Integrand size = 68, antiderivative size = 22 \[ \int \frac {e^{\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}} (8-2 x-4 \log (5 x))}{\left (48-24 x+3 x^2\right ) \log ^3(5 x)} \, dx=5 e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}} \]
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}} (8-2 x-4 \log (5 x))}{\left (48-24 x+3 x^2\right ) \log ^3(5 x)} \, dx=5 e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}} \]
Integrate[(E^((x + (24 - 6*x + (-12 + 3*x)*Log[5])*Log[5*x]^2)/((-12 + 3*x )*Log[5*x]^2))*(8 - 2*x - 4*Log[5*x]))/((48 - 24*x + 3*x^2)*Log[5*x]^3),x]
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.
| \(\displaystyle \int \frac {(-2 x-4 \log (5 x)+8) \exp \left (\frac {x+(-6 x+(3 x-12) \log (5)+24) \log ^2(5 x)}{(3 x-12) \log ^2(5 x)}\right )}{\left (3 x^2-24 x+48\right ) \log ^3(5 x)} \, dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 12 \int \frac {\exp \left (-\frac {3 (-\log (5) (4-x)-2 x+8) \log ^2(5 x)+x}{3 (4-x) \log ^2(5 x)}\right ) (-x-2 \log (5 x)+4)}{18 (4-x)^2 \log ^3(5 x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{3} \int \frac {\exp \left (-\frac {3 (-\log (5) (4-x)-2 x+8) \log ^2(5 x)+x}{3 (4-x) \log ^2(5 x)}\right ) (-x-2 \log (5 x)+4)}{(4-x)^2 \log ^3(5 x)}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \frac {2}{3} \int \frac {5 e^{\frac {x}{3 (x-4) \log ^2(5 x)}-2} (-x-2 \log (5 x)+4)}{(4-x)^2 \log ^3(5 x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {10}{3} \int \frac {e^{-\frac {x}{3 (4-x) \log ^2(5 x)}-2} (-x-2 \log (5 x)+4)}{(4-x)^2 \log ^3(5 x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {10}{3} \int \left (-\frac {e^{-\frac {x}{3 (4-x) \log ^2(5 x)}-2}}{\log ^3(5 x) (x-4)}-\frac {2 e^{-\frac {x}{3 (4-x) \log ^2(5 x)}-2}}{\log ^2(5 x) (x-4)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {10}{3} \left (-2 \int \frac {e^{-\frac {x}{3 (4-x) \log ^2(5 x)}-2}}{(x-4)^2 \log ^2(5 x)}dx-\int \frac {e^{-\frac {x}{3 (4-x) \log ^2(5 x)}-2}}{(x-4) \log ^3(5 x)}dx\right )\) |
Int[(E^((x + (24 - 6*x + (-12 + 3*x)*Log[5])*Log[5*x]^2)/((-12 + 3*x)*Log[ 5*x]^2))*(8 - 2*x - 4*Log[5*x]))/((48 - 24*x + 3*x^2)*Log[5*x]^3),x]
3.7.29.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 1.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73
| method | result | size |
| parallelrisch | \({\mathrm e}^{\frac {\left (\left (3 x -12\right ) \ln \left (5\right )-6 x +24\right ) \ln \left (5 x \right )^{2}+x}{\left (3 x -12\right ) \ln \left (5 x \right )^{2}}}\) | \(38\) |
| risch | \(5^{\frac {x}{x -4}} \left (\frac {1}{625}\right )^{\frac {1}{x -4}} {\mathrm e}^{-\frac {6 x \ln \left (5 x \right )^{2}-24 \ln \left (5 x \right )^{2}-x}{3 \left (x -4\right ) \ln \left (5 x \right )^{2}}}\) | \(53\) |
int((-4*ln(5*x)-2*x+8)*exp((((3*x-12)*ln(5)-6*x+24)*ln(5*x)^2+x)/(3*x-12)/ ln(5*x)^2)/(3*x^2-24*x+48)/ln(5*x)^3,x,method=_RETURNVERBOSE)
Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}} (8-2 x-4 \log (5 x))}{\left (48-24 x+3 x^2\right ) \log ^3(5 x)} \, dx=e^{\left (\frac {3 \, {\left ({\left (x - 4\right )} \log \left (5\right ) - 2 \, x + 8\right )} \log \left (5 \, x\right )^{2} + x}{3 \, {\left (x - 4\right )} \log \left (5 \, x\right )^{2}}\right )} \]
integrate((-4*log(5*x)-2*x+8)*exp((((3*x-12)*log(5)-6*x+24)*log(5*x)^2+x)/ (3*x-12)/log(5*x)^2)/(3*x^2-24*x+48)/log(5*x)^3,x, algorithm=\
Time = 0.40 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {e^{\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}} (8-2 x-4 \log (5 x))}{\left (48-24 x+3 x^2\right ) \log ^3(5 x)} \, dx=e^{\frac {x + \left (- 6 x + \left (3 x - 12\right ) \log {\left (5 \right )} + 24\right ) \log {\left (5 x \right )}^{2}}{\left (3 x - 12\right ) \log {\left (5 x \right )}^{2}}} \]
integrate((-4*ln(5*x)-2*x+8)*exp((((3*x-12)*ln(5)-6*x+24)*ln(5*x)**2+x)/(3 *x-12)/ln(5*x)**2)/(3*x**2-24*x+48)/ln(5*x)**3,x)
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (19) = 38\).
Time = 0.46 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.82 \[ \int \frac {e^{\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}} (8-2 x-4 \log (5 x))}{\left (48-24 x+3 x^2\right ) \log ^3(5 x)} \, dx=5 \, e^{\left (\frac {4}{3 \, {\left (x \log \left (5\right )^{2} + {\left (x - 4\right )} \log \left (x\right )^{2} - 4 \, \log \left (5\right )^{2} + 2 \, {\left (x \log \left (5\right ) - 4 \, \log \left (5\right )\right )} \log \left (x\right )\right )}} + \frac {1}{3 \, {\left (\log \left (5\right )^{2} + 2 \, \log \left (5\right ) \log \left (x\right ) + \log \left (x\right )^{2}\right )}} - 2\right )} \]
integrate((-4*log(5*x)-2*x+8)*exp((((3*x-12)*log(5)-6*x+24)*log(5*x)^2+x)/ (3*x-12)/log(5*x)^2)/(3*x^2-24*x+48)/log(5*x)^3,x, algorithm=\
5*e^(4/3/(x*log(5)^2 + (x - 4)*log(x)^2 - 4*log(5)^2 + 2*(x*log(5) - 4*log (5))*log(x)) + 1/3/(log(5)^2 + 2*log(5)*log(x) + log(x)^2) - 2)
Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (19) = 38\).
Time = 1.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 6.23 \[ \int \frac {e^{\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}} (8-2 x-4 \log (5 x))}{\left (48-24 x+3 x^2\right ) \log ^3(5 x)} \, dx=e^{\left (\frac {x \log \left (5\right ) \log \left (5 \, x\right )^{2}}{x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}} - \frac {2 \, x \log \left (5 \, x\right )^{2}}{x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}} - \frac {4 \, \log \left (5\right ) \log \left (5 \, x\right )^{2}}{x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}} + \frac {8 \, \log \left (5 \, x\right )^{2}}{x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}} + \frac {x}{3 \, {\left (x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}\right )}}\right )} \]
integrate((-4*log(5*x)-2*x+8)*exp((((3*x-12)*log(5)-6*x+24)*log(5*x)^2+x)/ (3*x-12)/log(5*x)^2)/(3*x^2-24*x+48)/log(5*x)^3,x, algorithm=\
e^(x*log(5)*log(5*x)^2/(x*log(5*x)^2 - 4*log(5*x)^2) - 2*x*log(5*x)^2/(x*l og(5*x)^2 - 4*log(5*x)^2) - 4*log(5)*log(5*x)^2/(x*log(5*x)^2 - 4*log(5*x) ^2) + 8*log(5*x)^2/(x*log(5*x)^2 - 4*log(5*x)^2) + 1/3*x/(x*log(5*x)^2 - 4 *log(5*x)^2))
Time = 12.01 (sec) , antiderivative size = 401, normalized size of antiderivative = 18.23 \[ \int \frac {e^{\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}} (8-2 x-4 \log (5 x))}{\left (48-24 x+3 x^2\right ) \log ^3(5 x)} \, dx=\frac {5^{\frac {{\ln \left (x\right )}^2}{{\ln \left (x\right )}^2+2\,\ln \left (5\right )\,\ln \left (x\right )+{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{\frac {3\,x\,{\ln \left (5\right )}^3}{3\,x\,{\ln \left (x\right )}^2-12\,{\ln \left (x\right )}^2+3\,x\,{\ln \left (5\right )}^2-24\,\ln \left (5\right )\,\ln \left (x\right )-12\,{\ln \left (5\right )}^2+6\,x\,\ln \left (5\right )\,\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {6\,x\,{\ln \left (5\right )}^2}{3\,x\,{\ln \left (x\right )}^2-12\,{\ln \left (x\right )}^2+3\,x\,{\ln \left (5\right )}^2-24\,\ln \left (5\right )\,\ln \left (x\right )-12\,{\ln \left (5\right )}^2+6\,x\,\ln \left (5\right )\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {24\,{\ln \left (x\right )}^2}{3\,x\,{\ln \left (x\right )}^2-12\,{\ln \left (x\right )}^2+3\,x\,{\ln \left (5\right )}^2-24\,\ln \left (5\right )\,\ln \left (x\right )-12\,{\ln \left (5\right )}^2+6\,x\,\ln \left (5\right )\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {x}{3\,x\,{\ln \left (x\right )}^2-12\,{\ln \left (x\right )}^2+3\,x\,{\ln \left (5\right )}^2-24\,\ln \left (5\right )\,\ln \left (x\right )-12\,{\ln \left (5\right )}^2+6\,x\,\ln \left (5\right )\,\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {6\,x\,{\ln \left (x\right )}^2}{3\,x\,{\ln \left (x\right )}^2-12\,{\ln \left (x\right )}^2+3\,x\,{\ln \left (5\right )}^2-24\,\ln \left (5\right )\,\ln \left (x\right )-12\,{\ln \left (5\right )}^2+6\,x\,\ln \left (5\right )\,\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {12\,{\ln \left (5\right )}^3}{3\,x\,{\ln \left (x\right )}^2-12\,{\ln \left (x\right )}^2+3\,x\,{\ln \left (5\right )}^2-24\,\ln \left (5\right )\,\ln \left (x\right )-12\,{\ln \left (5\right )}^2+6\,x\,\ln \left (5\right )\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {24\,{\ln \left (5\right )}^2}{3\,x\,{\ln \left (x\right )}^2-12\,{\ln \left (x\right )}^2+3\,x\,{\ln \left (5\right )}^2-24\,\ln \left (5\right )\,\ln \left (x\right )-12\,{\ln \left (5\right )}^2+6\,x\,\ln \left (5\right )\,\ln \left (x\right )}}}{x^{\frac {2\,\left (2\,\ln \left (5\right )-{\ln \left (5\right )}^2\right )}{{\ln \left (x\right )}^2+2\,\ln \left (5\right )\,\ln \left (x\right )+{\ln \left (5\right )}^2}}} \]
int(-(exp((x + log(5*x)^2*(log(5)*(3*x - 12) - 6*x + 24))/(log(5*x)^2*(3*x - 12)))*(2*x + 4*log(5*x) - 8))/(log(5*x)^3*(3*x^2 - 24*x + 48)),x)
(5^(log(x)^2/(log(x)^2 + 2*log(5)*log(x) + log(5)^2))*exp((3*x*log(5)^3)/( 3*x*log(x)^2 - 12*log(x)^2 + 3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp(-(6*x*log(5)^2)/(3*x*log(x)^2 - 12*log(x)^2 + 3 *x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp((24 *log(x)^2)/(3*x*log(x)^2 - 12*log(x)^2 + 3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp(x/(3*x*log(x)^2 - 12*log(x)^2 + 3*x *log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp(-(6*x *log(x)^2)/(3*x*log(x)^2 - 12*log(x)^2 + 3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp(-(12*log(5)^3)/(3*x*log(x)^2 - 12*l og(x)^2 + 3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x )))*exp((24*log(5)^2)/(3*x*log(x)^2 - 12*log(x)^2 + 3*x*log(5)^2 - 24*log( 5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x))))/x^((2*(2*log(5) - log(5)^2) )/(log(x)^2 + 2*log(5)*log(x) + log(5)^2))