\(\int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx\) [804]

Optimal result

Integrand size = 27, antiderivative size = 263 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=-\frac {4 a^4 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{5 x^4 \sqrt {1-a^2 x^2}}-\frac {3 a \sqrt {c-\frac {c}{a^2 x^2}}}{4 x^3 \sqrt {1-a^2 x^2}}-\frac {4 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{3 x^2 \sqrt {1-a^2 x^2}}-\frac {2 a^3 \sqrt {c-\frac {c}{a^2 x^2}}}{x \sqrt {1-a^2 x^2}}+\frac {4 a^5 \sqrt {c-\frac {c}{a^2 x^2}} x \log (x)}{\sqrt {1-a^2 x^2}}-\frac {4 a^5 \sqrt {c-\frac {c}{a^2 x^2}} x \log (1-a x)}{\sqrt {1-a^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.35 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left (-\frac {1}{5 x^5}-\frac {3 a}{4 x^4}-\frac {4 a^2}{3 x^3}-\frac {2 a^3}{x^2}-\frac {4 a^4}{x}+4 a^5 \log (x)-4 a^5 \log (1-a x)\right )}{\sqrt {1-a^2 x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 92, 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 = {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.

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

\(\Big \downarrow \) 6710

\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {1-a^2 x^2}}{x^6}dx}{\sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 6700

\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {(a x+1)^2}{x^6 (1-a x)}dx}{\sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \left (-\frac {4 a^6}{a x-1}+\frac {4 a^5}{x}+\frac {4 a^4}{x^2}+\frac {4 a^3}{x^3}+\frac {4 a^2}{x^4}+\frac {3 a}{x^5}+\frac {1}{x^6}\right )dx}{\sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(Sqrt[c - c/(a^2*x^2)]*x*(-1/5*1/x^5 - (3*a)/(4*x^4) - (4*a^2)/(3*x^3) - ( 
2*a^3)/x^2 - (4*a^4)/x + 4*a^5*Log[x] - 4*a^5*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 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 = 91, normalized size of antiderivative = 0.35

method	result	size
default	\(-\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (240 \ln \left (a x -1\right ) x^{5} a^{5}-240 a^{5} \ln \left (x \right ) x^{5}+240 a^{4} x^{4}+120 a^{3} x^{3}+80 a^{2} x^{2}+45 a x +12\right )}{60 x^{4} \sqrt {-a^{2} x^{2}+1}}\)	\(91\)

Input:

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

Output:

-1/60*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)/x^4*(240*ln(a*x-1)*x^5*a^5-240*a^5*ln( 
x)*x^5+240*a^4*x^4+120*a^3*x^3+80*a^2*x^2+45*a*x+12)/(-a^2*x^2+1)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 582, normalized size of antiderivative = 2.21 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\left [\frac {120 \, {\left (a^{6} x^{6} - a^{4} x^{4}\right )} \sqrt {-c} \log \left (-\frac {4 \, a^{5} c x^{5} - {\left (2 \, a^{6} - 4 \, a^{5} + 6 \, a^{4} - 4 \, a^{3} + a^{2}\right )} c x^{6} - {\left (4 \, a^{4} + 4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} c x^{4} + 5 \, a^{2} c x^{2} - 4 \, a c x - {\left (4 \, a^{4} x^{4} - 6 \, a^{3} x^{3} - {\left (4 \, a^{4} - 6 \, a^{3} + 4 \, a^{2} - a\right )} x^{5} + 4 \, a^{2} x^{2} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right ) + {\left (240 \, a^{4} x^{4} + 120 \, a^{3} x^{3} - {\left (240 \, a^{4} + 120 \, a^{3} + 80 \, a^{2} + 45 \, a + 12\right )} x^{5} + 80 \, a^{2} x^{2} + 45 \, a x + 12\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{60 \, {\left (a^{2} x^{6} - x^{4}\right )}}, \frac {240 \, {\left (a^{6} x^{6} - a^{4} x^{4}\right )} \sqrt {c} \arctan \left (\frac {{\left (2 \, a^{2} x^{2} - {\left (2 \, a^{3} - 2 \, a^{2} + a\right )} x^{3} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} - a^{2}\right )} c x^{4} - {\left (a^{2} - 2 \, a + 1\right )} c x^{2} - 2 \, a c x + c}\right ) + {\left (240 \, a^{4} x^{4} + 120 \, a^{3} x^{3} - {\left (240 \, a^{4} + 120 \, a^{3} + 80 \, a^{2} + 45 \, a + 12\right )} x^{5} + 80 \, a^{2} x^{2} + 45 \, a x + 12\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{60 \, {\left (a^{2} x^{6} - x^{4}\right )}}\right ] \] Input:

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

Output:

[1/60*(120*(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^4*x^4 - 6*a^3*x^3 - (4*a^4 - 6*a^3 + 4*a^2 - 
a)*x^5 + 4*a^2*x^2 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a^2*c*x^2 - c) 
/(a^2*x^2)) + c)/(a^4*x^6 - 2*a^3*x^5 + 2*a*x^3 - x^2)) + (240*a^4*x^4 + 1 
20*a^3*x^3 - (240*a^4 + 120*a^3 + 80*a^2 + 45*a + 12)*x^5 + 80*a^2*x^2 + 4 
5*a*x + 12)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^2*x^6 - 
 x^4), 1/60*(240*(a^6*x^6 - a^4*x^4)*sqrt(c)*arctan((2*a^2*x^2 - (2*a^3 - 
2*a^2 + a)*x^3 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2 
*x^2))/(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)) + (240*a^4*x^4 + 120*a^3*x^3 - (240*a^4 + 120*a^3 + 80*a^2 + 45*a 
+ 12)*x^5 + 80*a^2*x^2 + 45*a*x + 12)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - 
 c)/(a^2*x^2)))/(a^2*x^6 - x^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.91 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=-\frac {1}{2} i \, a^{4} \sqrt {c} \log \left (a x + 1\right ) + \frac {1}{2} i \, a^{4} \sqrt {c} \log \left (a x - 1\right ) - \frac {1}{2} \, {\left (-i \, a \sqrt {c} \log \left (a x + 1\right ) - i \, a \sqrt {c} \log \left (a x - 1\right ) + 2 i \, a \sqrt {c} \log \left (x\right ) - \frac {i \, \sqrt {c}}{a x^{2}}\right )} a^{3} - \frac {1}{2} \, {\left (3 i \, a^{2} \sqrt {c} \log \left (a x + 1\right ) - 3 i \, a^{2} \sqrt {c} \log \left (a x - 1\right ) - \frac {2 \, {\left (3 i \, a^{2} \sqrt {c} x^{2} + i \, \sqrt {c}\right )}}{a x^{3}}\right )} a^{2} - \frac {3}{4} \, {\left (-2 i \, a^{3} \sqrt {c} \log \left (a x + 1\right ) - 2 i \, a^{3} \sqrt {c} \log \left (a x - 1\right ) + 4 i \, a^{3} \sqrt {c} \log \left (x\right ) - \frac {2 i \, a^{2} \sqrt {c} x^{2} + i \, \sqrt {c}}{a x^{4}}\right )} a + \frac {i \, {\left (15 \, a^{4} \sqrt {c} x^{4} + 5 \, a^{2} \sqrt {c} x^{2} + 3 \, \sqrt {c}\right )}}{15 \, a x^{5}} \] Input:

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

Output:

-1/2*I*a^4*sqrt(c)*log(a*x + 1) + 1/2*I*a^4*sqrt(c)*log(a*x - 1) - 1/2*(-I 
*a*sqrt(c)*log(a*x + 1) - I*a*sqrt(c)*log(a*x - 1) + 2*I*a*sqrt(c)*log(x) 
- I*sqrt(c)/(a*x^2))*a^3 - 1/2*(3*I*a^2*sqrt(c)*log(a*x + 1) - 3*I*a^2*sqr 
t(c)*log(a*x - 1) - 2*(3*I*a^2*sqrt(c)*x^2 + I*sqrt(c))/(a*x^3))*a^2 - 3/4 
*(-2*I*a^3*sqrt(c)*log(a*x + 1) - 2*I*a^3*sqrt(c)*log(a*x - 1) + 4*I*a^3*s 
qrt(c)*log(x) - (2*I*a^2*sqrt(c)*x^2 + I*sqrt(c))/(a*x^4))*a + 1/15*I*(15* 
a^4*sqrt(c)*x^4 + 5*a^2*sqrt(c)*x^2 + 3*sqrt(c))/(a*x^5)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.31 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=-\frac {1}{60} \, {\left (240 \, a^{3} \log \left ({\left | a x - 1 \right |}\right ) \mathrm {sgn}\left (x\right ) - 240 \, a^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right ) + \frac {240 \, a^{4} x^{4} \mathrm {sgn}\left (x\right ) + 120 \, a^{3} x^{3} \mathrm {sgn}\left (x\right ) + 80 \, a^{2} x^{2} \mathrm {sgn}\left (x\right ) + 45 \, a x \mathrm {sgn}\left (x\right ) + 12 \, \mathrm {sgn}\left (x\right )}{a^{2} x^{5}}\right )} \sqrt {-c} {\left | a \right |} \] Input:

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

Output:

-1/60*(240*a^3*log(abs(a*x - 1))*sgn(x) - 240*a^3*log(abs(x))*sgn(x) + (24 
0*a^4*x^4*sgn(x) + 120*a^3*x^3*sgn(x) + 80*a^2*x^2*sgn(x) + 45*a*x*sgn(x) 
+ 12*sgn(x))/(a^2*x^5))*sqrt(-c)*abs(a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.25 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^5} \, dx=\frac {\sqrt {c}\, i \left (-240 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}+240 \,\mathrm {log}\left (x \right ) a^{5} x^{5}-240 a^{4} x^{4}-120 a^{3} x^{3}-80 a^{2} x^{2}-45 a x -12\right )}{60 a \,x^{5}} \] Input:

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

Output:

(sqrt(c)*i*( - 240*log(a*x - 1)*a**5*x**5 + 240*log(x)*a**5*x**5 - 240*a** 
4*x**4 - 120*a**3*x**3 - 80*a**2*x**2 - 45*a*x - 12))/(60*a*x**5)