\(\int \frac {e^{\frac {-10-2 x+(-3 x-x^2) \log (5)}{(5+x) \log (5)}} (15+10 x+x^2)}{25+10 x+x^2} \, dx\) [4157]

Optimal result

Integrand size = 48, antiderivative size = 24 \[ \int \frac {e^{\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}} \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx=1-e^{-x+\frac {2 x}{5+x}-\frac {2}{\log (5)}} \]

[Out]

1-exp(-2/ln(5)+2*x/(5+x)-x)

Rubi [F]

\[ \int \frac {e^{\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}} \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx=\int \frac {\exp \left (\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}\right ) \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx \]

[In]

Int[(E^((-10 - 2*x + (-3*x - x^2)*Log[5])/((5 + x)*Log[5]))*(15 + 10*x + x^2))/(25 + 10*x + x^2),x]

[Out]

Defer[Int][E^((-10 - x^2*Log[5] - x*(2 + Log[125]))/((5 + x)*Log[5])), x] - 10*Defer[Int][E^((-10 - x^2*Log[5]
- x*(2 + Log[125]))/((5 + x)*Log[5]))/(5 + x)^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}\right ) \left (15+10 x+x^2\right )}{(5+x)^2} \, dx \\ & = \int \frac {\exp \left (\frac {-10-x^2 \log (5)-x (2+\log (125))}{(5+x) \log (5)}\right ) \left (15+10 x+x^2\right )}{(5+x)^2} \, dx \\ & = \int \left (\exp \left (\frac {-10-x^2 \log (5)-x (2+\log (125))}{(5+x) \log (5)}\right )-\frac {10 \exp \left (\frac {-10-x^2 \log (5)-x (2+\log (125))}{(5+x) \log (5)}\right )}{(5+x)^2}\right ) \, dx \\ & = -\left (10 \int \frac {\exp \left (\frac {-10-x^2 \log (5)-x (2+\log (125))}{(5+x) \log (5)}\right )}{(5+x)^2} \, dx\right )+\int \exp \left (\frac {-10-x^2 \log (5)-x (2+\log (125))}{(5+x) \log (5)}\right ) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}} \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx=-\frac {2\ 25^{\frac {10+3 x}{(5+x) \log (5)}} e^{-4-x-\frac {2}{\log (5)}} \log (5)}{\log (25)} \]

[In]

Integrate[(E^((-10 - 2*x + (-3*x - x^2)*Log[5])/((5 + x)*Log[5]))*(15 + 10*x + x^2))/(25 + 10*x + x^2),x]

[Out]

(-2*25^((10 + 3*x)/((5 + x)*Log[5]))*E^(-4 - x - 2/Log[5])*Log[5])/Log[25]

Maple [A] (verified)

Time = 3.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29

method	result	size
gosper	\(-{\mathrm e}^{-\frac {x^{2} \ln \left (5\right )+3 x \ln \left (5\right )+2 x +10}{\left (5+x \right ) \ln \left (5\right )}}\)	\(31\)
risch	\(-{\mathrm e}^{-\frac {x^{2} \ln \left (5\right )+3 x \ln \left (5\right )+2 x +10}{\left (5+x \right ) \ln \left (5\right )}}\)	\(31\)
parallelrisch	\(-{\mathrm e}^{\frac {\left (-x^{2}-3 x \right ) \ln \left (5\right )-2 x -10}{\left (5+x \right ) \ln \left (5\right )}}\)	\(31\)
norman	\(\frac {-x \,{\mathrm e}^{\frac {\left (-x^{2}-3 x \right ) \ln \left (5\right )-2 x -10}{\left (5+x \right ) \ln \left (5\right )}}-5 \,{\mathrm e}^{\frac {\left (-x^{2}-3 x \right ) \ln \left (5\right )-2 x -10}{\left (5+x \right ) \ln \left (5\right )}}}{5+x}\)	\(69\)

[In]

int((x^2+10*x+15)*exp(((-x^2-3*x)*ln(5)-2*x-10)/(5+x)/ln(5))/(x^2+10*x+25),x,method=_RETURNVERBOSE)

[Out]

-exp(-(x^2*ln(5)+3*x*ln(5)+2*x+10)/(5+x)/ln(5))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}} \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx=-e^{\left (-\frac {{\left (x^{2} + 3 \, x\right )} \log \left (5\right ) + 2 \, x + 10}{{\left (x + 5\right )} \log \left (5\right )}\right )} \]

[In]

integrate((x^2+10*x+15)*exp(((-x^2-3*x)*log(5)-2*x-10)/(5+x)/log(5))/(x^2+10*x+25),x, algorithm="fricas")

[Out]

-e^(-((x^2 + 3*x)*log(5) + 2*x + 10)/((x + 5)*log(5)))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}} \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx=- e^{\frac {- 2 x + \left (- x^{2} - 3 x\right ) \log {\left (5 \right )} - 10}{\left (x + 5\right ) \log {\left (5 \right )}}} \]

[In]

integrate((x**2+10*x+15)*exp(((-x**2-3*x)*ln(5)-2*x-10)/(5+x)/ln(5))/(x**2+10*x+25),x)

[Out]

-exp((-2*x + (-x**2 - 3*x)*log(5) - 10)/((x + 5)*log(5)))

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}} \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx=-e^{\left (-x - \frac {10}{x + 5} - \frac {2}{\log \left (5\right )} + 2\right )} \]

[In]

integrate((x^2+10*x+15)*exp(((-x^2-3*x)*log(5)-2*x-10)/(5+x)/log(5))/(x^2+10*x+25),x, algorithm="maxima")

[Out]

-e^(-x - 10/(x + 5) - 2/log(5) + 2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (23) = 46\).

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.71 \[ \int \frac {e^{\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}} \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx=-e^{\left (-\frac {x^{2} \log \left (5\right )}{x \log \left (5\right ) + 5 \, \log \left (5\right )} - \frac {3 \, x \log \left (5\right )}{x \log \left (5\right ) + 5 \, \log \left (5\right )} - \frac {2 \, x}{x \log \left (5\right ) + 5 \, \log \left (5\right )} - \frac {10}{x \log \left (5\right ) + 5 \, \log \left (5\right )}\right )} \]

[In]

integrate((x^2+10*x+15)*exp(((-x^2-3*x)*log(5)-2*x-10)/(5+x)/log(5))/(x^2+10*x+25),x, algorithm="giac")

[Out]

-e^(-x^2*log(5)/(x*log(5) + 5*log(5)) - 3*x*log(5)/(x*log(5) + 5*log(5)) - 2*x/(x*log(5) + 5*log(5)) - 10/(x*l
og(5) + 5*log(5)))

Mupad [B] (verification not implemented)

Time = 10.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {e^{\frac {-10-2 x+\left (-3 x-x^2\right ) \log (5)}{(5+x) \log (5)}} \left (15+10 x+x^2\right )}{25+10 x+x^2} \, dx=-{\left (\frac {1}{5}\right )}^{\frac {x^2+3\,x}{5\,\ln \left (5\right )+x\,\ln \left (5\right )}}\,{\mathrm {e}}^{-\frac {2\,x}{5\,\ln \left (5\right )+x\,\ln \left (5\right )}}\,{\mathrm {e}}^{-\frac {10}{5\,\ln \left (5\right )+x\,\ln \left (5\right )}} \]

[In]

int((exp(-(2*x + log(5)*(3*x + x^2) + 10)/(log(5)*(x + 5)))*(10*x + x^2 + 15))/(10*x + x^2 + 25),x)

[Out]

-(1/5)^((3*x + x^2)/(5*log(5) + x*log(5)))*exp(-(2*x)/(5*log(5) + x*log(5)))*exp(-10/(5*log(5) + x*log(5)))