\(\int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^3} \, dx\) [842]

Optimal result

Integrand size = 27, antiderivative size = 186 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^3} \, dx=-\frac {4 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{3 x^2 \sqrt {1-a^2 x^2}}+\frac {3 a \sqrt {c-\frac {c}{a^2 x^2}}}{2 x \sqrt {1-a^2 x^2}}-\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}} x \log (x)}{\sqrt {1-a^2 x^2}}+\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}} x \log (1+a x)}{\sqrt {1-a^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.39 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^3} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left (-\frac {1}{3 x^3}+\frac {3 a}{2 x^2}-\frac {4 a^2}{x}-4 a^3 \log (x)+4 a^3 \log (1+a x)\right )}{\sqrt {1-a^2 x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.39, 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.

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

\(\Big \downarrow \) 6710

\(\displaystyle \frac {x \sqrt {c-\frac {c}{a^2 x^2}} \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {1-a^2 x^2}}{x^4}dx}{\sqrt {1-a^2 x^2}}\)

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(Sqrt[c - c/(a^2*x^2)]*x*(-1/3*1/x^3 + (3*a)/(2*x^2) - (4*a^2)/x - 4*a^3*L 
og[x] + 4*a^3*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 = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.40

method	result	size
default	\(\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (24 \ln \left (a x +1\right ) x^{3} a^{3}-24 \ln \left (x \right ) x^{3} a^{3}-24 a^{2} x^{2}+9 a x -2\right )}{6 x^{2} \sqrt {-a^{2} x^{2}+1}}\)	\(75\)

Input:

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

Output:

1/6*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)/x^2*(24*ln(a*x+1)*x^3*a^3-24*ln(x)*x^3*a 
^3-24*a^2*x^2+9*a*x-2)/(-a^2*x^2+1)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 523, normalized size of antiderivative = 2.81 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^3} \, dx=\left [\frac {12 \, {\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {-c} \log \left (\frac {4 \, a^{5} c x^{5} + {\left (2 \, a^{6} + 4 \, a^{5} + 6 \, a^{4} + 4 \, a^{3} + a^{2}\right )} c x^{6} + {\left (4 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} - 4 \, a - 1\right )} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, a c x + {\left (4 \, a^{4} x^{4} + 6 \, a^{3} x^{3} - {\left (4 \, a^{4} + 6 \, a^{3} + 4 \, a^{2} + a\right )} x^{5} + 4 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{4} x^{6} + 2 \, a^{3} x^{5} - 2 \, a x^{3} - x^{2}}\right ) + {\left (24 \, a^{2} x^{2} - {\left (24 \, a^{2} - 9 \, a + 2\right )} x^{3} - 9 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{6 \, {\left (a^{2} x^{4} - x^{2}\right )}}, \frac {24 \, {\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {c} \arctan \left (-\frac {{\left (2 \, a^{2} x^{2} + {\left (2 \, a^{3} + 2 \, a^{2} + a\right )} x^{3} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} + a^{2}\right )} c x^{4} + {\left (a^{2} + 2 \, a + 1\right )} c x^{2} - 2 \, a c x - c}\right ) + {\left (24 \, a^{2} x^{2} - {\left (24 \, a^{2} - 9 \, a + 2\right )} x^{3} - 9 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{6 \, {\left (a^{2} x^{4} - x^{2}\right )}}\right ] \] Input:

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

Output:

[1/6*(12*(a^4*x^4 - a^2*x^2)*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)) + (24*a^2*x^2 - (24*a 
^2 - 9*a + 2)*x^3 - 9*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^ 
2*x^2)))/(a^2*x^4 - x^2), 1/6*(24*(a^4*x^4 - a^2*x^2)*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)) + (24*a^2*x^2 - (24*a^2 - 9*a + 2)*x^3 - 9*a*x + 
 2)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^2*x^4 - x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.32 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^3} \, dx=\frac {1}{6} \, {\left (24 \, a \log \left ({\left | a x + 1 \right |}\right ) \mathrm {sgn}\left (x\right ) - 24 \, a \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right ) - \frac {24 \, a^{2} x^{2} \mathrm {sgn}\left (x\right ) - 9 \, a x \mathrm {sgn}\left (x\right ) + 2 \, \mathrm {sgn}\left (x\right )}{a^{2} x^{3}}\right )} \sqrt {-c} {\left | a \right |} \] Input:

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

Output:

1/6*(24*a*log(abs(a*x + 1))*sgn(x) - 24*a*log(abs(x))*sgn(x) - (24*a^2*x^2 
*sgn(x) - 9*a*x*sgn(x) + 2*sgn(x))/(a^2*x^3))*sqrt(-c)*abs(a)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.26 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^3} \, dx=\frac {\sqrt {c}\, i \left (-24 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+24 \,\mathrm {log}\left (x \right ) a^{3} x^{3}+24 a^{2} x^{2}-9 a x +2\right )}{6 a \,x^{3}} \] Input:

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

Output:

(sqrt(c)*i*( - 24*log(a*x + 1)*a**3*x**3 + 24*log(x)*a**3*x**3 + 24*a**2*x 
**2 - 9*a*x + 2))/(6*a*x**3)