Integrand size = 25, antiderivative size = 307 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^4}{\left (c-a^2 c x^2\right )^{5/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{4 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {23 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}-\frac {7 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}} \] Output:
a^5*(1-1/a^2/x^2)^(5/2)*x^4/(-a^2*c*x^2+c)^(5/2)-1/8*a^6*(1-1/a^2/x^2)^(5/ 2)*x^5/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)-3/4*a^6*(1-1/a^2/x^2)^(5/2)*x^5/(-a *x+1)/(-a^2*c*x^2+c)^(5/2)+1/8*a^6*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2*c *x^2+c)^(5/2)-a^6*(1-1/a^2/x^2)^(5/2)*x^5*ln(x)/(-a^2*c*x^2+c)^(5/2)+23/16 *a^6*(1-1/a^2/x^2)^(5/2)*x^5*ln(-a*x+1)/(-a^2*c*x^2+c)^(5/2)-7/16*a^6*(1-1 /a^2/x^2)^(5/2)*x^5*ln(a*x+1)/(-a^2*c*x^2+c)^(5/2)
Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.32 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (\frac {16}{x}-\frac {2 a}{(-1+a x)^2}+\frac {12 a}{-1+a x}+\frac {2 a}{1+a x}-16 a \log (x)+23 a \log (1-a x)-7 a \log (1+a x)\right )}{16 \left (c-a^2 c x^2\right )^{5/2}} \] Input:
Integrate[E^ArcCoth[a*x]/(x^2*(c - a^2*c*x^2)^(5/2)),x]
Output:
(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(16/x - (2*a)/(-1 + a*x)^2 + (12*a)/(-1 + a*x) + (2*a)/(1 + a*x) - 16*a*Log[x] + 23*a*Log[1 - a*x] - 7*a*Log[1 + a* x]))/(16*(c - a^2*c*x^2)^(5/2))
Time = 0.94 (sec) , antiderivative size = 108, 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.
| \(\displaystyle \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6746 |
| \(\displaystyle \frac {x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^7}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 6747 |
| \(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int -\frac {1}{x^2 (1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {1}{x^2 (1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \left (-\frac {23 a^2}{16 (a x-1)}+\frac {7 a^2}{16 (a x+1)}+\frac {3 a^2}{4 (a x-1)^2}+\frac {a^2}{8 (a x+1)^2}-\frac {a^2}{4 (a x-1)^3}+\frac {a}{x}+\frac {1}{x^2}\right )dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (\frac {3 a}{4 (1-a x)}-\frac {a}{8 (a x+1)}+\frac {a}{8 (1-a x)^2}+a \log (x)-\frac {23}{16} a \log (1-a x)+\frac {7}{16} a \log (a x+1)-\frac {1}{x}\right )}{\left (c-a^2 c x^2\right )^{5/2}}\) |
Input:
Int[E^ArcCoth[a*x]/(x^2*(c - a^2*c*x^2)^(5/2)),x]
Output:
-((a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(-x^(-1) + a/(8*(1 - a*x)^2) + (3*a)/(4 *(1 - a*x)) - a/(8*(1 + a*x)) + a*Log[x] - (23*a*Log[1 - a*x])/16 + (7*a*L og[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 = 225, normalized size of antiderivative = 0.73
| method | result | size |
| default | \(\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (7 \ln \left (a x +1\right ) x^{4} a^{4}-23 \ln \left (a x -1\right ) x^{4} a^{4}+16 \ln \left (x \right ) x^{4} a^{4}-7 \ln \left (a x +1\right ) x^{3} a^{3}+23 a^{3} \ln \left (a x -1\right ) x^{3}-16 \ln \left (x \right ) x^{3} a^{3}-30 a^{3} x^{3}-7 \ln \left (a x +1\right ) x^{2} a^{2}+23 a^{2} \ln \left (a x -1\right ) x^{2}-16 a^{2} \ln \left (x \right ) x^{2}+22 a^{2} x^{2}+7 \ln \left (a x +1\right ) x a -23 a \ln \left (a x -1\right ) x +16 a \ln \left (x \right ) x +28 a x -16\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} x \left (a x +1\right )}\) | \(225\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVER BOSE)
Output:
1/16/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(7*ln(a*x+1)*x ^4*a^4-23*ln(a*x-1)*x^4*a^4+16*ln(x)*x^4*a^4-7*ln(a*x+1)*x^3*a^3+23*a^3*ln (a*x-1)*x^3-16*ln(x)*x^3*a^3-30*a^3*x^3-7*ln(a*x+1)*x^2*a^2+23*a^2*ln(a*x- 1)*x^2-16*a^2*ln(x)*x^2+22*a^2*x^2+7*ln(a*x+1)*x*a-23*a*ln(a*x-1)*x+16*a*l n(x)*x+28*a*x-16)/(a^2*x^2-1)/c^3/x/(a*x+1)
Time = 0.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.57 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (30 \, a^{3} x^{3} - 22 \, a^{2} x^{2} - 28 \, a x - 7 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (a x + 1\right ) + 23 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (x\right ) + 16\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{4} c^{3} x^{4} - a^{3} c^{3} x^{3} - a^{2} c^{3} x^{2} + a c^{3} x\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm= "fricas")
Output:
-1/16*(30*a^3*x^3 - 22*a^2*x^2 - 28*a*x - 7*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(a*x + 1) + 23*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(a*x - 1) - 16*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(x) + 16)*sqrt(-a^2*c)/(a^4*c^3 *x^4 - a^3*c^3*x^3 - a^2*c^3*x^2 + a*c^3*x)
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)/x**2/(-a**2*c*x**2+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm= "maxima")
Output:
integrate(1/((-a^2*c*x^2 + c)^(5/2)*x^2*sqrt((a*x - 1)/(a*x + 1))), x)
Exception generated. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm= "giac")
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. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {1}{x^2\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \] Input:
int(1/(x^2*(c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
int(1/(x^2*(c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.34 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, i \left (16-10 a^{4} x^{4}-7 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{4} x^{4}-7 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{4} x^{4}+46 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{4} x^{4}-16 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a x -16 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a x +7 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}+7 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}+7 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}+7 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}-46 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}-46 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}-38 a x -12 a^{2} x^{2}+46 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x -7 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x -7 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x -16 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{4} x^{4}-16 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{4} x^{4}+16 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{3} x^{3}+16 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{3} x^{3}+40 a^{3} x^{3}+16 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{2} x^{2}+16 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{2} x^{2}\right )}{16 c^{3} x \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*i*( - 7*log(sqrt( - a*x + 1) - sqrt(2))*a**4*x**4 + 7*log(sqrt( - a*x + 1) - sqrt(2))*a**3*x**3 + 7*log(sqrt( - a*x + 1) - sqrt(2))*a**2*x* *2 - 7*log(sqrt( - a*x + 1) - sqrt(2))*a*x - 16*log(sqrt( - a*x + 1) - 1)* a**4*x**4 + 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 - 7*log(sqrt( - a*x + 1) + sqrt(2))*a**4*x**4 + 7*log(sqrt( - a*x + 1) + sqrt(2))*a**3*x**3 + 7 *log(sqrt( - a*x + 1) + sqrt(2))*a**2*x**2 - 7*log(sqrt( - a*x + 1) + sqrt (2))*a*x - 16*log(sqrt( - a*x + 1) + 1)*a**4*x**4 + 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 + 46*log(sqrt( - a*x + 1))*a**4*x**4 - 46*log(sqrt( - a *x + 1))*a**3*x**3 - 46*log(sqrt( - a*x + 1))*a**2*x**2 + 46*log(sqrt( - a *x + 1))*a*x - 10*a**4*x**4 + 40*a**3*x**3 - 12*a**2*x**2 - 38*a*x + 16))/ (16*c**3*x*(a**3*x**3 - a**2*x**2 - a*x + 1))