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

Optimal. Leaf size=24 \[ 1-e^{-x+\frac {2 x}{5+x}-\frac {2}{\log (5)}} \]

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Rubi [F]  time = 1.24, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [F]  time = 0.91, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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]

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]

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fricas [A]  time = 0.55, size = 29, normalized size = 1.21 \begin {gather*} -e^{\left (-\frac {{\left (x^{2} + 3 \, x\right )} \log \relax (5) + 2 \, x + 10}{{\left (x + 5\right )} \log \relax (5)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [B]  time = 0.15, size = 65, normalized size = 2.71 \begin {gather*} -e^{\left (-\frac {x^{2} \log \relax (5)}{x \log \relax (5) + 5 \, \log \relax (5)} - \frac {3 \, x \log \relax (5)}{x \log \relax (5) + 5 \, \log \relax (5)} - \frac {2 \, x}{x \log \relax (5) + 5 \, \log \relax (5)} - \frac {10}{x \log \relax (5) + 5 \, \log \relax (5)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [A]  time = 0.20, size = 31, normalized size = 1.29

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.83, size = 21, normalized size = 0.88 \begin {gather*} -e^{\left (-x - \frac {10}{x + 5} - \frac {2}{\log \relax (5)} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.97, size = 52, normalized size = 2.17 \begin {gather*} -{\left (\frac {1}{5}\right )}^{\frac {x^2+3\,x}{5\,\ln \relax (5)+x\,\ln \relax (5)}}\,{\mathrm {e}}^{-\frac {2\,x}{5\,\ln \relax (5)+x\,\ln \relax (5)}}\,{\mathrm {e}}^{-\frac {10}{5\,\ln \relax (5)+x\,\ln \relax (5)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.18, size = 26, normalized size = 1.08 \begin {gather*} - e^{\frac {- 2 x + \left (- x^{2} - 3 x\right ) \log {\relax (5 )} - 10}{\left (x + 5\right ) \log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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