Integrand size = 25, antiderivative size = 252 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x}{2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^2}{\left (c-a^2 c x^2\right )^{3/2}}+\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {7 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (1+a x)}{4 \left (c-a^2 c x^2\right )^{3/2}} \] Output:
-1/2*a^3*(1-1/a^2/x^2)^(3/2)*x/(-a^2*c*x^2+c)^(3/2)-a^4*(1-1/a^2/x^2)^(3/2 )*x^2/(-a^2*c*x^2+c)^(3/2)+1/2*a^5*(1-1/a^2/x^2)^(3/2)*x^3/(-a*x+1)/(-a^2* c*x^2+c)^(3/2)+2*a^5*(1-1/a^2/x^2)^(3/2)*x^3*ln(x)/(-a^2*c*x^2+c)^(3/2)-7/ 4*a^5*(1-1/a^2/x^2)^(3/2)*x^3*ln(-a*x+1)/(-a^2*c*x^2+c)^(3/2)-1/4*a^5*(1-1 /a^2/x^2)^(3/2)*x^3*ln(a*x+1)/(-a^2*c*x^2+c)^(3/2)
Time = 0.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.37 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \left (-\frac {2}{x^2}-\frac {4 a}{x}+\frac {2 a^2}{1-a x}+8 a^2 \log (x)-7 a^2 \log (1-a x)-a^2 \log (1+a x)\right )}{4 \left (c-a^2 c x^2\right )^{3/2}} \] Input:
Integrate[E^ArcCoth[a*x]/(x^3*(c - a^2*c*x^2)^(3/2)),x]
Output:
(a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3*(-2/x^2 - (4*a)/x + (2*a^2)/(1 - a*x) + 8 *a^2*Log[x] - 7*a^2*Log[1 - a*x] - a^2*Log[1 + a*x]))/(4*(c - a^2*c*x^2)^( 3/2))
Time = 0.89 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.39, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6746, 6747, 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.
\(\displaystyle \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^6}dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \frac {1}{x^3 (1-a x)^2 (a x+1)}dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \int \left (-\frac {7 a^3}{4 (a x-1)}-\frac {a^3}{4 (a x+1)}+\frac {a^3}{2 (a x-1)^2}+\frac {2 a^2}{x}+\frac {a}{x^2}+\frac {1}{x^3}\right )dx}{\left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (\frac {a^2}{2 (1-a x)}+2 a^2 \log (x)-\frac {7}{4} a^2 \log (1-a x)-\frac {1}{4} a^2 \log (a x+1)-\frac {a}{x}-\frac {1}{2 x^2}\right )}{\left (c-a^2 c x^2\right )^{3/2}}\) |
Input:
Int[E^ArcCoth[a*x]/(x^3*(c - a^2*c*x^2)^(3/2)),x]
Output:
(a^3*(1 - 1/(a^2*x^2))^(3/2)*x^3*(-1/2*1/x^2 - a/x + a^2/(2*(1 - a*x)) + 2 *a^2*Log[x] - (7*a^2*Log[1 - a*x])/4 - (a^2*Log[1 + a*x])/4))/(c - a^2*c*x ^2)^(3/2)
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 = 0.14 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.55
method | result | size |
default | \(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (\ln \left (a x +1\right ) x^{3} a^{3}+7 a^{3} \ln \left (a x -1\right ) x^{3}-8 \ln \left (x \right ) x^{3} a^{3}-\ln \left (a x +1\right ) x^{2} a^{2}-7 a^{2} \ln \left (a x -1\right ) x^{2}+8 a^{2} \ln \left (x \right ) x^{2}+6 a^{2} x^{2}-2 a x -2\right )}{4 \sqrt {\frac {a x -1}{a x +1}}\, \left (a^{2} x^{2}-1\right ) c^{2} x^{2}}\) | \(138\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/x^3/(-a^2*c*x^2+c)^(3/2),x,method=_RETURNVER BOSE)
Output:
-1/4/((a*x-1)/(a*x+1))^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(ln(a*x+1)*x^3*a^3+7*a ^3*ln(a*x-1)*x^3-8*ln(x)*x^3*a^3-ln(a*x+1)*x^2*a^2-7*a^2*ln(a*x-1)*x^2+8*a ^2*ln(x)*x^2+6*a^2*x^2-2*a*x-2)/(a^2*x^2-1)/c^2/x^2
Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.45 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {{\left (6 \, a^{2} x^{2} - 2 \, a x + {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 7 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 8 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (x\right ) - 2\right )} \sqrt {-a^{2} c}}{4 \, {\left (a^{2} c^{2} x^{3} - a c^{2} x^{2}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^3/(-a^2*c*x^2+c)^(3/2),x, algorithm= "fricas")
Output:
-1/4*(6*a^2*x^2 - 2*a*x + (a^3*x^3 - a^2*x^2)*log(a*x + 1) + 7*(a^3*x^3 - a^2*x^2)*log(a*x - 1) - 8*(a^3*x^3 - a^2*x^2)*log(x) - 2)*sqrt(-a^2*c)/(a^ 2*c^2*x^3 - a*c^2*x^2)
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)/x**3/(-a**2*c*x**2+c)**(3/2),x)
Output:
Timed out
\[ \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{3} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^3/(-a^2*c*x^2+c)^(3/2),x, algorithm= "maxima")
Output:
integrate(1/((-a^2*c*x^2 + c)^(3/2)*x^3*sqrt((a*x - 1)/(a*x + 1))), x)
Exception generated. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^3/(-a^2*c*x^2+c)^(3/2),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. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {1}{x^3\,{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \] Input:
int(1/(x^3*(c - a^2*c*x^2)^(3/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
int(1/(x^3*(c - a^2*c*x^2)^(3/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, i \left (\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}-\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}-8 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{3} x^{3}+8 \,\mathrm {log}\left (\sqrt {-a x +1}-1\right ) a^{2} x^{2}+\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}-\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}-8 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{3} x^{3}+8 \,\mathrm {log}\left (\sqrt {-a x +1}+1\right ) a^{2} x^{2}+14 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}-14 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}-3 a^{3} x^{3}+9 a^{2} x^{2}-2 a x -2\right )}{4 c^{2} x^{2} \left (a x -1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)/x^3/(-a^2*c*x^2+c)^(3/2),x)
Output:
(sqrt(c)*i*(log(sqrt( - a*x + 1) - sqrt(2))*a**3*x**3 - log(sqrt( - a*x + 1) - sqrt(2))*a**2*x**2 - 8*log(sqrt( - a*x + 1) - 1)*a**3*x**3 + 8*log(sq rt( - a*x + 1) - 1)*a**2*x**2 + log(sqrt( - a*x + 1) + sqrt(2))*a**3*x**3 - log(sqrt( - a*x + 1) + sqrt(2))*a**2*x**2 - 8*log(sqrt( - a*x + 1) + 1)* a**3*x**3 + 8*log(sqrt( - a*x + 1) + 1)*a**2*x**2 + 14*log(sqrt( - a*x + 1 ))*a**3*x**3 - 14*log(sqrt( - a*x + 1))*a**2*x**2 - 3*a**3*x**3 + 9*a**2*x **2 - 2*a*x - 2))/(4*c**2*x**2*(a*x - 1))