\(\int \frac {e^{\text {arctanh}(a x)}}{x^2 (c-a^2 c x^2)^{3/2}} \, dx\) [995]

Optimal result

Integrand size = 25, antiderivative size = 206 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{c x \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2}}{2 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} \log (x)}{c \sqrt {c-a^2 c x^2}}-\frac {5 a \sqrt {1-a^2 x^2} \log (1-a x)}{4 c \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} \log (1+a x)}{4 c \sqrt {c-a^2 c x^2}} \] Output:

-(-a^2*x^2+1)^(1/2)/c/x/(-a^2*c*x^2+c)^(1/2)+1/2*a*(-a^2*x^2+1)^(1/2)/c/(- 
a*x+1)/(-a^2*c*x^2+c)^(1/2)+a*(-a^2*x^2+1)^(1/2)*ln(x)/c/(-a^2*c*x^2+c)^(1 
/2)-5/4*a*(-a^2*x^2+1)^(1/2)*ln(-a*x+1)/c/(-a^2*c*x^2+c)^(1/2)+1/4*a*(-a^2 
*x^2+1)^(1/2)*ln(a*x+1)/c/(-a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.37 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-\frac {4}{x}+\frac {2 a}{1-a x}+4 a \log (x)-5 a \log (1-a x)+a \log (1+a x)\right )}{4 c \sqrt {c-a^2 c x^2}} \] Input:

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

Output:

(Sqrt[1 - a^2*x^2]*(-4/x + (2*a)/(1 - a*x) + 4*a*Log[x] - 5*a*Log[1 - a*x] 
 + a*Log[1 + a*x]))/(4*c*Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.38, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6703, 6700, 99, 2009}

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.

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

\(\Big \downarrow \) 6703

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

\(\Big \downarrow \) 6700

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {1}{x^2 (1-a x)^2 (a x+1)}dx}{c \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {5 a^2}{4 (a x-1)}+\frac {a^2}{4 (a x+1)}+\frac {a^2}{2 (a x-1)^2}+\frac {a}{x}+\frac {1}{x^2}\right )dx}{c \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {a}{2 (1-a x)}+a \log (x)-\frac {5}{4} a \log (1-a x)+\frac {1}{4} a \log (a x+1)-\frac {1}{x}\right )}{c \sqrt {c-a^2 c x^2}}\)

Input:

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

Output:

(Sqrt[1 - a^2*x^2]*(-x^(-1) + a/(2*(1 - a*x)) + a*Log[x] - (5*a*Log[1 - a* 
x])/4 + (a*Log[1 + a*x])/4))/(c*Sqrt[c - a^2*c*x^2])
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6700

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[c^p   Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], 
 x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || 
 GtQ[c, 0])
 

rule 6703

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar 
t[p])   Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, 
d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !I 
ntegerQ[n/2]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.54

method	result	size
default	\(\frac {\left (\ln \left (a x +1\right ) x^{2} a^{2}-5 a^{2} \ln \left (a x -1\right ) x^{2}+4 a^{2} \ln \left (x \right ) x^{2}-\ln \left (a x +1\right ) x a +5 a \ln \left (a x -1\right ) x -4 a \ln \left (x \right ) x -6 a x +4\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{4 \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right ) c^{2} x}\)	\(111\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^2/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVE 
RBOSE)
 

Output:

1/4*(ln(a*x+1)*x^2*a^2-5*a^2*ln(a*x-1)*x^2+4*a^2*ln(x)*x^2-ln(a*x+1)*x*a+5 
*a*ln(a*x-1)*x-4*a*ln(x)*x-6*a*x+4)/(-a^2*x^2+1)^(1/2)/(a*x-1)*(-c*(a^2*x^ 
2-1))^(1/2)/c^2/x
 

Fricas [F]

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

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

Output:

integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^5*c^2*x^7 - a^4*c^2*x 
^6 - 2*a^3*c^2*x^5 + 2*a^2*c^2*x^4 + a*c^2*x^3 - c^2*x^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.44 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (-5 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+5 \,\mathrm {log}\left (a x -1\right ) a x +\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-\mathrm {log}\left (a x +1\right ) a x +4 \,\mathrm {log}\left (x \right ) a^{2} x^{2}-4 \,\mathrm {log}\left (x \right ) a x -6 a^{2} x^{2}+4\right )}{4 c^{2} x \left (a x -1\right )} \] Input:

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

Output:

(sqrt(c)*( - 5*log(a*x - 1)*a**2*x**2 + 5*log(a*x - 1)*a*x + log(a*x + 1)* 
a**2*x**2 - log(a*x + 1)*a*x + 4*log(x)*a**2*x**2 - 4*log(x)*a*x - 6*a**2* 
x**2 + 4))/(4*c**2*x*(a*x - 1))