Integrand size = 171, antiderivative size = 30 \[ \int \frac {e^{\frac {2 \left (4-x^3 \log \left (-e^x-x+\log (-15+5 x)\right )\right )}{\log \left (-e^x-x+\log (-15+5 x)\right )}} \left (-32+8 x+e^x (-24+8 x)+\left (-18 x^3+6 x^4+e^x \left (-18 x^2+6 x^3\right )+\left (18 x^2-6 x^3\right ) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )\right )}{\left (e^x (3-x)+3 x-x^2+(-3+x) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )} \, dx=2+e^{-2 x^3+\frac {8}{\log \left (-e^x-x+\log (5 (-3+x))\right )}} \]
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {2 \left (4-x^3 \log \left (-e^x-x+\log (-15+5 x)\right )\right )}{\log \left (-e^x-x+\log (-15+5 x)\right )}} \left (-32+8 x+e^x (-24+8 x)+\left (-18 x^3+6 x^4+e^x \left (-18 x^2+6 x^3\right )+\left (18 x^2-6 x^3\right ) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )\right )}{\left (e^x (3-x)+3 x-x^2+(-3+x) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )} \, dx=e^{-2 x^3+\frac {8}{\log \left (-e^x-x+\log (5 (-3+x))\right )}} \]
Integrate[(E^((2*(4 - x^3*Log[-E^x - x + Log[-15 + 5*x]]))/Log[-E^x - x + Log[-15 + 5*x]])*(-32 + 8*x + E^x*(-24 + 8*x) + (-18*x^3 + 6*x^4 + E^x*(-1 8*x^2 + 6*x^3) + (18*x^2 - 6*x^3)*Log[-15 + 5*x])*Log[-E^x - x + Log[-15 + 5*x]]^2))/((E^x*(3 - x) + 3*x - x^2 + (-3 + x)*Log[-15 + 5*x])*Log[-E^x - x + Log[-15 + 5*x]]^2),x]
Time = 2.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {7239, 27, 25, 7257}
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.
| \(\displaystyle \int \frac {\left (\left (6 x^4-18 x^3+e^x \left (6 x^3-18 x^2\right )+\left (18 x^2-6 x^3\right ) \log (5 x-15)\right ) \log ^2\left (-x-e^x+\log (5 x-15)\right )+8 x+e^x (8 x-24)-32\right ) \exp \left (\frac {2 \left (4-x^3 \log \left (-x-e^x+\log (5 x-15)\right )\right )}{\log \left (-x-e^x+\log (5 x-15)\right )}\right )}{\left (-x^2+3 x+e^x (3-x)+(x-3) \log (5 x-15)\right ) \log ^2\left (-x-e^x+\log (5 x-15)\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int 2 e^{\frac {8}{\log \left (-x-e^x+\log (5 (x-3))\right )}-2 x^3} \left (-3 x^2-\frac {4 \left (e^x (x-3)+x-4\right )}{(x-3) \left (x+e^x-\log (5 (x-3))\right ) \log ^2\left (-x-e^x+\log (5 (x-3))\right )}\right )dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int -e^{\frac {8}{\log \left (-x-e^x+\log (-5 (3-x))\right )}-2 x^3} \left (3 x^2+\frac {4 \left (e^x (3-x)-x+4\right )}{(3-x) \left (x+e^x-\log (-5 (3-x))\right ) \log ^2\left (-x-e^x+\log (-5 (3-x))\right )}\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int e^{\frac {8}{\log \left (-x-e^x+\log (-5 (3-x))\right )}-2 x^3} \left (3 x^2+\frac {4 \left (e^x (3-x)-x+4\right )}{(3-x) \left (x+e^x-\log (-5 (3-x))\right ) \log ^2\left (-x-e^x+\log (-5 (3-x))\right )}\right )dx\) |
| \(\Big \downarrow \) 7257 |
| \(\displaystyle \exp \left (\frac {8}{\log \left (-x-e^x+\log (-5 (3-x))\right )}-2 x^3\right )\) |
Int[(E^((2*(4 - x^3*Log[-E^x - x + Log[-15 + 5*x]]))/Log[-E^x - x + Log[-1 5 + 5*x]])*(-32 + 8*x + E^x*(-24 + 8*x) + (-18*x^3 + 6*x^4 + E^x*(-18*x^2 + 6*x^3) + (18*x^2 - 6*x^3)*Log[-15 + 5*x])*Log[-E^x - x + Log[-15 + 5*x]] ^2))/((E^x*(3 - x) + 3*x - x^2 + (-3 + x)*Log[-15 + 5*x])*Log[-E^x - x + L og[-15 + 5*x]]^2),x]
3.8.87.3.1 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[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40
\[{\mathrm e}^{-\frac {2 \left (x^{3} \ln \left (\ln \left (5 x -15\right )-{\mathrm e}^{x}-x \right )-4\right )}{\ln \left (\ln \left (5 x -15\right )-{\mathrm e}^{x}-x \right )}}\]
int((((-6*x^3+18*x^2)*ln(5*x-15)+(6*x^3-18*x^2)*exp(x)+6*x^4-18*x^3)*ln(ln (5*x-15)-exp(x)-x)^2+(8*x-24)*exp(x)+8*x-32)*exp((-x^3*ln(ln(5*x-15)-exp(x )-x)+4)/ln(ln(5*x-15)-exp(x)-x))^2/((-3+x)*ln(5*x-15)+(-x+3)*exp(x)-x^2+3* x)/ln(ln(5*x-15)-exp(x)-x)^2,x)
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {e^{\frac {2 \left (4-x^3 \log \left (-e^x-x+\log (-15+5 x)\right )\right )}{\log \left (-e^x-x+\log (-15+5 x)\right )}} \left (-32+8 x+e^x (-24+8 x)+\left (-18 x^3+6 x^4+e^x \left (-18 x^2+6 x^3\right )+\left (18 x^2-6 x^3\right ) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )\right )}{\left (e^x (3-x)+3 x-x^2+(-3+x) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )} \, dx=e^{\left (-\frac {2 \, {\left (x^{3} \log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right ) - 4\right )}}{\log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right )}\right )} \]
integrate((((-6*x^3+18*x^2)*log(5*x-15)+(6*x^3-18*x^2)*exp(x)+6*x^4-18*x^3 )*log(log(5*x-15)-exp(x)-x)^2+(8*x-24)*exp(x)+8*x-32)*exp((-x^3*log(log(5* x-15)-exp(x)-x)+4)/log(log(5*x-15)-exp(x)-x))^2/((-3+x)*log(5*x-15)+(-x+3) *exp(x)-x^2+3*x)/log(log(5*x-15)-exp(x)-x)^2,x, algorithm=\
Timed out. \[ \int \frac {e^{\frac {2 \left (4-x^3 \log \left (-e^x-x+\log (-15+5 x)\right )\right )}{\log \left (-e^x-x+\log (-15+5 x)\right )}} \left (-32+8 x+e^x (-24+8 x)+\left (-18 x^3+6 x^4+e^x \left (-18 x^2+6 x^3\right )+\left (18 x^2-6 x^3\right ) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )\right )}{\left (e^x (3-x)+3 x-x^2+(-3+x) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )} \, dx=\text {Timed out} \]
integrate((((-6*x**3+18*x**2)*ln(5*x-15)+(6*x**3-18*x**2)*exp(x)+6*x**4-18 *x**3)*ln(ln(5*x-15)-exp(x)-x)**2+(8*x-24)*exp(x)+8*x-32)*exp((-x**3*ln(ln (5*x-15)-exp(x)-x)+4)/ln(ln(5*x-15)-exp(x)-x))**2/((-3+x)*ln(5*x-15)+(-x+3 )*exp(x)-x**2+3*x)/ln(ln(5*x-15)-exp(x)-x)**2,x)
\[ \int \frac {e^{\frac {2 \left (4-x^3 \log \left (-e^x-x+\log (-15+5 x)\right )\right )}{\log \left (-e^x-x+\log (-15+5 x)\right )}} \left (-32+8 x+e^x (-24+8 x)+\left (-18 x^3+6 x^4+e^x \left (-18 x^2+6 x^3\right )+\left (18 x^2-6 x^3\right ) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )\right )}{\left (e^x (3-x)+3 x-x^2+(-3+x) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )} \, dx=\int { -\frac {2 \, {\left (3 \, {\left (x^{4} - 3 \, x^{3} + {\left (x^{3} - 3 \, x^{2}\right )} e^{x} - {\left (x^{3} - 3 \, x^{2}\right )} \log \left (5 \, x - 15\right )\right )} \log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right )^{2} + 4 \, {\left (x - 3\right )} e^{x} + 4 \, x - 16\right )} e^{\left (-\frac {2 \, {\left (x^{3} \log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right ) - 4\right )}}{\log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right )}\right )}}{{\left (x^{2} + {\left (x - 3\right )} e^{x} - {\left (x - 3\right )} \log \left (5 \, x - 15\right ) - 3 \, x\right )} \log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right )^{2}} \,d x } \]
integrate((((-6*x^3+18*x^2)*log(5*x-15)+(6*x^3-18*x^2)*exp(x)+6*x^4-18*x^3 )*log(log(5*x-15)-exp(x)-x)^2+(8*x-24)*exp(x)+8*x-32)*exp((-x^3*log(log(5* x-15)-exp(x)-x)+4)/log(log(5*x-15)-exp(x)-x))^2/((-3+x)*log(5*x-15)+(-x+3) *exp(x)-x^2+3*x)/log(log(5*x-15)-exp(x)-x)^2,x, algorithm=\
x*e^(-2*x^3 + x + 8/log(-x - e^x + log(5) + log(x - 3)))/((x - 3)*e^x + x - 4) + x*e^(-2*x^3 + 8/log(-x - e^x + log(5) + log(x - 3)))/((x - 3)*e^x + x - 4) - 3*e^(-2*x^3 + x + 8/log(-x - e^x + log(5) + log(x - 3)))/((x - 3 )*e^x + x - 4) - 4*e^(-2*x^3 + 8/log(-x - e^x + log(5) + log(x - 3)))/((x - 3)*e^x + x - 4) - 6*integrate(x^2*e^(-2*x^3 + 8/log(-x - e^x + log(5) + log(x - 3))), x)
Time = 13.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {2 \left (4-x^3 \log \left (-e^x-x+\log (-15+5 x)\right )\right )}{\log \left (-e^x-x+\log (-15+5 x)\right )}} \left (-32+8 x+e^x (-24+8 x)+\left (-18 x^3+6 x^4+e^x \left (-18 x^2+6 x^3\right )+\left (18 x^2-6 x^3\right ) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )\right )}{\left (e^x (3-x)+3 x-x^2+(-3+x) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )} \, dx=e^{\left (-2 \, x^{3} + \frac {8}{\log \left (-x - e^{x} + \log \left (5 \, x - 15\right )\right )}\right )} \]
integrate((((-6*x^3+18*x^2)*log(5*x-15)+(6*x^3-18*x^2)*exp(x)+6*x^4-18*x^3 )*log(log(5*x-15)-exp(x)-x)^2+(8*x-24)*exp(x)+8*x-32)*exp((-x^3*log(log(5* x-15)-exp(x)-x)+4)/log(log(5*x-15)-exp(x)-x))^2/((-3+x)*log(5*x-15)+(-x+3) *exp(x)-x^2+3*x)/log(log(5*x-15)-exp(x)-x)^2,x, algorithm=\
Time = 8.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {2 \left (4-x^3 \log \left (-e^x-x+\log (-15+5 x)\right )\right )}{\log \left (-e^x-x+\log (-15+5 x)\right )}} \left (-32+8 x+e^x (-24+8 x)+\left (-18 x^3+6 x^4+e^x \left (-18 x^2+6 x^3\right )+\left (18 x^2-6 x^3\right ) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )\right )}{\left (e^x (3-x)+3 x-x^2+(-3+x) \log (-15+5 x)\right ) \log ^2\left (-e^x-x+\log (-15+5 x)\right )} \, dx={\mathrm {e}}^{-2\,x^3}\,{\mathrm {e}}^{\frac {8}{\ln \left (\ln \left (5\,x-15\right )-x-{\mathrm {e}}^x\right )}} \]
int((exp(-(2*(x^3*log(log(5*x - 15) - x - exp(x)) - 4))/log(log(5*x - 15) - x - exp(x)))*(8*x + exp(x)*(8*x - 24) - log(log(5*x - 15) - x - exp(x))^ 2*(exp(x)*(18*x^2 - 6*x^3) - log(5*x - 15)*(18*x^2 - 6*x^3) + 18*x^3 - 6*x ^4) - 32))/(log(log(5*x - 15) - x - exp(x))^2*(3*x - exp(x)*(x - 3) - x^2 + log(5*x - 15)*(x - 3))),x)