Integrand size = 23, antiderivative size = 137 \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \text {arctanh}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \] Output:
-1/8*a^3*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)-1/8*a^3*( 1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)+1/8*a^3*(1-1/a^2/x^2)^ (5/2)*x^5*arctanh(a*x)/(-a^2*c*x^2+c)^(5/2)
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.45 \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (-\frac {1}{(-1+a x)^2}-\frac {1}{1+a x}+\text {arctanh}(a x)\right )}{8 \left (c-a^2 c x^2\right )^{5/2}} \] Input:
Integrate[(E^ArcCoth[a*x]*x)/(c - a^2*c*x^2)^(5/2),x]
Output:
(a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5*(-(-1 + a*x)^(-2) - (1 + a*x)^(-1) + ArcT anh[a*x]))/(8*(c - a^2*c*x^2)^(5/2))
Time = 0.81 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.57, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6746, 6747, 25, 86, 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.
\(\displaystyle \int \frac {x e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^4}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int -\frac {x}{(1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {x}{(1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \left (\frac {1}{8 \left (a^2 x^2-1\right ) a}-\frac {1}{8 (a x+1)^2 a}-\frac {1}{4 (a x-1)^3 a}\right )dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (-\frac {\text {arctanh}(a x)}{8 a^2}+\frac {1}{8 a^2 (a x+1)}+\frac {1}{8 a^2 (1-a x)^2}\right )}{\left (c-a^2 c x^2\right )^{5/2}}\) |
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*a^2*(1 - a*x)^2) + 1/(8*a^2*(1 + a*x)) - ArcTanh[a*x]/(8*a^2)))/(c - a^2*c*x^2)^(5/2))
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 164, normalized size of antiderivative = 1.20
method | result | size |
default | \(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (\ln \left (a x +1\right ) x^{3} a^{3}-a^{3} \ln \left (a x -1\right ) x^{3}-\ln \left (a x +1\right ) x^{2} a^{2}+a^{2} \ln \left (a x -1\right ) x^{2}-2 a^{2} x^{2}-\ln \left (a x +1\right ) x a +a \ln \left (a x -1\right ) x +2 a x +\ln \left (a x +1\right )-\ln \left (a x -1\right )-4\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a^{2} \left (a x +1\right )}\) | \(164\) |
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)*(ln(a*x+1)*x^ 3*a^3-a^3*ln(a*x-1)*x^3-ln(a*x+1)*x^2*a^2+a^2*ln(a*x-1)*x^2-2*a^2*x^2-ln(a *x+1)*x*a+a*ln(a*x-1)*x+2*a*x+ln(a*x+1)-ln(a*x-1)-4)/(a^2*x^2-1)/c^3/a^2/( a*x+1)
Time = 0.12 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (a^{4} x^{3} - a^{3} x^{2} - a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, {\left (a^{2} x^{2} - a x + 2\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{6} c^{3} x^{3} - a^{5} c^{3} x^{2} - a^{4} c^{3} x + a^{3} c^{3}\right )}} \] 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*((a^4*x^3 - a^3*x^2 - a^2*x + a)*sqrt(-c)*log((a^2*c*x^2 - 2*sqrt(-a ^2*c)*sqrt(-c)*x + c)/(a^2*x^2 - 1)) - 2*(a^2*x^2 - a*x + 2)*sqrt(-a^2*c)) /(a^6*c^3*x^3 - a^5*c^3*x^2 - a^4*c^3*x + a^3*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
\[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] 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(x/((-a^2*c*x^2 + c)^(5/2)*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. \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \] Input:
int(x/((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
int(x/((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.77 \[ \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, i \left (\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}-\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}-\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x +\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )+\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}-\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}-\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x +\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )-2 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}+2 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}+2 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x -2 \,\mathrm {log}\left (\sqrt {-a x +1}\right )+a^{3} x^{3}-3 a^{2} x^{2}+a x -3\right )}{16 a^{2} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] 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*(log(sqrt( - a*x + 1) - sqrt(2))*a**3*x**3 - log(sqrt( - a*x + 1) - sqrt(2))*a**2*x**2 - log(sqrt( - a*x + 1) - sqrt(2))*a*x + log(sqrt( - a*x + 1) - sqrt(2)) + log(sqrt( - a*x + 1) + sqrt(2))*a**3*x**3 - log(sq rt( - a*x + 1) + sqrt(2))*a**2*x**2 - log(sqrt( - a*x + 1) + sqrt(2))*a*x + log(sqrt( - a*x + 1) + sqrt(2)) - 2*log(sqrt( - a*x + 1))*a**3*x**3 + 2* log(sqrt( - a*x + 1))*a**2*x**2 + 2*log(sqrt( - a*x + 1))*a*x - 2*log(sqrt ( - a*x + 1)) + a**3*x**3 - 3*a**2*x**2 + a*x - 3))/(16*a**2*c**3*(a**3*x* *3 - a**2*x**2 - a*x + 1))