3.5.27 \(\int x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\) [427]

Optimal. Leaf size=136 \[ \frac {x \sqrt {1-a^2 x^2}}{24 a^3}+\frac {x^3 \sqrt {1-a^2 x^2}}{20 a}+\frac {11 \text {ArcSin}(a x)}{120 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \]

[Out]

11/120*arcsin(a*x)/a^4+1/24*x*(-a^2*x^2+1)^(1/2)/a^3+1/20*x^3*(-a^2*x^2+1)^(1/2)/a-2/15*arctanh(a*x)*(-a^2*x^2
+1)^(1/2)/a^4-1/15*x^2*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^2+1/5*x^4*arctanh(a*x)*(-a^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6157, 6163, 327, 222, 6141} \begin {gather*} \frac {11 \text {ArcSin}(a x)}{120 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {x^3 \sqrt {1-a^2 x^2}}{20 a}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}+\frac {x \sqrt {1-a^2 x^2}}{24 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[1 - a^2*x^2])/(24*a^3) + (x^3*Sqrt[1 - a^2*x^2])/(20*a) + (11*ArcSin[a*x])/(120*a^4) - (2*Sqrt[1 - a^2
*x^2]*ArcTanh[a*x])/(15*a^4) - (x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(15*a^2) + (x^4*Sqrt[1 - a^2*x^2]*ArcTanh[
a*x])/5

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 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 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) \, dx &=\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{5} \int \frac {x^3 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{5} a \int \frac {x^4}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {x^3 \sqrt {1-a^2 x^2}}{20 a}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {2 \int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{15 a^2}+\frac {\int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{15 a}-\frac {3 \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{20 a}\\ &=\frac {x \sqrt {1-a^2 x^2}}{24 a^3}+\frac {x^3 \sqrt {1-a^2 x^2}}{20 a}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{30 a^3}-\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{40 a^3}+\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{15 a^3}\\ &=\frac {x \sqrt {1-a^2 x^2}}{24 a^3}+\frac {x^3 \sqrt {1-a^2 x^2}}{20 a}+\frac {11 \sin ^{-1}(a x)}{120 a^4}-\frac {2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac {x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^2}+\frac {1}{5} x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 79, normalized size = 0.58 \begin {gather*} \frac {a x \sqrt {1-a^2 x^2} \left (5+6 a^2 x^2\right )+11 \text {ArcSin}(a x)+8 \sqrt {1-a^2 x^2} \left (-2-a^2 x^2+3 a^4 x^4\right ) \tanh ^{-1}(a x)}{120 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*Sqrt[1 - a^2*x^2]*(5 + 6*a^2*x^2) + 11*ArcSin[a*x] + 8*Sqrt[1 - a^2*x^2]*(-2 - a^2*x^2 + 3*a^4*x^4)*ArcTa
nh[a*x])/(120*a^4)

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Maple [C] Result contains complex when optimal does not.
time = 1.37, size = 120, normalized size = 0.88

method	result	size
default	\(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (24 a^{4} x^{4} \arctanh \left (a x \right )+6 a^{3} x^{3}-8 a^{2} x^{2} \arctanh \left (a x \right )+5 a x -16 \arctanh \left (a x \right )\right )}{120 a^{4}}+\frac {11 i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}+i\right )}{120 a^{4}}-\frac {11 i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-i\right )}{120 a^{4}}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120/a^4*(-(a*x-1)*(a*x+1))^(1/2)*(24*a^4*x^4*arctanh(a*x)+6*a^3*x^3-8*a^2*x^2*arctanh(a*x)+5*a*x-16*arctanh(
a*x))+11/120*I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)+I)/a^4-11/120*I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-I)/a^4

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Maxima [A]
time = 0.46, size = 128, normalized size = 0.94 \begin {gather*} -\frac {1}{120} \, a {\left (\frac {3 \, {\left (\frac {2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{2}} - \frac {\arcsin \left (a x\right )}{a^{3}}\right )}}{a^{2}} - \frac {8 \, {\left (\sqrt {-a^{2} x^{2} + 1} x + \frac {\arcsin \left (a x\right )}{a}\right )}}{a^{4}}\right )} - \frac {1}{15} \, {\left (\frac {3 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{a^{2}} + \frac {2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4}}\right )} \operatorname {artanh}\left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*a*(3*(2*(-a^2*x^2 + 1)^(3/2)*x/a^2 - sqrt(-a^2*x^2 + 1)*x/a^2 - arcsin(a*x)/a^3)/a^2 - 8*(sqrt(-a^2*x^2
+ 1)*x + arcsin(a*x)/a)/a^4) - 1/15*(3*(-a^2*x^2 + 1)^(3/2)*x^2/a^2 + 2*(-a^2*x^2 + 1)^(3/2)/a^4)*arctanh(a*x
)

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Fricas [A]
time = 0.41, size = 91, normalized size = 0.67 \begin {gather*} \frac {{\left (6 \, a^{3} x^{3} + 5 \, a x + 4 \, {\left (3 \, a^{4} x^{4} - a^{2} x^{2} - 2\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )} \sqrt {-a^{2} x^{2} + 1} - 22 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{120 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*((6*a^3*x^3 + 5*a*x + 4*(3*a^4*x^4 - a^2*x^2 - 2)*log(-(a*x + 1)/(a*x - 1)))*sqrt(-a^2*x^2 + 1) - 22*arc
tan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^4

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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}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x), 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)*(-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 = -0.01 \begin {gather*} \int x^3\,\mathrm {atanh}\left (a\,x\right )\,\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)*(1 - a^2*x^2)^(1/2),x)

[Out]

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

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