Integrand size = 203, antiderivative size = 30 \[ \int \frac {-e^{2 x}+3 x-3 x^2-3 e^x \log (4)+\left (e (3-6 x)-3 x+6 x^2+e^{2 x} (-2 e+2 x)+e^x (-3 e+3 x) \log (4)\right ) \log (e-x)}{e^{4 x} (e-x)-9 x^3+18 x^4-9 x^5+e \left (9 x^2-18 x^3+9 x^4\right )+e^{3 x} (6 e-6 x) \log (4)+e^x \left (18 x^2-18 x^3+e \left (-18 x+18 x^2\right )\right ) \log (4)+e^{2 x} \left (6 x^2-6 x^3+e \left (-6 x+6 x^2\right )+(9 e-9 x) \log ^2(4)\right )} \, dx=\frac {\log (e-x)}{e^{2 x}+3 \left (-x+x^2+e^x \log (4)\right )} \]
Time = 0.96 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-e^{2 x}+3 x-3 x^2-3 e^x \log (4)+\left (e (3-6 x)-3 x+6 x^2+e^{2 x} (-2 e+2 x)+e^x (-3 e+3 x) \log (4)\right ) \log (e-x)}{e^{4 x} (e-x)-9 x^3+18 x^4-9 x^5+e \left (9 x^2-18 x^3+9 x^4\right )+e^{3 x} (6 e-6 x) \log (4)+e^x \left (18 x^2-18 x^3+e \left (-18 x+18 x^2\right )\right ) \log (4)+e^{2 x} \left (6 x^2-6 x^3+e \left (-6 x+6 x^2\right )+(9 e-9 x) \log ^2(4)\right )} \, dx=\frac {\log (e-x)}{e^{2 x}-3 x+3 x^2+e^x \log (64)} \]
Integrate[(-E^(2*x) + 3*x - 3*x^2 - 3*E^x*Log[4] + (E*(3 - 6*x) - 3*x + 6* x^2 + E^(2*x)*(-2*E + 2*x) + E^x*(-3*E + 3*x)*Log[4])*Log[E - x])/(E^(4*x) *(E - x) - 9*x^3 + 18*x^4 - 9*x^5 + E*(9*x^2 - 18*x^3 + 9*x^4) + E^(3*x)*( 6*E - 6*x)*Log[4] + E^x*(18*x^2 - 18*x^3 + E*(-18*x + 18*x^2))*Log[4] + E^ (2*x)*(6*x^2 - 6*x^3 + E*(-6*x + 6*x^2) + (9*E - 9*x)*Log[4]^2)),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.
\(\displaystyle \int \frac {-3 x^2+\left (6 x^2-3 x+e (3-6 x)+e^{2 x} (2 x-2 e)+e^x (3 x-3 e) \log (4)\right ) \log (e-x)+3 x-e^{2 x}-3 e^x \log (4)}{-9 x^5+18 x^4-9 x^3+e^{2 x} \left (-6 x^3+6 x^2+e \left (6 x^2-6 x\right )+(9 e-9 x) \log ^2(4)\right )+e^x \left (-18 x^3+18 x^2+e \left (18 x^2-18 x\right )\right ) \log (4)+e \left (9 x^4-18 x^3+9 x^2\right )+e^{4 x} (e-x)+e^{3 x} (6 e-6 x) \log (4)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-3 (x-1) x-e^{2 x}-(e-x) \left (6 x+2 e^{2 x}+e^x \log (64)-3\right ) \log (e-x)-e^x \log (64)}{(e-x) \left (3 (x-1) x+e^{2 x}+e^x \log (64)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (6 x^2-12 x+e^x \log (64)+3\right ) \log (e-x)}{\left (3 x^2-3 x+e^{2 x}+e^x \log (64)\right )^2}-\frac {-2 x \log (e-x)+2 e \log (e-x)+1}{(e-x) \left (3 x^2-3 x+e^{2 x}+e^x \log (64)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \log (e-x) \int \frac {1}{\left (3 x^2-3 x+e^{2 x}+e^x \log (64)\right )^2}dx+\log (64) \log (e-x) \int \frac {e^x}{\left (3 x^2-3 x+e^{2 x}+e^x \log (64)\right )^2}dx-12 \log (e-x) \int \frac {x}{\left (3 x^2-3 x+e^{2 x}+e^x \log (64)\right )^2}dx+6 \log (e-x) \int \frac {x^2}{\left (3 x^2-3 x+e^{2 x}+e^x \log (64)\right )^2}dx-2 \log (e-x) \int \frac {1}{3 x^2-3 x+e^{2 x}+e^x \log (64)}dx-\int \frac {1}{(e-x) \left (3 x^2-3 x+e^{2 x}+e^x \log (64)\right )}dx+6 \int \frac {\int \frac {x^2}{\left (3 (x-1) x+e^{2 x}+e^x \log (64)\right )^2}dx}{e-x}dx+3 \int \frac {\int \frac {1}{\left (3 (x-1) x+e^{2 x}+e^x \log (64)\right )^2}dx}{e-x}dx+\log (64) \int \frac {\int \frac {e^x}{\left (3 (x-1) x+e^{2 x}+e^x \log (64)\right )^2}dx}{e-x}dx-12 \int \frac {\int \frac {x}{\left (3 (x-1) x+e^{2 x}+e^x \log (64)\right )^2}dx}{e-x}dx-2 \int \frac {\int \frac {1}{3 (x-1) x+e^{2 x}+e^x \log (64)}dx}{e-x}dx+2 e \int \frac {\int \frac {1}{(e-x) \left (3 (x-1) x+e^{2 x}+e^x \log (64)\right )}dx}{e-x}dx+2 e \int \frac {\int \frac {1}{(e-x) \left (3 (x-1) x+e^{2 x}+e^x \log (64)\right )}dx}{x-e}dx\) |
Int[(-E^(2*x) + 3*x - 3*x^2 - 3*E^x*Log[4] + (E*(3 - 6*x) - 3*x + 6*x^2 + E^(2*x)*(-2*E + 2*x) + E^x*(-3*E + 3*x)*Log[4])*Log[E - x])/(E^(4*x)*(E - x) - 9*x^3 + 18*x^4 - 9*x^5 + E*(9*x^2 - 18*x^3 + 9*x^4) + E^(3*x)*(6*E - 6*x)*Log[4] + E^x*(18*x^2 - 18*x^3 + E*(-18*x + 18*x^2))*Log[4] + E^(2*x)* (6*x^2 - 6*x^3 + E*(-6*x + 6*x^2) + (9*E - 9*x)*Log[4]^2)),x]
3.11.65.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.94 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\ln \left ({\mathrm e}-x \right )}{{\mathrm e}^{2 x}+3 x^{2}+6 \,{\mathrm e}^{x} \ln \left (2\right )-3 x}\) | \(30\) |
parallelrisch | \(\frac {\ln \left ({\mathrm e}-x \right )}{{\mathrm e}^{2 x}+3 x^{2}+6 \,{\mathrm e}^{x} \ln \left (2\right )-3 x}\) | \(30\) |
int((((-2*exp(1)+2*x)*exp(x)^2+2*(-3*exp(1)+3*x)*ln(2)*exp(x)+(-6*x+3)*exp (1)+6*x^2-3*x)*ln(exp(1)-x)-exp(x)^2-6*exp(x)*ln(2)-3*x^2+3*x)/((exp(1)-x) *exp(x)^4+2*(6*exp(1)-6*x)*ln(2)*exp(x)^3+(4*(9*exp(1)-9*x)*ln(2)^2+(6*x^2 -6*x)*exp(1)-6*x^3+6*x^2)*exp(x)^2+2*((18*x^2-18*x)*exp(1)-18*x^3+18*x^2)* ln(2)*exp(x)+(9*x^4-18*x^3+9*x^2)*exp(1)-9*x^5+18*x^4-9*x^3),x,method=_RET URNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-e^{2 x}+3 x-3 x^2-3 e^x \log (4)+\left (e (3-6 x)-3 x+6 x^2+e^{2 x} (-2 e+2 x)+e^x (-3 e+3 x) \log (4)\right ) \log (e-x)}{e^{4 x} (e-x)-9 x^3+18 x^4-9 x^5+e \left (9 x^2-18 x^3+9 x^4\right )+e^{3 x} (6 e-6 x) \log (4)+e^x \left (18 x^2-18 x^3+e \left (-18 x+18 x^2\right )\right ) \log (4)+e^{2 x} \left (6 x^2-6 x^3+e \left (-6 x+6 x^2\right )+(9 e-9 x) \log ^2(4)\right )} \, dx=\frac {\log \left (-x + e\right )}{3 \, x^{2} + 6 \, e^{x} \log \left (2\right ) - 3 \, x + e^{\left (2 \, x\right )}} \]
integrate((((-2*exp(1)+2*x)*exp(x)^2+2*(-3*exp(1)+3*x)*log(2)*exp(x)+(-6*x +3)*exp(1)+6*x^2-3*x)*log(exp(1)-x)-exp(x)^2-6*exp(x)*log(2)-3*x^2+3*x)/(( exp(1)-x)*exp(x)^4+2*(6*exp(1)-6*x)*log(2)*exp(x)^3+(4*(9*exp(1)-9*x)*log( 2)^2+(6*x^2-6*x)*exp(1)-6*x^3+6*x^2)*exp(x)^2+2*((18*x^2-18*x)*exp(1)-18*x ^3+18*x^2)*log(2)*exp(x)+(9*x^4-18*x^3+9*x^2)*exp(1)-9*x^5+18*x^4-9*x^3),x , algorithm=\
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-e^{2 x}+3 x-3 x^2-3 e^x \log (4)+\left (e (3-6 x)-3 x+6 x^2+e^{2 x} (-2 e+2 x)+e^x (-3 e+3 x) \log (4)\right ) \log (e-x)}{e^{4 x} (e-x)-9 x^3+18 x^4-9 x^5+e \left (9 x^2-18 x^3+9 x^4\right )+e^{3 x} (6 e-6 x) \log (4)+e^x \left (18 x^2-18 x^3+e \left (-18 x+18 x^2\right )\right ) \log (4)+e^{2 x} \left (6 x^2-6 x^3+e \left (-6 x+6 x^2\right )+(9 e-9 x) \log ^2(4)\right )} \, dx=\frac {\log {\left (e - x \right )}}{3 x^{2} - 3 x + e^{2 x} + 6 e^{x} \log {\left (2 \right )}} \]
integrate((((-2*exp(1)+2*x)*exp(x)**2+2*(-3*exp(1)+3*x)*ln(2)*exp(x)+(-6*x +3)*exp(1)+6*x**2-3*x)*ln(exp(1)-x)-exp(x)**2-6*exp(x)*ln(2)-3*x**2+3*x)/( (exp(1)-x)*exp(x)**4+2*(6*exp(1)-6*x)*ln(2)*exp(x)**3+(4*(9*exp(1)-9*x)*ln (2)**2+(6*x**2-6*x)*exp(1)-6*x**3+6*x**2)*exp(x)**2+2*((18*x**2-18*x)*exp( 1)-18*x**3+18*x**2)*ln(2)*exp(x)+(9*x**4-18*x**3+9*x**2)*exp(1)-9*x**5+18* x**4-9*x**3),x)
Time = 0.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-e^{2 x}+3 x-3 x^2-3 e^x \log (4)+\left (e (3-6 x)-3 x+6 x^2+e^{2 x} (-2 e+2 x)+e^x (-3 e+3 x) \log (4)\right ) \log (e-x)}{e^{4 x} (e-x)-9 x^3+18 x^4-9 x^5+e \left (9 x^2-18 x^3+9 x^4\right )+e^{3 x} (6 e-6 x) \log (4)+e^x \left (18 x^2-18 x^3+e \left (-18 x+18 x^2\right )\right ) \log (4)+e^{2 x} \left (6 x^2-6 x^3+e \left (-6 x+6 x^2\right )+(9 e-9 x) \log ^2(4)\right )} \, dx=\frac {\log \left (-x + e\right )}{3 \, x^{2} + 6 \, e^{x} \log \left (2\right ) - 3 \, x + e^{\left (2 \, x\right )}} \]
integrate((((-2*exp(1)+2*x)*exp(x)^2+2*(-3*exp(1)+3*x)*log(2)*exp(x)+(-6*x +3)*exp(1)+6*x^2-3*x)*log(exp(1)-x)-exp(x)^2-6*exp(x)*log(2)-3*x^2+3*x)/(( exp(1)-x)*exp(x)^4+2*(6*exp(1)-6*x)*log(2)*exp(x)^3+(4*(9*exp(1)-9*x)*log( 2)^2+(6*x^2-6*x)*exp(1)-6*x^3+6*x^2)*exp(x)^2+2*((18*x^2-18*x)*exp(1)-18*x ^3+18*x^2)*log(2)*exp(x)+(9*x^4-18*x^3+9*x^2)*exp(1)-9*x^5+18*x^4-9*x^3),x , algorithm=\
Time = 0.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-e^{2 x}+3 x-3 x^2-3 e^x \log (4)+\left (e (3-6 x)-3 x+6 x^2+e^{2 x} (-2 e+2 x)+e^x (-3 e+3 x) \log (4)\right ) \log (e-x)}{e^{4 x} (e-x)-9 x^3+18 x^4-9 x^5+e \left (9 x^2-18 x^3+9 x^4\right )+e^{3 x} (6 e-6 x) \log (4)+e^x \left (18 x^2-18 x^3+e \left (-18 x+18 x^2\right )\right ) \log (4)+e^{2 x} \left (6 x^2-6 x^3+e \left (-6 x+6 x^2\right )+(9 e-9 x) \log ^2(4)\right )} \, dx=\frac {\log \left (-x + e\right )}{3 \, x^{2} + 6 \, e^{x} \log \left (2\right ) - 3 \, x + e^{\left (2 \, x\right )}} \]
integrate((((-2*exp(1)+2*x)*exp(x)^2+2*(-3*exp(1)+3*x)*log(2)*exp(x)+(-6*x +3)*exp(1)+6*x^2-3*x)*log(exp(1)-x)-exp(x)^2-6*exp(x)*log(2)-3*x^2+3*x)/(( exp(1)-x)*exp(x)^4+2*(6*exp(1)-6*x)*log(2)*exp(x)^3+(4*(9*exp(1)-9*x)*log( 2)^2+(6*x^2-6*x)*exp(1)-6*x^3+6*x^2)*exp(x)^2+2*((18*x^2-18*x)*exp(1)-18*x ^3+18*x^2)*log(2)*exp(x)+(9*x^4-18*x^3+9*x^2)*exp(1)-9*x^5+18*x^4-9*x^3),x , algorithm=\
Time = 14.51 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-e^{2 x}+3 x-3 x^2-3 e^x \log (4)+\left (e (3-6 x)-3 x+6 x^2+e^{2 x} (-2 e+2 x)+e^x (-3 e+3 x) \log (4)\right ) \log (e-x)}{e^{4 x} (e-x)-9 x^3+18 x^4-9 x^5+e \left (9 x^2-18 x^3+9 x^4\right )+e^{3 x} (6 e-6 x) \log (4)+e^x \left (18 x^2-18 x^3+e \left (-18 x+18 x^2\right )\right ) \log (4)+e^{2 x} \left (6 x^2-6 x^3+e \left (-6 x+6 x^2\right )+(9 e-9 x) \log ^2(4)\right )} \, dx=\frac {\ln \left (\mathrm {e}-x\right )}{{\mathrm {e}}^{2\,x}-3\,x+6\,{\mathrm {e}}^x\,\ln \left (2\right )+3\,x^2} \]
int((exp(2*x) - 3*x - log(exp(1) - x)*(exp(2*x)*(2*x - 2*exp(1)) - 3*x + 6 *x^2 - exp(1)*(6*x - 3) + 2*exp(x)*log(2)*(3*x - 3*exp(1))) + 6*exp(x)*log (2) + 3*x^2)/(exp(4*x)*(x - exp(1)) + exp(2*x)*(4*log(2)^2*(9*x - 9*exp(1) ) + exp(1)*(6*x - 6*x^2) - 6*x^2 + 6*x^3) - exp(1)*(9*x^2 - 18*x^3 + 9*x^4 ) + 9*x^3 - 18*x^4 + 9*x^5 + 2*exp(3*x)*log(2)*(6*x - 6*exp(1)) + 2*exp(x) *log(2)*(exp(1)*(18*x - 18*x^2) - 18*x^2 + 18*x^3)),x)