\(\int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx\) [1201]

Optimal result

Integrand size = 27, antiderivative size = 223 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=-\frac {\sqrt {c-a^2 c x^2}}{4 x^4 \sqrt {1-a^2 x^2}}-\frac {a \sqrt {c-a^2 c x^2}}{x^3 \sqrt {1-a^2 x^2}}-\frac {2 a^2 \sqrt {c-a^2 c x^2}}{x^2 \sqrt {1-a^2 x^2}}-\frac {4 a^3 \sqrt {c-a^2 c x^2}}{x \sqrt {1-a^2 x^2}}+\frac {4 a^4 \sqrt {c-a^2 c x^2} \log (x)}{\sqrt {1-a^2 x^2}}-\frac {4 a^4 \sqrt {c-a^2 c x^2} \log (1-a x)}{\sqrt {1-a^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.35 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (-\frac {1}{4 x^4}-\frac {a}{x^3}-\frac {2 a^2}{x^2}-\frac {4 a^3}{x}+4 a^4 \log (x)-4 a^4 \log (1-a x)\right )}{\sqrt {1-a^2 x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.35, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, 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.

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

\(\Big \downarrow \) 6703

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(Sqrt[c - a^2*c*x^2]*(-1/4*1/x^4 - a/x^3 - (2*a^2)/x^2 - (4*a^3)/x + 4*a^4 
*Log[x] - 4*a^4*Log[1 - a*x]))/Sqrt[1 - a^2*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 = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.35

method	result	size
default	\(-\frac {\left (16 \ln \left (a x -1\right ) x^{4} a^{4}-16 \ln \left (x \right ) x^{4} a^{4}+16 a^{3} x^{3}+8 a^{2} x^{2}+4 a x +1\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{4 \sqrt {-a^{2} x^{2}+1}\, x^{4}}\)	\(78\)

Input:

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

Output:

-1/4*(16*ln(a*x-1)*x^4*a^4-16*ln(x)*x^4*a^4+16*a^3*x^3+8*a^2*x^2+4*a*x+1)/ 
(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.26 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=\left [\frac {8 \, {\left (a^{6} x^{6} - a^{4} x^{4}\right )} \sqrt {c} \log \left (-\frac {4 \, a^{5} c x^{5} - {\left (2 \, a^{6} - 4 \, a^{5} + 6 \, a^{4} - 4 \, a^{3} + a^{2}\right )} c x^{6} - {\left (4 \, a^{4} + 4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} c x^{4} + 5 \, a^{2} c x^{2} - 4 \, a c x + {\left (4 \, a^{3} x^{3} - {\left (4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} x^{4} - 6 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right ) + {\left (16 \, a^{3} x^{3} - {\left (16 \, a^{3} + 8 \, a^{2} + 4 \, a + 1\right )} x^{4} + 8 \, a^{2} x^{2} + 4 \, a x + 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{4 \, {\left (a^{2} x^{6} - x^{4}\right )}}, -\frac {16 \, {\left (a^{6} x^{6} - a^{4} x^{4}\right )} \sqrt {-c} \arctan \left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a^{2} - 2 \, a + 1\right )} x^{2} - 2 \, a x + 1\right )} \sqrt {-c}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} - a^{2}\right )} c x^{4} - {\left (a^{2} - 2 \, a + 1\right )} c x^{2} - 2 \, a c x + c}\right ) - {\left (16 \, a^{3} x^{3} - {\left (16 \, a^{3} + 8 \, a^{2} + 4 \, a + 1\right )} x^{4} + 8 \, a^{2} x^{2} + 4 \, a x + 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{4 \, {\left (a^{2} x^{6} - x^{4}\right )}}\right ] \] Input:

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

Output:

[1/4*(8*(a^6*x^6 - a^4*x^4)*sqrt(c)*log(-(4*a^5*c*x^5 - (2*a^6 - 4*a^5 + 6 
*a^4 - 4*a^3 + a^2)*c*x^6 - (4*a^4 + 4*a^3 - 6*a^2 + 4*a - 1)*c*x^4 + 5*a^ 
2*c*x^2 - 4*a*c*x + (4*a^3*x^3 - (4*a^3 - 6*a^2 + 4*a - 1)*x^4 - 6*a^2*x^2 
 + 4*a*x - 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) + c)/(a^4*x^ 
6 - 2*a^3*x^5 + 2*a*x^3 - x^2)) + (16*a^3*x^3 - (16*a^3 + 8*a^2 + 4*a + 1) 
*x^4 + 8*a^2*x^2 + 4*a*x + 1)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^ 
2*x^6 - x^4), -1/4*(16*(a^6*x^6 - a^4*x^4)*sqrt(-c)*arctan(-sqrt(-a^2*c*x^ 
2 + c)*sqrt(-a^2*x^2 + 1)*((2*a^2 - 2*a + 1)*x^2 - 2*a*x + 1)*sqrt(-c)/(2* 
a^3*c*x^3 - (2*a^3 - a^2)*c*x^4 - (a^2 - 2*a + 1)*c*x^2 - 2*a*c*x + c)) - 
(16*a^3*x^3 - (16*a^3 + 8*a^2 + 4*a + 1)*x^4 + 8*a^2*x^2 + 4*a*x + 1)*sqrt 
(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^2*x^6 - x^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.37 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=-\frac {1}{2} \, \left (-1\right )^{-2 \, a^{2} c x^{2} + 2 \, c} a^{4} \sqrt {c} \log \left (-2 \, a^{2} c + \frac {2 \, c}{x^{2}}\right ) + \frac {a^{4} c}{2 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {1}{2} \, {\left (a \sqrt {c} \log \left (a x + 1\right ) - a \sqrt {c} \log \left (a x - 1\right ) - \frac {2 \, \sqrt {c}}{x}\right )} a^{3} - \frac {3}{2} \, {\left (\left (-1\right )^{-2 \, a^{2} c x^{2} + 2 \, c} a^{2} \sqrt {c} \log \left (-2 \, a^{2} c + \frac {2 \, c}{x^{2}}\right ) - \frac {a^{2} c}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c}} + \frac {c}{\sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} x^{2}}\right )} a^{2} + \frac {1}{2} \, {\left (3 \, a^{3} \sqrt {c} \log \left (a x + 1\right ) - 3 \, a^{3} \sqrt {c} \log \left (a x - 1\right ) - \frac {2 \, {\left (3 \, a^{2} \sqrt {c} x^{2} + \sqrt {c}\right )}}{x^{3}}\right )} a - \frac {a^{2} c}{4 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} x^{2}} - \frac {c}{4 \, \sqrt {a^{4} c x^{4} - 2 \, a^{2} c x^{2} + c} x^{4}} \] Input:

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

Output:

-1/2*(-1)^(-2*a^2*c*x^2 + 2*c)*a^4*sqrt(c)*log(-2*a^2*c + 2*c/x^2) + 1/2*a 
^4*c/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c) + 1/2*(a*sqrt(c)*log(a*x + 1) - a*s 
qrt(c)*log(a*x - 1) - 2*sqrt(c)/x)*a^3 - 3/2*((-1)^(-2*a^2*c*x^2 + 2*c)*a^ 
2*sqrt(c)*log(-2*a^2*c + 2*c/x^2) - a^2*c/sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c 
) + c/(sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)*x^2))*a^2 + 1/2*(3*a^3*sqrt(c)*lo 
g(a*x + 1) - 3*a^3*sqrt(c)*log(a*x - 1) - 2*(3*a^2*sqrt(c)*x^2 + sqrt(c))/ 
x^3)*a - 1/4*a^2*c/(sqrt(a^4*c*x^4 - 2*a^2*c*x^2 + c)*x^2) - 1/4*c/(sqrt(a 
^4*c*x^4 - 2*a^2*c*x^2 + c)*x^4)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.23 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=-\frac {1}{4} \, {\left (16 \, a^{4} \log \left ({\left | a x - 1 \right |}\right ) - 16 \, a^{4} \log \left ({\left | x \right |}\right ) + \frac {16 \, a^{3} x^{3} + 8 \, a^{2} x^{2} + 4 \, a x + 1}{x^{4}}\right )} \sqrt {c} \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^(1/2)/x^5,x, algorit 
hm="giac")
 

Output:

-1/4*(16*a^4*log(abs(a*x - 1)) - 16*a^4*log(abs(x)) + (16*a^3*x^3 + 8*a^2* 
x^2 + 4*a*x + 1)/x^4)*sqrt(c)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.24 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=\frac {\sqrt {c}\, \left (-16 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+16 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-16 a^{3} x^{3}-8 a^{2} x^{2}-4 a x -1\right )}{4 x^{4}} \] Input:

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

Output:

(sqrt(c)*( - 16*log(a*x - 1)*a**4*x**4 + 16*log(x)*a**4*x**4 - 16*a**3*x** 
3 - 8*a**2*x**2 - 4*a*x - 1))/(4*x**4)