\(\int e^{3 \text {arctanh}(a x)} (c-\frac {c}{a^2 x^2})^{5/2} \, dx\) [729]

Optimal result

Integrand size = 24, antiderivative size = 220 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=-\frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}{4 \left (1-a^2 x^2\right )^{5/2}}-\frac {a \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}{\left (1-a^2 x^2\right )^{5/2}}-\frac {a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}{\left (1-a^2 x^2\right )^{5/2}}+\frac {2 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^4}{\left (1-a^2 x^2\right )^{5/2}}-\frac {a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^6}{\left (1-a^2 x^2\right )^{5/2}}-\frac {3 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (1-a^2 x^2\right )^{5/2}} \] Output:

-1/4*(c-c/a^2/x^2)^(5/2)*x/(-a^2*x^2+1)^(5/2)-a*(c-c/a^2/x^2)^(5/2)*x^2/(- 
a^2*x^2+1)^(5/2)-a^2*(c-c/a^2/x^2)^(5/2)*x^3/(-a^2*x^2+1)^(5/2)+2*a^3*(c-c 
/a^2/x^2)^(5/2)*x^4/(-a^2*x^2+1)^(5/2)-a^5*(c-c/a^2/x^2)^(5/2)*x^6/(-a^2*x 
^2+1)^(5/2)-3*a^4*(c-c/a^2/x^2)^(5/2)*x^5*ln(x)/(-a^2*x^2+1)^(5/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.41 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \left (1+4 a x+4 a^2 x^2-8 a^3 x^3+5 a^4 x^4+4 a^5 x^5+12 a^4 x^4 \log (x)\right )}{4 a^4 x^3 \sqrt {1-a^2 x^2}} \] Input:

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

Output:

-1/4*(c^2*Sqrt[c - c/(a^2*x^2)]*(1 + 4*a*x + 4*a^2*x^2 - 8*a^3*x^3 + 5*a^4 
*x^4 + 4*a^5*x^5 + 12*a^4*x^4*Log[x]))/(a^4*x^3*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.35, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6710, 6700, 84, 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.

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

\(\Big \downarrow \) 6710

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 84

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \left (a^5 (-x)-3 a^4 \log (x)+\frac {2 a^3}{x}-\frac {a^2}{x^2}-\frac {a}{x^3}-\frac {1}{4 x^4}\right )}{\left (1-a^2 x^2\right )^{5/2}}\)

Input:

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

Output:

((c - c/(a^2*x^2))^(5/2)*x^5*(-1/4*1/x^4 - a/x^3 - a^2/x^2 + (2*a^3)/x - a 
^5*x - 3*a^4*Log[x]))/(1 - a^2*x^2)^(5/2)
 

Defintions of rubi rules used

rule 84

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : 
> Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, 
 d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n + p + 2, 0 
] && GtQ[n + 2*p, 0])
 

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 = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.39

method	result	size
default	\(-\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}} x \left (4 a^{5} x^{5}+12 \ln \left (x \right ) x^{4} a^{4}-8 a^{3} x^{3}+4 a^{2} x^{2}+4 a x +1\right )}{4 \left (a^{2} x^{2}-1\right )^{2} \sqrt {-a^{2} x^{2}+1}}\)	\(86\)

Input:

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

Output:

-1/4*(c*(a^2*x^2-1)/a^2/x^2)^(5/2)*x/(a^2*x^2-1)^2/(-a^2*x^2+1)^(1/2)*(4*a 
^5*x^5+12*ln(x)*x^4*a^4-8*a^3*x^3+4*a^2*x^2+4*a*x+1)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 476, normalized size of antiderivative = 2.16 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\left [\frac {6 \, {\left (a^{5} c^{2} x^{5} - a^{3} c^{2} x^{3}\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} - {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) + {\left (4 \, a^{5} c^{2} x^{5} - 8 \, a^{3} c^{2} x^{3} - {\left (4 \, a^{5} - 8 \, a^{3} + 4 \, a^{2} + 4 \, a + 1\right )} c^{2} x^{4} + 4 \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x + c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{6} x^{5} - a^{4} x^{3}\right )}}, \frac {12 \, {\left (a^{5} c^{2} x^{5} - a^{3} c^{2} x^{3}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} - a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} + {\left (a^{2} - 1\right )} c x^{2} - c}\right ) + {\left (4 \, a^{5} c^{2} x^{5} - 8 \, a^{3} c^{2} x^{3} - {\left (4 \, a^{5} - 8 \, a^{3} + 4 \, a^{2} + 4 \, a + 1\right )} c^{2} x^{4} + 4 \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x + c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{4 \, {\left (a^{6} x^{5} - a^{4} x^{3}\right )}}\right ] \] Input:

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

Output:

[1/4*(6*(a^5*c^2*x^5 - a^3*c^2*x^3)*sqrt(-c)*log((a^2*c*x^6 + a^2*c*x^2 - 
c*x^4 - (a*x^5 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c)/(a^ 
2*x^2)) - c)/(a^2*x^4 - x^2)) + (4*a^5*c^2*x^5 - 8*a^3*c^2*x^3 - (4*a^5 - 
8*a^3 + 4*a^2 + 4*a + 1)*c^2*x^4 + 4*a^2*c^2*x^2 + 4*a*c^2*x + c^2)*sqrt(- 
a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^6*x^5 - a^4*x^3), 1/4*(12 
*(a^5*c^2*x^5 - a^3*c^2*x^3)*sqrt(c)*arctan(sqrt(-a^2*x^2 + 1)*(a*x^3 - a* 
x)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^4 + (a^2 - 1)*c*x^2 - 
c)) + (4*a^5*c^2*x^5 - 8*a^3*c^2*x^3 - (4*a^5 - 8*a^3 + 4*a^2 + 4*a + 1)*c 
^2*x^4 + 4*a^2*c^2*x^2 + 4*a*c^2*x + c^2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x 
^2 - c)/(a^2*x^2)))/(a^6*x^5 - a^4*x^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.39 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=-\frac {1}{4} \, {\left (\frac {4 \, c^{2} x \mathrm {sgn}\left (x\right )}{a} + \frac {12 \, c^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{2}} - \frac {8 \, a^{3} c^{2} x^{3} \mathrm {sgn}\left (x\right ) - 4 \, a^{2} c^{2} x^{2} \mathrm {sgn}\left (x\right ) - 4 \, a c^{2} x \mathrm {sgn}\left (x\right ) - c^{2} \mathrm {sgn}\left (x\right )}{a^{6} x^{4}}\right )} \sqrt {-c} {\left | a \right |} \] Input:

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

Output:

-1/4*(4*c^2*x*sgn(x)/a + 12*c^2*log(abs(x))*sgn(x)/a^2 - (8*a^3*c^2*x^3*sg 
n(x) - 4*a^2*c^2*x^2*sgn(x) - 4*a*c^2*x*sgn(x) - c^2*sgn(x))/(a^6*x^4))*sq 
rt(-c)*abs(a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.25 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} i \left (-12 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-4 a^{5} x^{5}+8 a^{3} x^{3}-4 a^{2} x^{2}-4 a x -1\right )}{4 a^{5} x^{4}} \] Input:

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

Output:

(sqrt(c)*c**2*i*( - 12*log(x)*a**4*x**4 - 4*a**5*x**5 + 8*a**3*x**3 - 4*a* 
*2*x**2 - 4*a*x - 1))/(4*a**5*x**4)