3.451 \(\int (1-a^2 x^2)^{3/2} \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=189 \[ \frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {3 \sqrt {1-a^2 x^2}}{8 a}+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {3}{8} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{8 a}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{8 a}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{4 a} \]

[Out]

1/12*(-a^2*x^2+1)^(3/2)/a+1/4*x*(-a^2*x^2+1)^(3/2)*arctanh(a*x)-3/4*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arcta
nh(a*x)/a-3/8*I*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a+3/8*I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a
+3/8*(-a^2*x^2+1)^(1/2)/a+3/8*x*arctanh(a*x)*(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5942, 5950} \[ -\frac {3 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{8 a}+\frac {3 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{8 a}+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {3 \sqrt {1-a^2 x^2}}{8 a}+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {3}{8} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[1 - a^2*x^2])/(8*a) + (1 - a^2*x^2)^(3/2)/(12*a) + (3*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/8 + (x*(1 - a^
2*x^2)^(3/2)*ArcTanh[a*x])/4 - (3*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(4*a) - (((3*I)/8)*PolyLog
[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a + (((3*I)/8)*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 )^{3/2} \tanh ^{-1}(a x) \, dx &=\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {3}{4} \int \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\\ &=\frac {3 \sqrt {1-a^2 x^2}}{8 a}+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {3}{8} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {3}{8} \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {3 \sqrt {1-a^2 x^2}}{8 a}+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {3}{8} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{4 a}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 176, normalized size = 0.93 \[ \frac {-2 a^2 x^2 \sqrt {1-a^2 x^2}+11 \sqrt {1-a^2 x^2}+15 a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-6 a^3 x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-9 i \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )+9 i \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )-9 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+9 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )}{24 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(11*Sqrt[1 - a^2*x^2] - 2*a^2*x^2*Sqrt[1 - a^2*x^2] + 15*a*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 6*a^3*x^3*Sqrt[1
- a^2*x^2]*ArcTanh[a*x] - (9*I)*ArcTanh[a*x]*Log[1 - I/E^ArcTanh[a*x]] + (9*I)*ArcTanh[a*x]*Log[1 + I/E^ArcTa
nh[a*x]] - (9*I)*PolyLog[2, (-I)/E^ArcTanh[a*x]] + (9*I)*PolyLog[2, I/E^ArcTanh[a*x]])/(24*a)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} x^{2} - 1\right )} \sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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maple [A]  time = 0.96, size = 173, normalized size = 0.92 \[ -\frac {\left (6 a^{3} x^{3} \arctanh \left (a x \right )+2 a^{2} x^{2}-15 a x \arctanh \left (a x \right )-11\right ) \sqrt {-a^{2} x^{2}+1}}{24 a}-\frac {3 i \arctanh \left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a}+\frac {3 i \arctanh \left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a}-\frac {3 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a}+\frac {3 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*a^3*x^3*arctanh(a*x)+2*a^2*x^2-15*a*x*arctanh(a*x)-11)*(-a^2*x^2+1)^(1/2)/a-3/8*I/a*arctanh(a*x)*ln(1
+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+3/8*I/a*arctanh(a*x)*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))-3/8*I/a*dilog(1+I*(a*x+
1)/(-a^2*x^2+1)^(1/2))+3/8*I/a*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)*(1 - a^2*x^2)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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