Integrand size = 90, antiderivative size = 34 \[ \int \frac {e^{-2 e^3} \left (-e^{2 e^3} x \log (4)+e^{2 e^3} (2+x) \log (4) \log (2+x)+\left (8 x+4 x^2+e^{2 e^3} (-4-2 x+(-2-x) \log (4))\right ) \log ^2(2+x)\right )}{(2+x) \log (4) \log ^2(2+x)} \, dx=2-x+\frac {2 \left (-x+e^{-2 e^3} x^2\right )}{\log (4)}+\frac {x}{\log (2+x)} \]
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-2 e^3} \left (-e^{2 e^3} x \log (4)+e^{2 e^3} (2+x) \log (4) \log (2+x)+\left (8 x+4 x^2+e^{2 e^3} (-4-2 x+(-2-x) \log (4))\right ) \log ^2(2+x)\right )}{(2+x) \log (4) \log ^2(2+x)} \, dx=\frac {2 e^{-2 e^3} x^2}{\log (4)}-\frac {x (2+\log (4))}{\log (4)}+\frac {x}{\log (2+x)} \]
Integrate[(-(E^(2*E^3)*x*Log[4]) + E^(2*E^3)*(2 + x)*Log[4]*Log[2 + x] + ( 8*x + 4*x^2 + E^(2*E^3)*(-4 - 2*x + (-2 - x)*Log[4]))*Log[2 + x]^2)/(E^(2* E^3)*(2 + x)*Log[4]*Log[2 + x]^2),x]
Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(34)=68\).
Time = 0.75 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {27, 25, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{-2 e^3} \left (\left (4 x^2+8 x+e^{2 e^3} (-2 x+(-x-2) \log (4)-4)\right ) \log ^2(x+2)+e^{2 e^3} (x+2) \log (4) \log (x+2)-e^{2 e^3} x \log (4)\right )}{(x+2) \log (4) \log ^2(x+2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{-2 e^3} \int -\frac {-\left (\left (4 x^2+8 x-e^{2 e^3} (2 x+(x+2) \log (4)+4)\right ) \log ^2(x+2)\right )-e^{2 e^3} (x+2) \log (4) \log (x+2)+e^{2 e^3} x \log (4)}{(x+2) \log ^2(x+2)}dx}{\log (4)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {e^{-2 e^3} \int \frac {-\left (\left (4 x^2+8 x-e^{2 e^3} (2 x+(x+2) \log (4)+4)\right ) \log ^2(x+2)\right )-e^{2 e^3} (x+2) \log (4) \log (x+2)+e^{2 e^3} x \log (4)}{(x+2) \log ^2(x+2)}dx}{\log (4)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {e^{-2 e^3} \int \left (\frac {e^{2 e^3} \log (4) x}{(x+2) \log ^2(x+2)}-4 x-\frac {e^{2 e^3} \log (4)}{\log (x+2)}+2 e^{2 e^3} (1+\log (2))\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{-2 e^3} \left (-2 x^2+2 e^{2 e^3} x (1+\log (2))-\frac {e^{2 e^3} (x+2) \log (4)}{\log (x+2)}+\frac {2 e^{2 e^3} \log (4)}{\log (x+2)}\right )}{\log (4)}\) |
Int[(-(E^(2*E^3)*x*Log[4]) + E^(2*E^3)*(2 + x)*Log[4]*Log[2 + x] + (8*x + 4*x^2 + E^(2*E^3)*(-4 - 2*x + (-2 - x)*Log[4]))*Log[2 + x]^2)/(E^(2*E^3)*( 2 + x)*Log[4]*Log[2 + x]^2),x]
-((-2*x^2 + 2*E^(2*E^3)*x*(1 + Log[2]) + (2*E^(2*E^3)*Log[4])/Log[2 + x] - (E^(2*E^3)*(2 + x)*Log[4])/Log[2 + x])/(E^(2*E^3)*Log[4]))
3.8.79.3.1 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 = 2.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97
method | result | size |
risch | \(-x -\frac {x}{\ln \left (2\right )}+\frac {{\mathrm e}^{-2 \,{\mathrm e}^{3}} x^{2}}{\ln \left (2\right )}+\frac {x}{\ln \left (2+x \right )}\) | \(33\) |
parts | \(-\frac {2}{\ln \left (2+x \right )}+\frac {2+x}{\ln \left (2+x \right )}-x -\frac {x}{\ln \left (2\right )}+\frac {{\mathrm e}^{-2 \,{\mathrm e}^{3}} x^{2}}{\ln \left (2\right )}\) | \(43\) |
norman | \(\frac {\left (x \,{\mathrm e}^{{\mathrm e}^{3}}+\frac {{\mathrm e}^{-{\mathrm e}^{3}} x^{2} \ln \left (2+x \right )}{\ln \left (2\right )}-\frac {{\mathrm e}^{{\mathrm e}^{3}} \left (1+\ln \left (2\right )\right ) x \ln \left (2+x \right )}{\ln \left (2\right )}\right ) {\mathrm e}^{-{\mathrm e}^{3}}}{\ln \left (2+x \right )}\) | \(54\) |
parallelrisch | \(-\frac {{\mathrm e}^{-2 \,{\mathrm e}^{3}} \left (2 \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \left (2+x \right ) x -2 x \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}}-8 \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \left (2+x \right )+2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \left (2+x \right ) x -8 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \left (2+x \right )-2 \ln \left (2+x \right ) x^{2}+8 \ln \left (2+x \right )\right )}{2 \ln \left (2\right ) \ln \left (2+x \right )}\) | \(94\) |
derivativedivides | \(\frac {{\mathrm e}^{-2 \,{\mathrm e}^{3}} \left (-2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \left (2\right ) \left (2+x \right )-2 \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}} \operatorname {Ei}_{1}\left (-\ln \left (2+x \right )\right )-2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \left (2+x \right )-2 \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}} \left (-\frac {2+x}{\ln \left (2+x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (2+x \right )\right )\right )+2 \left (2+x \right )^{2}-\frac {4 \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}}}{\ln \left (2+x \right )}-16-8 x \right )}{2 \ln \left (2\right )}\) | \(109\) |
default | \(\frac {{\mathrm e}^{-2 \,{\mathrm e}^{3}} \left (-2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \left (2\right ) \left (2+x \right )-2 \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}} \operatorname {Ei}_{1}\left (-\ln \left (2+x \right )\right )-2 \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \left (2+x \right )-2 \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}} \left (-\frac {2+x}{\ln \left (2+x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (2+x \right )\right )\right )+2 \left (2+x \right )^{2}-\frac {4 \ln \left (2\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}}}{\ln \left (2+x \right )}-16-8 x \right )}{2 \ln \left (2\right )}\) | \(109\) |
int(1/2*(((2*(-2-x)*ln(2)-2*x-4)*exp(exp(3))^2+4*x^2+8*x)*ln(2+x)^2+2*(2+x )*ln(2)*exp(exp(3))^2*ln(2+x)-2*x*ln(2)*exp(exp(3))^2)/(2+x)/ln(2)/exp(exp (3))^2/ln(2+x)^2,x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-2 e^3} \left (-e^{2 e^3} x \log (4)+e^{2 e^3} (2+x) \log (4) \log (2+x)+\left (8 x+4 x^2+e^{2 e^3} (-4-2 x+(-2-x) \log (4))\right ) \log ^2(2+x)\right )}{(2+x) \log (4) \log ^2(2+x)} \, dx=\frac {{\left (x e^{\left (2 \, e^{3}\right )} \log \left (2\right ) + {\left (x^{2} - {\left (x \log \left (2\right ) + x\right )} e^{\left (2 \, e^{3}\right )}\right )} \log \left (x + 2\right )\right )} e^{\left (-2 \, e^{3}\right )}}{\log \left (2\right ) \log \left (x + 2\right )} \]
integrate(1/2*(((2*(-2-x)*log(2)-2*x-4)*exp(exp(3))^2+4*x^2+8*x)*log(2+x)^ 2+2*(2+x)*log(2)*exp(exp(3))^2*log(2+x)-2*x*log(2)*exp(exp(3))^2)/(2+x)/lo g(2)/exp(exp(3))^2/log(2+x)^2,x, algorithm=\
(x*e^(2*e^3)*log(2) + (x^2 - (x*log(2) + x)*e^(2*e^3))*log(x + 2))*e^(-2*e ^3)/(log(2)*log(x + 2))
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-2 e^3} \left (-e^{2 e^3} x \log (4)+e^{2 e^3} (2+x) \log (4) \log (2+x)+\left (8 x+4 x^2+e^{2 e^3} (-4-2 x+(-2-x) \log (4))\right ) \log ^2(2+x)\right )}{(2+x) \log (4) \log ^2(2+x)} \, dx=\frac {x^{2}}{e^{2 e^{3}} \log {\left (2 \right )}} + \frac {x \left (-1 - \log {\left (2 \right )}\right )}{\log {\left (2 \right )}} + \frac {x}{\log {\left (x + 2 \right )}} \]
integrate(1/2*(((2*(-2-x)*ln(2)-2*x-4)*exp(exp(3))**2+4*x**2+8*x)*ln(2+x)* *2+2*(2+x)*ln(2)*exp(exp(3))**2*ln(2+x)-2*x*ln(2)*exp(exp(3))**2)/(2+x)/ln (2)/exp(exp(3))**2/ln(2+x)**2,x)
Time = 0.32 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-2 e^3} \left (-e^{2 e^3} x \log (4)+e^{2 e^3} (2+x) \log (4) \log (2+x)+\left (8 x+4 x^2+e^{2 e^3} (-4-2 x+(-2-x) \log (4))\right ) \log ^2(2+x)\right )}{(2+x) \log (4) \log ^2(2+x)} \, dx=\frac {{\left (x e^{\left (2 \, e^{3}\right )} \log \left (2\right ) - {\left (x {\left (\log \left (2\right ) + 1\right )} e^{\left (2 \, e^{3}\right )} - x^{2}\right )} \log \left (x + 2\right )\right )} e^{\left (-2 \, e^{3}\right )}}{\log \left (2\right ) \log \left (x + 2\right )} \]
integrate(1/2*(((2*(-2-x)*log(2)-2*x-4)*exp(exp(3))^2+4*x^2+8*x)*log(2+x)^ 2+2*(2+x)*log(2)*exp(exp(3))^2*log(2+x)-2*x*log(2)*exp(exp(3))^2)/(2+x)/lo g(2)/exp(exp(3))^2/log(2+x)^2,x, algorithm=\
(x*e^(2*e^3)*log(2) - (x*(log(2) + 1)*e^(2*e^3) - x^2)*log(x + 2))*e^(-2*e ^3)/(log(2)*log(x + 2))
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.79 \[ \int \frac {e^{-2 e^3} \left (-e^{2 e^3} x \log (4)+e^{2 e^3} (2+x) \log (4) \log (2+x)+\left (8 x+4 x^2+e^{2 e^3} (-4-2 x+(-2-x) \log (4))\right ) \log ^2(2+x)\right )}{(2+x) \log (4) \log ^2(2+x)} \, dx=-\frac {{\left (x e^{\left (2 \, e^{3}\right )} \log \left (2\right ) \log \left (x + 2\right ) - x e^{\left (2 \, e^{3}\right )} \log \left (2\right ) - x^{2} \log \left (x + 2\right ) + x e^{\left (2 \, e^{3}\right )} \log \left (x + 2\right )\right )} e^{\left (-2 \, e^{3}\right )}}{\log \left (2\right ) \log \left (x + 2\right )} \]
integrate(1/2*(((2*(-2-x)*log(2)-2*x-4)*exp(exp(3))^2+4*x^2+8*x)*log(2+x)^ 2+2*(2+x)*log(2)*exp(exp(3))^2*log(2+x)-2*x*log(2)*exp(exp(3))^2)/(2+x)/lo g(2)/exp(exp(3))^2/log(2+x)^2,x, algorithm=\
-(x*e^(2*e^3)*log(2)*log(x + 2) - x*e^(2*e^3)*log(2) - x^2*log(x + 2) + x* e^(2*e^3)*log(x + 2))*e^(-2*e^3)/(log(2)*log(x + 2))
Time = 9.43 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-2 e^3} \left (-e^{2 e^3} x \log (4)+e^{2 e^3} (2+x) \log (4) \log (2+x)+\left (8 x+4 x^2+e^{2 e^3} (-4-2 x+(-2-x) \log (4))\right ) \log ^2(2+x)\right )}{(2+x) \log (4) \log ^2(2+x)} \, dx=\frac {x}{\ln \left (x+2\right )}-x-\frac {x}{\ln \left (2\right )}+\frac {x^2\,{\mathrm {e}}^{-2\,{\mathrm {e}}^3}}{\ln \left (2\right )} \]