Integrand size = 66, antiderivative size = 26 \begin {dmath*} \int \frac {1}{5} e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx=e^{4+\frac {4 e^x}{5 x}} \left (4+e^x\right ) x^2 (3+x) \end {dmath*}
Time = 5.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {1}{5} e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx=e^{4+\frac {4 e^x}{5 x}} \left (4+e^x\right ) x^2 (3+x) \end {dmath*}
Integrate[(E^((4*E^x + 20*x)/(5*x))*(120*x + 60*x^2 + E^(2*x)*(-12 + 8*x + 4*x^2) + E^x*(-48 + 62*x + 46*x^2 + 5*x^3)))/5,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.
\(\displaystyle \int \frac {1}{5} e^{\frac {20 x+4 e^x}{5 x}} \left (60 x^2+e^{2 x} \left (4 x^2+8 x-12\right )+e^x \left (5 x^3+46 x^2+62 x-48\right )+120 x\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int e^{\frac {4 \left (5 x+e^x\right )}{5 x}} \left (60 x^2+120 x-4 e^{2 x} \left (-x^2-2 x+3\right )-e^x \left (-5 x^3-46 x^2-62 x+48\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (60 e^{\frac {4 \left (5 x+e^x\right )}{5 x}} x^2+120 e^{\frac {4 \left (5 x+e^x\right )}{5 x}} x+4 e^{2 x+\frac {4 \left (5 x+e^x\right )}{5 x}} \left (x^2+2 x-3\right )+e^{x+\frac {4 \left (5 x+e^x\right )}{5 x}} \left (5 x^3+46 x^2+62 x-48\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (5 \int e^{x+\frac {4 \left (5 x+e^x\right )}{5 x}} x^3dx+60 \int e^{\frac {4 \left (5 x+e^x\right )}{5 x}} x^2dx+46 \int e^{x+\frac {4 \left (5 x+e^x\right )}{5 x}} x^2dx+4 \int e^{2 x+\frac {4 \left (5 x+e^x\right )}{5 x}} x^2dx-48 \int e^{x+\frac {4 \left (5 x+e^x\right )}{5 x}}dx-12 \int e^{2 x+\frac {4 \left (5 x+e^x\right )}{5 x}}dx+120 \int e^{\frac {4 \left (5 x+e^x\right )}{5 x}} xdx+62 \int e^{x+\frac {4 \left (5 x+e^x\right )}{5 x}} xdx+8 \int e^{2 x+\frac {4 \left (5 x+e^x\right )}{5 x}} xdx\right )\) |
Int[(E^((4*E^x + 20*x)/(5*x))*(120*x + 60*x^2 + E^(2*x)*(-12 + 8*x + 4*x^2 ) + E^x*(-48 + 62*x + 46*x^2 + 5*x^3)))/5,x]
3.9.97.3.1 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 = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\left ({\mathrm e}^{x}+4\right ) x^{2} \left (3+x \right ) {\mathrm e}^{\frac {4 x +\frac {4 \,{\mathrm e}^{x}}{5}}{x}}\) | \(24\) |
parallelrisch | \({\mathrm e}^{\frac {4 x +\frac {4 \,{\mathrm e}^{x}}{5}}{x}} {\mathrm e}^{x} x^{3}+4 x^{3} {\mathrm e}^{\frac {4 x +\frac {4 \,{\mathrm e}^{x}}{5}}{x}}+3 \,{\mathrm e}^{\frac {4 x +\frac {4 \,{\mathrm e}^{x}}{5}}{x}} x^{2} {\mathrm e}^{x}+12 \,{\mathrm e}^{\frac {4 x +\frac {4 \,{\mathrm e}^{x}}{5}}{x}} x^{2}\) | \(73\) |
norman | \({\mathrm e}^{x} x^{3} {\mathrm e}^{\frac {4 \,{\mathrm e}^{x}+20 x}{5 x}}+12 x^{2} {\mathrm e}^{\frac {4 \,{\mathrm e}^{x}+20 x}{5 x}}+4 x^{3} {\mathrm e}^{\frac {4 \,{\mathrm e}^{x}+20 x}{5 x}}+3 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {4 \,{\mathrm e}^{x}+20 x}{5 x}}\) | \(81\) |
int(1/5*((4*x^2+8*x-12)*exp(x)^2+(5*x^3+46*x^2+62*x-48)*exp(x)+60*x^2+120* x)*exp(1/5*(4*exp(x)+20*x)/x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \begin {dmath*} \int \frac {1}{5} e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx={\left (4 \, x^{3} + 12 \, x^{2} + {\left (x^{3} + 3 \, x^{2}\right )} e^{x}\right )} e^{\left (\frac {4 \, {\left (5 \, x + e^{x}\right )}}{5 \, x}\right )} \end {dmath*}
integrate(1/5*((4*x^2+8*x-12)*exp(x)^2+(5*x^3+46*x^2+62*x-48)*exp(x)+60*x^ 2+120*x)*exp(1/5*(4*exp(x)+20*x)/x),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \begin {dmath*} \int \frac {1}{5} e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx=\left (x^{3} e^{x} + 4 x^{3} + 3 x^{2} e^{x} + 12 x^{2}\right ) e^{\frac {4 x + \frac {4 e^{x}}{5}}{x}} \end {dmath*}
integrate(1/5*((4*x**2+8*x-12)*exp(x)**2+(5*x**3+46*x**2+62*x-48)*exp(x)+6 0*x**2+120*x)*exp(1/5*(4*exp(x)+20*x)/x),x)
Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \begin {dmath*} \int \frac {1}{5} e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx={\left (4 \, x^{3} e^{4} + 12 \, x^{2} e^{4} + {\left (x^{3} e^{4} + 3 \, x^{2} e^{4}\right )} e^{x}\right )} e^{\left (\frac {4 \, e^{x}}{5 \, x}\right )} \end {dmath*}
integrate(1/5*((4*x^2+8*x-12)*exp(x)^2+(5*x^3+46*x^2+62*x-48)*exp(x)+60*x^ 2+120*x)*exp(1/5*(4*exp(x)+20*x)/x),x, algorithm=\
\begin {dmath*} \int \frac {1}{5} e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx=\int { \frac {1}{5} \, {\left (60 \, x^{2} + 4 \, {\left (x^{2} + 2 \, x - 3\right )} e^{\left (2 \, x\right )} + {\left (5 \, x^{3} + 46 \, x^{2} + 62 \, x - 48\right )} e^{x} + 120 \, x\right )} e^{\left (\frac {4 \, {\left (5 \, x + e^{x}\right )}}{5 \, x}\right )} \,d x } \end {dmath*}
integrate(1/5*((4*x^2+8*x-12)*exp(x)^2+(5*x^3+46*x^2+62*x-48)*exp(x)+60*x^ 2+120*x)*exp(1/5*(4*exp(x)+20*x)/x),x, algorithm=\
integrate(1/5*(60*x^2 + 4*(x^2 + 2*x - 3)*e^(2*x) + (5*x^3 + 46*x^2 + 62*x - 48)*e^x + 120*x)*e^(4/5*(5*x + e^x)/x), x)
Time = 16.47 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \begin {dmath*} \int \frac {1}{5} e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx=x^2\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^x}{5\,x}+4}\,\left ({\mathrm {e}}^x+4\right )\,\left (x+3\right ) \end {dmath*}