\(\int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a^2 c x^2)^{3/2}} \, dx\) [1012]

Optimal result

Integrand size = 25, antiderivative size = 297 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{3 c x^3 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{2 c x^2 \sqrt {c-a^2 c x^2}}-\frac {2 a^2 \sqrt {1-a^2 x^2}}{c x \sqrt {c-a^2 c x^2}}+\frac {a^3 \sqrt {1-a^2 x^2}}{2 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {2 a^3 \sqrt {1-a^2 x^2} \log (x)}{c \sqrt {c-a^2 c x^2}}-\frac {9 a^3 \sqrt {1-a^2 x^2} \log (1-a x)}{4 c \sqrt {c-a^2 c x^2}}+\frac {a^3 \sqrt {1-a^2 x^2} \log (1+a x)}{4 c \sqrt {c-a^2 c x^2}} \] Output:

-1/3*(-a^2*x^2+1)^(1/2)/c/x^3/(-a^2*c*x^2+c)^(1/2)-1/2*a*(-a^2*x^2+1)^(1/2 
)/c/x^2/(-a^2*c*x^2+c)^(1/2)-2*a^2*(-a^2*x^2+1)^(1/2)/c/x/(-a^2*c*x^2+c)^( 
1/2)+1/2*a^3*(-a^2*x^2+1)^(1/2)/c/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+2*a^3*(-a^ 
2*x^2+1)^(1/2)*ln(x)/c/(-a^2*c*x^2+c)^(1/2)-9/4*a^3*(-a^2*x^2+1)^(1/2)*ln( 
-a*x+1)/c/(-a^2*c*x^2+c)^(1/2)+1/4*a^3*(-a^2*x^2+1)^(1/2)*ln(a*x+1)/c/(-a^ 
2*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.33 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-\frac {4}{x^3}-\frac {6 a}{x^2}-\frac {24 a^2}{x}+\frac {6 a^3}{1-a x}+24 a^3 \log (x)-27 a^3 \log (1-a x)+3 a^3 \log (1+a x)\right )}{12 c \sqrt {c-a^2 c x^2}} \] Input:

Integrate[E^ArcTanh[a*x]/(x^4*(c - a^2*c*x^2)^(3/2)),x]
 

Output:

(Sqrt[1 - a^2*x^2]*(-4/x^3 - (6*a)/x^2 - (24*a^2)/x + (6*a^3)/(1 - a*x) + 
24*a^3*Log[x] - 27*a^3*Log[1 - a*x] + 3*a^3*Log[1 + a*x]))/(12*c*Sqrt[c - 
a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.36, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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.

\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 6703

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {9 a^4}{4 (a x-1)}+\frac {a^4}{4 (a x+1)}+\frac {a^4}{2 (a x-1)^2}+\frac {2 a^3}{x}+\frac {2 a^2}{x^2}+\frac {a}{x^3}+\frac {1}{x^4}\right )dx}{c \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {a^3}{2 (1-a x)}+2 a^3 \log (x)-\frac {9}{4} a^3 \log (1-a x)+\frac {1}{4} a^3 \log (a x+1)-\frac {2 a^2}{x}-\frac {a}{2 x^2}-\frac {1}{3 x^3}\right )}{c \sqrt {c-a^2 c x^2}}\)

Input:

Int[E^ArcTanh[a*x]/(x^4*(c - a^2*c*x^2)^(3/2)),x]
 

Output:

(Sqrt[1 - a^2*x^2]*(-1/3*1/x^3 - a/(2*x^2) - (2*a^2)/x + a^3/(2*(1 - a*x)) 
 + 2*a^3*Log[x] - (9*a^3*Log[1 - a*x])/4 + (a^3*Log[1 + a*x])/4))/(c*Sqrt[ 
c - a^2*c*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 = 0.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.47

method	result	size
default	\(\frac {\left (3 \ln \left (a x +1\right ) x^{4} a^{4}-27 \ln \left (a x -1\right ) x^{4} a^{4}+24 \ln \left (x \right ) x^{4} a^{4}-3 \ln \left (a x +1\right ) x^{3} a^{3}+27 a^{3} \ln \left (a x -1\right ) x^{3}-24 \ln \left (x \right ) x^{3} a^{3}-30 a^{3} x^{3}+18 a^{2} x^{2}+2 a x +4\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{12 \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right ) c^{2} x^{3}}\)	\(140\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVE 
RBOSE)
 

Output:

1/12*(3*ln(a*x+1)*x^4*a^4-27*ln(a*x-1)*x^4*a^4+24*ln(x)*x^4*a^4-3*ln(a*x+1 
)*x^3*a^3+27*a^3*ln(a*x-1)*x^3-24*ln(x)*x^3*a^3-30*a^3*x^3+18*a^2*x^2+2*a* 
x+4)/(-a^2*x^2+1)^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)/c^2/x^3
 

Fricas [F]

\[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1} x^{4}} \,d x } \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c)^(3/2),x, algorithm 
="fricas")
 

Output:

integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^5*c^2*x^9 - a^4*c^2*x 
^8 - 2*a^3*c^2*x^7 + 2*a^2*c^2*x^6 + a*c^2*x^5 - c^2*x^4), x)
 

Sympy [F]

\[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {a x + 1}{x^{4} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x**4/(-a**2*c*x**2+c)**(3/2),x)
 

Output:

Integral((a*x + 1)/(x**4*sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1 
))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1} x^{4}} \,d x } \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c)^(3/2),x, algorithm 
="maxima")
 

Output:

integrate((a*x + 1)/((-a^2*c*x^2 + c)^(3/2)*sqrt(-a^2*x^2 + 1)*x^4), x)
 

Giac [F]

\[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {-a^{2} x^{2} + 1} x^{4}} \,d x } \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c)^(3/2),x, algorithm 
="giac")
 

Output:

integrate((a*x + 1)/((-a^2*c*x^2 + c)^(3/2)*sqrt(-a^2*x^2 + 1)*x^4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {a\,x+1}{x^4\,{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.39 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (-27 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+27 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+3 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}-3 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+24 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-24 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-30 a^{4} x^{4}+18 a^{2} x^{2}+2 a x +4\right )}{12 c^{2} x^{3} \left (a x -1\right )} \] Input:

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

Output:

(sqrt(c)*( - 27*log(a*x - 1)*a**4*x**4 + 27*log(a*x - 1)*a**3*x**3 + 3*log 
(a*x + 1)*a**4*x**4 - 3*log(a*x + 1)*a**3*x**3 + 24*log(x)*a**4*x**4 - 24* 
log(x)*a**3*x**3 - 30*a**4*x**4 + 18*a**2*x**2 + 2*a*x + 4))/(12*c**2*x**3 
*(a*x - 1))