Integrand size = 25, antiderivative size = 252 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{2 c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {11 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {5 \sqrt {1-a^2 x^2} \log (1+a x)}{16 c^2 \sqrt {c-a^2 c x^2}} \] Output:
1/8*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)+1/2*(-a^2*x^2+1 )^(1/2)/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+1/8*(-a^2*x^2+1)^(1/2)/c^2/(a*x+ 1)/(-a^2*c*x^2+c)^(1/2)+(-a^2*x^2+1)^(1/2)*ln(x)/c^2/(-a^2*c*x^2+c)^(1/2)- 11/16*(-a^2*x^2+1)^(1/2)*ln(-a*x+1)/c^2/(-a^2*c*x^2+c)^(1/2)-5/16*(-a^2*x^ 2+1)^(1/2)*ln(a*x+1)/c^2/(-a^2*c*x^2+c)^(1/2)
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.34 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (\frac {8}{1-a x}+\frac {2}{(-1+a x)^2}+\frac {2}{1+a x}+16 \log (x)-11 \log (1-a x)-5 \log (1+a x)\right )}{16 c^2 \sqrt {c-a^2 c x^2}} \] Input:
Integrate[E^ArcTanh[a*x]/(x*(c - a^2*c*x^2)^(5/2)),x]
Output:
(Sqrt[1 - a^2*x^2]*(8/(1 - a*x) + 2/(-1 + a*x)^2 + 2/(1 + a*x) + 16*Log[x] - 11*Log[1 - a*x] - 5*Log[1 + a*x]))/(16*c^2*Sqrt[c - a^2*c*x^2])
Time = 0.84 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.37, 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.
| \(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6703 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)}}{x \left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 6700 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {1}{x (1-a x)^3 (a x+1)^2}dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (-\frac {11 a}{16 (a x-1)}-\frac {5 a}{16 (a x+1)}+\frac {a}{2 (a x-1)^2}-\frac {a}{8 (a x+1)^2}-\frac {a}{4 (a x-1)^3}+\frac {1}{x}\right )dx}{c^2 \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {1}{2 (1-a x)}+\frac {1}{8 (a x+1)}+\frac {1}{8 (1-a x)^2}-\frac {11}{16} \log (1-a x)-\frac {5}{16} \log (a x+1)+\log (x)\right )}{c^2 \sqrt {c-a^2 c x^2}}\) |
Input:
Int[E^ArcTanh[a*x]/(x*(c - a^2*c*x^2)^(5/2)),x]
Output:
(Sqrt[1 - a^2*x^2]*(1/(8*(1 - a*x)^2) + 1/(2*(1 - a*x)) + 1/(8*(1 + a*x)) + Log[x] - (11*Log[1 - a*x])/16 - (5*Log[1 + a*x])/16))/(c^2*Sqrt[c - a^2* c*x^2])
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]))
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])
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]
Time = 0.12 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(-\frac {\left (5 \ln \left (a x +1\right ) x^{3} a^{3}+11 a^{3} \ln \left (a x -1\right ) x^{3}-16 \ln \left (x \right ) x^{3} a^{3}-5 \ln \left (a x +1\right ) x^{2} a^{2}-11 a^{2} \ln \left (a x -1\right ) x^{2}+16 a^{2} \ln \left (x \right ) x^{2}+6 a^{2} x^{2}-5 \ln \left (a x +1\right ) x a -11 a \ln \left (a x -1\right ) x +16 a \ln \left (x \right ) x +2 a x +5 \ln \left (a x +1\right )+11 \ln \left (a x -1\right )-16 \ln \left (x \right )-12\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{16 \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (a x -1\right )^{2} c^{3}}\) | \(182\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERB OSE)
Output:
-1/16*(5*ln(a*x+1)*x^3*a^3+11*a^3*ln(a*x-1)*x^3-16*ln(x)*x^3*a^3-5*ln(a*x+ 1)*x^2*a^2-11*a^2*ln(a*x-1)*x^2+16*a^2*ln(x)*x^2+6*a^2*x^2-5*ln(a*x+1)*x*a -11*a*ln(a*x-1)*x+16*a*ln(x)*x+2*a*x+5*ln(a*x+1)+11*ln(a*x-1)-16*ln(x)-12) /(-a^2*x^2+1)^(1/2)/(a*x+1)/(a*x-1)^2*(-c*(a^2*x^2-1))^(1/2)/c^3
\[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1} x} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-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^8 - a^6*c^3*x^ 7 - 3*a^5*c^3*x^6 + 3*a^4*c^3*x^5 + 3*a^3*c^3*x^4 - 3*a^2*c^3*x^3 - a*c^3* x^2 + c^3*x), x)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {a x + 1}{x \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x/(-a**2*c*x**2+c)**(5/2),x)
Output:
Integral((a*x + 1)/(x*sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))* *(5/2)), x)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1} x} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-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), x)
\[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1} x} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-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), x)
Timed out. \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {a\,x+1}{x\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((a*x + 1)/(x*(c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)),x)
Output:
int((a*x + 1)/(x*(c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)), x)
Time = 0.16 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.66 \[ \int \frac {e^{\text {arctanh}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-11 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+11 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+11 \,\mathrm {log}\left (a x -1\right ) a x -11 \,\mathrm {log}\left (a x -1\right )-5 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}+5 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}+5 \,\mathrm {log}\left (a x +1\right ) a x -5 \,\mathrm {log}\left (a x +1\right )+16 \,\mathrm {log}\left (x \right ) a^{3} x^{3}-16 \,\mathrm {log}\left (x \right ) a^{2} x^{2}-16 \,\mathrm {log}\left (x \right ) a x +16 \,\mathrm {log}\left (x \right )-6 a^{3} x^{3}+4 a x +6\right )}{16 c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*( - 11*log(a*x - 1)*a**3*x**3 + 11*log(a*x - 1)*a**2*x**2 + 11*lo g(a*x - 1)*a*x - 11*log(a*x - 1) - 5*log(a*x + 1)*a**3*x**3 + 5*log(a*x + 1)*a**2*x**2 + 5*log(a*x + 1)*a*x - 5*log(a*x + 1) + 16*log(x)*a**3*x**3 - 16*log(x)*a**2*x**2 - 16*log(x)*a*x + 16*log(x) - 6*a**3*x**3 + 4*a*x + 6 ))/(16*c**3*(a**3*x**3 - a**2*x**2 - a*x + 1))