\(\int (1-a^2 x^2)^{3/2} \text {arctanh}(a x) \, dx\) [451]

Optimal result

Integrand size = 19, antiderivative size = 189 \[ \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx=\frac {3 \sqrt {1-a^2 x^2}}{8 a}+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}+\frac {3}{8} x \sqrt {1-a^2 x^2} \text {arctanh}(a x)+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)-\frac {3 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \text {arctanh}(a x)}{4 a}-\frac {3 i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a}+\frac {3 i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a} \] Output:

3/8*(-a^2*x^2+1)^(1/2)/a+1/12*(-a^2*x^2+1)^(3/2)/a+3/8*x*(-a^2*x^2+1)^(1/2 
)*arctanh(a*x)+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))*arctanh(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
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93 \[ \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx=\frac {11 \sqrt {1-a^2 x^2}-2 a^2 x^2 \sqrt {1-a^2 x^2}+15 a x \sqrt {1-a^2 x^2} \text {arctanh}(a x)-6 a^3 x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)-9 i \text {arctanh}(a x) \log \left (1-i e^{-\text {arctanh}(a x)}\right )+9 i \text {arctanh}(a x) \log \left (1+i e^{-\text {arctanh}(a x)}\right )-9 i \operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )+9 i \operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )}{24 a} \] Input:

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

Output:

(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)*ArcTa 
nh[a*x]*Log[1 - I/E^ArcTanh[a*x]] + (9*I)*ArcTanh[a*x]*Log[1 + I/E^ArcTanh 
[a*x]] - (9*I)*PolyLog[2, (-I)/E^ArcTanh[a*x]] + (9*I)*PolyLog[2, I/E^ArcT 
anh[a*x]])/(24*a)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6504, 6504, 6512}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx\)

\(\Big \downarrow \) 6504

\(\displaystyle \frac {3}{4} \int \sqrt {1-a^2 x^2} \text {arctanh}(a x)dx+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}\)

\(\Big \downarrow \) 6504

\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx+\frac {1}{2} x \sqrt {1-a^2 x^2} \text {arctanh}(a x)+\frac {\sqrt {1-a^2 x^2}}{2 a}\right )+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}\)

\(\Big \downarrow \) 6512

\(\displaystyle \frac {3}{4} \left (\frac {1}{2} x \sqrt {1-a^2 x^2} \text {arctanh}(a x)+\frac {\sqrt {1-a^2 x^2}}{2 a}+\frac {1}{2} \left (-\frac {2 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \text {arctanh}(a x)}{a}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}\right )\right )+\frac {1}{4} x \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a}\)

Input:

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

Output:

(1 - a^2*x^2)^(3/2)/(12*a) + (x*(1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/4 + (3*( 
Sqrt[1 - a^2*x^2]/(2*a) + (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/2 + ((-2*ArcT 
an[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/a - (I*PolyLog[2, ((-I)*Sqrt 
[1 - a*x])/Sqrt[1 + a*x]])/a + (I*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a* 
x]])/a)/2))/4
 

Defintions of rubi rules used

rule 6504

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

rule 6512

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*S 
qrt[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]
 

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {\left (6 a^{3} x^{3} \operatorname {arctanh}\left (a x \right )+2 a^{2} x^{2}-15 a x \,\operatorname {arctanh}\left (a x \right )-11\right ) \sqrt {-a^{2} x^{2}+1}}{24 a}-\frac {3 i \operatorname {arctanh}\left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a}+\frac {3 i \operatorname {arctanh}\left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a}-\frac {3 i \operatorname {dilog}\left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a}+\frac {3 i \operatorname {dilog}\left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a}\)	\(173\)

Input:

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

Output:

-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))
 

Fricas [F]

\[ \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx=\int { {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right ) \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx=\int \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}{\left (a x \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx=\int { {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right ) \,d x } \] Input:

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx=\int \mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x) \, dx=-\left (\int \sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right ) x^{2}d x \right ) a^{2}+\int \sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right )d x \] Input:

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

Output:

 - int(sqrt( - a**2*x**2 + 1)*atanh(a*x)*x**2,x)*a**2 + int(sqrt( - a**2*x 
**2 + 1)*atanh(a*x),x)