\(\int \frac {e^{-3 \text {arctanh}(a x)}}{(c-\frac {c}{a^2 x^2})^{5/2}} \, dx\) [758]

Optimal result

Integrand size = 24, antiderivative size = 267 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=-\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^4}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)^3}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)^2}+\frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (1+a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5} \] Output:

-(-a^2*x^2+1)^(5/2)/a^5/(c-c/a^2/x^2)^(5/2)/x^4+1/6*(-a^2*x^2+1)^(5/2)/a^6 
/(c-c/a^2/x^2)^(5/2)/x^5/(a*x+1)^3-9/8*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2 
)^(5/2)/x^5/(a*x+1)^2+31/8*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/ 
(a*x+1)-1/16*(-a^2*x^2+1)^(5/2)*ln(-a*x+1)/a^6/(c-c/a^2/x^2)^(5/2)/x^5+49/ 
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 = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.41 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (140+270 a x+42 a^2 x^2-144 a^3 x^3-48 a^4 x^4-3 (1+a x)^3 \log (1-a x)+147 (1+a x)^3 \log (1+a x)\right )}{48 a^2 c^2 \sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^3} \] Input:

Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))^(5/2)),x]
 

Output:

(Sqrt[1 - a^2*x^2]*(140 + 270*a*x + 42*a^2*x^2 - 144*a^3*x^3 - 48*a^4*x^4 
- 3*(1 + a*x)^3*Log[1 - a*x] + 147*(1 + a*x)^3*Log[1 + a*x]))/(48*a^2*c^2* 
Sqrt[c - c/(a^2*x^2)]*x*(1 + a*x)^3)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.41, 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.

\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6710

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {\left (1-a^2 x^2\right )^{5/2} \int \left (\frac {49}{16 a^5 (a x+1)}-\frac {31}{8 a^5 (a x+1)^2}+\frac {9}{4 a^5 (a x+1)^3}-\frac {1}{2 a^5 (a x+1)^4}-\frac {1}{a^5}-\frac {1}{16 a^5 (a x-1)}\right )dx}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {31}{8 a^6 (a x+1)}-\frac {9}{8 a^6 (a x+1)^2}+\frac {1}{6 a^6 (a x+1)^3}-\frac {\log (1-a x)}{16 a^6}+\frac {49 \log (a x+1)}{16 a^6}-\frac {x}{a^5}\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\)

Input:

Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))^(5/2)),x]
 

Output:

((1 - a^2*x^2)^(5/2)*(-(x/a^5) + 1/(6*a^6*(1 + a*x)^3) - 9/(8*a^6*(1 + a*x 
)^2) + 31/(8*a^6*(1 + a*x)) - Log[1 - a*x]/(16*a^6) + (49*Log[1 + a*x])/(1 
6*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	\(\frac {\left (-48 a^{4} x^{4}+147 \ln \left (a x +1\right ) x^{3} a^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}-144 a^{3} x^{3}+441 \ln \left (a x +1\right ) x^{2} a^{2}-9 a^{2} \ln \left (a x -1\right ) x^{2}+42 a^{2} x^{2}+441 \ln \left (a x +1\right ) x a -9 a \ln \left (a x -1\right ) x +270 a x +147 \ln \left (a x +1\right )-3 \ln \left (a x -1\right )+140\right ) \left (a x -1\right )^{2} \sqrt {-a^{2} x^{2}+1}}{48 \left (a x +1\right ) a^{6} x^{5} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\)	\(176\)

Input:

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x,method=_RETURNVER 
BOSE)
 

Output:

1/48*(-48*a^4*x^4+147*ln(a*x+1)*x^3*a^3-3*a^3*ln(a*x-1)*x^3-144*a^3*x^3+44 
1*ln(a*x+1)*x^2*a^2-9*a^2*ln(a*x-1)*x^2+42*a^2*x^2+441*ln(a*x+1)*x*a-9*a*l 
n(a*x-1)*x+270*a*x+147*ln(a*x+1)-3*ln(a*x-1)+140)*(a*x-1)^2*(-a^2*x^2+1)^( 
1/2)/(a*x+1)/a^6/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)
 

Fricas [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm= 
"fricas")
 

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 + 3*a^6*c^3*x^6 + a^5*c^3*x^5 - 5*a^4*c^3*x^4 - 5*a^3*c^3*x^3 + a^ 
2*c^3*x^2 + 3*a*c^3*x + c^3), x)
 

Sympy [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )^{3}}\, dx \] Input:

integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(c-c/a**2/x**2)**(5/2),x)
 

Output:

Integral((-(a*x - 1)*(a*x + 1))**(3/2)/((-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))* 
*(5/2)*(a*x + 1)**3), x)
 

Maxima [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm= 
"maxima")
 

Output:

integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a^2*x^2))^(5/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),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. \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,{\left (a\,x+1\right )}^3} \,d x \] Input:

int((1 - a^2*x^2)^(3/2)/((c - c/(a^2*x^2))^(5/2)*(a*x + 1)^3),x)
 

Output:

int((1 - a^2*x^2)^(3/2)/((c - c/(a^2*x^2))^(5/2)*(a*x + 1)^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {c}\, i \left (3 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+9 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+9 \,\mathrm {log}\left (a x -1\right ) a x +3 \,\mathrm {log}\left (a x -1\right )-147 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-441 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-441 \,\mathrm {log}\left (a x +1\right ) a x -147 \,\mathrm {log}\left (a x +1\right )+48 a^{4} x^{4}+158 a^{3} x^{3}-228 a x -126\right )}{48 a \,c^{3} \left (a^{3} x^{3}+3 a^{2} x^{2}+3 a x +1\right )} \] Input:

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x)
 

Output:

(sqrt(c)*i*(3*log(a*x - 1)*a**3*x**3 + 9*log(a*x - 1)*a**2*x**2 + 9*log(a* 
x - 1)*a*x + 3*log(a*x - 1) - 147*log(a*x + 1)*a**3*x**3 - 441*log(a*x + 1 
)*a**2*x**2 - 441*log(a*x + 1)*a*x - 147*log(a*x + 1) + 48*a**4*x**4 + 158 
*a**3*x**3 - 228*a*x - 126))/(48*a*c**3*(a**3*x**3 + 3*a**2*x**2 + 3*a*x + 
 1))