Integrand size = 90, antiderivative size = 24 \[ \int e^{-4+e^{-4 x^2+2 x^3}+e^{-4+e^{-4 x^2+2 x^3}+x^2-2 x \log (5)+\log ^2(5)}+x^2-2 x \log (5)+\log ^2(5)} \left (2 x+e^{-4 x^2+2 x^3} \left (-8 x+6 x^2\right )-2 \log (5)\right ) \, dx=e^{e^{-4+e^{2 (-2+x) x^2}+(x-\log (5))^2}} \]
\[ \int e^{-4+e^{-4 x^2+2 x^3}+e^{-4+e^{-4 x^2+2 x^3}+x^2-2 x \log (5)+\log ^2(5)}+x^2-2 x \log (5)+\log ^2(5)} \left (2 x+e^{-4 x^2+2 x^3} \left (-8 x+6 x^2\right )-2 \log (5)\right ) \, dx=\int e^{-4+e^{-4 x^2+2 x^3}+e^{-4+e^{-4 x^2+2 x^3}+x^2-2 x \log (5)+\log ^2(5)}+x^2-2 x \log (5)+\log ^2(5)} \left (2 x+e^{-4 x^2+2 x^3} \left (-8 x+6 x^2\right )-2 \log (5)\right ) \, dx \]
Integrate[E^(-4 + E^(-4*x^2 + 2*x^3) + E^(-4 + E^(-4*x^2 + 2*x^3) + x^2 - 2*x*Log[5] + Log[5]^2) + x^2 - 2*x*Log[5] + Log[5]^2)*(2*x + E^(-4*x^2 + 2 *x^3)*(-8*x + 6*x^2) - 2*Log[5]),x]
Integrate[E^(-4 + E^(-4*x^2 + 2*x^3) + E^(-4 + E^(-4*x^2 + 2*x^3) + x^2 - 2*x*Log[5] + Log[5]^2) + x^2 - 2*x*Log[5] + Log[5]^2)*(2*x + E^(-4*x^2 + 2 *x^3)*(-8*x + 6*x^2) - 2*Log[5]), 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.
\(\displaystyle \int \left (e^{2 x^3-4 x^2} \left (6 x^2-8 x\right )+2 x-2 \log (5)\right ) \exp \left (x^2+e^{2 x^3-4 x^2}+e^{x^2+e^{2 x^3-4 x^2}-2 x \log (5)-4+\log ^2(5)}-2 x \log (5)-4+\log ^2(5)\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \left (e^{2 x^3-4 x^2} \left (6 x^2-8 x\right )+2 x-2 \log (5)\right ) \exp \left (x^2+e^{2 x^3-4 x^2}+e^{x^2+e^{2 x^3-4 x^2}-2 x \log (5)-4+\log ^2(5)}-2 x \log (5)-4 \left (1-\frac {\log ^2(5)}{4}\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 x \exp \left (x^2+e^{2 x^3-4 x^2}+e^{x^2+e^{2 x^3-4 x^2}-2 x \log (5)-4+\log ^2(5)}-2 x \log (5)-4 \left (1-\frac {\log ^2(5)}{4}\right )\right )+2 (3 x-4) x \exp \left (2 (x-2) x^2+x^2+e^{2 x^3-4 x^2}+e^{x^2+e^{2 x^3-4 x^2}-2 x \log (5)-4+\log ^2(5)}-2 x \log (5)-4 \left (1-\frac {\log ^2(5)}{4}\right )\right )-2 \log (5) \exp \left (x^2+e^{2 x^3-4 x^2}+e^{x^2+e^{2 x^3-4 x^2}-2 x \log (5)-4+\log ^2(5)}-2 x \log (5)-4 \left (1-\frac {\log ^2(5)}{4}\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \log (5) \int \exp \left (x^2-2 \log (5) x+e^{2 x^3-4 x^2}+e^{x^2-2 \log (5) x+e^{2 x^3-4 x^2}+\log ^2(5)-4}-4 \left (1-\frac {\log ^2(5)}{4}\right )\right )dx+2 \int \exp \left (x^2-2 \log (5) x+e^{2 x^3-4 x^2}+e^{x^2-2 \log (5) x+e^{2 x^3-4 x^2}+\log ^2(5)-4}-4 \left (1-\frac {\log ^2(5)}{4}\right )\right ) xdx-8 \int \exp \left (2 (x-2) x^2+x^2-2 \log (5) x+e^{2 x^3-4 x^2}+e^{x^2-2 \log (5) x+e^{2 x^3-4 x^2}+\log ^2(5)-4}-4 \left (1-\frac {\log ^2(5)}{4}\right )\right ) xdx+6 \int \exp \left (2 (x-2) x^2+x^2-2 \log (5) x+e^{2 x^3-4 x^2}+e^{x^2-2 \log (5) x+e^{2 x^3-4 x^2}+\log ^2(5)-4}-4 \left (1-\frac {\log ^2(5)}{4}\right )\right ) x^2dx\) |
Int[E^(-4 + E^(-4*x^2 + 2*x^3) + E^(-4 + E^(-4*x^2 + 2*x^3) + x^2 - 2*x*Lo g[5] + Log[5]^2) + x^2 - 2*x*Log[5] + Log[5]^2)*(2*x + E^(-4*x^2 + 2*x^3)* (-8*x + 6*x^2) - 2*Log[5]),x]
3.15.59.3.1 Defintions of rubi rules used
Time = 0.99 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
risch | \({\mathrm e}^{\left (\frac {1}{25}\right )^{x} {\mathrm e}^{{\mathrm e}^{2 \left (-2+x \right ) x^{2}}+\ln \left (5\right )^{2}-4+x^{2}}}\) | \(25\) |
derivativedivides | \({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x^{3}-4 x^{2}}+\ln \left (5\right )^{2}-2 x \ln \left (5\right )+x^{2}-4}}\) | \(29\) |
default | \({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x^{3}-4 x^{2}}+\ln \left (5\right )^{2}-2 x \ln \left (5\right )+x^{2}-4}}\) | \(29\) |
norman | \({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x^{3}-4 x^{2}}+\ln \left (5\right )^{2}-2 x \ln \left (5\right )+x^{2}-4}}\) | \(29\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{2 x^{3}-4 x^{2}}+\ln \left (5\right )^{2}-2 x \ln \left (5\right )+x^{2}-4}}\) | \(29\) |
int(((6*x^2-8*x)*exp(2*x^3-4*x^2)-2*ln(5)+2*x)*exp(exp(2*x^3-4*x^2)+ln(5)^ 2-2*x*ln(5)+x^2-4)*exp(exp(exp(2*x^3-4*x^2)+ln(5)^2-2*x*ln(5)+x^2-4)),x,me thod=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int e^{-4+e^{-4 x^2+2 x^3}+e^{-4+e^{-4 x^2+2 x^3}+x^2-2 x \log (5)+\log ^2(5)}+x^2-2 x \log (5)+\log ^2(5)} \left (2 x+e^{-4 x^2+2 x^3} \left (-8 x+6 x^2\right )-2 \log (5)\right ) \, dx=e^{\left (e^{\left (x^{2} - 2 \, x \log \left (5\right ) + \log \left (5\right )^{2} + e^{\left (2 \, x^{3} - 4 \, x^{2}\right )} - 4\right )}\right )} \]
integrate(((6*x^2-8*x)*exp(2*x^3-4*x^2)-2*log(5)+2*x)*exp(exp(2*x^3-4*x^2) +log(5)^2-2*x*log(5)+x^2-4)*exp(exp(exp(2*x^3-4*x^2)+log(5)^2-2*x*log(5)+x ^2-4)),x, algorithm=\
Time = 0.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int e^{-4+e^{-4 x^2+2 x^3}+e^{-4+e^{-4 x^2+2 x^3}+x^2-2 x \log (5)+\log ^2(5)}+x^2-2 x \log (5)+\log ^2(5)} \left (2 x+e^{-4 x^2+2 x^3} \left (-8 x+6 x^2\right )-2 \log (5)\right ) \, dx=e^{e^{x^{2} - 2 x \log {\left (5 \right )} + e^{2 x^{3} - 4 x^{2}} - 4 + \log {\left (5 \right )}^{2}}} \]
integrate(((6*x**2-8*x)*exp(2*x**3-4*x**2)-2*ln(5)+2*x)*exp(exp(2*x**3-4*x **2)+ln(5)**2-2*x*ln(5)+x**2-4)*exp(exp(exp(2*x**3-4*x**2)+ln(5)**2-2*x*ln (5)+x**2-4)),x)
Time = 0.84 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int e^{-4+e^{-4 x^2+2 x^3}+e^{-4+e^{-4 x^2+2 x^3}+x^2-2 x \log (5)+\log ^2(5)}+x^2-2 x \log (5)+\log ^2(5)} \left (2 x+e^{-4 x^2+2 x^3} \left (-8 x+6 x^2\right )-2 \log (5)\right ) \, dx=e^{\left (e^{\left (x^{2} - 2 \, x \log \left (5\right ) + \log \left (5\right )^{2} + e^{\left (2 \, x^{3} - 4 \, x^{2}\right )} - 4\right )}\right )} \]
integrate(((6*x^2-8*x)*exp(2*x^3-4*x^2)-2*log(5)+2*x)*exp(exp(2*x^3-4*x^2) +log(5)^2-2*x*log(5)+x^2-4)*exp(exp(exp(2*x^3-4*x^2)+log(5)^2-2*x*log(5)+x ^2-4)),x, algorithm=\
\[ \int e^{-4+e^{-4 x^2+2 x^3}+e^{-4+e^{-4 x^2+2 x^3}+x^2-2 x \log (5)+\log ^2(5)}+x^2-2 x \log (5)+\log ^2(5)} \left (2 x+e^{-4 x^2+2 x^3} \left (-8 x+6 x^2\right )-2 \log (5)\right ) \, dx=\int { 2 \, {\left ({\left (3 \, x^{2} - 4 \, x\right )} e^{\left (2 \, x^{3} - 4 \, x^{2}\right )} + x - \log \left (5\right )\right )} e^{\left (x^{2} - 2 \, x \log \left (5\right ) + \log \left (5\right )^{2} + e^{\left (2 \, x^{3} - 4 \, x^{2}\right )} + e^{\left (x^{2} - 2 \, x \log \left (5\right ) + \log \left (5\right )^{2} + e^{\left (2 \, x^{3} - 4 \, x^{2}\right )} - 4\right )} - 4\right )} \,d x } \]
integrate(((6*x^2-8*x)*exp(2*x^3-4*x^2)-2*log(5)+2*x)*exp(exp(2*x^3-4*x^2) +log(5)^2-2*x*log(5)+x^2-4)*exp(exp(exp(2*x^3-4*x^2)+log(5)^2-2*x*log(5)+x ^2-4)),x, algorithm=\
integrate(2*((3*x^2 - 4*x)*e^(2*x^3 - 4*x^2) + x - log(5))*e^(x^2 - 2*x*lo g(5) + log(5)^2 + e^(2*x^3 - 4*x^2) + e^(x^2 - 2*x*log(5) + log(5)^2 + e^( 2*x^3 - 4*x^2) - 4) - 4), x)
Time = 13.60 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int e^{-4+e^{-4 x^2+2 x^3}+e^{-4+e^{-4 x^2+2 x^3}+x^2-2 x \log (5)+\log ^2(5)}+x^2-2 x \log (5)+\log ^2(5)} \left (2 x+e^{-4 x^2+2 x^3} \left (-8 x+6 x^2\right )-2 \log (5)\right ) \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{{\ln \left (5\right )}^2}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x^3}\,{\mathrm {e}}^{-4\,x^2}}}{5^{2\,x}}} \]