Integrand size = 107, antiderivative size = 24 \[ \int \frac {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx=\log \left (\frac {8 \left (\frac {x}{e^{(5-x) x}+x}+\log (5)\right )}{x}\right ) \]
Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx=-\log (x)-\log \left (e^{5 x}+e^{x^2} x\right )+\log \left (e^{x^2} x+e^{5 x} \log (5)+e^{x^2} x \log (5)\right ) \]
Integrate[(-x^2 - E^(10*x - 2*x^2)*Log[5] - x^2*Log[5] + E^(5*x - x^2)*(-5 *x^2 + 2*x^3 - 2*x*Log[5]))/(x^3 + E^(10*x - 2*x^2)*x*Log[5] + x^3*Log[5] + E^(5*x - x^2)*(x^2 + 2*x^2*Log[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.
\(\displaystyle \int \frac {-x^2+x^2 (-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (2 x^3-5 x^2-2 x \log (5)\right )}{x^3+x^3 \log (5)+e^{10 x-2 x^2} x \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (2 x^3-5 x^2-2 x \log (5)\right )}{x^3+x^3 \log (5)+e^{10 x-2 x^2} x \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (2 x^3-5 x^2-2 x \log (5)\right )}{x^3 (1+\log (5))+e^{10 x-2 x^2} x \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x^2} \left (x^2 (-1-\log (5))-e^{10 x-2 x^2} \log (5)+e^{5 x-x^2} \left (2 x^3-5 x^2-2 x \log (5)\right )\right )}{x \left (e^{x^2} x+e^{5 x}\right ) \left (e^{x^2} x (1+\log (5))+e^{5 x} \log (5)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{2 x^2} x \left (2 x^2-5 x+1\right )}{e^{x (x+5)} x+e^{10 x}}+\frac {e^{2 x^2-5 x} x \left (-2 x^2+5 x-1\right ) (1+\log (5))^2}{\log (5) \left (e^{x^2} x (1+\log (5))+e^{5 x} \log (5)\right )}+\frac {e^{x^2-5 x} \left (2 x^2-5 x+1\right )}{\log (5)}-\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^{2 x^2} x}{e^{x (x+5)} x+e^{10 x}}dx-5 \int \frac {e^{2 x^2} x^2}{e^{x (x+5)} x+e^{10 x}}dx+\frac {(1+\log (5))^2 \int \frac {e^{2 x^2-5 x} x}{-e^{x^2} (1+\log (5)) x-e^{5 x} \log (5)}dx}{\log (5)}+\frac {5 (1+\log (5))^2 \int \frac {e^{2 x^2-5 x} x^2}{e^{x^2} (1+\log (5)) x+e^{5 x} \log (5)}dx}{\log (5)}+2 \int \frac {e^{2 x^2} x^3}{e^{x (x+5)} x+e^{10 x}}dx+\frac {2 (1+\log (5))^2 \int \frac {e^{2 x^2-5 x} x^3}{-e^{x^2} (1+\log (5)) x-e^{5 x} \log (5)}dx}{\log (5)}+\frac {e^{x^2-5 x} \left (5 x-2 x^2\right )}{(5-2 x) \log (5)}-\log (x)\) |
Int[(-x^2 - E^(10*x - 2*x^2)*Log[5] - x^2*Log[5] + E^(5*x - x^2)*(-5*x^2 + 2*x^3 - 2*x*Log[5]))/(x^3 + E^(10*x - 2*x^2)*x*Log[5] + x^3*Log[5] + E^(5 *x - x^2)*(x^2 + 2*x^2*Log[5])),x]
3.26.14.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54
method | result | size |
risch | \(-\ln \left (x \right )+\ln \left ({\mathrm e}^{-\left (-5+x \right ) x}+\frac {\left (\ln \left (5\right )+1\right ) x}{\ln \left (5\right )}\right )-\ln \left (x +{\mathrm e}^{-\left (-5+x \right ) x}\right )\) | \(37\) |
norman | \(-\ln \left (x \right )-\ln \left (x +{\mathrm e}^{-x^{2}+5 x}\right )+\ln \left (x \ln \left (5\right )+{\mathrm e}^{-x^{2}+5 x} \ln \left (5\right )+x \right )\) | \(41\) |
parallelrisch | \(-\ln \left (x \right )+\ln \left (\frac {x \ln \left (5\right )+{\mathrm e}^{-x^{2}+5 x} \ln \left (5\right )+x}{\ln \left (5\right )+1}\right )-\ln \left (x +{\mathrm e}^{-x^{2}+5 x}\right )\) | \(48\) |
int((-ln(5)*exp(-x^2+5*x)^2+(-2*x*ln(5)+2*x^3-5*x^2)*exp(-x^2+5*x)-x^2*ln( 5)-x^2)/(x*ln(5)*exp(-x^2+5*x)^2+(2*x^2*ln(5)+x^2)*exp(-x^2+5*x)+x^3*ln(5) +x^3),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx=\log \left (x \log \left (5\right ) + e^{\left (-x^{2} + 5 \, x\right )} \log \left (5\right ) + x\right ) - \log \left (x + e^{\left (-x^{2} + 5 \, x\right )}\right ) - \log \left (x\right ) \]
integrate((-log(5)*exp(-x^2+5*x)^2+(-2*x*log(5)+2*x^3-5*x^2)*exp(-x^2+5*x) -x^2*log(5)-x^2)/(x*log(5)*exp(-x^2+5*x)^2+(2*x^2*log(5)+x^2)*exp(-x^2+5*x )+x^3*log(5)+x^3),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx=- \log {\left (x \right )} - \log {\left (x + e^{- x^{2} + 5 x} \right )} + \log {\left (\frac {2 x + 2 x \log {\left (5 \right )}}{2 \log {\left (5 \right )}} + e^{- x^{2} + 5 x} \right )} \]
integrate((-ln(5)*exp(-x**2+5*x)**2+(-2*x*ln(5)+2*x**3-5*x**2)*exp(-x**2+5 *x)-x**2*ln(5)-x**2)/(x*ln(5)*exp(-x**2+5*x)**2+(2*x**2*ln(5)+x**2)*exp(-x **2+5*x)+x**3*ln(5)+x**3),x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (22) = 44\).
Time = 0.36 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx=-\log \left (x\right ) - \log \left (\frac {x e^{\left (x^{2}\right )} + e^{\left (5 \, x\right )}}{x}\right ) + \log \left (\frac {x {\left (\log \left (5\right ) + 1\right )} e^{\left (x^{2}\right )} + e^{\left (5 \, x\right )} \log \left (5\right )}{x {\left (\log \left (5\right ) + 1\right )}}\right ) \]
integrate((-log(5)*exp(-x^2+5*x)^2+(-2*x*log(5)+2*x^3-5*x^2)*exp(-x^2+5*x) -x^2*log(5)-x^2)/(x*log(5)*exp(-x^2+5*x)^2+(2*x^2*log(5)+x^2)*exp(-x^2+5*x )+x^3*log(5)+x^3),x, algorithm=\
-log(x) - log((x*e^(x^2) + e^(5*x))/x) + log((x*(log(5) + 1)*e^(x^2) + e^( 5*x)*log(5))/(x*(log(5) + 1)))
Time = 0.39 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx=\log \left (-x \log \left (5\right ) - e^{\left (-x^{2} + 5 \, x\right )} \log \left (5\right ) - x\right ) - \log \left (x + e^{\left (-x^{2} + 5 \, x\right )}\right ) - \log \left (x\right ) \]
integrate((-log(5)*exp(-x^2+5*x)^2+(-2*x*log(5)+2*x^3-5*x^2)*exp(-x^2+5*x) -x^2*log(5)-x^2)/(x*log(5)*exp(-x^2+5*x)^2+(2*x^2*log(5)+x^2)*exp(-x^2+5*x )+x^3*log(5)+x^3),x, algorithm=\
Time = 1.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-x^2-e^{10 x-2 x^2} \log (5)-x^2 \log (5)+e^{5 x-x^2} \left (-5 x^2+2 x^3-2 x \log (5)\right )}{x^3+e^{10 x-2 x^2} x \log (5)+x^3 \log (5)+e^{5 x-x^2} \left (x^2+2 x^2 \log (5)\right )} \, dx=\ln \left (x+x\,\ln \left (5\right )+{\mathrm {e}}^{-x\,\left (x-5\right )}\,\ln \left (5\right )\right )-\ln \left (x+{\mathrm {e}}^{-x\,\left (x-5\right )}\right )-\ln \left (x\right ) \]