Integrand size = 100, antiderivative size = 35 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-4 x^3 \log (3)+e^x \left (12+9 x^2\right ) \log (3)-4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^2}} \left (e^x (-24+4 x) \log (3)+\left (-4 x^3+4 x^4\right ) \log (3)+4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^3} \, dx=3^{3+\frac {4 \left (\frac {3}{x}-e^{-x} x^2-\log \left (\frac {4}{x}\right )\right )}{3 x}} \]
Time = 2.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-4 x^3 \log (3)+e^x \left (12+9 x^2\right ) \log (3)-4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^2}} \left (e^x (-24+4 x) \log (3)+\left (-4 x^3+4 x^4\right ) \log (3)+4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^3} \, dx=\frac {4\ 3^{3+\frac {4}{x^2}-\frac {4 e^{-x} x}{3}-\frac {4 \log \left (\frac {4}{x}\right )}{3 x}} \log (3)}{\log (81)} \]
Integrate[(E^(-x + (-4*x^3*Log[3] + E^x*(12 + 9*x^2)*Log[3] - 4*E^x*x*Log[ 3]*Log[4/x])/(3*E^x*x^2))*(E^x*(-24 + 4*x)*Log[3] + (-4*x^3 + 4*x^4)*Log[3 ] + 4*E^x*x*Log[3]*Log[4/x]))/(3*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.
\(\displaystyle \int \frac {\left (\left (4 x^4-4 x^3\right ) \log (3)+e^x (4 x-24) \log (3)+4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right ) \exp \left (\frac {e^{-x} \left (-4 x^3 \log (3)+e^x \left (9 x^2+12\right ) \log (3)-4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^2}-x\right )}{3 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {4 \exp \left (-x-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right ) \left (e^x \log (3) (6-x)-e^x x \log (3) \log \left (\frac {4}{x}\right )+\left (x^3-x^4\right ) \log (3)\right )}{x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{3} \int \frac {\exp \left (-x-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right ) \left (e^x \log (3) (6-x)-e^x x \log (3) \log \left (\frac {4}{x}\right )+\left (x^3-x^4\right ) \log (3)\right )}{x^3}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {4}{3} \int \frac {\exp \left (-x-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right ) \log (3) \left (-x^4+x^3-e^x x-e^x \log \left (\frac {4}{x}\right ) x+6 e^x\right )}{x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{3} \log (3) \int \frac {\exp \left (-x-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right ) \left (-x^4+x^3-e^x x-e^x \log \left (\frac {4}{x}\right ) x+6 e^x\right )}{x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {4}{3} \log (3) \int \left (-\exp \left (-x-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right ) x+\exp \left (-x-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right )-\frac {\exp \left (-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right ) \left (\log \left (\frac {4}{x}\right ) x+x-6\right )}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4}{3} \log (3) \left (\int \exp \left (-x-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right )dx-\int \exp \left (-x-\frac {e^{-x} \left (4 \log (3) x^3+4 e^x \log (3) \log \left (\frac {4}{x}\right ) x-3 e^x \left (3 x^2+4\right ) \log (3)\right )}{3 x^2}\right ) xdx+2 \int \frac {3^{\frac {4 e^{-x} \left (-x^3+3 e^x x^2-e^x \log \left (\frac {4}{x}\right ) x+3 e^x\right )}{3 x^2}}}{x^3}dx-\int \frac {3^{\frac {e^{-x} \left (-4 x^3+9 e^x x^2-4 e^x \log \left (\frac {4}{x}\right ) x+12 e^x\right )}{3 x^2}}}{x^2}dx-\int \frac {3^{\frac {e^{-x} \left (-4 x^3+9 e^x x^2-4 e^x \log \left (\frac {4}{x}\right ) x+12 e^x\right )}{3 x^2}} \log \left (\frac {4}{x}\right )}{x^2}dx\right )\) |
Int[(E^(-x + (-4*x^3*Log[3] + E^x*(12 + 9*x^2)*Log[3] - 4*E^x*x*Log[3]*Log [4/x])/(3*E^x*x^2))*(E^x*(-24 + 4*x)*Log[3] + (-4*x^3 + 4*x^4)*Log[3] + 4* E^x*x*Log[3]*Log[4/x]))/(3*x^3),x]
3.4.17.3.1 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 = 17.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97
method | result | size |
risch | \(3^{-\frac {4 x^{3} {\mathrm e}^{-x}-4 x \ln \left (x \right )+8 x \ln \left (2\right )-9 x^{2}-12}{3 x^{2}}}\) | \(34\) |
parallelrisch | \({\mathrm e}^{-\frac {\ln \left (3\right ) \left (4 x \,{\mathrm e}^{x} \ln \left (\frac {4}{x}\right )-9 \,{\mathrm e}^{x} x^{2}+4 x^{3}-12 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{3 x^{2}}}\) | \(41\) |
int(1/3*(4*x*ln(3)*exp(x)*ln(4/x)+(4*x-24)*ln(3)*exp(x)+(4*x^4-4*x^3)*ln(3 ))*exp(1/3*(-4*x*ln(3)*exp(x)*ln(4/x)+(9*x^2+12)*ln(3)*exp(x)-4*x^3*ln(3)) /exp(x)/x^2)/exp(x)/x^3,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-4 x^3 \log (3)+e^x \left (12+9 x^2\right ) \log (3)-4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^2}} \left (e^x (-24+4 x) \log (3)+\left (-4 x^3+4 x^4\right ) \log (3)+4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^3} \, dx=e^{\left (x - \frac {{\left (4 \, x^{3} \log \left (3\right ) + 4 \, x e^{x} \log \left (3\right ) \log \left (\frac {4}{x}\right ) + 3 \, {\left (x^{3} - {\left (3 \, x^{2} + 4\right )} \log \left (3\right )\right )} e^{x}\right )} e^{\left (-x\right )}}{3 \, x^{2}}\right )} \]
integrate(1/3*(4*x*log(3)*exp(x)*log(4/x)+(4*x-24)*log(3)*exp(x)+(4*x^4-4* x^3)*log(3))*exp(1/3*(-4*x*log(3)*exp(x)*log(4/x)+(9*x^2+12)*log(3)*exp(x) -4*x^3*log(3))/exp(x)/x^2)/exp(x)/x^3,x, algorithm=\
e^(x - 1/3*(4*x^3*log(3) + 4*x*e^x*log(3)*log(4/x) + 3*(x^3 - (3*x^2 + 4)* log(3))*e^x)*e^(-x)/x^2)
Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-4 x^3 \log (3)+e^x \left (12+9 x^2\right ) \log (3)-4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^2}} \left (e^x (-24+4 x) \log (3)+\left (-4 x^3+4 x^4\right ) \log (3)+4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^3} \, dx=e^{\frac {\left (- \frac {4 x^{3} \log {\left (3 \right )}}{3} - \frac {4 x e^{x} \log {\left (3 \right )} \log {\left (\frac {4}{x} \right )}}{3} + \frac {\left (9 x^{2} + 12\right ) e^{x} \log {\left (3 \right )}}{3}\right ) e^{- x}}{x^{2}}} \]
integrate(1/3*(4*x*ln(3)*exp(x)*ln(4/x)+(4*x-24)*ln(3)*exp(x)+(4*x**4-4*x* *3)*ln(3))*exp(1/3*(-4*x*ln(3)*exp(x)*ln(4/x)+(9*x**2+12)*ln(3)*exp(x)-4*x **3*ln(3))/exp(x)/x**2)/exp(x)/x**3,x)
exp((-4*x**3*log(3)/3 - 4*x*exp(x)*log(3)*log(4/x)/3 + (9*x**2 + 12)*exp(x )*log(3)/3)*exp(-x)/x**2)
Time = 0.46 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-4 x^3 \log (3)+e^x \left (12+9 x^2\right ) \log (3)-4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^2}} \left (e^x (-24+4 x) \log (3)+\left (-4 x^3+4 x^4\right ) \log (3)+4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^3} \, dx=27 \, e^{\left (-\frac {4}{3} \, x e^{\left (-x\right )} \log \left (3\right ) - \frac {8 \, \log \left (3\right ) \log \left (2\right )}{3 \, x} + \frac {4 \, \log \left (3\right ) \log \left (x\right )}{3 \, x} + \frac {4 \, \log \left (3\right )}{x^{2}}\right )} \]
integrate(1/3*(4*x*log(3)*exp(x)*log(4/x)+(4*x-24)*log(3)*exp(x)+(4*x^4-4* x^3)*log(3))*exp(1/3*(-4*x*log(3)*exp(x)*log(4/x)+(9*x^2+12)*log(3)*exp(x) -4*x^3*log(3))/exp(x)/x^2)/exp(x)/x^3,x, algorithm=\
\[ \int \frac {e^{-x+\frac {e^{-x} \left (-4 x^3 \log (3)+e^x \left (12+9 x^2\right ) \log (3)-4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^2}} \left (e^x (-24+4 x) \log (3)+\left (-4 x^3+4 x^4\right ) \log (3)+4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^3} \, dx=\int { \frac {4 \, {\left (x e^{x} \log \left (3\right ) \log \left (\frac {4}{x}\right ) + {\left (x - 6\right )} e^{x} \log \left (3\right ) + {\left (x^{4} - x^{3}\right )} \log \left (3\right )\right )} e^{\left (-x - \frac {{\left (4 \, x^{3} \log \left (3\right ) + 4 \, x e^{x} \log \left (3\right ) \log \left (\frac {4}{x}\right ) - 3 \, {\left (3 \, x^{2} + 4\right )} e^{x} \log \left (3\right )\right )} e^{\left (-x\right )}}{3 \, x^{2}}\right )}}{3 \, x^{3}} \,d x } \]
integrate(1/3*(4*x*log(3)*exp(x)*log(4/x)+(4*x-24)*log(3)*exp(x)+(4*x^4-4* x^3)*log(3))*exp(1/3*(-4*x*log(3)*exp(x)*log(4/x)+(9*x^2+12)*log(3)*exp(x) -4*x^3*log(3))/exp(x)/x^2)/exp(x)/x^3,x, algorithm=\
integrate(4/3*(x*e^x*log(3)*log(4/x) + (x - 6)*e^x*log(3) + (x^4 - x^3)*lo g(3))*e^(-x - 1/3*(4*x^3*log(3) + 4*x*e^x*log(3)*log(4/x) - 3*(3*x^2 + 4)* e^x*log(3))*e^(-x)/x^2)/x^3, x)
Time = 7.92 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-4 x^3 \log (3)+e^x \left (12+9 x^2\right ) \log (3)-4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^2}} \left (e^x (-24+4 x) \log (3)+\left (-4 x^3+4 x^4\right ) \log (3)+4 e^x x \log (3) \log \left (\frac {4}{x}\right )\right )}{3 x^3} \, dx=\frac {27\,3^{\frac {4}{x^2}}\,{\mathrm {e}}^{-\frac {4\,\ln \left (\frac {1}{x}\right )\,\ln \left (3\right )}{3\,x}}}{2^{\frac {8\,\ln \left (3\right )}{3\,x}}\,3^{\frac {4\,x\,{\mathrm {e}}^{-x}}{3}}} \]