Integrand size = 25, antiderivative size = 271 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {11 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {5 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}} \] Output:
-1/8*a^5*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)-1/2*a^5*( 1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)/(-a^2*c*x^2+c)^(5/2)-1/8*a^5*(1-1/a^2/x^2) ^(5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)-a^5*(1-1/a^2/x^2)^(5/2)*x^5*ln(x)/ (-a^2*c*x^2+c)^(5/2)+11/16*a^5*(1-1/a^2/x^2)^(5/2)*x^5*ln(-a*x+1)/(-a^2*c* x^2+c)^(5/2)+5/16*a^5*(1-1/a^2/x^2)^(5/2)*x^5*ln(a*x+1)/(-a^2*c*x^2+c)^(5/ 2)
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.32 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (-\frac {2}{(-1+a x)^2}+\frac {8}{-1+a x}-\frac {2}{1+a x}-16 \log (x)+11 \log (1-a x)+5 \log (1+a x)\right )}{16 \left (c-a^2 c x^2\right )^{5/2}} \] Input:
Integrate[E^ArcCoth[a*x]/(x*(c - a^2*c*x^2)^(5/2)),x]
Output:
(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(-2/(-1 + a*x)^2 + 8/(-1 + a*x) - 2/(1 + a*x) - 16*Log[x] + 11*Log[1 - a*x] + 5*Log[1 + a*x]))/(16*(c - a^2*c*x^2)^ (5/2))
Time = 0.89 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.35, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6746, 6747, 25, 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.
| \(\displaystyle \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6746 |
| \(\displaystyle \frac {x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^6}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 6747 |
| \(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int -\frac {1}{x (1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {1}{x (1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \left (-\frac {11 a}{16 (a x-1)}-\frac {5 a}{16 (a x+1)}+\frac {a}{2 (a x-1)^2}-\frac {a}{8 (a x+1)^2}-\frac {a}{4 (a x-1)^3}+\frac {1}{x}\right )dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (\frac {1}{2 (1-a x)}+\frac {1}{8 (a x+1)}+\frac {1}{8 (1-a x)^2}-\frac {11}{16} \log (1-a x)-\frac {5}{16} \log (a x+1)+\log (x)\right )}{\left (c-a^2 c x^2\right )^{5/2}}\) |
Input:
Int[E^ArcCoth[a*x]/(x*(c - a^2*c*x^2)^(5/2)),x]
Output:
-((a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(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 - a^2*c*x^2)^(5/2))
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^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(c + d*x^2)^p/(x^(2*p)*(1 - 1/(a^2*x^2))^p) Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && !IntegerQ[n/2] && !IntegerQ[p]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[c^p/a^(2*p) Int[(u/x^(2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] && !Inte gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
Time = 0.15 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (5 \ln \left (a x +1\right ) x^{3} a^{3}+11 a^{3} \ln \left (a x -1\right ) x^{3}-16 \ln \left (x \right ) x^{3} a^{3}-5 \ln \left (a x +1\right ) x^{2} a^{2}-11 a^{2} \ln \left (a x -1\right ) x^{2}+16 a^{2} \ln \left (x \right ) x^{2}+6 a^{2} x^{2}-5 \ln \left (a x +1\right ) x a -11 a \ln \left (a x -1\right ) x +16 a \ln \left (x \right ) x +2 a x +5 \ln \left (a x +1\right )+11 \ln \left (a x -1\right )-16 \ln \left (x \right )-12\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} \left (a x +1\right )}\) | \(196\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBO SE)
Output:
-1/16/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(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*l n(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+1 6*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)/c^3/(a *x+1)
Time = 0.11 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.54 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (6 \, a^{2} x^{2} + 2 \, a x + 5 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 11 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (x\right ) - 12\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{4} c^{3} x^{3} - a^{3} c^{3} x^{2} - a^{2} c^{3} x + a c^{3}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x, algorithm="f ricas")
Output:
-1/16*(6*a^2*x^2 + 2*a*x + 5*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) + 11*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x - 1) - 16*(a^3*x^3 - a^2*x^2 - a* x + 1)*log(x) - 12)*sqrt(-a^2*c)/(a^4*c^3*x^3 - a^3*c^3*x^2 - a^2*c^3*x + a*c^3)
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)/x/(-a**2*c*x**2+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x, algorithm="m axima")
Output:
integrate(1/((-a^2*c*x^2 + c)^(5/2)*x*sqrt((a*x - 1)/(a*x + 1))), x)
Exception generated. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x, algorithm="g iac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \] Input:
int(1/(x*(c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
int(1/(x*(c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)
Time = 0.16 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.37 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, i \left (5 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}-5 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}-5 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +5 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )-16 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{3} x^{3}+16 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{2} x^{2}+16 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a x -16 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right )+5 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}-5 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}-5 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x +5 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )-16 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{3} x^{3}+16 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{2} x^{2}+16 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a x -16 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right )+22 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}-22 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}-22 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x +22 \,\mathrm {log}\left (\sqrt {-a x +1}\right )-3 a^{3} x^{3}+9 a^{2} x^{2}+5 a x -15\right )}{16 c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/x/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*i*(5*log(sqrt( - a*x + 1) - sqrt(2))*a**3*x**3 - 5*log(sqrt( - a* x + 1) - sqrt(2))*a**2*x**2 - 5*log(sqrt( - a*x + 1) - sqrt(2))*a*x + 5*lo g(sqrt( - a*x + 1) - sqrt(2)) - 16*log(sqrt( - a*x + 1) - 1)*a**3*x**3 + 1 6*log(sqrt( - a*x + 1) - 1)*a**2*x**2 + 16*log(sqrt( - a*x + 1) - 1)*a*x - 16*log(sqrt( - a*x + 1) - 1) + 5*log(sqrt( - a*x + 1) + sqrt(2))*a**3*x** 3 - 5*log(sqrt( - a*x + 1) + sqrt(2))*a**2*x**2 - 5*log(sqrt( - a*x + 1) + sqrt(2))*a*x + 5*log(sqrt( - a*x + 1) + sqrt(2)) - 16*log(sqrt( - a*x + 1 ) + 1)*a**3*x**3 + 16*log(sqrt( - a*x + 1) + 1)*a**2*x**2 + 16*log(sqrt( - a*x + 1) + 1)*a*x - 16*log(sqrt( - a*x + 1) + 1) + 22*log(sqrt( - a*x + 1 ))*a**3*x**3 - 22*log(sqrt( - a*x + 1))*a**2*x**2 - 22*log(sqrt( - a*x + 1 ))*a*x + 22*log(sqrt( - a*x + 1)) - 3*a**3*x**3 + 9*a**2*x**2 + 5*a*x - 15 ))/(16*c**3*(a**3*x**3 - a**2*x**2 - a*x + 1))