Integrand size = 106, antiderivative size = 24 \[ \int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx=\frac {e^{e^x-x+x^2} \left (x+\log \left ((4+x)^2\right )\right )}{x} \]
Time = 5.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx=\frac {e^{e^x+(-1+x) x} \left (x+\log \left ((4+x)^2\right )\right )}{x} \]
Integrate[(E^(E^x - 2*x + x^2)*(E^(2*x)*(4*x^2 + x^3) + E^x*(2*x - 4*x^2 + 7*x^3 + 2*x^4) + (E^(2*x)*(4*x + x^2) + E^x*(-4 - 5*x + 7*x^2 + 2*x^3))*L og[16 + 8*x + x^2]))/(4*x^2 + x^3),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.
\(\displaystyle \int \frac {e^{x^2-2 x+e^x} \left (e^{2 x} \left (x^3+4 x^2\right )+\left (e^{2 x} \left (x^2+4 x\right )+e^x \left (2 x^3+7 x^2-5 x-4\right )\right ) \log \left (x^2+8 x+16\right )+e^x \left (2 x^4+7 x^3-4 x^2+2 x\right )\right )}{x^3+4 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{x^2-2 x+e^x} \left (e^{2 x} \left (x^3+4 x^2\right )+\left (e^{2 x} \left (x^2+4 x\right )+e^x \left (2 x^3+7 x^2-5 x-4\right )\right ) \log \left (x^2+8 x+16\right )+e^x \left (2 x^4+7 x^3-4 x^2+2 x\right )\right )}{x^2 (x+4)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x^2+e^x} \left (x+\log \left ((x+4)^2\right )\right )}{x}+\frac {e^{x^2-x+e^x} \left (2 x^4+7 x^3+2 x^3 \log \left ((x+4)^2\right )-4 x^2+7 x^2 \log \left ((x+4)^2\right )+2 x-5 x \log \left ((x+4)^2\right )-4 \log \left ((x+4)^2\right )\right )}{x^2 (x+4)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{x^2+e^x}dx-\int e^{x^2-x+e^x}dx+\frac {1}{2} \int \frac {e^{x^2-x+e^x}}{x}dx+2 \int e^{x^2-x+e^x} xdx-\frac {1}{2} \int \frac {e^{x^2-x+e^x}}{x+4}dx+2 \int \frac {\int \frac {e^{(x-1) x+e^x}}{x^2}dx}{x+4}dx-2 \int \frac {\int \frac {e^{x^2+e^x}}{x}dx}{x+4}dx+2 \log \left ((x+4)^2\right ) \int e^{x^2-x+e^x}dx-\log \left ((x+4)^2\right ) \int \frac {e^{x^2-x+e^x}}{x^2}dx+\log \left ((x+4)^2\right ) \int \frac {e^{x^2+e^x}}{x}dx-\log \left ((x+4)^2\right ) \int \frac {e^{x^2-x+e^x}}{x}dx-4 \int \frac {\int e^{(x-1) x+e^x}dx}{x+4}dx+2 \int \frac {\int \frac {e^{(x-1) x+e^x}}{x}dx}{x+4}dx\) |
Int[(E^(E^x - 2*x + x^2)*(E^(2*x)*(4*x^2 + x^3) + E^x*(2*x - 4*x^2 + 7*x^3 + 2*x^4) + (E^(2*x)*(4*x + x^2) + E^x*(-4 - 5*x + 7*x^2 + 2*x^3))*Log[16 + 8*x + x^2]))/(4*x^2 + x^3),x]
3.12.42.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.78 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71
method | result | size |
parallelrisch | \(\frac {\left (8 \,{\mathrm e}^{x} x +8 \,{\mathrm e}^{x} \ln \left (x^{2}+8 x +16\right )\right ) {\mathrm e}^{{\mathrm e}^{x}+x^{2}-2 x}}{8 x}\) | \(41\) |
risch | \(\frac {\left (-i \pi \operatorname {csgn}\left (i \left (4+x \right )\right )^{2} \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{3}+2 x +4 \ln \left (4+x \right )\right ) {\mathrm e}^{x^{2}+{\mathrm e}^{x}-x}}{2 x}\) | \(85\) |
int((((x^2+4*x)*exp(x)^2+(2*x^3+7*x^2-5*x-4)*exp(x))*ln(x^2+8*x+16)+(x^3+4 *x^2)*exp(x)^2+(2*x^4+7*x^3-4*x^2+2*x)*exp(x))/(x^3+4*x^2)/exp(-exp(x)-x^2 +2*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx=\frac {{\left (x e^{x} + e^{x} \log \left (x^{2} + 8 \, x + 16\right )\right )} e^{\left (x^{2} - 2 \, x + e^{x}\right )}}{x} \]
integrate((((x^2+4*x)*exp(x)^2+(2*x^3+7*x^2-5*x-4)*exp(x))*log(x^2+8*x+16) +(x^3+4*x^2)*exp(x)^2+(2*x^4+7*x^3-4*x^2+2*x)*exp(x))/(x^3+4*x^2)/exp(-exp (x)-x^2+2*x),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx=\frac {\left (x e^{x} + e^{x} \log {\left (x^{2} + 8 x + 16 \right )}\right ) e^{x^{2} - 2 x + e^{x}}}{x} \]
integrate((((x**2+4*x)*exp(x)**2+(2*x**3+7*x**2-5*x-4)*exp(x))*ln(x**2+8*x +16)+(x**3+4*x**2)*exp(x)**2+(2*x**4+7*x**3-4*x**2+2*x)*exp(x))/(x**3+4*x* *2)/exp(-exp(x)-x**2+2*x),x)
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx=\frac {{\left (x + 2 \, \log \left (x + 4\right )\right )} e^{\left (x^{2} - x + e^{x}\right )}}{x} \]
integrate((((x^2+4*x)*exp(x)^2+(2*x^3+7*x^2-5*x-4)*exp(x))*log(x^2+8*x+16) +(x^3+4*x^2)*exp(x)^2+(2*x^4+7*x^3-4*x^2+2*x)*exp(x))/(x^3+4*x^2)/exp(-exp (x)-x^2+2*x),x, algorithm=\
Time = 0.36 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx=\frac {{\left (x e^{\left (x^{2} + e^{x}\right )} + e^{\left (x^{2} + e^{x}\right )} \log \left (x^{2} + 8 \, x + 16\right )\right )} e^{\left (-x\right )}}{x} \]
integrate((((x^2+4*x)*exp(x)^2+(2*x^3+7*x^2-5*x-4)*exp(x))*log(x^2+8*x+16) +(x^3+4*x^2)*exp(x)^2+(2*x^4+7*x^3-4*x^2+2*x)*exp(x))/(x^3+4*x^2)/exp(-exp (x)-x^2+2*x),x, algorithm=\
Timed out. \[ \int \frac {e^{e^x-2 x+x^2} \left (e^{2 x} \left (4 x^2+x^3\right )+e^x \left (2 x-4 x^2+7 x^3+2 x^4\right )+\left (e^{2 x} \left (4 x+x^2\right )+e^x \left (-4-5 x+7 x^2+2 x^3\right )\right ) \log \left (16+8 x+x^2\right )\right )}{4 x^2+x^3} \, dx=\int \frac {{\mathrm {e}}^{{\mathrm {e}}^x-2\,x+x^2}\,\left ({\mathrm {e}}^{2\,x}\,\left (x^3+4\,x^2\right )+{\mathrm {e}}^x\,\left (2\,x^4+7\,x^3-4\,x^2+2\,x\right )+\ln \left (x^2+8\,x+16\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (x^2+4\,x\right )-{\mathrm {e}}^x\,\left (-2\,x^3-7\,x^2+5\,x+4\right )\right )\right )}{x^3+4\,x^2} \,d x \]
int((exp(exp(x) - 2*x + x^2)*(exp(2*x)*(4*x^2 + x^3) + exp(x)*(2*x - 4*x^2 + 7*x^3 + 2*x^4) + log(8*x + x^2 + 16)*(exp(2*x)*(4*x + x^2) - exp(x)*(5* x - 7*x^2 - 2*x^3 + 4))))/(4*x^2 + x^3),x)