\(\int \frac {\text {arctanh}(a x)^3}{(1-a^2 x^2)^{3/2}} \, dx\) [407]

Optimal result

Integrand size = 21, antiderivative size = 88 \[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx=-\frac {6}{a \sqrt {1-a^2 x^2}}+\frac {6 x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 \text {arctanh}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}} \] Output:

-6/a/(-a^2*x^2+1)^(1/2)+6*x*arctanh(a*x)/(-a^2*x^2+1)^(1/2)-3*arctanh(a*x) 
^2/a/(-a^2*x^2+1)^(1/2)+x*arctanh(a*x)^3/(-a^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.51 \[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {-6+6 a x \text {arctanh}(a x)-3 \text {arctanh}(a x)^2+a x \text {arctanh}(a x)^3}{a \sqrt {1-a^2 x^2}} \] Input:

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

Output:

(-6 + 6*a*x*ArcTanh[a*x] - 3*ArcTanh[a*x]^2 + a*x*ArcTanh[a*x]^3)/(a*Sqrt[ 
1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6524, 6520}

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.

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

\(\Big \downarrow \) 6524

\(\displaystyle 6 \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}}-\frac {3 \text {arctanh}(a x)^2}{a \sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 6520

\(\displaystyle \frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}}-\frac {3 \text {arctanh}(a x)^2}{a \sqrt {1-a^2 x^2}}+6 \left (\frac {x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}-\frac {1}{a \sqrt {1-a^2 x^2}}\right )\)

Input:

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

Output:

(-3*ArcTanh[a*x]^2)/(a*Sqrt[1 - a^2*x^2]) + (x*ArcTanh[a*x]^3)/Sqrt[1 - a^ 
2*x^2] + 6*(-(1/(a*Sqrt[1 - a^2*x^2])) + (x*ArcTanh[a*x])/Sqrt[1 - a^2*x^2 
])
 

Defintions of rubi rules used

rule 6520

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symb 
ol] :> Simp[-b/(c*d*Sqrt[d + e*x^2]), x] + Simp[x*((a + b*ArcTanh[c*x])/(d* 
Sqrt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0]
 

rule 6524

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x 
_Symbol] :> Simp[(-b)*p*((a + b*ArcTanh[c*x])^(p - 1)/(c*d*Sqrt[d + e*x^2]) 
), x] + (Simp[x*((a + b*ArcTanh[c*x])^p/(d*Sqrt[d + e*x^2])), x] + Simp[b^2 
*p*(p - 1)   Int[(a + b*ArcTanh[c*x])^(p - 2)/(d + e*x^2)^(3/2), x], x]) /; 
 FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 1]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (\operatorname {arctanh}\left (a x \right )^{3} a x +6 a x \,\operatorname {arctanh}\left (a x \right )-3 \operatorname {arctanh}\left (a x \right )^{2}-6\right )}{a \left (a^{2} x^{2}-1\right )}\)	\(56\)
orering	\(-\frac {8 \left (a x -1\right ) \left (a x +1\right ) x \left (9 a^{2} x^{2}-4\right ) \operatorname {arctanh}\left (a x \right )^{3}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {2 \left (a x +1\right )^{2} \left (a x -1\right )^{2} \left (34 a^{2} x^{2}-5\right ) \left (\frac {3 \operatorname {arctanh}\left (a x \right )^{2} a}{\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}+\frac {3 \operatorname {arctanh}\left (a x \right )^{3} a^{2} x}{\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}\right )}{a^{2}}-\frac {16 x \left (a x +1\right )^{3} \left (a x -1\right )^{3} \left (\frac {6 \,\operatorname {arctanh}\left (a x \right ) a^{2}}{\left (-a^{2} x^{2}+1\right )^{\frac {7}{2}}}+\frac {24 \operatorname {arctanh}\left (a x \right )^{2} a^{3} x}{\left (-a^{2} x^{2}+1\right )^{\frac {7}{2}}}+\frac {15 \operatorname {arctanh}\left (a x \right )^{3} a^{4} x^{2}}{\left (-a^{2} x^{2}+1\right )^{\frac {7}{2}}}+\frac {3 \operatorname {arctanh}\left (a x \right )^{3} a^{2}}{\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}\right )}{a^{2}}-\frac {\left (a x +1\right )^{4} \left (a x -1\right )^{4} \left (\frac {6 a^{3}}{\left (-a^{2} x^{2}+1\right )^{\frac {9}{2}}}+\frac {90 \,\operatorname {arctanh}\left (a x \right ) a^{4} x}{\left (-a^{2} x^{2}+1\right )^{\frac {9}{2}}}+\frac {213 \operatorname {arctanh}\left (a x \right )^{2} a^{5} x^{2}}{\left (-a^{2} x^{2}+1\right )^{\frac {9}{2}}}+\frac {33 \operatorname {arctanh}\left (a x \right )^{2} a^{3}}{\left (-a^{2} x^{2}+1\right )^{\frac {7}{2}}}+\frac {105 \operatorname {arctanh}\left (a x \right )^{3} a^{6} x^{3}}{\left (-a^{2} x^{2}+1\right )^{\frac {9}{2}}}+\frac {45 \operatorname {arctanh}\left (a x \right )^{3} a^{4} x}{\left (-a^{2} x^{2}+1\right )^{\frac {7}{2}}}\right )}{a^{4}}\)	\(391\)

Input:

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

Output:

-1/a*(-a^2*x^2+1)^(1/2)*(arctanh(a*x)^3*a*x+6*a*x*arctanh(a*x)-3*arctanh(a 
*x)^2-6)/(a^2*x^2-1)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99 \[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx=-\frac {{\left (a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 24 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 6 \, \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 48\right )} \sqrt {-a^{2} x^{2} + 1}}{8 \, {\left (a^{3} x^{2} - a\right )}} \] Input:

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

Output:

-1/8*(a*x*log(-(a*x + 1)/(a*x - 1))^3 + 24*a*x*log(-(a*x + 1)/(a*x - 1)) - 
 6*log(-(a*x + 1)/(a*x - 1))^2 - 48)*sqrt(-a^2*x^2 + 1)/(a^3*x^2 - a)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98 \[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx=\frac {x \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}} + 6 \, a {\left (\frac {x \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} a} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} a^{2}}\right )} - \frac {3 \, \operatorname {artanh}\left (a x\right )^{2}}{\sqrt {-a^{2} x^{2} + 1} a} \] Input:

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

Output:

x*arctanh(a*x)^3/sqrt(-a^2*x^2 + 1) + 6*a*(x*arctanh(a*x)/(sqrt(-a^2*x^2 + 
 1)*a) - 1/(sqrt(-a^2*x^2 + 1)*a^2)) - 3*arctanh(a*x)^2/(sqrt(-a^2*x^2 + 1 
)*a)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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