Integrand size = 25, antiderivative size = 262 \[ \int \frac {e^{\coth ^{-1}(a x)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^6}{\left (c-a^2 c x^2\right )^{5/2}}-\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 a (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{a (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 a (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {23 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 a \left (c-a^2 c x^2\right )^{5/2}}-\frac {7 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 a \left (c-a^2 c x^2\right )^{5/2}} \] Output:
(1-1/a^2/x^2)^(5/2)*x^6/(-a^2*c*x^2+c)^(5/2)-1/8*(1-1/a^2/x^2)^(5/2)*x^5/a /(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)+(1-1/a^2/x^2)^(5/2)*x^5/a/(-a*x+1)/(-a^2* c*x^2+c)^(5/2)-1/8*(1-1/a^2/x^2)^(5/2)*x^5/a/(a*x+1)/(-a^2*c*x^2+c)^(5/2)+ 23/16*(1-1/a^2/x^2)^(5/2)*x^5*ln(-a*x+1)/a/(-a^2*c*x^2+c)^(5/2)-7/16*(1-1/ a^2/x^2)^(5/2)*x^5*ln(a*x+1)/a/(-a^2*c*x^2+c)^(5/2)
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.38 \[ \int \frac {e^{\coth ^{-1}(a x)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (x-\frac {1}{8 a (-1+a x)^2}+\frac {1}{a-a^2 x}-\frac {1}{8 a+8 a^2 x}+\frac {23 \log (1-a x)}{16 a}-\frac {7 \log (1+a x)}{16 a}\right )}{\left (c-a^2 c x^2\right )^{5/2}} \] Input:
Integrate[(E^ArcCoth[a*x]*x^5)/(c - a^2*c*x^2)^(5/2),x]
Output:
((1 - 1/(a^2*x^2))^(5/2)*x^5*(x - 1/(8*a*(-1 + a*x)^2) + (a - a^2*x)^(-1) - (8*a + 8*a^2*x)^(-1) + (23*Log[1 - a*x])/(16*a) - (7*Log[1 + a*x])/(16*a )))/(c - a^2*c*x^2)^(5/2)
Time = 0.83 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.43, 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.
\(\displaystyle \int \frac {x^5 e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int -\frac {x^5}{(1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \frac {x^5}{(1-a x)^3 (a x+1)^2}dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \int \left (\frac {7}{16 a^5 (a x+1)}-\frac {1}{8 a^5 (a x+1)^2}-\frac {1}{a^5}-\frac {23}{16 a^5 (a x-1)}-\frac {1}{a^5 (a x-1)^2}-\frac {1}{4 a^5 (a x-1)^3}\right )dx}{\left (c-a^2 c x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (-\frac {1}{a^6 (1-a x)}+\frac {1}{8 a^6 (a x+1)}+\frac {1}{8 a^6 (1-a x)^2}-\frac {23 \log (1-a x)}{16 a^6}+\frac {7 \log (a x+1)}{16 a^6}-\frac {x}{a^5}\right )}{\left (c-a^2 c x^2\right )^{5/2}}\) |
Input:
Int[(E^ArcCoth[a*x]*x^5)/(c - a^2*c*x^2)^(5/2),x]
Output:
-((a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(-(x/a^5) + 1/(8*a^6*(1 - a*x)^2) - 1/( a^6*(1 - a*x)) + 1/(8*a^6*(1 + a*x)) - (23*Log[1 - a*x])/(16*a^6) + (7*Log [1 + a*x])/(16*a^6)))/(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 = 185, normalized size of antiderivative = 0.71
method | result | size |
default | \(\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (-16 a^{4} x^{4}+7 \ln \left (a x +1\right ) x^{3} a^{3}-23 a^{3} \ln \left (a x -1\right ) x^{3}+16 a^{3} x^{3}-7 \ln \left (a x +1\right ) x^{2} a^{2}+23 a^{2} \ln \left (a x -1\right ) x^{2}+34 a^{2} x^{2}-7 \ln \left (a x +1\right ) x a +23 a \ln \left (a x -1\right ) x -18 a x +7 \ln \left (a x +1\right )-23 \ln \left (a x -1\right )-12\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a^{6} \left (a x +1\right )}\) | \(185\) |
Input:
int(1/((a*x-1)/(a*x+1))^(1/2)*x^5/(-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)*(-16*a^4*x^4+7 *ln(a*x+1)*x^3*a^3-23*a^3*ln(a*x-1)*x^3+16*a^3*x^3-7*ln(a*x+1)*x^2*a^2+23* a^2*ln(a*x-1)*x^2+34*a^2*x^2-7*ln(a*x+1)*x*a+23*a*ln(a*x-1)*x-18*a*x+7*ln( a*x+1)-23*ln(a*x-1)-12)/(a^2*x^2-1)/c^3/a^6/(a*x+1)
Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.53 \[ \int \frac {e^{\coth ^{-1}(a x)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (16 \, a^{4} x^{4} - 16 \, a^{3} x^{3} - 34 \, a^{2} x^{2} + 18 \, a x - 7 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 23 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) + 12\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{10} c^{3} x^{3} - a^{9} c^{3} x^{2} - a^{8} c^{3} x + a^{7} c^{3}\right )}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^5/(-a^2*c*x^2+c)^(5/2),x, algorithm= "fricas")
Output:
-1/16*(16*a^4*x^4 - 16*a^3*x^3 - 34*a^2*x^2 + 18*a*x - 7*(a^3*x^3 - a^2*x^ 2 - a*x + 1)*log(a*x + 1) + 23*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x - 1) + 12)*sqrt(-a^2*c)/(a^10*c^3*x^3 - a^9*c^3*x^2 - a^8*c^3*x + a^7*c^3)
Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/((a*x-1)/(a*x+1))**(1/2)*x**5/(-a**2*c*x**2+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {e^{\coth ^{-1}(a x)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{5}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^5/(-a^2*c*x^2+c)^(5/2),x, algorithm= "maxima")
Output:
integrate(x^5/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)
Exception generated. \[ \int \frac {e^{\coth ^{-1}(a x)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^5/(-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. \[ \int \frac {e^{\coth ^{-1}(a x)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^5}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \] Input:
int(x^5/((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)
Output:
int(x^5/((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)
Time = 0.16 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.98 \[ \int \frac {e^{\coth ^{-1}(a x)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, i \left (-7 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{3} x^{3}+7 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a^{2} x^{2}+7 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right ) a x -7 \,\mathrm {log}\left (\sqrt {-a x +1}-\sqrt {2}\right )-7 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{3} x^{3}+7 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a^{2} x^{2}+7 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right ) a x -7 \,\mathrm {log}\left (\sqrt {-a x +1}+\sqrt {2}\right )+46 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{3} x^{3}-46 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a^{2} x^{2}-46 \,\mathrm {log}\left (\sqrt {-a x +1}\right ) a x +46 \,\mathrm {log}\left (\sqrt {-a x +1}\right )+16 a^{4} x^{4}-23 a^{3} x^{3}-27 a^{2} x^{2}+25 a x +5\right )}{16 a^{6} c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:
int(1/((a*x-1)/(a*x+1))^(1/2)*x^5/(-a^2*c*x^2+c)^(5/2),x)
Output:
(sqrt(c)*i*( - 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 - 7 *log(sqrt( - a*x + 1) - sqrt(2)) - 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 - 7*log(sqrt( - a*x + 1) + sqrt(2)) + 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 + 46*log(sqrt( - a*x + 1)) + 16*a**4*x**4 - 23*a**3*x**3 - 27*a* *2*x**2 + 25*a*x + 5))/(16*a**6*c**3*(a**3*x**3 - a**2*x**2 - a*x + 1))