\(\int \frac {1}{(1-a^2 x^2)^{3/2} \text {arctanh}(a x)^3} \, dx\) [424]

Optimal result

Integrand size = 21, antiderivative size = 65 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=-\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}-\frac {x}{2 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}+\frac {\text {Chi}(\text {arctanh}(a x))}{2 a} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^3} \, dx=\frac {-\frac {1+a x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}+\text {Chi}(\text {arctanh}(a x))}{2 a} \] Input:

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

Output:

(-((1 + a*x*ArcTanh[a*x])/(Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)) + CoshIntegr 
al[ArcTanh[a*x]])/(2*a)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6528, 6568, 6530, 3042, 3782}

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.

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

\(\Big \downarrow \) 6528

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

\(\Big \downarrow \) 6568

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

\(\Big \downarrow \) 6530

\(\displaystyle \frac {1}{2} a \left (\frac {\int \frac {1}{\sqrt {1-a^2 x^2} \text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}-\frac {x}{a \sqrt {1-a^2 x^2} \text {arctanh}(a x)}\right )-\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {1}{2 a \sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}+\frac {1}{2} a \left (-\frac {x}{a \sqrt {1-a^2 x^2} \text {arctanh}(a x)}+\frac {\int \frac {\sin \left (i \text {arctanh}(a x)+\frac {\pi }{2}\right )}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a^2}\right )\)

\(\Big \downarrow \) 3782

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3782

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo 
l] :> Simp[CoshIntegral[c*f*(fz/d) + f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz 
}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]
 

rule 6528

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

rule 6530

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x 
_Symbol] :> Simp[d^q/c   Subst[Int[(a + b*x)^p/Cosh[x]^(2*(q + 1)), x], x, 
ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && I 
LtQ[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])
 

rule 6568

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x^2)^(q + 1)*((a + b*Arc 
Tanh[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Simp[f*(m/(b*c*(p + 1)))   Int[(f 
*x)^(m - 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x] /; FreeQ[{a, 
 b, c, d, e, f, m, q}, x] && EqQ[c^2*d + e, 0] && EqQ[m + 2*q + 2, 0] && Lt 
Q[p, -1]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.32

method	result	size
default	\(\frac {\operatorname {arctanh}\left (a x \right )^{2} \operatorname {Chi}\left (\operatorname {arctanh}\left (a x \right )\right ) a^{2} x^{2}+x \,\operatorname {arctanh}\left (a x \right ) a \sqrt {-a^{2} x^{2}+1}-\operatorname {Chi}\left (\operatorname {arctanh}\left (a x \right )\right ) \operatorname {arctanh}\left (a x \right )^{2}+\sqrt {-a^{2} x^{2}+1}}{2 a \operatorname {arctanh}\left (a x \right )^{2} \left (a^{2} x^{2}-1\right )}\)	\(86\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(-a^2*x^2 + 1)/((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)^3), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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