\(\int \frac {e^{\frac {1}{27} (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}})+\frac {2+2 x+e^x (1+x)}{x}} (-50+e^x (-25+25 x+25 x^2))}{27 x^2} \, dx\) [7706]

Optimal result

Integrand size = 68, antiderivative size = 26 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\frac {25}{9} \left (-8+\frac {1}{3} e^{\frac {\left (2+e^x\right ) (1+x)}{x}}\right )} \]

[Out]

exp(25/27*exp((exp(x)+2)/x*(1+x))-200/9)

Rubi [F]

\[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=\int \frac {\exp \left (\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}\right ) \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx \]

[In]

Int[(E^((-600 + 25*E^((2 + 2*x + E^x*(1 + x))/x))/27 + (2 + 2*x + E^x*(1 + x))/x)*(-50 + E^x*(-25 + 25*x + 25*
x^2)))/(27*x^2),x]

[Out]

(25*Defer[Int][E^(x + (54 + 27*E^x - 546*x + 27*E^x*x + 25*E^(((2 + E^x)*(1 + x))/x)*x)/(27*x)), x])/27 - (50*
Defer[Int][E^((54 + 27*E^x - 546*x + 27*E^x*x + 25*E^(((2 + E^x)*(1 + x))/x)*x)/(27*x))/x^2, x])/27 - (25*Defe
r[Int][E^(x + (54 + 27*E^x - 546*x + 27*E^x*x + 25*E^(((2 + E^x)*(1 + x))/x)*x)/(27*x))/x^2, x])/27 + (25*Defe
r[Int][E^(x + (54 + 27*E^x - 546*x + 27*E^x*x + 25*E^(((2 + E^x)*(1 + x))/x)*x)/(27*x))/x, x])/27

Rubi steps \begin{align*} \text {integral}& = \frac {1}{27} \int \frac {\exp \left (\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}\right ) \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{x^2} \, dx \\ & = \frac {1}{27} \int \frac {\exp \left (\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right ) \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{x^2} \, dx \\ & = \frac {1}{27} \int \left (-\frac {50 \exp \left (\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )}{x^2}+\frac {25 \exp \left (x+\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right ) \left (-1+x+x^2\right )}{x^2}\right ) \, dx \\ & = \frac {25}{27} \int \frac {\exp \left (x+\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right ) \left (-1+x+x^2\right )}{x^2} \, dx-\frac {50}{27} \int \frac {\exp \left (\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )}{x^2} \, dx \\ & = \frac {25}{27} \int \left (\exp \left (x+\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )-\frac {\exp \left (x+\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )}{x^2}+\frac {\exp \left (x+\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )}{x}\right ) \, dx-\frac {50}{27} \int \frac {\exp \left (\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )}{x^2} \, dx \\ & = \frac {25}{27} \int \exp \left (x+\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right ) \, dx-\frac {25}{27} \int \frac {\exp \left (x+\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )}{x^2} \, dx+\frac {25}{27} \int \frac {\exp \left (x+\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )}{x} \, dx-\frac {50}{27} \int \frac {\exp \left (\frac {54+27 e^x-546 x+27 e^x x+25 e^{\frac {\left (2+e^x\right ) (1+x)}{x}} x}{27 x}\right )}{x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\frac {25}{27} \left (-24+e^{\frac {\left (2+e^x\right ) (1+x)}{x}}\right )} \]

[In]

Integrate[(E^((-600 + 25*E^((2 + 2*x + E^x*(1 + x))/x))/27 + (2 + 2*x + E^x*(1 + x))/x)*(-50 + E^x*(-25 + 25*x
+ 25*x^2)))/(27*x^2),x]

[Out]

E^((25*(-24 + E^(((2 + E^x)*(1 + x))/x)))/27)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69

method	result	size
risch	\({\mathrm e}^{\frac {25 \,{\mathrm e}^{\frac {\left ({\mathrm e}^{x}+2\right ) \left (1+x \right )}{x}}}{27}-\frac {200}{9}}\)	\(18\)
norman	\({\mathrm e}^{\frac {25 \,{\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{x}+2 x +2}{x}}}{27}-\frac {200}{9}}\)	\(22\)
parallelrisch	\({\mathrm e}^{\frac {25 \,{\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{x}+2 x +2}{x}}}{27}-\frac {200}{9}}\)	\(22\)

[In]

int(1/27*((25*x^2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*exp(25/27*exp(((1+x)*exp(x)+2*x+2)/x)-200/9)
/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(25/27*exp((exp(x)+2)/x*(1+x))-200/9)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (17) = 34\).

Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\left (\frac {27 \, {\left (x + 1\right )} e^{x} + 25 \, x e^{\left (\frac {{\left (x + 1\right )} e^{x} + 2 \, x + 2}{x}\right )} - 546 \, x + 54}{27 \, x} - \frac {{\left (x + 1\right )} e^{x} + 2 \, x + 2}{x}\right )} \]

[In]

integrate(1/27*((25*x^2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*exp(25/27*exp(((1+x)*exp(x)+2*x+2)/x)-
200/9)/x^2,x, algorithm="fricas")

[Out]

e^(1/27*(27*(x + 1)*e^x + 25*x*e^(((x + 1)*e^x + 2*x + 2)/x) - 546*x + 54)/x - ((x + 1)*e^x + 2*x + 2)/x)

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\frac {25 e^{\frac {2 x + \left (x + 1\right ) e^{x} + 2}{x}}}{27} - \frac {200}{9}} \]

[In]

integrate(1/27*((25*x**2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*exp(25/27*exp(((1+x)*exp(x)+2*x+2)/x)
-200/9)/x**2,x)

[Out]

exp(25*exp((2*x + (x + 1)*exp(x) + 2)/x)/27 - 200/9)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\left (\frac {25}{27} \, e^{\left (\frac {e^{x}}{x} + \frac {2}{x} + e^{x} + 2\right )} - \frac {200}{9}\right )} \]

[In]

integrate(1/27*((25*x^2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*exp(25/27*exp(((1+x)*exp(x)+2*x+2)/x)-
200/9)/x^2,x, algorithm="maxima")

[Out]

e^(25/27*e^(e^x/x + 2/x + e^x + 2) - 200/9)

Giac [F]

\[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=\int { \frac {25 \, {\left ({\left (x^{2} + x - 1\right )} e^{x} - 2\right )} e^{\left (\frac {{\left (x + 1\right )} e^{x} + 2 \, x + 2}{x} + \frac {25}{27} \, e^{\left (\frac {{\left (x + 1\right )} e^{x} + 2 \, x + 2}{x}\right )} - \frac {200}{9}\right )}}{27 \, x^{2}} \,d x } \]

[In]

integrate(1/27*((25*x^2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*exp(25/27*exp(((1+x)*exp(x)+2*x+2)/x)-
200/9)/x^2,x, algorithm="giac")

[Out]

integrate(25/27*((x^2 + x - 1)*e^x - 2)*e^(((x + 1)*e^x + 2*x + 2)/x + 25/27*e^(((x + 1)*e^x + 2*x + 2)/x) - 2
00/9)/x^2, x)

Mupad [B] (verification not implemented)

Time = 13.40 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx={\mathrm {e}}^{-\frac {200}{9}}\,{\mathrm {e}}^{\frac {25\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{2/x}}{27}} \]

[In]

int((exp((25*exp((2*x + exp(x)*(x + 1) + 2)/x))/27 - 200/9)*exp((2*x + exp(x)*(x + 1) + 2)/x)*(exp(x)*(25*x +
25*x^2 - 25) - 50))/(27*x^2),x)

[Out]

exp(-200/9)*exp((25*exp(exp(x))*exp(2)*exp(exp(x)/x)*exp(2/x))/27)