\(\int \frac {e^{\text {arctanh}(a x)}}{x^3 (c-a^2 c x^2)^{5/2}} \, dx\) [1007]

Optimal result

Integrand size = 25, antiderivative size = 345 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{2 c^2 x^2 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {3 a^2 \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {39 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {9 a^2 \sqrt {1-a^2 x^2} \log (1+a x)}{16 c^2 \sqrt {c-a^2 c x^2}} \] Output:

-1/2*(-a^2*x^2+1)^(1/2)/c^2/x^2/(-a^2*c*x^2+c)^(1/2)-a*(-a^2*x^2+1)^(1/2)/ 
c^2/x/(-a^2*c*x^2+c)^(1/2)+1/8*a^2*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)^2/(-a^2 
*c*x^2+c)^(1/2)+a^2*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+1 
/8*a^2*(-a^2*x^2+1)^(1/2)/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)+3*a^2*(-a^2*x^2 
+1)^(1/2)*ln(x)/c^2/(-a^2*c*x^2+c)^(1/2)-39/16*a^2*(-a^2*x^2+1)^(1/2)*ln(- 
a*x+1)/c^2/(-a^2*c*x^2+c)^(1/2)-9/16*a^2*(-a^2*x^2+1)^(1/2)*ln(a*x+1)/c^2/ 
(-a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.33 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-\frac {8}{x^2}-\frac {16 a}{x}+\frac {16 a^2}{1-a x}+\frac {2 a^2}{(-1+a x)^2}+\frac {2 a^2}{1+a x}+48 a^2 \log (x)-39 a^2 \log (1-a x)-9 a^2 \log (1+a x)\right )}{16 c^2 \sqrt {c-a^2 c x^2}} \] Input:

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

Output:

(Sqrt[1 - a^2*x^2]*(-8/x^2 - (16*a)/x + (16*a^2)/(1 - a*x) + (2*a^2)/(-1 + 
 a*x)^2 + (2*a^2)/(1 + a*x) + 48*a^2*Log[x] - 39*a^2*Log[1 - a*x] - 9*a^2* 
Log[1 + a*x]))/(16*c^2*Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.35, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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.

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

\(\Big \downarrow \) 6703

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

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {39 a^3}{16 (a x-1)}-\frac {9 a^3}{16 (a x+1)}+\frac {a^3}{(a x-1)^2}-\frac {a^3}{8 (a x+1)^2}-\frac {a^3}{4 (a x-1)^3}+\frac {3 a^2}{x}+\frac {a}{x^2}+\frac {1}{x^3}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {a^2}{1-a x}+\frac {a^2}{8 (a x+1)}+\frac {a^2}{8 (1-a x)^2}+3 a^2 \log (x)-\frac {39}{16} a^2 \log (1-a x)-\frac {9}{16} a^2 \log (a x+1)-\frac {a}{x}-\frac {1}{2 x^2}\right )}{c^2 \sqrt {c-a^2 c x^2}}\)

Input:

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

Output:

(Sqrt[1 - a^2*x^2]*(-1/2*1/x^2 - a/x + a^2/(8*(1 - a*x)^2) + a^2/(1 - a*x) 
 + a^2/(8*(1 + a*x)) + 3*a^2*Log[x] - (39*a^2*Log[1 - a*x])/16 - (9*a^2*Lo 
g[1 + a*x])/16))/(c^2*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 = 0.21 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.67

method	result	size
default	\(-\frac {\left (9 \ln \left (a x +1\right ) x^{5} a^{5}+39 \ln \left (a x -1\right ) x^{5} a^{5}-48 a^{5} \ln \left (x \right ) x^{5}-9 \ln \left (a x +1\right ) x^{4} a^{4}-39 \ln \left (a x -1\right ) x^{4} a^{4}+48 \ln \left (x \right ) x^{4} a^{4}+30 a^{4} x^{4}-9 \ln \left (a x +1\right ) x^{3} a^{3}-39 a^{3} \ln \left (a x -1\right ) x^{3}+48 \ln \left (x \right ) x^{3} a^{3}-6 a^{3} x^{3}+9 \ln \left (a x +1\right ) x^{2} a^{2}+39 a^{2} \ln \left (a x -1\right ) x^{2}-48 a^{2} \ln \left (x \right ) x^{2}-44 a^{2} x^{2}+8 a x +8\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{16 \sqrt {-a^{2} x^{2}+1}\, \left (a x -1\right )^{2} \left (a x +1\right ) c^{3} x^{2}}\)	\(231\)

Input:

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

Output:

-1/16*(9*ln(a*x+1)*x^5*a^5+39*ln(a*x-1)*x^5*a^5-48*a^5*ln(x)*x^5-9*ln(a*x+ 
1)*x^4*a^4-39*ln(a*x-1)*x^4*a^4+48*ln(x)*x^4*a^4+30*a^4*x^4-9*ln(a*x+1)*x^ 
3*a^3-39*a^3*ln(a*x-1)*x^3+48*ln(x)*x^3*a^3-6*a^3*x^3+9*ln(a*x+1)*x^2*a^2+ 
39*a^2*ln(a*x-1)*x^2-48*a^2*ln(x)*x^2-44*a^2*x^2+8*a*x+8)/(-a^2*x^2+1)^(1/ 
2)/(a*x-1)^2/(a*x+1)*(-c*(a^2*x^2-1))^(1/2)/c^3/x^2
 

Fricas [F]

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

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

Output:

integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^7*c^3*x^10 - a^6*c^3*x 
^9 - 3*a^5*c^3*x^8 + 3*a^4*c^3*x^7 + 3*a^3*c^3*x^6 - 3*a^2*c^3*x^5 - a*c^3 
*x^4 + c^3*x^3), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.62 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-39 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}+39 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}+39 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}-39 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}-9 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}+9 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+9 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-9 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+48 \,\mathrm {log}\left (x \right ) a^{5} x^{5}-48 \,\mathrm {log}\left (x \right ) a^{4} x^{4}-48 \,\mathrm {log}\left (x \right ) a^{3} x^{3}+48 \,\mathrm {log}\left (x \right ) a^{2} x^{2}-30 a^{5} x^{5}+36 a^{3} x^{3}+14 a^{2} x^{2}-8 a x -8\right )}{16 c^{3} x^{2} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:

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

Output:

(sqrt(c)*( - 39*log(a*x - 1)*a**5*x**5 + 39*log(a*x - 1)*a**4*x**4 + 39*lo 
g(a*x - 1)*a**3*x**3 - 39*log(a*x - 1)*a**2*x**2 - 9*log(a*x + 1)*a**5*x** 
5 + 9*log(a*x + 1)*a**4*x**4 + 9*log(a*x + 1)*a**3*x**3 - 9*log(a*x + 1)*a 
**2*x**2 + 48*log(x)*a**5*x**5 - 48*log(x)*a**4*x**4 - 48*log(x)*a**3*x**3 
 + 48*log(x)*a**2*x**2 - 30*a**5*x**5 + 36*a**3*x**3 + 14*a**2*x**2 - 8*a* 
x - 8))/(16*c**3*x**2*(a**3*x**3 - a**2*x**2 - a*x + 1))