Integrand size = 133, antiderivative size = 29 \[ \int \frac {e^{\frac {2 (-x+e \log (x))}{e}} \left (-2 x^2+e \left (4 x+x^2\right )\right )+e \left (8 x+4 x^2\right ) \log (5)+\left (e^{\frac {2 (-x+e \log (x))}{e}} (-2 x+e (2+x))+4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (-x+e \log (x))}{e}}+4 \log (5)\right )\right )}{2 e^{1+\frac {2 (-x+e \log (x))}{e}} x+8 e x \log (5)} \, dx=x+\frac {1}{4} \left (x+\log \left (\frac {1}{4} e^{-\frac {2 x}{e}} x^2+\log (5)\right )\right )^2 \]
1/2*(x+ln(ln(5)+1/4*exp(ln(x)-x/exp(1))^2))*(1/2*x+1/2*ln(ln(5)+1/4*exp(ln (x)-x/exp(1))^2))+x
Leaf count is larger than twice the leaf count of optimal. \(114\) vs. \(2(29)=58\).
Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 3.93 \[ \int \frac {e^{\frac {2 (-x+e \log (x))}{e}} \left (-2 x^2+e \left (4 x+x^2\right )\right )+e \left (8 x+4 x^2\right ) \log (5)+\left (e^{\frac {2 (-x+e \log (x))}{e}} (-2 x+e (2+x))+4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (-x+e \log (x))}{e}}+4 \log (5)\right )\right )}{2 e^{1+\frac {2 (-x+e \log (x))}{e}} x+8 e x \log (5)} \, dx=\frac {-\frac {2 x^2}{e}+\frac {1}{2} e x (4+x)+2 x \log \left (x^2+4 e^{\frac {2 x}{e}} \log (5)\right )-\frac {1}{2} e \log ^2\left (x^2+4 e^{\frac {2 x}{e}} \log (5)\right )+\log \left (\frac {1}{4} e^{-\frac {2 x}{e}} x^2+\log (5)\right ) \left ((-2+e) x+e \log \left (x^2+4 e^{\frac {2 x}{e}} \log (5)\right )\right )}{2 e} \]
Integrate[(E^((2*(-x + E*Log[x]))/E)*(-2*x^2 + E*(4*x + x^2)) + E*(8*x + 4 *x^2)*Log[5] + (E^((2*(-x + E*Log[x]))/E)*(-2*x + E*(2 + x)) + 4*E*x*Log[5 ])*Log[(E^((2*(-x + E*Log[x]))/E) + 4*Log[5])/4])/(2*E^(1 + (2*(-x + E*Log [x]))/E)*x + 8*E*x*Log[5]),x]
((-2*x^2)/E + (E*x*(4 + x))/2 + 2*x*Log[x^2 + 4*E^((2*x)/E)*Log[5]] - (E*L og[x^2 + 4*E^((2*x)/E)*Log[5]]^2)/2 + Log[x^2/(4*E^((2*x)/E)) + Log[5]]*(( -2 + E)*x + E*Log[x^2 + 4*E^((2*x)/E)*Log[5]]))/(2*E)
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.
| \(\displaystyle \int \frac {e \left (4 x^2+8 x\right ) \log (5)+\left (e \left (x^2+4 x\right )-2 x^2\right ) e^{\frac {2 (e \log (x)-x)}{e}}+\left (4 e x \log (5)+(e (x+2)-2 x) e^{\frac {2 (e \log (x)-x)}{e}}\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (e \log (x)-x)}{e}}+4 \log (5)\right )\right )}{2 x e^{\frac {2 (e \log (x)-x)}{e}+1}+8 e x \log (5)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{\frac {2 x}{e}-1} \left (e \left (4 x^2+8 x\right ) \log (5)+\left (e \left (x^2+4 x\right )-2 x^2\right ) e^{\frac {2 (e \log (x)-x)}{e}}+\left (4 e x \log (5)+(e (x+2)-2 x) e^{\frac {2 (e \log (x)-x)}{e}}\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (e \log (x)-x)}{e}}+4 \log (5)\right )\right )\right )}{2 x \left (x^2+e^{\frac {2 x}{e}} \log (625)\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int -\frac {e^{\frac {2 x}{e}-1} \left (e^{-\frac {2 x}{e}} \left (2 x^2-e \left (x^2+4 x\right )\right ) x^2+\left (e^{-\frac {2 x}{e}} x^2 (2 x-e (x+2))-4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{-\frac {2 x}{e}} x^2+\log (625)\right )\right )-4 e \left (x^2+2 x\right ) \log (5)\right )}{x \left (x^2+e^{\frac {2 x}{e}} \log (625)\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {e^{\frac {2 x}{e}-1} \left (e^{-\frac {2 x}{e}} \left (2 x^2-e \left (x^2+4 x\right )\right ) x^2+\left (e^{-\frac {2 x}{e}} x^2 (2 x-e (x+2))-4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{-\frac {2 x}{e}} x^2+\log (625)\right )\right )-4 e \left (x^2+2 x\right ) \log (5)\right )}{x \left (x^2+e^{\frac {2 x}{e}} \log (625)\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{2} \int \left (\frac {2 \left (1-\frac {e}{2}\right ) x^2+2 \left (1-\frac {e}{2}\right ) \log \left (\frac {1}{4} e^{-\frac {2 x}{e}} x^2+\log (5)\right ) x-4 e x-2 e \log \left (\frac {1}{4} e^{-\frac {2 x}{e}} x^2+\log (5)\right )}{e x}+\frac {e^{\frac {2 x}{e}-1} (e-x) \log (390625) \left (x+\log \left (e^{-\frac {2 x}{e}} x^2+\log (625)\right )-\log (4)\right )}{x \left (x^2+e^{\frac {2 x}{e}} \log (625)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {2 (2-e) \int \frac {x^2}{x^2+4 e^{\frac {2 x}{e}} \log (5)}dx}{e}-\log (390625) \int \frac {e^{\frac {2 x}{e}}}{x^2+e^{\frac {2 x}{e}} \log (625)}dx+\log (390625) \log \left (e^{-\frac {2 x}{e}} x^2+\log (625)\right ) \int \frac {e^{\frac {2 x}{e}-1}}{x^2+e^{\frac {2 x}{e}} \log (625)}dx-\log (4) \log (390625) \int \frac {e^{\frac {2 x}{e}-1}}{x^2+e^{\frac {2 x}{e}} \log (625)}dx-\log (390625) \log \left (e^{-\frac {2 x}{e}} x^2+\log (625)\right ) \int \frac {e^{\frac {2 x}{e}}}{x \left (x^2+e^{\frac {2 x}{e}} \log (625)\right )}dx+\log (4) \log (390625) \int \frac {e^{\frac {2 x}{e}}}{x \left (x^2+e^{\frac {2 x}{e}} \log (625)\right )}dx+\log (390625) \int \frac {e^{\frac {2 x}{e}-1} x}{x^2+e^{\frac {2 x}{e}} \log (625)}dx+2 \int \frac {\log \left (\frac {1}{4} e^{-\frac {2 x}{e}} x^2+\log (5)\right )}{x}dx-2 \log (390625) \int \frac {x \int \frac {e^{\frac {2 x}{e}-1}}{x^2+e^{\frac {2 x}{e}} \log (625)}dx}{x^2+e^{\frac {2 x}{e}} \log (625)}dx+\frac {2 \log (390625) \int \frac {x^2 \int \frac {e^{\frac {2 x}{e}-1}}{x^2+e^{\frac {2 x}{e}} \log (625)}dx}{x^2+e^{\frac {2 x}{e}} \log (625)}dx}{e}-\frac {2 (2-e) \int \frac {x^3}{x^2+4 e^{\frac {2 x}{e}} \log (5)}dx}{e^2}+2 \log (390625) \int \frac {x \int \frac {e^{\frac {2 x}{e}}}{x^3+e^{\frac {2 x}{e}} \log (625) x}dx}{x^2+e^{\frac {2 x}{e}} \log (625)}dx-\frac {2 \log (390625) \int \frac {x^2 \int \frac {e^{\frac {2 x}{e}}}{x^3+e^{\frac {2 x}{e}} \log (625) x}dx}{x^2+e^{\frac {2 x}{e}} \log (625)}dx}{e}-\frac {(2-e) x^2}{2 e}-\frac {(2-e) x \log \left (\frac {1}{4} e^{-\frac {2 x}{e}} x^2+\log (5)\right )}{e}+4 x\right )\) |
Int[(E^((2*(-x + E*Log[x]))/E)*(-2*x^2 + E*(4*x + x^2)) + E*(8*x + 4*x^2)* Log[5] + (E^((2*(-x + E*Log[x]))/E)*(-2*x + E*(2 + x)) + 4*E*x*Log[5])*Log [(E^((2*(-x + E*Log[x]))/E) + 4*Log[5])/4])/(2*E^(1 + (2*(-x + E*Log[x]))/ E)*x + 8*E*x*Log[5]),x]
3.19.100.3.1 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 = 4.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55
| method | result | size |
| risch | \(\frac {\ln \left (\frac {x^{2} {\mathrm e}^{-2 \,{\mathrm e}^{-1} x}}{4}+\ln \left (5\right )\right )^{2}}{4}+\frac {x \ln \left (\frac {x^{2} {\mathrm e}^{-2 \,{\mathrm e}^{-1} x}}{4}+\ln \left (5\right )\right )}{2}+\frac {x^{2}}{4}+x\) | \(45\) |
| parallelrisch | \(\frac {\left (x^{2} {\mathrm e}+2 \,{\mathrm e} x \ln \left (\frac {{\mathrm e}^{2 \left ({\mathrm e} \ln \left (x \right )-x \right ) {\mathrm e}^{-1}}}{4}+\ln \left (5\right )\right )+{\mathrm e} \ln \left (\frac {{\mathrm e}^{2 \left ({\mathrm e} \ln \left (x \right )-x \right ) {\mathrm e}^{-1}}}{4}+\ln \left (5\right )\right )^{2}+4 x \,{\mathrm e}\right ) {\mathrm e}^{-1}}{4}\) | \(75\) |
int(((((2+x)*exp(1)-2*x)*exp((exp(1)*ln(x)-x)/exp(1))^2+4*x*exp(1)*ln(5))* ln(1/4*exp((exp(1)*ln(x)-x)/exp(1))^2+ln(5))+((x^2+4*x)*exp(1)-2*x^2)*exp( (exp(1)*ln(x)-x)/exp(1))^2+(4*x^2+8*x)*exp(1)*ln(5))/(2*x*exp(1)*exp((exp( 1)*ln(x)-x)/exp(1))^2+8*x*exp(1)*ln(5)),x,method=_RETURNVERBOSE)
1/4*ln(1/4*x^2*exp(-2*exp(-1)*x)+ln(5))^2+1/2*x*ln(1/4*x^2*exp(-2*exp(-1)* x)+ln(5))+1/4*x^2+x
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.41 \[ \int \frac {e^{\frac {2 (-x+e \log (x))}{e}} \left (-2 x^2+e \left (4 x+x^2\right )\right )+e \left (8 x+4 x^2\right ) \log (5)+\left (e^{\frac {2 (-x+e \log (x))}{e}} (-2 x+e (2+x))+4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (-x+e \log (x))}{e}}+4 \log (5)\right )\right )}{2 e^{1+\frac {2 (-x+e \log (x))}{e}} x+8 e x \log (5)} \, dx=\frac {1}{4} \, x^{2} + \frac {1}{2} \, x \log \left (\frac {1}{4} \, {\left (4 \, e \log \left (5\right ) + e^{\left ({\left (2 \, e \log \left (x\right ) - 2 \, x + e\right )} e^{\left (-1\right )}\right )}\right )} e^{\left (-1\right )}\right ) + \frac {1}{4} \, \log \left (\frac {1}{4} \, {\left (4 \, e \log \left (5\right ) + e^{\left ({\left (2 \, e \log \left (x\right ) - 2 \, x + e\right )} e^{\left (-1\right )}\right )}\right )} e^{\left (-1\right )}\right )^{2} + x \]
integrate(((((2+x)*exp(1)-2*x)*exp((exp(1)*log(x)-x)/exp(1))^2+4*x*exp(1)* log(5))*log(1/4*exp((exp(1)*log(x)-x)/exp(1))^2+log(5))+((x^2+4*x)*exp(1)- 2*x^2)*exp((exp(1)*log(x)-x)/exp(1))^2+(4*x^2+8*x)*exp(1)*log(5))/(2*x*exp (1)*exp((exp(1)*log(x)-x)/exp(1))^2+8*x*exp(1)*log(5)),x, algorithm=\
1/4*x^2 + 1/2*x*log(1/4*(4*e*log(5) + e^((2*e*log(x) - 2*x + e)*e^(-1)))*e ^(-1)) + 1/4*log(1/4*(4*e*log(5) + e^((2*e*log(x) - 2*x + e)*e^(-1)))*e^(- 1))^2 + x
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (26) = 52\).
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.07 \[ \int \frac {e^{\frac {2 (-x+e \log (x))}{e}} \left (-2 x^2+e \left (4 x+x^2\right )\right )+e \left (8 x+4 x^2\right ) \log (5)+\left (e^{\frac {2 (-x+e \log (x))}{e}} (-2 x+e (2+x))+4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (-x+e \log (x))}{e}}+4 \log (5)\right )\right )}{2 e^{1+\frac {2 (-x+e \log (x))}{e}} x+8 e x \log (5)} \, dx=\frac {x^{2}}{4} + \frac {x \log {\left (\frac {e^{\frac {2 \left (- x + e \log {\left (x \right )}\right )}{e}}}{4} + \log {\left (5 \right )} \right )}}{2} + x + \frac {\log {\left (\frac {e^{\frac {2 \left (- x + e \log {\left (x \right )}\right )}{e}}}{4} + \log {\left (5 \right )} \right )}^{2}}{4} \]
integrate(((((2+x)*exp(1)-2*x)*exp((exp(1)*ln(x)-x)/exp(1))**2+4*x*exp(1)* ln(5))*ln(1/4*exp((exp(1)*ln(x)-x)/exp(1))**2+ln(5))+((x**2+4*x)*exp(1)-2* x**2)*exp((exp(1)*ln(x)-x)/exp(1))**2+(4*x**2+8*x)*exp(1)*ln(5))/(2*x*exp( 1)*exp((exp(1)*ln(x)-x)/exp(1))**2+8*x*exp(1)*ln(5)),x)
x**2/4 + x*log(exp(2*(-x + E*log(x))*exp(-1))/4 + log(5))/2 + x + log(exp( 2*(-x + E*log(x))*exp(-1))/4 + log(5))**2/4
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (25) = 50\).
Time = 0.41 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.00 \[ \int \frac {e^{\frac {2 (-x+e \log (x))}{e}} \left (-2 x^2+e \left (4 x+x^2\right )\right )+e \left (8 x+4 x^2\right ) \log (5)+\left (e^{\frac {2 (-x+e \log (x))}{e}} (-2 x+e (2+x))+4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (-x+e \log (x))}{e}}+4 \log (5)\right )\right )}{2 e^{1+\frac {2 (-x+e \log (x))}{e}} x+8 e x \log (5)} \, dx=\frac {1}{4} \, {\left (x^{2} {\left (e^{2} - 4 \, e + 4\right )} + e^{2} \log \left (x^{2} + 4 \, e^{\left (2 \, x e^{\left (-1\right )}\right )} \log \left (5\right )\right )^{2} - 4 \, {\left ({\left (\log \left (2\right ) - 1\right )} e^{2} - 2 \, e \log \left (2\right )\right )} x + 2 \, {\left (x {\left (e^{2} - 2 \, e\right )} - 2 \, e^{2} \log \left (2\right )\right )} \log \left (x^{2} + 4 \, e^{\left (2 \, x e^{\left (-1\right )}\right )} \log \left (5\right )\right )\right )} e^{\left (-2\right )} \]
integrate(((((2+x)*exp(1)-2*x)*exp((exp(1)*log(x)-x)/exp(1))^2+4*x*exp(1)* log(5))*log(1/4*exp((exp(1)*log(x)-x)/exp(1))^2+log(5))+((x^2+4*x)*exp(1)- 2*x^2)*exp((exp(1)*log(x)-x)/exp(1))^2+(4*x^2+8*x)*exp(1)*log(5))/(2*x*exp (1)*exp((exp(1)*log(x)-x)/exp(1))^2+8*x*exp(1)*log(5)),x, algorithm=\
1/4*(x^2*(e^2 - 4*e + 4) + e^2*log(x^2 + 4*e^(2*x*e^(-1))*log(5))^2 - 4*(( log(2) - 1)*e^2 - 2*e*log(2))*x + 2*(x*(e^2 - 2*e) - 2*e^2*log(2))*log(x^2 + 4*e^(2*x*e^(-1))*log(5)))*e^(-2)
\[ \int \frac {e^{\frac {2 (-x+e \log (x))}{e}} \left (-2 x^2+e \left (4 x+x^2\right )\right )+e \left (8 x+4 x^2\right ) \log (5)+\left (e^{\frac {2 (-x+e \log (x))}{e}} (-2 x+e (2+x))+4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (-x+e \log (x))}{e}}+4 \log (5)\right )\right )}{2 e^{1+\frac {2 (-x+e \log (x))}{e}} x+8 e x \log (5)} \, dx=\int { \frac {4 \, {\left (x^{2} + 2 \, x\right )} e \log \left (5\right ) - {\left (2 \, x^{2} - {\left (x^{2} + 4 \, x\right )} e\right )} e^{\left (2 \, {\left (e \log \left (x\right ) - x\right )} e^{\left (-1\right )}\right )} + {\left (4 \, x e \log \left (5\right ) + {\left ({\left (x + 2\right )} e - 2 \, x\right )} e^{\left (2 \, {\left (e \log \left (x\right ) - x\right )} e^{\left (-1\right )}\right )}\right )} \log \left (\frac {1}{4} \, e^{\left (2 \, {\left (e \log \left (x\right ) - x\right )} e^{\left (-1\right )}\right )} + \log \left (5\right )\right )}{2 \, {\left (4 \, x e \log \left (5\right ) + x e^{\left (2 \, {\left (e \log \left (x\right ) - x\right )} e^{\left (-1\right )} + 1\right )}\right )}} \,d x } \]
integrate(((((2+x)*exp(1)-2*x)*exp((exp(1)*log(x)-x)/exp(1))^2+4*x*exp(1)* log(5))*log(1/4*exp((exp(1)*log(x)-x)/exp(1))^2+log(5))+((x^2+4*x)*exp(1)- 2*x^2)*exp((exp(1)*log(x)-x)/exp(1))^2+(4*x^2+8*x)*exp(1)*log(5))/(2*x*exp (1)*exp((exp(1)*log(x)-x)/exp(1))^2+8*x*exp(1)*log(5)),x, algorithm=\
integrate(1/2*(4*(x^2 + 2*x)*e*log(5) - (2*x^2 - (x^2 + 4*x)*e)*e^(2*(e*lo g(x) - x)*e^(-1)) + (4*x*e*log(5) + ((x + 2)*e - 2*x)*e^(2*(e*log(x) - x)* e^(-1)))*log(1/4*e^(2*(e*log(x) - x)*e^(-1)) + log(5)))/(4*x*e*log(5) + x* e^(2*(e*log(x) - x)*e^(-1) + 1)), x)
Timed out. \[ \int \frac {e^{\frac {2 (-x+e \log (x))}{e}} \left (-2 x^2+e \left (4 x+x^2\right )\right )+e \left (8 x+4 x^2\right ) \log (5)+\left (e^{\frac {2 (-x+e \log (x))}{e}} (-2 x+e (2+x))+4 e x \log (5)\right ) \log \left (\frac {1}{4} \left (e^{\frac {2 (-x+e \log (x))}{e}}+4 \log (5)\right )\right )}{2 e^{1+\frac {2 (-x+e \log (x))}{e}} x+8 e x \log (5)} \, dx=\int \frac {{\mathrm {e}}^{-2\,{\mathrm {e}}^{-1}\,\left (x-\mathrm {e}\,\ln \left (x\right )\right )}\,\left (\mathrm {e}\,\left (x^2+4\,x\right )-2\,x^2\right )-\ln \left (\ln \left (5\right )+\frac {{\mathrm {e}}^{-2\,{\mathrm {e}}^{-1}\,\left (x-\mathrm {e}\,\ln \left (x\right )\right )}}{4}\right )\,\left ({\mathrm {e}}^{-2\,{\mathrm {e}}^{-1}\,\left (x-\mathrm {e}\,\ln \left (x\right )\right )}\,\left (2\,x-\mathrm {e}\,\left (x+2\right )\right )-4\,x\,\mathrm {e}\,\ln \left (5\right )\right )+\mathrm {e}\,\ln \left (5\right )\,\left (4\,x^2+8\,x\right )}{8\,x\,\mathrm {e}\,\ln \left (5\right )+2\,x\,\mathrm {e}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{-1}\,\left (x-\mathrm {e}\,\ln \left (x\right )\right )}} \,d x \]
int((exp(-2*exp(-1)*(x - exp(1)*log(x)))*(exp(1)*(4*x + x^2) - 2*x^2) - lo g(log(5) + exp(-2*exp(-1)*(x - exp(1)*log(x)))/4)*(exp(-2*exp(-1)*(x - exp (1)*log(x)))*(2*x - exp(1)*(x + 2)) - 4*x*exp(1)*log(5)) + exp(1)*log(5)*( 8*x + 4*x^2))/(8*x*exp(1)*log(5) + 2*x*exp(1)*exp(-2*exp(-1)*(x - exp(1)*l og(x)))),x)
int((exp(-2*exp(-1)*(x - exp(1)*log(x)))*(exp(1)*(4*x + x^2) - 2*x^2) - lo g(log(5) + exp(-2*exp(-1)*(x - exp(1)*log(x)))/4)*(exp(-2*exp(-1)*(x - exp (1)*log(x)))*(2*x - exp(1)*(x + 2)) - 4*x*exp(1)*log(5)) + exp(1)*log(5)*( 8*x + 4*x^2))/(8*x*exp(1)*log(5) + 2*x*exp(1)*exp(-2*exp(-1)*(x - exp(1)*l og(x)))), x)