Integrand size = 75, antiderivative size = 33 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\left (5+e^{2 x}\right ) \left (-x+e^4 x-\log (2+x)\right ) \log (\log (i \pi +\log (2))) \]
Time = 2.57 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\left (\left (-1+e^4\right ) \left (10+\left (5+e^{2 x}\right ) x\right )-\left (5+e^{2 x}\right ) \log (2+x)\right ) \log (\log (i \pi +\log (2))) \]
Integrate[((-15 - 5*x + E^4*(10 + 5*x) + E^(2*x)*(-3 - 5*x - 2*x^2 + E^4*( 2 + 5*x + 2*x^2)) + E^(2*x)*(-4 - 2*x)*Log[2 + x])*Log[Log[I*Pi + Log[2]]] )/(2 + x),x]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.73 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.24, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, 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.
\(\displaystyle \int \frac {\log (\log (\log (2)+i \pi )) \left (e^{2 x} \left (-2 x^2+e^4 \left (2 x^2+5 x+2\right )-5 x-3\right )-5 x+e^4 (5 x+10)+e^{2 x} (-2 x-4) \log (x+2)-15\right )}{x+2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (\log (\log (2)+i \pi )) \int -\frac {5 x-5 e^4 (x+2)+e^{2 x} \left (2 x^2+5 x-e^4 \left (2 x^2+5 x+2\right )+3\right )+2 e^{2 x} (x+2) \log (x+2)+15}{x+2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\log (\log (\log (2)+i \pi )) \int \frac {5 x-5 e^4 (x+2)+e^{2 x} \left (2 x^2+5 x-e^4 \left (2 x^2+5 x+2\right )+3\right )+2 e^{2 x} (x+2) \log (x+2)+15}{x+2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\log (\log (\log (2)+i \pi )) \int \left (\frac {5 \left (\left (1-e^4\right ) x-2 e^4+3\right )}{x+2}+\frac {e^{2 x} \left (2 \left (1-e^4\right ) x^2+2 \log (x+2) x+5 \left (1-e^4\right ) x+4 \log (x+2)+3 \left (1-\frac {2 e^4}{3}\right )\right )}{x+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\left (\log (\log (\log (2)+i \pi )) \left (\frac {\operatorname {ExpIntegralEi}(2 (x+2))}{e^4}-\frac {\operatorname {ExpIntegralEi}(2 x+4)}{e^4}+\left (1-e^4\right ) e^{2 x} x+5 \left (1-e^4\right ) x+e^{2 x} \log (x+2)+5 \log (x+2)\right )\right )\) |
Int[((-15 - 5*x + E^4*(10 + 5*x) + E^(2*x)*(-3 - 5*x - 2*x^2 + E^4*(2 + 5* x + 2*x^2)) + E^(2*x)*(-4 - 2*x)*Log[2 + x])*Log[Log[I*Pi + Log[2]]])/(2 + x),x]
-((5*(1 - E^4)*x + E^(2*x)*(1 - E^4)*x + ExpIntegralEi[2*(2 + x)]/E^4 - Ex pIntegralEi[4 + 2*x]/E^4 + 5*Log[2 + x] + E^(2*x)*Log[2 + x])*Log[Log[I*Pi + Log[2]]])
3.8.98.3.1 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 = 2.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39
method | result | size |
default | \(\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left (\left ({\mathrm e}^{4}-1\right ) x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (2+x \right )+5 x \,{\mathrm e}^{4}-5 x -5 \ln \left (2+x \right )\right )\) | \(46\) |
parallelrisch | \(\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left ({\mathrm e}^{4} {\mathrm e}^{2 x} x +5 x \,{\mathrm e}^{4}-x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (2+x \right )-20 \,{\mathrm e}^{4}-5 x -5 \ln \left (2+x \right )+20\right )\) | \(56\) |
parts | \(\left (\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x \,{\mathrm e}^{2 x}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )+5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \left (x \,{\mathrm e}^{4}-x -\ln \left (2+x \right )\right )\) | \(76\) |
norman | \(\left (5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x -5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \ln \left (2+x \right )+\left (\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right )\right ) x \,{\mathrm e}^{2 x}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )\) | \(93\) |
risch | \(-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) {\mathrm e}^{2 x} \ln \left (2+x \right )+\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{4+2 x}+5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{4}-\ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x \,{\mathrm e}^{2 x}-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) \ln \left (2+x \right )-5 \ln \left (\ln \left (\ln \left (2\right )+i \pi \right )\right ) x\) | \(95\) |
int(((-2*x-4)*exp(2*x)*ln(2+x)+((2*x^2+5*x+2)*exp(4)-2*x^2-5*x-3)*exp(2*x) +(5*x+10)*exp(4)-5*x-15)*ln(ln(ln(2)+I*Pi))/(2+x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx={\left (5 \, x e^{4} + {\left (x e^{4} - x\right )} e^{\left (2 \, x\right )} - {\left (e^{\left (2 \, x\right )} + 5\right )} \log \left (x + 2\right ) - 5 \, x\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right ) \]
integrate(((-2*x-4)*exp(2*x)*log(2+x)+((2*x^2+5*x+2)*exp(4)-2*x^2-5*x-3)*e xp(2*x)+(5*x+10)*exp(4)-5*x-15)*log(log(log(2)+I*pi))/(2+x),x, algorithm=\
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (27) = 54\).
Time = 0.51 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.79 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=x \left (- 5 \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} + 5 e^{4} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )}\right ) + \left (- x \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} + x e^{4} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} - \log {\left (x + 2 \right )} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )}\right ) e^{2 x} - 5 \log {\left (x + 2 \right )} \log {\left (\log {\left (\log {\left (2 \right )} + i \pi \right )} \right )} \]
integrate(((-2*x-4)*exp(2*x)*ln(2+x)+((2*x**2+5*x+2)*exp(4)-2*x**2-5*x-3)* exp(2*x)+(5*x+10)*exp(4)-5*x-15)*ln(ln(ln(2)+I*pi))/(2+x),x)
x*(-5*log(log(log(2) + I*pi)) + 5*exp(4)*log(log(log(2) + I*pi))) + (-x*lo g(log(log(2) + I*pi)) + x*exp(4)*log(log(log(2) + I*pi)) - log(x + 2)*log( log(log(2) + I*pi)))*exp(2*x) - 5*log(x + 2)*log(log(log(2) + I*pi))
\[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=\int { -\frac {{\left (2 \, {\left (x + 2\right )} e^{\left (2 \, x\right )} \log \left (x + 2\right ) - 5 \, {\left (x + 2\right )} e^{4} + {\left (2 \, x^{2} - {\left (2 \, x^{2} + 5 \, x + 2\right )} e^{4} + 5 \, x + 3\right )} e^{\left (2 \, x\right )} + 5 \, x + 15\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right )}{x + 2} \,d x } \]
integrate(((-2*x-4)*exp(2*x)*log(2+x)+((2*x^2+5*x+2)*exp(4)-2*x^2-5*x-3)*e xp(2*x)+(5*x+10)*exp(4)-5*x-15)*log(log(log(2)+I*pi))/(2+x),x, algorithm=\
(x*(e^4 - 1)*e^(2*x) + 5*(x - 2*log(x + 2))*e^4 + 3*e^(-4)*exp_integral_e( 1, -2*x - 4) + 10*e^4*log(x + 2) - e^(2*x)*log(x + 2) - 5*x + 3*integrate( e^(2*x)/(x + 2), x) - 5*log(x + 2))*log(log(I*pi + log(2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (29) = 58\).
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.52 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx={\left (5 \, {\left (x + 2\right )} e^{8} - 5 \, {\left (x + 2\right )} e^{4} + {\left (x + 2\right )} e^{\left (2 \, x + 8\right )} - {\left (x + 2\right )} e^{\left (2 \, x + 4\right )} - 5 \, e^{4} \log \left (x + 2\right ) - e^{\left (2 \, x + 4\right )} \log \left (x + 2\right ) - 2 \, e^{\left (2 \, x + 8\right )} + 2 \, e^{\left (2 \, x + 4\right )}\right )} e^{\left (-4\right )} \log \left (\log \left (i \, \pi + \log \left (2\right )\right )\right ) \]
integrate(((-2*x-4)*exp(2*x)*log(2+x)+((2*x^2+5*x+2)*exp(4)-2*x^2-5*x-3)*e xp(2*x)+(5*x+10)*exp(4)-5*x-15)*log(log(log(2)+I*pi))/(2+x),x, algorithm=\
(5*(x + 2)*e^8 - 5*(x + 2)*e^4 + (x + 2)*e^(2*x + 8) - (x + 2)*e^(2*x + 4) - 5*e^4*log(x + 2) - e^(2*x + 4)*log(x + 2) - 2*e^(2*x + 8) + 2*e^(2*x + 4))*e^(-4)*log(log(I*pi + log(2)))
Time = 9.78 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \frac {\left (-15-5 x+e^4 (10+5 x)+e^{2 x} \left (-3-5 x-2 x^2+e^4 \left (2+5 x+2 x^2\right )\right )+e^{2 x} (-4-2 x) \log (2+x)\right ) \log (\log (i \pi +\log (2)))}{2+x} \, dx=-\ln \left (\ln \left (\ln \left (2\right )+\Pi \,1{}\mathrm {i}\right )\right )\,\left (5\,x+5\,\ln \left (x+2\right )+x\,{\mathrm {e}}^{2\,x}-5\,x\,{\mathrm {e}}^4-x\,{\mathrm {e}}^{2\,x+4}+\ln \left (x+2\right )\,{\mathrm {e}}^{2\,x}\right ) \]
int(-(log(log(Pi*1i + log(2)))*(5*x + exp(2*x)*(5*x - exp(4)*(5*x + 2*x^2 + 2) + 2*x^2 + 3) - exp(4)*(5*x + 10) + log(x + 2)*exp(2*x)*(2*x + 4) + 15 ))/(x + 2),x)