Integrand size = 88, antiderivative size = 24 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=x+e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} \log (4+2 x) \] Output:
exp(2-x*(1+4/ln(3)^2))*ln(4+2*x)+x
Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=x+e^{2+x \left (-1-\frac {4}{\log ^2(3)}\right )} \log (2 (2+x)) \] Input:
Integrate[(E^((-4*x + (2 - x)*Log[3]^2)/Log[3]^2)*Log[3]^2 + (2 + x)*Log[3 ]^2 + E^((-4*x + (2 - x)*Log[3]^2)/Log[3]^2)*(-8 - 4*x + (-2 - x)*Log[3]^2 )*Log[4 + 2*x])/((2 + x)*Log[3]^2),x]
Output:
x + E^(2 + x*(-1 - 4/Log[3]^2))*Log[2*(2 + x)]
Leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(24)=48\).
Time = 0.67 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {27, 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.
| \(\displaystyle \int \frac {(x+2) \log ^2(3)+e^{\frac {(2-x) \log ^2(3)-4 x}{\log ^2(3)}} \left (-4 x+(-x-2) \log ^2(3)-8\right ) \log (2 x+4)+\log ^2(3) e^{\frac {(2-x) \log ^2(3)-4 x}{\log ^2(3)}}}{(x+2) \log ^2(3)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\log ^2(3) (x+2)-e^{-\frac {4 x-(2-x) \log ^2(3)}{\log ^2(3)}} \left (4 x+(x+2) \log ^2(3)+8\right ) \log (2 x+4)+e^{-\frac {4 x-(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)}{x+2}dx}{\log ^2(3)}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {\int \left (\frac {e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} \left (-4 x \left (1+\frac {\log ^2(3)}{4}\right ) \log (2 (x+2))-8 \left (1+\frac {\log ^2(3)}{4}\right ) \log (2 (x+2))+\log ^2(3)\right )}{x+2}+\log ^2(3)\right )dx}{\log ^2(3)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x \log ^2(3)+\frac {e^{2-x \left (1+\frac {4}{\log ^2(3)}\right )} \left (x \left (4+\log ^2(3)\right ) \log (2 (x+2))+2 \left (4+\log ^2(3)\right ) \log (2 (x+2))\right )}{(x+2) \left (1+\frac {4}{\log ^2(3)}\right )}}{\log ^2(3)}\) |
Input:
Int[(E^((-4*x + (2 - x)*Log[3]^2)/Log[3]^2)*Log[3]^2 + (2 + x)*Log[3]^2 + E^((-4*x + (2 - x)*Log[3]^2)/Log[3]^2)*(-8 - 4*x + (-2 - x)*Log[3]^2)*Log[ 4 + 2*x])/((2 + x)*Log[3]^2),x]
Output:
(x*Log[3]^2 + (E^(2 - x*(1 + 4/Log[3]^2))*(2*(4 + Log[3]^2)*Log[2*(2 + x)] + x*(4 + Log[3]^2)*Log[2*(2 + x)]))/((2 + x)*(1 + 4/Log[3]^2)))/Log[3]^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.91 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25
| method | result | size |
| parts | \(x +{\mathrm e}^{\frac {\left (2-x \right ) \ln \left (3\right )^{2}-4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )\) | \(30\) |
| risch | \(x +{\mathrm e}^{-\frac {x \ln \left (3\right )^{2}-2 \ln \left (3\right )^{2}+4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )\) | \(33\) |
| norman | \(\frac {x \ln \left (3\right )+\ln \left (3\right ) {\mathrm e}^{\frac {\left (2-x \right ) \ln \left (3\right )^{2}-4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )}{\ln \left (3\right )}\) | \(40\) |
| default | \(\frac {x \ln \left (3\right )^{2}+\ln \left (3\right )^{2} {\mathrm e}^{\frac {\left (2-x \right ) \ln \left (3\right )^{2}-4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )}{\ln \left (3\right )^{2}}\) | \(44\) |
| parallelrisch | \(\frac {\ln \left (3\right )^{2} {\mathrm e}^{\frac {\left (2-x \right ) \ln \left (3\right )^{2}-4 x}{\ln \left (3\right )^{2}}} \ln \left (4+2 x \right )+x \ln \left (3\right )^{2}-4 \ln \left (3\right )^{2}}{\ln \left (3\right )^{2}}\) | \(50\) |
Input:
int((((-2-x)*ln(3)^2-4*x-8)*exp(((2-x)*ln(3)^2-4*x)/ln(3)^2)*ln(4+2*x)+ln( 3)^2*exp(((2-x)*ln(3)^2-4*x)/ln(3)^2)+(2+x)*ln(3)^2)/(2+x)/ln(3)^2,x,metho d=_RETURNVERBOSE)
Output:
x+exp(((2-x)*ln(3)^2-4*x)/ln(3)^2)*ln(4+2*x)
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=e^{\left (-\frac {{\left (x - 2\right )} \log \left (3\right )^{2} + 4 \, x}{\log \left (3\right )^{2}}\right )} \log \left (2 \, x + 4\right ) + x \] Input:
integrate((((-2-x)*log(3)^2-4*x-8)*exp(((2-x)*log(3)^2-4*x)/log(3)^2)*log( 4+2*x)+log(3)^2*exp(((2-x)*log(3)^2-4*x)/log(3)^2)+(2+x)*log(3)^2)/(2+x)/l og(3)^2,x, algorithm="fricas")
Output:
e^(-((x - 2)*log(3)^2 + 4*x)/log(3)^2)*log(2*x + 4) + x
Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=x + e^{\frac {- 4 x + \left (2 - x\right ) \log {\left (3 \right )}^{2}}{\log {\left (3 \right )}^{2}}} \log {\left (2 x + 4 \right )} \] Input:
integrate((((-2-x)*ln(3)**2-4*x-8)*exp(((2-x)*ln(3)**2-4*x)/ln(3)**2)*ln(4 +2*x)+ln(3)**2*exp(((2-x)*ln(3)**2-4*x)/ln(3)**2)+(2+x)*ln(3)**2)/(2+x)/ln (3)**2,x)
Output:
x + exp((-4*x + (2 - x)*log(3)**2)/log(3)**2)*log(2*x + 4)
\[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=\int { \frac {{\left (x + 2\right )} \log \left (3\right )^{2} + e^{\left (-\frac {{\left (x - 2\right )} \log \left (3\right )^{2} + 4 \, x}{\log \left (3\right )^{2}}\right )} \log \left (3\right )^{2} - {\left ({\left (x + 2\right )} \log \left (3\right )^{2} + 4 \, x + 8\right )} e^{\left (-\frac {{\left (x - 2\right )} \log \left (3\right )^{2} + 4 \, x}{\log \left (3\right )^{2}}\right )} \log \left (2 \, x + 4\right )}{{\left (x + 2\right )} \log \left (3\right )^{2}} \,d x } \] Input:
integrate((((-2-x)*log(3)^2-4*x-8)*exp(((2-x)*log(3)^2-4*x)/log(3)^2)*log( 4+2*x)+log(3)^2*exp(((2-x)*log(3)^2-4*x)/log(3)^2)+(2+x)*log(3)^2)/(2+x)/l og(3)^2,x, algorithm="maxima")
Output:
-(e^(2*(log(3)^2 + 4)/log(3)^2 + 2)*exp_integral_e(1, (log(3)^2 + 4)*(x + 2)/log(3)^2)*log(3)^2 - e^(-x - 4*x/log(3)^2 + 2)*log(3)^2*log(x + 2) - (x - 2*log(x + 2))*log(3)^2 - 2*log(3)^2*log(x + 2) + integrate(((log(3)^2 + 4)*x*e^2*log(2) + (log(3)^2 + 2*(log(3)^2 + 4)*log(2))*e^2)*e^(-x - 4*x/l og(3)^2)/(x + 2), x))/log(3)^2
\[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=\int { \frac {{\left (x + 2\right )} \log \left (3\right )^{2} + e^{\left (-\frac {{\left (x - 2\right )} \log \left (3\right )^{2} + 4 \, x}{\log \left (3\right )^{2}}\right )} \log \left (3\right )^{2} - {\left ({\left (x + 2\right )} \log \left (3\right )^{2} + 4 \, x + 8\right )} e^{\left (-\frac {{\left (x - 2\right )} \log \left (3\right )^{2} + 4 \, x}{\log \left (3\right )^{2}}\right )} \log \left (2 \, x + 4\right )}{{\left (x + 2\right )} \log \left (3\right )^{2}} \,d x } \] Input:
integrate((((-2-x)*log(3)^2-4*x-8)*exp(((2-x)*log(3)^2-4*x)/log(3)^2)*log( 4+2*x)+log(3)^2*exp(((2-x)*log(3)^2-4*x)/log(3)^2)+(2+x)*log(3)^2)/(2+x)/l og(3)^2,x, algorithm="giac")
Output:
undef
Timed out. \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=\int \frac {{\ln \left (3\right )}^2\,\left (x+2\right )+{\mathrm {e}}^{-\frac {4\,x+{\ln \left (3\right )}^2\,\left (x-2\right )}{{\ln \left (3\right )}^2}}\,{\ln \left (3\right )}^2-{\mathrm {e}}^{-\frac {4\,x+{\ln \left (3\right )}^2\,\left (x-2\right )}{{\ln \left (3\right )}^2}}\,\ln \left (2\,x+4\right )\,\left (4\,x+{\ln \left (3\right )}^2\,\left (x+2\right )+8\right )}{{\ln \left (3\right )}^2\,\left (x+2\right )} \,d x \] Input:
int((log(3)^2*(x + 2) + exp(-(4*x + log(3)^2*(x - 2))/log(3)^2)*log(3)^2 - exp(-(4*x + log(3)^2*(x - 2))/log(3)^2)*log(2*x + 4)*(4*x + log(3)^2*(x + 2) + 8))/(log(3)^2*(x + 2)),x)
Output:
int((log(3)^2*(x + 2) + exp(-(4*x + log(3)^2*(x - 2))/log(3)^2)*log(3)^2 - exp(-(4*x + log(3)^2*(x - 2))/log(3)^2)*log(2*x + 4)*(4*x + log(3)^2*(x + 2) + 8))/(log(3)^2*(x + 2)), x)
Time = 0.17 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \log ^2(3)+(2+x) \log ^2(3)+e^{\frac {-4 x+(2-x) \log ^2(3)}{\log ^2(3)}} \left (-8-4 x+(-2-x) \log ^2(3)\right ) \log (4+2 x)}{(2+x) \log ^2(3)} \, dx=\frac {e^{\frac {\mathrm {log}\left (3\right )^{2} x +4 x}{\mathrm {log}\left (3\right )^{2}}} x +\mathrm {log}\left (2 x +4\right ) e^{2}}{e^{\frac {\mathrm {log}\left (3\right )^{2} x +4 x}{\mathrm {log}\left (3\right )^{2}}}} \] Input:
int((((-2-x)*log(3)^2-4*x-8)*exp(((2-x)*log(3)^2-4*x)/log(3)^2)*log(4+2*x) +log(3)^2*exp(((2-x)*log(3)^2-4*x)/log(3)^2)+(2+x)*log(3)^2)/(2+x)/log(3)^ 2,x)
Output:
(e**((log(3)**2*x + 4*x)/log(3)**2)*x + log(2*x + 4)*e**2)/e**((log(3)**2* x + 4*x)/log(3)**2)