Integrand size = 100, antiderivative size = 27 \[ \int \frac {1}{3} e^{-e^x+\frac {1}{3} e^{-e^x} \left (12 x^3+100 x^4+\left (-12 x^2-100 x^3\right ) \log (4)\right )} \left (36 x^2+400 x^3+\left (-24 x-300 x^2\right ) \log (4)+e^x \left (-12 x^3-100 x^4+\left (12 x^2+100 x^3\right ) \log (4)\right )\right ) \, dx=e^{\frac {4}{3} e^{-e^x} x^2 (3+25 x) (x-\log (4))} \]
\[ \int \frac {1}{3} e^{-e^x+\frac {1}{3} e^{-e^x} \left (12 x^3+100 x^4+\left (-12 x^2-100 x^3\right ) \log (4)\right )} \left (36 x^2+400 x^3+\left (-24 x-300 x^2\right ) \log (4)+e^x \left (-12 x^3-100 x^4+\left (12 x^2+100 x^3\right ) \log (4)\right )\right ) \, dx=\int \frac {1}{3} e^{-e^x+\frac {1}{3} e^{-e^x} \left (12 x^3+100 x^4+\left (-12 x^2-100 x^3\right ) \log (4)\right )} \left (36 x^2+400 x^3+\left (-24 x-300 x^2\right ) \log (4)+e^x \left (-12 x^3-100 x^4+\left (12 x^2+100 x^3\right ) \log (4)\right )\right ) \, dx \]
Integrate[(E^(-E^x + (12*x^3 + 100*x^4 + (-12*x^2 - 100*x^3)*Log[4])/(3*E^ E^x))*(36*x^2 + 400*x^3 + (-24*x - 300*x^2)*Log[4] + E^x*(-12*x^3 - 100*x^ 4 + (12*x^2 + 100*x^3)*Log[4])))/3,x]
Integrate[E^(-E^x + (12*x^3 + 100*x^4 + (-12*x^2 - 100*x^3)*Log[4])/(3*E^E ^x))*(36*x^2 + 400*x^3 + (-24*x - 300*x^2)*Log[4] + E^x*(-12*x^3 - 100*x^4 + (12*x^2 + 100*x^3)*Log[4])), x]/3
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.
\(\displaystyle \int \frac {1}{3} \left (400 x^3+36 x^2+\left (-300 x^2-24 x\right ) \log (4)+e^x \left (-100 x^4-12 x^3+\left (100 x^3+12 x^2\right ) \log (4)\right )\right ) \exp \left (\frac {1}{3} e^{-e^x} \left (100 x^4+12 x^3+\left (-100 x^3-12 x^2\right ) \log (4)\right )-e^x\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int 4 \exp \left (\frac {4}{3} e^{-e^x} \left (25 x^4+3 x^3-\left (25 x^3+3 x^2\right ) \log (4)\right )-e^x\right ) \left (100 x^3+9 x^2-e^x \left (25 x^4+3 x^3-\left (25 x^3+3 x^2\right ) \log (4)\right )-3 \left (25 x^2+2 x\right ) \log (4)\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{3} \int \exp \left (\frac {4}{3} e^{-e^x} \left (25 x^4+3 x^3-\left (25 x^3+3 x^2\right ) \log (4)\right )-e^x\right ) \left (100 x^3+9 x^2-e^x \left (25 x^4+3 x^3-\left (25 x^3+3 x^2\right ) \log (4)\right )-3 \left (25 x^2+2 x\right ) \log (4)\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {4}{3} \int \exp \left (-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) \left (100 x^3+9 x^2-e^x \left (25 x^4+3 x^3-\left (25 x^3+3 x^2\right ) \log (4)\right )-3 \left (25 x^2+2 x\right ) \log (4)\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {4}{3} \int \left (100 \exp \left (-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) x^3+9 \exp \left (-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) x^2-\exp \left (x-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) (25 x+3) (x-\log (4)) x^2-3 \exp \left (-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) (25 x+2) \log (4) x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{3} \left (-6 \log (4) \int \exp \left (-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) xdx-75 \log (4) \int \exp \left (-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) x^2dx+9 \int \exp \left (-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) x^2dx+3 \log (4) \int \exp \left (x-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) x^2dx+100 \int \exp \left (-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) x^3dx-(3-25 \log (4)) \int \exp \left (x-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) x^3dx-25 \int \exp \left (x-\frac {1}{3} e^{-e^x} \left (-100 x^4-12 \left (1-\frac {25 \log (4)}{3}\right ) x^3+12 \log (4) x^2+3 e^{x+e^x}\right )\right ) x^4dx\right )\) |
Int[(E^(-E^x + (12*x^3 + 100*x^4 + (-12*x^2 - 100*x^3)*Log[4])/(3*E^E^x))* (36*x^2 + 400*x^3 + (-24*x - 300*x^2)*Log[4] + E^x*(-12*x^3 - 100*x^4 + (1 2*x^2 + 100*x^3)*Log[4])))/3,x]
3.2.5.3.1 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]]
Time = 4.78 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \({\mathrm e}^{-\frac {4 x^{2} \left (3+25 x \right ) \left (2 \ln \left (2\right )-x \right ) {\mathrm e}^{-{\mathrm e}^{x}}}{3}}\) | \(25\) |
parallelrisch | \({\mathrm e}^{-\frac {4 x^{2} \left (50 x \ln \left (2\right )-25 x^{2}+6 \ln \left (2\right )-3 x \right ) {\mathrm e}^{-{\mathrm e}^{x}}}{3}}\) | \(30\) |
int(1/3*((2*(100*x^3+12*x^2)*ln(2)-100*x^4-12*x^3)*exp(x)+2*(-300*x^2-24*x )*ln(2)+400*x^3+36*x^2)*exp(1/3*(2*(-100*x^3-12*x^2)*ln(2)+100*x^4+12*x^3) /exp(exp(x)))/exp(exp(x)),x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {1}{3} e^{-e^x+\frac {1}{3} e^{-e^x} \left (12 x^3+100 x^4+\left (-12 x^2-100 x^3\right ) \log (4)\right )} \left (36 x^2+400 x^3+\left (-24 x-300 x^2\right ) \log (4)+e^x \left (-12 x^3-100 x^4+\left (12 x^2+100 x^3\right ) \log (4)\right )\right ) \, dx=e^{\left (\frac {4}{3} \, {\left (25 \, x^{4} + 3 \, x^{3} - 2 \, {\left (25 \, x^{3} + 3 \, x^{2}\right )} \log \left (2\right )\right )} e^{\left (-e^{x}\right )}\right )} \]
integrate(1/3*((2*(100*x^3+12*x^2)*log(2)-100*x^4-12*x^3)*exp(x)+2*(-300*x ^2-24*x)*log(2)+400*x^3+36*x^2)*exp(1/3*(2*(-100*x^3-12*x^2)*log(2)+100*x^ 4+12*x^3)/exp(exp(x)))/exp(exp(x)),x, algorithm=\
Time = 0.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {1}{3} e^{-e^x+\frac {1}{3} e^{-e^x} \left (12 x^3+100 x^4+\left (-12 x^2-100 x^3\right ) \log (4)\right )} \left (36 x^2+400 x^3+\left (-24 x-300 x^2\right ) \log (4)+e^x \left (-12 x^3-100 x^4+\left (12 x^2+100 x^3\right ) \log (4)\right )\right ) \, dx=e^{\left (\frac {100 x^{4}}{3} + 4 x^{3} + \frac {\left (- 200 x^{3} - 24 x^{2}\right ) \log {\left (2 \right )}}{3}\right ) e^{- e^{x}}} \]
integrate(1/3*((2*(100*x**3+12*x**2)*ln(2)-100*x**4-12*x**3)*exp(x)+2*(-30 0*x**2-24*x)*ln(2)+400*x**3+36*x**2)*exp(1/3*(2*(-100*x**3-12*x**2)*ln(2)+ 100*x**4+12*x**3)/exp(exp(x)))/exp(exp(x)),x)
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.55 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {1}{3} e^{-e^x+\frac {1}{3} e^{-e^x} \left (12 x^3+100 x^4+\left (-12 x^2-100 x^3\right ) \log (4)\right )} \left (36 x^2+400 x^3+\left (-24 x-300 x^2\right ) \log (4)+e^x \left (-12 x^3-100 x^4+\left (12 x^2+100 x^3\right ) \log (4)\right )\right ) \, dx=e^{\left (\frac {100}{3} \, x^{4} e^{\left (-e^{x}\right )} - \frac {200}{3} \, x^{3} e^{\left (-e^{x}\right )} \log \left (2\right ) + 4 \, x^{3} e^{\left (-e^{x}\right )} - 8 \, x^{2} e^{\left (-e^{x}\right )} \log \left (2\right )\right )} \]
integrate(1/3*((2*(100*x^3+12*x^2)*log(2)-100*x^4-12*x^3)*exp(x)+2*(-300*x ^2-24*x)*log(2)+400*x^3+36*x^2)*exp(1/3*(2*(-100*x^3-12*x^2)*log(2)+100*x^ 4+12*x^3)/exp(exp(x)))/exp(exp(x)),x, algorithm=\
\[ \int \frac {1}{3} e^{-e^x+\frac {1}{3} e^{-e^x} \left (12 x^3+100 x^4+\left (-12 x^2-100 x^3\right ) \log (4)\right )} \left (36 x^2+400 x^3+\left (-24 x-300 x^2\right ) \log (4)+e^x \left (-12 x^3-100 x^4+\left (12 x^2+100 x^3\right ) \log (4)\right )\right ) \, dx=\int { \frac {4}{3} \, {\left (100 \, x^{3} + 9 \, x^{2} - {\left (25 \, x^{4} + 3 \, x^{3} - 2 \, {\left (25 \, x^{3} + 3 \, x^{2}\right )} \log \left (2\right )\right )} e^{x} - 6 \, {\left (25 \, x^{2} + 2 \, x\right )} \log \left (2\right )\right )} e^{\left (\frac {4}{3} \, {\left (25 \, x^{4} + 3 \, x^{3} - 2 \, {\left (25 \, x^{3} + 3 \, x^{2}\right )} \log \left (2\right )\right )} e^{\left (-e^{x}\right )} - e^{x}\right )} \,d x } \]
integrate(1/3*((2*(100*x^3+12*x^2)*log(2)-100*x^4-12*x^3)*exp(x)+2*(-300*x ^2-24*x)*log(2)+400*x^3+36*x^2)*exp(1/3*(2*(-100*x^3-12*x^2)*log(2)+100*x^ 4+12*x^3)/exp(exp(x)))/exp(exp(x)),x, algorithm=\
integrate(4/3*(100*x^3 + 9*x^2 - (25*x^4 + 3*x^3 - 2*(25*x^3 + 3*x^2)*log( 2))*e^x - 6*(25*x^2 + 2*x)*log(2))*e^(4/3*(25*x^4 + 3*x^3 - 2*(25*x^3 + 3* x^2)*log(2))*e^(-e^x) - e^x), x)
Timed out. \[ \int \frac {1}{3} e^{-e^x+\frac {1}{3} e^{-e^x} \left (12 x^3+100 x^4+\left (-12 x^2-100 x^3\right ) \log (4)\right )} \left (36 x^2+400 x^3+\left (-24 x-300 x^2\right ) \log (4)+e^x \left (-12 x^3-100 x^4+\left (12 x^2+100 x^3\right ) \log (4)\right )\right ) \, dx=\int -\frac {{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^{{\mathrm {e}}^{-{\mathrm {e}}^x}\,\left (4\,x^3-\frac {2\,\ln \left (2\right )\,\left (100\,x^3+12\,x^2\right )}{3}+\frac {100\,x^4}{3}\right )}\,\left (2\,\ln \left (2\right )\,\left (300\,x^2+24\,x\right )+{\mathrm {e}}^x\,\left (12\,x^3-2\,\ln \left (2\right )\,\left (100\,x^3+12\,x^2\right )+100\,x^4\right )-36\,x^2-400\,x^3\right )}{3} \,d x \]
int(-(exp(-exp(x))*exp(exp(-exp(x))*(4*x^3 - (2*log(2)*(12*x^2 + 100*x^3)) /3 + (100*x^4)/3))*(2*log(2)*(24*x + 300*x^2) + exp(x)*(12*x^3 - 2*log(2)* (12*x^2 + 100*x^3) + 100*x^4) - 36*x^2 - 400*x^3))/3,x)