3.459 \(\int (1-a^2 x^2)^{5/2} \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=233 \[ \frac {\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac {5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac {5 \sqrt {1-a^2 x^2}}{16 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac {5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {5}{16} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {5 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a}+\frac {5 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a}-\frac {5 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{8 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5942, 5950} \[ -\frac {5 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a}+\frac {5 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac {5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac {5 \sqrt {1-a^2 x^2}}{16 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac {5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {5}{16} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {5 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Sqrt[1 - a^2*x^2])/(16*a) + (5*(1 - a^2*x^2)^(3/2))/(72*a) + (1 - a^2*x^2)^(5/2)/(30*a) + (5*x*Sqrt[1 - a^2
*x^2]*ArcTanh[a*x])/16 + (5*x*(1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/24 + (x*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x])/6 -
(5*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

Rule 5942

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(b*(d + e*x^2)^q)/(2*c*
q*(2*q + 1)), x] + (Dist[(2*d*q)/(2*q + 1), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x]), x], x] + Simp[(x*(d
+ e*x^2)^q*(a + b*ArcTanh[c*x]))/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[q, 0]

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 1.18, size = 224, normalized size = 0.96 \[ \frac {-98 a^2 x^2 \sqrt {1-a^2 x^2}+299 \sqrt {1-a^2 x^2}+495 a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+120 a^5 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+24 a^4 x^4 \sqrt {1-a^2 x^2}-390 a^3 x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-225 i \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )+225 i \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )-225 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+225 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )}{720 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(299*Sqrt[1 - a^2*x^2] - 98*a^2*x^2*Sqrt[1 - a^2*x^2] + 24*a^4*x^4*Sqrt[1 - a^2*x^2] + 495*a*x*Sqrt[1 - a^2*x^
2]*ArcTanh[a*x] - 390*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] - (2
25*I)*ArcTanh[a*x]*Log[1 - I/E^ArcTanh[a*x]] + (225*I)*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] - (225*I)*PolyLo
g[2, (-I)/E^ArcTanh[a*x]] + (225*I)*PolyLog[2, I/E^ArcTanh[a*x]])/(720*a)

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

________________________________________________________________________________________

maple [A]  time = 1.14, size = 193, normalized size = 0.83 \[ \frac {\left (120 \arctanh \left (a x \right ) x^{5} a^{5}+24 x^{4} a^{4}-390 a^{3} x^{3} \arctanh \left (a x \right )-98 a^{2} x^{2}+495 a x \arctanh \left (a x \right )+299\right ) \sqrt {-a^{2} x^{2}+1}}{720 a}-\frac {5 i \arctanh \left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a}+\frac {5 i \arctanh \left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a}-\frac {5 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a}+\frac {5 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/720*(120*arctanh(a*x)*x^5*a^5+24*x^4*a^4-390*a^3*x^3*arctanh(a*x)-98*a^2*x^2+495*a*x*arctanh(a*x)+299)*(-a^2
*x^2+1)^(1/2)/a-5/16*I/a*arctanh(a*x)*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+5/16*I/a*arctanh(a*x)*ln(1-I*(a*x+1)/
(-a^2*x^2+1)^(1/2))-5/16*I/a*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+5/16*I/a*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/
2))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)*(1 - a^2*x^2)^(5/2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \operatorname {atanh}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________