3.364 \(\int \frac {x^5 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=139 \[ \frac {89 \sin ^{-1}(a x)}{120 a^6}-\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{5 a^2}-\frac {8 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^6}-\frac {5 x \sqrt {1-a^2 x^2}}{24 a^5}-\frac {4 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac {x^3 \sqrt {1-a^2 x^2}}{20 a^3} \]

[Out]

89/120*arcsin(a*x)/a^6-5/24*x*(-a^2*x^2+1)^(1/2)/a^5-1/20*x^3*(-a^2*x^2+1)^(1/2)/a^3-8/15*arctanh(a*x)*(-a^2*x
^2+1)^(1/2)/a^6-4/15*x^2*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^4-1/5*x^4*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.23, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6016, 321, 216, 5994} \[ -\frac {x^3 \sqrt {1-a^2 x^2}}{20 a^3}-\frac {5 x \sqrt {1-a^2 x^2}}{24 a^5}-\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{5 a^2}-\frac {4 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac {8 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^6}+\frac {89 \sin ^{-1}(a x)}{120 a^6} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*ArcTanh[a*x])/Sqrt[1 - a^2*x^2],x]

[Out]

(-5*x*Sqrt[1 - a^2*x^2])/(24*a^5) - (x^3*Sqrt[1 - a^2*x^2])/(20*a^3) + (89*ArcSin[a*x])/(120*a^6) - (8*Sqrt[1
- a^2*x^2]*ArcTanh[a*x])/(15*a^6) - (4*x^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(15*a^4) - (x^4*Sqrt[1 - a^2*x^2]*A
rcTanh[a*x])/(5*a^2)

Rule 216

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 321

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 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^5 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{5 a^2}+\frac {4 \int \frac {x^3 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{5 a^2}+\frac {\int \frac {x^4}{\sqrt {1-a^2 x^2}} \, dx}{5 a}\\ &=-\frac {x^3 \sqrt {1-a^2 x^2}}{20 a^3}-\frac {4 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{5 a^2}+\frac {8 \int \frac {x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{15 a^4}+\frac {3 \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{20 a^3}+\frac {4 \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{15 a^3}\\ &=-\frac {5 x \sqrt {1-a^2 x^2}}{24 a^5}-\frac {x^3 \sqrt {1-a^2 x^2}}{20 a^3}-\frac {8 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^6}-\frac {4 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{5 a^2}+\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{40 a^5}+\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{15 a^5}+\frac {8 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{15 a^5}\\ &=-\frac {5 x \sqrt {1-a^2 x^2}}{24 a^5}-\frac {x^3 \sqrt {1-a^2 x^2}}{20 a^3}+\frac {89 \sin ^{-1}(a x)}{120 a^6}-\frac {8 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^6}-\frac {4 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{15 a^4}-\frac {x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{5 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 79, normalized size = 0.57 \[ -\frac {a x \sqrt {1-a^2 x^2} \left (6 a^2 x^2+25\right )+8 \sqrt {1-a^2 x^2} \left (3 a^4 x^4+4 a^2 x^2+8\right ) \tanh ^{-1}(a x)-89 \sin ^{-1}(a x)}{120 a^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*ArcTanh[a*x])/Sqrt[1 - a^2*x^2],x]

[Out]

-1/120*(a*x*Sqrt[1 - a^2*x^2]*(25 + 6*a^2*x^2) - 89*ArcSin[a*x] + 8*Sqrt[1 - a^2*x^2]*(8 + 4*a^2*x^2 + 3*a^4*x
^4)*ArcTanh[a*x])/a^6

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fricas [A]  time = 0.60, size = 91, normalized size = 0.65 \[ -\frac {{\left (6 \, a^{3} x^{3} + 25 \, a x + 4 \, {\left (3 \, a^{4} x^{4} + 4 \, a^{2} x^{2} + 8\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )} \sqrt {-a^{2} x^{2} + 1} + 178 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{120 \, a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctanh(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-1/120*((6*a^3*x^3 + 25*a*x + 4*(3*a^4*x^4 + 4*a^2*x^2 + 8)*log(-(a*x + 1)/(a*x - 1)))*sqrt(-a^2*x^2 + 1) + 17
8*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^6

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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^5*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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.44, size = 120, normalized size = 0.86 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (24 a^{4} x^{4} \arctanh \left (a x \right )+6 x^{3} a^{3}+32 a^{2} x^{2} \arctanh \left (a x \right )+25 a x +64 \arctanh \left (a x \right )\right )}{120 a^{6}}+\frac {89 i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}+i\right )}{120 a^{6}}-\frac {89 i \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}-i\right )}{120 a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*arctanh(a*x)/(-a^2*x^2+1)^(1/2),x)

[Out]

-1/120/a^6*(-(a*x-1)*(a*x+1))^(1/2)*(24*a^4*x^4*arctanh(a*x)+6*x^3*a^3+32*a^2*x^2*arctanh(a*x)+25*a*x+64*arcta
nh(a*x))+89/120*I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)+I)/a^6-89/120*I*ln((a*x+1)/(-a^2*x^2+1)^(1/2)-I)/a^6

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maxima [A]  time = 0.41, size = 163, normalized size = 1.17 \[ -\frac {1}{120} \, a {\left (\frac {3 \, {\left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} x^{3}}{a^{2}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{a^{4}} - \frac {3 \, \arcsin \left (a x\right )}{a^{5}}\right )}}{a^{2}} + \frac {16 \, {\left (\frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{2}} - \frac {\arcsin \left (a x\right )}{a^{3}}\right )}}{a^{4}} - \frac {64 \, \arcsin \left (a x\right )}{a^{7}}\right )} - \frac {1}{15} \, {\left (\frac {3 \, \sqrt {-a^{2} x^{2} + 1} x^{4}}{a^{2}} + \frac {4 \, \sqrt {-a^{2} x^{2} + 1} x^{2}}{a^{4}} + \frac {8 \, \sqrt {-a^{2} x^{2} + 1}}{a^{6}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctanh(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-1/120*a*(3*(2*sqrt(-a^2*x^2 + 1)*x^3/a^2 + 3*sqrt(-a^2*x^2 + 1)*x/a^4 - 3*arcsin(a*x)/a^5)/a^2 + 16*(sqrt(-a^
2*x^2 + 1)*x/a^2 - arcsin(a*x)/a^3)/a^4 - 64*arcsin(a*x)/a^7) - 1/15*(3*sqrt(-a^2*x^2 + 1)*x^4/a^2 + 4*sqrt(-a
^2*x^2 + 1)*x^2/a^4 + 8*sqrt(-a^2*x^2 + 1)/a^6)*arctanh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5\,\mathrm {atanh}\left (a\,x\right )}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*atanh(a*x))/(1 - a^2*x^2)^(1/2),x)

[Out]

int((x^5*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 \[ \int \frac {x^{5} \operatorname {atanh}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*atanh(a*x)/(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**5*atanh(a*x)/sqrt(-(a*x - 1)*(a*x + 1)), x)

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