\(\int \frac {e^{-\text {arctanh}(a x)}}{(c-\frac {c}{a^2 x^2})^{5/2}} \, dx\) [742]

Optimal result

Integrand size = 24, antiderivative size = 265 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^4}+\frac {\left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)^2}-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)}+\frac {7 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}-\frac {23 \left (1-a^2 x^2\right )^{5/2} \log (1+a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5} \] Output:

(-a^2*x^2+1)^(5/2)/a^5/(c-c/a^2/x^2)^(5/2)/x^4+1/8*(-a^2*x^2+1)^(5/2)/a^6/ 
(c-c/a^2/x^2)^(5/2)/x^5/(-a*x+1)+1/8*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^ 
(5/2)/x^5/(a*x+1)^2-(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/(a*x+1) 
+7/16*(-a^2*x^2+1)^(5/2)*ln(-a*x+1)/a^6/(c-c/a^2/x^2)^(5/2)/x^5-23/16*(-a^ 
2*x^2+1)^(5/2)*ln(a*x+1)/a^6/(c-c/a^2/x^2)^(5/2)/x^5
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.33 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\left (1-a^2 x^2\right )^{5/2} \left (2 \left (8 a x+\frac {1}{1-a x}+\frac {1}{(1+a x)^2}-\frac {8}{1+a x}\right )+7 \log (1-a x)-23 \log (1+a x)\right )}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5} \] Input:

Integrate[1/(E^ArcTanh[a*x]*(c - c/(a^2*x^2))^(5/2)),x]
 

Output:

((1 - a^2*x^2)^(5/2)*(2*(8*a*x + (1 - a*x)^(-1) + (1 + a*x)^(-2) - 8/(1 + 
a*x)) + 7*Log[1 - a*x] - 23*Log[1 + a*x]))/(16*a^6*(c - c/(a^2*x^2))^(5/2) 
*x^5)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.40, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6710, 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.

\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6710

\(\displaystyle \frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {e^{-\text {arctanh}(a x)} x^5}{\left (1-a^2 x^2\right )^{5/2}}dx}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\)

\(\Big \downarrow \) 6700

\(\displaystyle \frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {x^5}{(1-a x)^2 (a x+1)^3}dx}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\left (1-a^2 x^2\right )^{5/2} \int \left (-\frac {23}{16 a^5 (a x+1)}+\frac {1}{a^5 (a x+1)^2}-\frac {1}{4 a^5 (a x+1)^3}+\frac {1}{a^5}+\frac {7}{16 a^5 (a x-1)}+\frac {1}{8 a^5 (a x-1)^2}\right )dx}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {1}{8 a^6 (1-a x)}-\frac {1}{a^6 (a x+1)}+\frac {1}{8 a^6 (a x+1)^2}+\frac {7 \log (1-a x)}{16 a^6}-\frac {23 \log (a x+1)}{16 a^6}+\frac {x}{a^5}\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\)

Input:

Int[1/(E^ArcTanh[a*x]*(c - c/(a^2*x^2))^(5/2)),x]
 

Output:

((1 - a^2*x^2)^(5/2)*(x/a^5 + 1/(8*a^6*(1 - a*x)) + 1/(8*a^6*(1 + a*x)^2) 
- 1/(a^6*(1 + a*x)) + (7*Log[1 - a*x])/(16*a^6) - (23*Log[1 + a*x])/(16*a^ 
6)))/((c - c/(a^2*x^2))^(5/2)*x^5)
 

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 6710

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo 
l] :> Simp[x^(2*p)*((c + d/x^2)^p/(1 - a^2*x^2)^p)   Int[(u/x^(2*p))*(1 - a 
^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c 
+ a^2*d, 0] &&  !IntegerQ[p] &&  !IntegerQ[n/2]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {\left (-16 a^{4} x^{4}+23 \ln \left (a x +1\right ) x^{3} a^{3}-7 a^{3} \ln \left (a x -1\right ) x^{3}-16 a^{3} x^{3}+23 \ln \left (a x +1\right ) x^{2} a^{2}-7 a^{2} \ln \left (a x -1\right ) x^{2}+34 a^{2} x^{2}-23 \ln \left (a x +1\right ) x a +7 a \ln \left (a x -1\right ) x +18 a x -23 \ln \left (a x +1\right )+7 \ln \left (a x -1\right )-12\right ) \left (a x -1\right ) \left (a^{2} x^{2}-1\right )}{16 \sqrt {-a^{2} x^{2}+1}\, a^{6} x^{5} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\)	\(176\)

Input:

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x,method=_RETURNVERBO 
SE)
 

Output:

1/16*(-16*a^4*x^4+23*ln(a*x+1)*x^3*a^3-7*a^3*ln(a*x-1)*x^3-16*a^3*x^3+23*l 
n(a*x+1)*x^2*a^2-7*a^2*ln(a*x-1)*x^2+34*a^2*x^2-23*ln(a*x+1)*x*a+7*a*ln(a* 
x-1)*x+18*a*x-23*ln(a*x+1)+7*ln(a*x-1)-12)*(a*x-1)*(a^2*x^2-1)/(-a^2*x^2+1 
)^(1/2)/a^6/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)
 

Fricas [F]

\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="f 
ricas")
 

Output:

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

Sympy [F]

\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}\, dx \] Input:

integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(c-c/a**2/x**2)**(5/2),x)
 

Output:

Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))**(5 
/2)*(a*x + 1)), x)
 

Maxima [F]

\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="m 
axima")
 

Output:

integrate(sqrt(-a^2*x^2 + 1)/((a*x + 1)*(c - c/(a^2*x^2))^(5/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="g 
iac")
 

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. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,\left (a\,x+1\right )} \,d x \] Input:

int((1 - a^2*x^2)^(1/2)/((c - c/(a^2*x^2))^(5/2)*(a*x + 1)),x)
 

Output:

int((1 - a^2*x^2)^(1/2)/((c - c/(a^2*x^2))^(5/2)*(a*x + 1)), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {c}\, i \left (-7 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-7 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+7 \,\mathrm {log}\left (a x -1\right ) a x +7 \,\mathrm {log}\left (a x -1\right )+23 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+23 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-23 \,\mathrm {log}\left (a x +1\right ) a x -23 \,\mathrm {log}\left (a x +1\right )-16 a^{4} x^{4}-50 a^{3} x^{3}+52 a x +22\right )}{16 a \,c^{3} \left (a^{3} x^{3}+a^{2} x^{2}-a x -1\right )} \] Input:

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x)
 

Output:

(sqrt(c)*i*( - 7*log(a*x - 1)*a**3*x**3 - 7*log(a*x - 1)*a**2*x**2 + 7*log 
(a*x - 1)*a*x + 7*log(a*x - 1) + 23*log(a*x + 1)*a**3*x**3 + 23*log(a*x + 
1)*a**2*x**2 - 23*log(a*x + 1)*a*x - 23*log(a*x + 1) - 16*a**4*x**4 - 50*a 
**3*x**3 + 52*a*x + 22))/(16*a*c**3*(a**3*x**3 + a**2*x**2 - a*x - 1))