Integrand size = 99, antiderivative size = 29 \[ \int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2+4 x^3+2 x^4} \, dx=x-\frac {e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}}}{2+2 x} \]
Time = 3.43 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2+4 x^3+2 x^4} \, dx=x-\frac {2^{-1+\frac {4 e^{-x}}{x}} e^{\frac {2 e^{16-x}}{x}}}{1+x} \]
Integrate[(E^x*(2*x^2 + 4*x^3 + 2*x^4) + E^((2*(E^16 + Log[4]))/(E^x*x))*( E^x*x^2 + E^16*(2 + 4*x + 2*x^2) + (2 + 4*x + 2*x^2)*Log[4]))/(E^x*(2*x^2 + 4*x^3 + 2*x^4)),x]
Leaf count is larger than twice the leaf count of optimal. \(83\) vs. \(2(29)=58\).
Time = 5.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2026, 2007, 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.
\(\displaystyle \int \frac {e^{-x} \left (e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2 x^2+4 x+2\right )+\left (2 x^2+4 x+2\right ) \log (4)\right )+e^x \left (2 x^4+4 x^3+2 x^2\right )\right )}{2 x^4+4 x^3+2 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2 x^2+4 x+2\right )+\left (2 x^2+4 x+2\right ) \log (4)\right )+e^x \left (2 x^4+4 x^3+2 x^2\right )\right )}{x^2 \left (2 x^2+4 x+2\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-x} \left (e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2 x^2+4 x+2\right )+\left (2 x^2+4 x+2\right ) \log (4)\right )+e^x \left (2 x^4+4 x^3+2 x^2\right )\right )}{x^2 \left (\sqrt {2} x+\sqrt {2}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2^{\frac {4 e^{-x}}{x}-1} e^{\frac {2 e^{16-x}}{x}-x} \left (e^x x^2+2 e^{16} x^2 \left (1+\frac {\log (16)}{2 e^{16}}\right )+4 e^{16} x \left (1+\frac {\log (4)}{e^{16}}\right )+2 e^{16} \left (1+\frac {\log (16)}{2 e^{16}}\right )\right )}{x^2 (x+1)^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x-\frac {2^{\frac {4 e^{-x}}{x}-3} e^{\frac {2 e^{16-x}}{x}-x} \left (x^2 \log (16)+4 x \log (4)+\log (16)\right )}{\left (\frac {e^{-x}}{x^2}+\frac {e^{-x}}{x}\right ) x^2 (x+1)^2 \log (2)}\) |
Int[(E^x*(2*x^2 + 4*x^3 + 2*x^4) + E^((2*(E^16 + Log[4]))/(E^x*x))*(E^x*x^ 2 + E^16*(2 + 4*x + 2*x^2) + (2 + 4*x + 2*x^2)*Log[4]))/(E^x*(2*x^2 + 4*x^ 3 + 2*x^4)),x]
x - (2^(-3 + 4/(E^x*x))*E^((2*E^(16 - x))/x - x)*(4*x*Log[4] + Log[16] + x ^2*Log[16]))/((1/(E^x*x^2) + 1/(E^x*x))*x^2*(1 + x)^2*Log[2])
3.4.74.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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 = 0.92 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
risch | \(x -\frac {{\mathrm e}^{\frac {2 \left (2 \ln \left (2\right )+{\mathrm e}^{16}\right ) {\mathrm e}^{-x}}{x}}}{2 \left (1+x \right )}\) | \(27\) |
parallelrisch | \(\frac {-2+2 x^{2}-{\mathrm e}^{\frac {2 \left (2 \ln \left (2\right )+{\mathrm e}^{16}\right ) {\mathrm e}^{-x}}{x}}}{2+2 x}\) | \(35\) |
norman | \(\frac {\left ({\mathrm e}^{x} x^{3}-{\mathrm e}^{x} x -\frac {{\mathrm e}^{x} x \,{\mathrm e}^{\frac {2 \left (2 \ln \left (2\right )+{\mathrm e}^{16}\right ) {\mathrm e}^{-x}}{x}}}{2}\right ) {\mathrm e}^{-x}}{x \left (1+x \right )}\) | \(49\) |
int(((exp(x)*x^2+2*(2*x^2+4*x+2)*ln(2)+(2*x^2+4*x+2)*exp(16))*exp((2*ln(2) +exp(16))/exp(x)/x)^2+(2*x^4+4*x^3+2*x^2)*exp(x))/(2*x^4+4*x^3+2*x^2)/exp( x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2+4 x^3+2 x^4} \, dx=\frac {2 \, x^{2} + 2 \, x - e^{\left (\frac {2 \, {\left (e^{16} + 2 \, \log \left (2\right )\right )} e^{\left (-x\right )}}{x}\right )}}{2 \, {\left (x + 1\right )}} \]
integrate(((exp(x)*x^2+2*(2*x^2+4*x+2)*log(2)+(2*x^2+4*x+2)*exp(16))*exp(( 2*log(2)+exp(16))/exp(x)/x)^2+(2*x^4+4*x^3+2*x^2)*exp(x))/(2*x^4+4*x^3+2*x ^2)/exp(x),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2+4 x^3+2 x^4} \, dx=x - \frac {e^{\frac {2 \cdot \left (2 \log {\left (2 \right )} + e^{16}\right ) e^{- x}}{x}}}{2 x + 2} \]
integrate(((exp(x)*x**2+2*(2*x**2+4*x+2)*ln(2)+(2*x**2+4*x+2)*exp(16))*exp ((2*ln(2)+exp(16))/exp(x)/x)**2+(2*x**4+4*x**3+2*x**2)*exp(x))/(2*x**4+4*x **3+2*x**2)/exp(x),x)
\[ \int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2+4 x^3+2 x^4} \, dx=\int { \frac {{\left (2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} e^{x} + {\left (x^{2} e^{x} + 2 \, {\left (x^{2} + 2 \, x + 1\right )} e^{16} + 4 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (2\right )\right )} e^{\left (\frac {2 \, {\left (e^{16} + 2 \, \log \left (2\right )\right )} e^{\left (-x\right )}}{x}\right )}\right )} e^{\left (-x\right )}}{2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}} \,d x } \]
integrate(((exp(x)*x^2+2*(2*x^2+4*x+2)*log(2)+(2*x^2+4*x+2)*exp(16))*exp(( 2*log(2)+exp(16))/exp(x)/x)^2+(2*x^4+4*x^3+2*x^2)*exp(x))/(2*x^4+4*x^3+2*x ^2)/exp(x),x, algorithm=\
(x^2 + x - 1)/(x + 1) + 1/(x + 1) + 1/2*integrate((2*x^2*(e^16 + 2*log(2)) + x^2*e^x + 4*x*(e^16 + 2*log(2)) + 2*e^16 + 4*log(2))*e^(-x + 4*e^(-x)*l og(2)/x + 2*e^(-x + 16)/x)/(x^4 + 2*x^3 + x^2), x)
\[ \int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2+4 x^3+2 x^4} \, dx=\int { \frac {{\left (2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} e^{x} + {\left (x^{2} e^{x} + 2 \, {\left (x^{2} + 2 \, x + 1\right )} e^{16} + 4 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (2\right )\right )} e^{\left (\frac {2 \, {\left (e^{16} + 2 \, \log \left (2\right )\right )} e^{\left (-x\right )}}{x}\right )}\right )} e^{\left (-x\right )}}{2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}} \,d x } \]
integrate(((exp(x)*x^2+2*(2*x^2+4*x+2)*log(2)+(2*x^2+4*x+2)*exp(16))*exp(( 2*log(2)+exp(16))/exp(x)/x)^2+(2*x^4+4*x^3+2*x^2)*exp(x))/(2*x^4+4*x^3+2*x ^2)/exp(x),x, algorithm=\
integrate(1/2*(2*(x^4 + 2*x^3 + x^2)*e^x + (x^2*e^x + 2*(x^2 + 2*x + 1)*e^ 16 + 4*(x^2 + 2*x + 1)*log(2))*e^(2*(e^16 + 2*log(2))*e^(-x)/x))*e^(-x)/(x ^4 + 2*x^3 + x^2), x)
Timed out. \[ \int \frac {e^{-x} \left (e^x \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2 e^{-x} \left (e^{16}+\log (4)\right )}{x}} \left (e^x x^2+e^{16} \left (2+4 x+2 x^2\right )+\left (2+4 x+2 x^2\right ) \log (4)\right )\right )}{2 x^2+4 x^3+2 x^4} \, dx=\int \frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^x\,\left (2\,x^4+4\,x^3+2\,x^2\right )+{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{16}+2\,\ln \left (2\right )\right )}{x}}\,\left (x^2\,{\mathrm {e}}^x+{\mathrm {e}}^{16}\,\left (2\,x^2+4\,x+2\right )+2\,\ln \left (2\right )\,\left (2\,x^2+4\,x+2\right )\right )\right )}{2\,x^4+4\,x^3+2\,x^2} \,d x \]
int((exp(-x)*(exp(x)*(2*x^2 + 4*x^3 + 2*x^4) + exp((2*exp(-x)*(exp(16) + 2 *log(2)))/x)*(x^2*exp(x) + exp(16)*(4*x + 2*x^2 + 2) + 2*log(2)*(4*x + 2*x ^2 + 2))))/(2*x^2 + 4*x^3 + 2*x^4),x)