Integrand size = 71, antiderivative size = 28 \[ \int \frac {e^3 (-1-x)-x^2-3 x^3+(-1-x) \log (4)-2 x^2 \log (x)+e^{e^x} \left (x+2 x^2+e^x x^3+\left (x+e^x x^2\right ) \log (x)\right )}{x} \, dx=-3+\left (-e^3+\left (e^{e^x}-x\right ) x-\log (4)\right ) (x+\log (x)) \]
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^3 (-1-x)-x^2-3 x^3+(-1-x) \log (4)-2 x^2 \log (x)+e^{e^x} \left (x+2 x^2+e^x x^3+\left (x+e^x x^2\right ) \log (x)\right )}{x} \, dx=-\left (\left (e^3-e^{e^x} x+x^2+\log (4)\right ) (x+\log (x))\right ) \]
Integrate[(E^3*(-1 - x) - x^2 - 3*x^3 + (-1 - x)*Log[4] - 2*x^2*Log[x] + E ^E^x*(x + 2*x^2 + E^x*x^3 + (x + E^x*x^2)*Log[x]))/x,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.
\(\displaystyle \int \frac {-3 x^3-x^2-2 x^2 \log (x)+e^{e^x} \left (e^x x^3+2 x^2+\left (e^x x^2+x\right ) \log (x)+x\right )+e^3 (-x-1)+(-x-1) \log (4)}{x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-3 x^3-x^2-2 x^2 \log (x)+e^{e^x} \left (e^x x^3+2 x^2+\left (e^x x^2+x\right ) \log (x)+x\right )+(-x-1) \left (e^3+\log (4)\right )}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {-3 x^3+2 e^{e^x} x^2-x^2-2 x^2 \log (x)+e^{e^x} x+e^{e^x} x \log (x)-e^3 x \left (1+\frac {\log (4)}{e^3}\right )-e^3 \left (1+\frac {\log (4)}{e^3}\right )}{x}+e^{x+e^x} x (x+\log (x))\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\operatorname {ExpIntegralEi}\left (e^x\right )}{x}dx+\int e^{x+e^x} x^2dx+2 \int e^{e^x} xdx-\int \frac {\int e^{x+e^x} xdx}{x}dx+\log (x) \int e^{x+e^x} xdx+\operatorname {ExpIntegralEi}\left (e^x\right )+\operatorname {ExpIntegralEi}\left (e^x\right ) \log (x)-x^3-x^2 \log (x)-x \left (e^3+\log (4)\right )-\left (e^3+\log (4)\right ) \log (x)\) |
Int[(E^3*(-1 - x) - x^2 - 3*x^3 + (-1 - x)*Log[4] - 2*x^2*Log[x] + E^E^x*( x + 2*x^2 + E^x*x^3 + (x + E^x*x^2)*Log[x]))/x,x]
3.17.90.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 1.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71
method | result | size |
risch | \(-x^{2} \ln \left (x \right )-x^{3}-2 \ln \left (2\right ) \ln \left (x \right )-2 x \ln \left (2\right )-\ln \left (x \right ) {\mathrm e}^{3}-x \,{\mathrm e}^{3}+\left (x \ln \left (x \right )+x^{2}\right ) {\mathrm e}^{{\mathrm e}^{x}}\) | \(48\) |
parallelrisch | \(x \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x}}-x^{2} \ln \left (x \right )+{\mathrm e}^{{\mathrm e}^{x}} x^{2}-x^{3}-2 \ln \left (2\right ) \ln \left (x \right )-2 x \ln \left (2\right )-\ln \left (x \right ) {\mathrm e}^{3}-x \,{\mathrm e}^{3}\) | \(50\) |
int((((exp(x)*x^2+x)*ln(x)+exp(x)*x^3+2*x^2+x)*exp(exp(x))-2*x^2*ln(x)+2*( -1-x)*ln(2)+(-1-x)*exp(3)-3*x^3-x^2)/x,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {e^3 (-1-x)-x^2-3 x^3+(-1-x) \log (4)-2 x^2 \log (x)+e^{e^x} \left (x+2 x^2+e^x x^3+\left (x+e^x x^2\right ) \log (x)\right )}{x} \, dx=-x^{3} - x e^{3} + {\left (x^{2} + x \log \left (x\right )\right )} e^{\left (e^{x}\right )} - 2 \, x \log \left (2\right ) - {\left (x^{2} + e^{3} + 2 \, \log \left (2\right )\right )} \log \left (x\right ) \]
integrate((((exp(x)*x^2+x)*log(x)+exp(x)*x^3+2*x^2+x)*exp(exp(x))-2*x^2*lo g(x)+2*(-1-x)*log(2)+(-1-x)*exp(3)-3*x^3-x^2)/x,x, algorithm=\
Time = 11.96 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^3 (-1-x)-x^2-3 x^3+(-1-x) \log (4)-2 x^2 \log (x)+e^{e^x} \left (x+2 x^2+e^x x^3+\left (x+e^x x^2\right ) \log (x)\right )}{x} \, dx=- x^{3} - x^{2} \log {\left (x \right )} - x \left (2 \log {\left (2 \right )} + e^{3}\right ) + \left (x^{2} + x \log {\left (x \right )}\right ) e^{e^{x}} - \left (2 \log {\left (2 \right )} + e^{3}\right ) \log {\left (x \right )} \]
integrate((((exp(x)*x**2+x)*ln(x)+exp(x)*x**3+2*x**2+x)*exp(exp(x))-2*x**2 *ln(x)+2*(-1-x)*ln(2)+(-1-x)*exp(3)-3*x**3-x**2)/x,x)
-x**3 - x**2*log(x) - x*(2*log(2) + exp(3)) + (x**2 + x*log(x))*exp(exp(x) ) - (2*log(2) + exp(3))*log(x)
\[ \int \frac {e^3 (-1-x)-x^2-3 x^3+(-1-x) \log (4)-2 x^2 \log (x)+e^{e^x} \left (x+2 x^2+e^x x^3+\left (x+e^x x^2\right ) \log (x)\right )}{x} \, dx=\int { -\frac {3 \, x^{3} + 2 \, x^{2} \log \left (x\right ) + x^{2} + {\left (x + 1\right )} e^{3} - {\left (x^{3} e^{x} + 2 \, x^{2} + {\left (x^{2} e^{x} + x\right )} \log \left (x\right ) + x\right )} e^{\left (e^{x}\right )} + 2 \, {\left (x + 1\right )} \log \left (2\right )}{x} \,d x } \]
integrate((((exp(x)*x^2+x)*log(x)+exp(x)*x^3+2*x^2+x)*exp(exp(x))-2*x^2*lo g(x)+2*(-1-x)*log(2)+(-1-x)*exp(3)-3*x^3-x^2)/x,x, algorithm=\
-x^3 - x^2*log(x) - x*e^3 + (x^2 + x*log(x))*e^(e^x) - 2*x*log(2) - e^3*lo g(x) - 2*log(2)*log(x) + Ei(e^x) - integrate(e^(e^x), x)
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (24) = 48\).
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.46 \[ \int \frac {e^3 (-1-x)-x^2-3 x^3+(-1-x) \log (4)-2 x^2 \log (x)+e^{e^x} \left (x+2 x^2+e^x x^3+\left (x+e^x x^2\right ) \log (x)\right )}{x} \, dx=-{\left (x^{3} e^{x} + x^{2} e^{x} \log \left (x\right ) - x^{2} e^{\left (x + e^{x}\right )} + 2 \, x e^{x} \log \left (2\right ) - x e^{\left (x + e^{x}\right )} \log \left (x\right ) + 2 \, e^{x} \log \left (2\right ) \log \left (x\right ) + x e^{\left (x + 3\right )} + e^{\left (x + 3\right )} \log \left (x\right )\right )} e^{\left (-x\right )} \]
integrate((((exp(x)*x^2+x)*log(x)+exp(x)*x^3+2*x^2+x)*exp(exp(x))-2*x^2*lo g(x)+2*(-1-x)*log(2)+(-1-x)*exp(3)-3*x^3-x^2)/x,x, algorithm=\
-(x^3*e^x + x^2*e^x*log(x) - x^2*e^(x + e^x) + 2*x*e^x*log(2) - x*e^(x + e ^x)*log(x) + 2*e^x*log(2)*log(x) + x*e^(x + 3) + e^(x + 3)*log(x))*e^(-x)
Time = 11.80 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {e^3 (-1-x)-x^2-3 x^3+(-1-x) \log (4)-2 x^2 \log (x)+e^{e^x} \left (x+2 x^2+e^x x^3+\left (x+e^x x^2\right ) \log (x)\right )}{x} \, dx={\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x\,\ln \left (x\right )+x^2\right )-x^2\,\ln \left (x\right )-\ln \left (x\right )\,\left ({\mathrm {e}}^3+\ln \left (4\right )\right )-x\,\left ({\mathrm {e}}^3+\ln \left (4\right )\right )-x^3 \]
int(-(2*x^2*log(x) + exp(3)*(x + 1) + 2*log(2)*(x + 1) + x^2 + 3*x^3 - exp (exp(x))*(x + x^3*exp(x) + log(x)*(x + x^2*exp(x)) + 2*x^2))/x,x)