20} \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 {11 \sqrt {1-a^2 x^2}}{60 a^4}-\frac {\left (1-a^2 x^2\right )^{3/2}}{30 a^4}+\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{12 a^3}+\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{10 a}-\frac {11 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{30 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-\frac {11 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{60 a^4}+\frac {11 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{60 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 175, normalized size = 0.62 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (11+11 a x \tanh ^{-1}(a x)+6 a x \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)+12 \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)^2+2 \left (-1+a^2 x^2\right ) \left (1+10 \tanh ^{-1}(a x)^2\right )-\frac {11 i \left (\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 )+\text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-\text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}\right )}{60 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(11 + 11*a*x*ArcTanh[a*x] + 6*a*x*(-1 + a^2*x^2)*ArcTanh[a*x] + 12*(-1 + a^2*x^2)^2*ArcTanh
[a*x]^2 + 2*(-1 + a^2*x^2)*(1 + 10*ArcTanh[a*x]^2) - ((11*I)*(ArcTanh[a*x]*(Log[1 - I/E^ArcTanh[a*x]] - Log[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]))/(6
0*a^4)

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Maple [A]
time = 0.70, size = 211, normalized size = 0.75

method	result	size
default	\(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (12 a^{4} x^{4} \arctanh \left (a x \right )^{2}+6 a^{3} x^{3} \arctanh \left (a x \right )-4 a^{2} x^{2} \arctanh \left (a x \right )^{2}+2 a^{2} x^{2}+5 a x \arctanh \left (a x \right )-8 \arctanh \left (a x \right )^{2}+9\right )}{60 a^{4}}-\frac {11 i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{60 a^{4}}+\frac {11 i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{60 a^{4}}-\frac {11 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{60 a^{4}}+\frac {11 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{60 a^{4}}\)	\(211\)

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,method=_RETURNVERBOSE)

[Out]

1/60/a^4*(-(a*x-1)*(a*x+1))^(1/2)*(12*a^4*x^4*arctanh(a*x)^2+6*a^3*x^3*arctanh(a*x)-4*a^2*x^2*arctanh(a*x)^2+2
*a^2*x^2+5*a*x*arctanh(a*x)-8*arctanh(a*x)^2+9)-11/60*I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^4+11
/60*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^4-11/60*I*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^4+11
/60*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)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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

[Out]

Integral(x**3*sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x)**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)^(1/2),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 = -0.00 \begin {gather*} \int x^3\,{\mathrm {atanh}\left (a\,x\right )}^2\,\sqrt {1-a^2\,x^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)^(1/2),x)

[Out]

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

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