Integrand size = 132, antiderivative size = 28 \[ \int \frac {e^{-3+e^{e^x}} \left (e^x \left (-4 x-4 x^2\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{-6+2 e^{e^x}} \left (-8 \log (4)+4 e^x x \log (4)+e^x \left (-4 x-4 x^2\right ) \log (4) \log (x)\right )}{x+2 e^{-3+e^{e^x}} x \log (x)+e^{-6+2 e^{e^x}} x \log ^2(x)} \, dx=\frac {4 \left (2-e^x x\right ) \log (4)}{e^{3-e^{e^x}}+\log (x)} \]
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-3+e^{e^x}} \left (e^x \left (-4 x-4 x^2\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{-6+2 e^{e^x}} \left (-8 \log (4)+4 e^x x \log (4)+e^x \left (-4 x-4 x^2\right ) \log (4) \log (x)\right )}{x+2 e^{-3+e^{e^x}} x \log (x)+e^{-6+2 e^{e^x}} x \log ^2(x)} \, dx=-\frac {4 e^{e^{e^x}} \left (-2+e^x x\right ) \log (4)}{e^3+e^{e^{e^x}} \log (x)} \]
Integrate[(E^(-3 + E^E^x)*(E^x*(-4*x - 4*x^2)*Log[4] + E^E^x*(8*E^x*x*Log[ 4] - 4*E^(2*x)*x^2*Log[4])) + E^(-6 + 2*E^E^x)*(-8*Log[4] + 4*E^x*x*Log[4] + E^x*(-4*x - 4*x^2)*Log[4]*Log[x]))/(x + 2*E^(-3 + E^E^x)*x*Log[x] + E^( -6 + 2*E^E^x)*x*Log[x]^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.
| \(\displaystyle \int \frac {e^{e^{e^x}-3} \left (e^x \left (-4 x^2-4 x\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{2 e^{e^x}-6} \left (e^x \left (-4 x^2-4 x\right ) \log (4) \log (x)+4 e^x x \log (4)-8 \log (4)\right )}{x+e^{2 e^{e^x}-6} x \log ^2(x)+2 e^{e^{e^x}-3} x \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^6 \left (e^{e^{e^x}-3} \left (e^x \left (-4 x^2-4 x\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{2 e^{e^x}-6} \left (e^x \left (-4 x^2-4 x\right ) \log (4) \log (x)+4 e^x x \log (4)-8 \log (4)\right )\right )}{x \left (e^{e^{e^x}} \log (x)+e^3\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^6 \int -\frac {4 \left (e^{-3+e^{e^x}} \left (e^x \left (x^2+x\right ) \log (4)-e^{e^x} \left (2 e^x x \log (4)-e^{2 x} x^2 \log (4)\right )\right )+e^{-6+2 e^{e^x}} \left (-e^x \log (4) x+e^x \left (x^2+x\right ) \log (4) \log (x)+2 \log (4)\right )\right )}{x \left (e^{e^{e^x}} \log (x)+e^3\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -4 e^6 \int \frac {e^{-3+e^{e^x}} \left (e^x \left (x^2+x\right ) \log (4)-e^{e^x} \left (2 e^x x \log (4)-e^{2 x} x^2 \log (4)\right )\right )-e^{-6+2 e^{e^x}} \left (e^x \log (4) x-e^x \left (x^2+x\right ) \log (4) \log (x)-\log (16)\right )}{x \left (e^{e^{e^x}} \log (x)+e^3\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -4 e^6 \int \left (\frac {e^{2 x+e^{e^x}+e^x-3} \log (4) x}{\left (e^{e^{e^x}} \log (x)+e^3\right )^2}+\frac {e^{x+e^{e^x}-6} \log (4) \left (e^{e^{e^x}} \log (x) x+e^3 x-e^{e^{e^x}}-2 e^{3+e^x}+e^{e^{e^x}} \log (x)+e^3\right )}{\left (e^{e^{e^x}} \log (x)+e^3\right )^2}+\frac {e^{2 \left (-3+e^{e^x}\right )} \log (16)}{\left (e^{e^{e^x}} \log (x)+e^3\right )^2 x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 e^6 \left (-2 \log (4) \int \frac {e^{x+e^{e^x}+e^x-3}}{\left (e^{e^{e^x}} \log (x)+e^3\right )^2}dx+\log (16) \int \frac {e^{2 \left (-3+e^{e^x}\right )}}{x \left (e^{e^{e^x}} \log (x)+e^3\right )^2}dx+\log (4) \int \frac {e^{2 x+e^{e^x}+e^x-3} x}{\left (e^{e^{e^x}} \log (x)+e^3\right )^2}dx+\log (4) \int \frac {e^{x+e^{e^x}-3}}{\log (x) \left (e^{e^{e^x}} \log (x)+e^3\right )^2}dx+\log (4) \int \frac {e^{x+e^{e^x}-6}}{e^{e^{e^x}} \log (x)+e^3}dx+\log (4) \int \frac {e^{x+e^{e^x}-6} x}{e^{e^{e^x}} \log (x)+e^3}dx-\log (4) \int \frac {e^{x+e^{e^x}-6}}{\log (x) \left (e^{e^{e^x}} \log (x)+e^3\right )}dx\right )\) |
Int[(E^(-3 + E^E^x)*(E^x*(-4*x - 4*x^2)*Log[4] + E^E^x*(8*E^x*x*Log[4] - 4 *E^(2*x)*x^2*Log[4])) + E^(-6 + 2*E^E^x)*(-8*Log[4] + 4*E^x*x*Log[4] + E^x *(-4*x - 4*x^2)*Log[4]*Log[x]))/(x + 2*E^(-3 + E^E^x)*x*Log[x] + E^(-6 + 2 *E^E^x)*x*Log[x]^2),x]
3.20.27.3.1 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 = 20.72 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43
| method | result | size |
| parallelrisch | \(-\frac {8 x \ln \left (2\right ) {\mathrm e}^{x} {\mathrm e}^{-3+{\mathrm e}^{{\mathrm e}^{x}}}-16 \ln \left (2\right ) {\mathrm e}^{-3+{\mathrm e}^{{\mathrm e}^{x}}}}{\ln \left (x \right ) {\mathrm e}^{-3+{\mathrm e}^{{\mathrm e}^{x}}}+1}\) | \(40\) |
| risch | \(-\frac {8 \left ({\mathrm e}^{x} x -2\right ) \ln \left (2\right )}{\ln \left (x \right )}+\frac {8 \left ({\mathrm e}^{x} x -2\right ) \ln \left (2\right )}{\ln \left (x \right ) \left (\ln \left (x \right ) {\mathrm e}^{-3+{\mathrm e}^{{\mathrm e}^{x}}}+1\right )}\) | \(43\) |
int(((2*(-4*x^2-4*x)*ln(2)*exp(x)*ln(x)+8*x*ln(2)*exp(x)-16*ln(2))*exp(-3+ exp(exp(x)))^2+((-8*x^2*ln(2)*exp(x)^2+16*x*ln(2)*exp(x))*exp(exp(x))+2*(- 4*x^2-4*x)*ln(2)*exp(x))*exp(-3+exp(exp(x))))/(x*ln(x)^2*exp(-3+exp(exp(x) ))^2+2*x*ln(x)*exp(-3+exp(exp(x)))+x),x,method=_RETURNVERBOSE)
-(8*x*ln(2)*exp(x)*exp(-3+exp(exp(x)))-16*ln(2)*exp(-3+exp(exp(x))))/(ln(x )*exp(-3+exp(exp(x)))+1)
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-3+e^{e^x}} \left (e^x \left (-4 x-4 x^2\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{-6+2 e^{e^x}} \left (-8 \log (4)+4 e^x x \log (4)+e^x \left (-4 x-4 x^2\right ) \log (4) \log (x)\right )}{x+2 e^{-3+e^{e^x}} x \log (x)+e^{-6+2 e^{e^x}} x \log ^2(x)} \, dx=-\frac {8 \, {\left (x e^{x} \log \left (2\right ) - 2 \, \log \left (2\right )\right )} e^{\left (e^{\left (e^{x}\right )} - 3\right )}}{e^{\left (e^{\left (e^{x}\right )} - 3\right )} \log \left (x\right ) + 1} \]
integrate(((2*(-4*x^2-4*x)*log(2)*exp(x)*log(x)+8*x*log(2)*exp(x)-16*log(2 ))*exp(-3+exp(exp(x)))^2+((-8*x^2*log(2)*exp(x)^2+16*x*log(2)*exp(x))*exp( exp(x))+2*(-4*x^2-4*x)*log(2)*exp(x))*exp(-3+exp(exp(x))))/(x*log(x)^2*exp (-3+exp(exp(x)))^2+2*x*log(x)*exp(-3+exp(exp(x)))+x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {e^{-3+e^{e^x}} \left (e^x \left (-4 x-4 x^2\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{-6+2 e^{e^x}} \left (-8 \log (4)+4 e^x x \log (4)+e^x \left (-4 x-4 x^2\right ) \log (4) \log (x)\right )}{x+2 e^{-3+e^{e^x}} x \log (x)+e^{-6+2 e^{e^x}} x \log ^2(x)} \, dx=- \frac {8 x e^{x} \log {\left (2 \right )}}{\log {\left (x \right )}} + \frac {16 \log {\left (2 \right )}}{\log {\left (x \right )}} + \frac {8 x e^{x} \log {\left (2 \right )} - 16 \log {\left (2 \right )}}{e^{e^{e^{x}} - 3} \log {\left (x \right )}^{2} + \log {\left (x \right )}} \]
integrate(((2*(-4*x**2-4*x)*ln(2)*exp(x)*ln(x)+8*x*ln(2)*exp(x)-16*ln(2))* exp(-3+exp(exp(x)))**2+((-8*x**2*ln(2)*exp(x)**2+16*x*ln(2)*exp(x))*exp(ex p(x))+2*(-4*x**2-4*x)*ln(2)*exp(x))*exp(-3+exp(exp(x))))/(x*ln(x)**2*exp(- 3+exp(exp(x)))**2+2*x*ln(x)*exp(-3+exp(exp(x)))+x),x)
-8*x*exp(x)*log(2)/log(x) + 16*log(2)/log(x) + (8*x*exp(x)*log(2) - 16*log (2))/(exp(exp(exp(x)) - 3)*log(x)**2 + log(x))
Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-3+e^{e^x}} \left (e^x \left (-4 x-4 x^2\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{-6+2 e^{e^x}} \left (-8 \log (4)+4 e^x x \log (4)+e^x \left (-4 x-4 x^2\right ) \log (4) \log (x)\right )}{x+2 e^{-3+e^{e^x}} x \log (x)+e^{-6+2 e^{e^x}} x \log ^2(x)} \, dx=-\frac {8 \, {\left (x e^{x} \log \left (2\right ) - 2 \, \log \left (2\right )\right )} e^{\left (e^{\left (e^{x}\right )}\right )}}{e^{\left (e^{\left (e^{x}\right )}\right )} \log \left (x\right ) + e^{3}} \]
integrate(((2*(-4*x^2-4*x)*log(2)*exp(x)*log(x)+8*x*log(2)*exp(x)-16*log(2 ))*exp(-3+exp(exp(x)))^2+((-8*x^2*log(2)*exp(x)^2+16*x*log(2)*exp(x))*exp( exp(x))+2*(-4*x^2-4*x)*log(2)*exp(x))*exp(-3+exp(exp(x))))/(x*log(x)^2*exp (-3+exp(exp(x)))^2+2*x*log(x)*exp(-3+exp(exp(x)))+x),x, algorithm=\
\[ \int \frac {e^{-3+e^{e^x}} \left (e^x \left (-4 x-4 x^2\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{-6+2 e^{e^x}} \left (-8 \log (4)+4 e^x x \log (4)+e^x \left (-4 x-4 x^2\right ) \log (4) \log (x)\right )}{x+2 e^{-3+e^{e^x}} x \log (x)+e^{-6+2 e^{e^x}} x \log ^2(x)} \, dx=\int { -\frac {8 \, {\left ({\left ({\left (x^{2} + x\right )} e^{x} \log \left (2\right ) \log \left (x\right ) - x e^{x} \log \left (2\right ) + 2 \, \log \left (2\right )\right )} e^{\left (2 \, e^{\left (e^{x}\right )} - 6\right )} + {\left ({\left (x^{2} + x\right )} e^{x} \log \left (2\right ) + {\left (x^{2} e^{\left (2 \, x\right )} \log \left (2\right ) - 2 \, x e^{x} \log \left (2\right )\right )} e^{\left (e^{x}\right )}\right )} e^{\left (e^{\left (e^{x}\right )} - 3\right )}\right )}}{x e^{\left (2 \, e^{\left (e^{x}\right )} - 6\right )} \log \left (x\right )^{2} + 2 \, x e^{\left (e^{\left (e^{x}\right )} - 3\right )} \log \left (x\right ) + x} \,d x } \]
integrate(((2*(-4*x^2-4*x)*log(2)*exp(x)*log(x)+8*x*log(2)*exp(x)-16*log(2 ))*exp(-3+exp(exp(x)))^2+((-8*x^2*log(2)*exp(x)^2+16*x*log(2)*exp(x))*exp( exp(x))+2*(-4*x^2-4*x)*log(2)*exp(x))*exp(-3+exp(exp(x))))/(x*log(x)^2*exp (-3+exp(exp(x)))^2+2*x*log(x)*exp(-3+exp(exp(x)))+x),x, algorithm=\
integrate(-8*(((x^2 + x)*e^x*log(2)*log(x) - x*e^x*log(2) + 2*log(2))*e^(2 *e^(e^x) - 6) + ((x^2 + x)*e^x*log(2) + (x^2*e^(2*x)*log(2) - 2*x*e^x*log( 2))*e^(e^x))*e^(e^(e^x) - 3))/(x*e^(2*e^(e^x) - 6)*log(x)^2 + 2*x*e^(e^(e^ x) - 3)*log(x) + x), x)
Time = 10.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.89 \[ \int \frac {e^{-3+e^{e^x}} \left (e^x \left (-4 x-4 x^2\right ) \log (4)+e^{e^x} \left (8 e^x x \log (4)-4 e^{2 x} x^2 \log (4)\right )\right )+e^{-6+2 e^{e^x}} \left (-8 \log (4)+4 e^x x \log (4)+e^x \left (-4 x-4 x^2\right ) \log (4) \log (x)\right )}{x+2 e^{-3+e^{e^x}} x \log (x)+e^{-6+2 e^{e^x}} x \log ^2(x)} \, dx=\frac {8\,\ln \left (2\right )\,\left (2\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}-3}+x^2\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^x}-x^2\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\,{\mathrm {e}}^x-x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}-3}\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}-3}\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\,\ln \left (x\right )-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}-3}\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\ln \left (x\right )\right )}{\left (x\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\,\ln \left (x\right )+1\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}-3}\,\ln \left (x\right )+1\right )} \]
int(-(exp(2*exp(exp(x)) - 6)*(16*log(2) - 8*x*exp(x)*log(2) + 2*exp(x)*log (2)*log(x)*(4*x + 4*x^2)) - exp(exp(exp(x)) - 3)*(exp(exp(x))*(16*x*exp(x) *log(2) - 8*x^2*exp(2*x)*log(2)) - 2*exp(x)*log(2)*(4*x + 4*x^2)))/(x + x* exp(2*exp(exp(x)) - 6)*log(x)^2 + 2*x*exp(exp(exp(x)) - 3)*log(x)),x)
(8*log(2)*(2*exp(exp(exp(x)) - 3) + x^2*exp(2*x + exp(x)) - x^2*exp(x + ex p(x))*exp(x) - x*exp(exp(exp(x)) - 3)*exp(x) + 2*x*exp(exp(exp(x)) - 3)*ex p(x + exp(x))*log(x) - x^2*exp(exp(exp(x)) - 3)*exp(x + exp(x))*exp(x)*log (x)))/((x*exp(x + exp(x))*log(x) + 1)*(exp(exp(exp(x)) - 3)*log(x) + 1))