Integrand size = 71, antiderivative size = 32 \[ \int \frac {-19-e^6+e^3 (-8-2 x)-9 x-x^2+\left (20+19 x+5 x^2\right ) \log (5)+\left (-9 x-2 e^3 x-2 x^2+\left (19 x+10 x^2\right ) \log (5)\right ) \log (x)}{x} \, dx=\left (-3-x-\left (4+e^3+x\right )^2+\left (-x+5 (2+x)^2\right ) \log (5)\right ) \log (x) \]
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.91 \[ \int \frac {-19-e^6+e^3 (-8-2 x)-9 x-x^2+\left (20+19 x+5 x^2\right ) \log (5)+\left (-9 x-2 e^3 x-2 x^2+\left (19 x+10 x^2\right ) \log (5)\right ) \log (x)}{x} \, dx=-19 \log (x)-8 e^3 \log (x)-e^6 \log (x)-9 x \log (x)-2 e^3 x \log (x)-x^2 \log (x)+20 \log (5) \log (x)+19 x \log (5) \log (x)+5 x^2 \log (5) \log (x) \]
Integrate[(-19 - E^6 + E^3*(-8 - 2*x) - 9*x - x^2 + (20 + 19*x + 5*x^2)*Lo g[5] + (-9*x - 2*E^3*x - 2*x^2 + (19*x + 10*x^2)*Log[5])*Log[x])/x,x]
-19*Log[x] - 8*E^3*Log[x] - E^6*Log[x] - 9*x*Log[x] - 2*E^3*x*Log[x] - x^2 *Log[x] + 20*Log[5]*Log[x] + 19*x*Log[5]*Log[x] + 5*x^2*Log[5]*Log[x]
Leaf count is larger than twice the leaf count of optimal. \(82\) vs. \(2(32)=64\).
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.56, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {2010, 2009}
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.
\(\displaystyle \int \frac {-x^2+\left (-2 x^2+\left (10 x^2+19 x\right ) \log (5)-2 e^3 x-9 x\right ) \log (x)+\left (5 x^2+19 x+20\right ) \log (5)-9 x+e^3 (-2 x-8)-e^6-19}{x} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {-\left (x^2 (1-5 \log (5))\right )-x \left (9+2 e^3-19 \log (5)\right )-e^6-8 e^3-19+20 \log (5)}{x}+\left (-2 x (1-5 \log (5))-2 e^3-9+19 \log (5)\right ) \log (x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log (x) \left (2 x (1-5 \log (5))+2 e^3+9-19 \log (5)\right )^2}{4 (1-5 \log (5))}+\frac {\left (9+2 e^3-19 \log (5)\right )^2 \log (x)}{4 (1-5 \log (5))}-\left (19+8 e^3+e^6-20 \log (5)\right ) \log (x)\) |
Int[(-19 - E^6 + E^3*(-8 - 2*x) - 9*x - x^2 + (20 + 19*x + 5*x^2)*Log[5] + (-9*x - 2*E^3*x - 2*x^2 + (19*x + 10*x^2)*Log[5])*Log[x])/x,x]
-((19 + 8*E^3 + E^6 - 20*Log[5])*Log[x]) + ((9 + 2*E^3 - 19*Log[5])^2*Log[ x])/(4*(1 - 5*Log[5])) - ((9 + 2*E^3 + 2*x*(1 - 5*Log[5]) - 19*Log[5])^2*L og[x])/(4*(1 - 5*Log[5]))
3.23.27.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47
method | result | size |
norman | \(\left (-{\mathrm e}^{6}+20 \ln \left (5\right )-8 \,{\mathrm e}^{3}-19\right ) \ln \left (x \right )+\left (5 \ln \left (5\right )-1\right ) x^{2} \ln \left (x \right )+\left (19 \ln \left (5\right )-2 \,{\mathrm e}^{3}-9\right ) x \ln \left (x \right )\) | \(47\) |
risch | \(\left (5 x^{2} \ln \left (5\right )+19 x \ln \left (5\right )-2 x \,{\mathrm e}^{3}-x^{2}-9 x \right ) \ln \left (x \right )-\ln \left (x \right ) {\mathrm e}^{6}+20 \ln \left (5\right ) \ln \left (x \right )-8 \ln \left (x \right ) {\mathrm e}^{3}-19 \ln \left (x \right )\) | \(53\) |
parallelrisch | \(5 x^{2} \ln \left (5\right ) \ln \left (x \right )+19 x \ln \left (5\right ) \ln \left (x \right )-2 x \,{\mathrm e}^{3} \ln \left (x \right )-x^{2} \ln \left (x \right )-\ln \left (x \right ) {\mathrm e}^{6}+20 \ln \left (5\right ) \ln \left (x \right )-8 \ln \left (x \right ) {\mathrm e}^{3}-9 x \ln \left (x \right )-19 \ln \left (x \right )\) | \(61\) |
parts | \(10 \ln \left (5\right ) \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+19 \ln \left (5\right ) \left (x \ln \left (x \right )-x \right )-2 \,{\mathrm e}^{3} \left (x \ln \left (x \right )-x \right )-x^{2} \ln \left (x \right )-9 x \ln \left (x \right )+\frac {5 x^{2} \ln \left (5\right )}{2}+19 x \ln \left (5\right )-2 x \,{\mathrm e}^{3}+\left (-{\mathrm e}^{6}+20 \ln \left (5\right )-8 \,{\mathrm e}^{3}-19\right ) \ln \left (x \right )\) | \(89\) |
default | \(10 \ln \left (5\right ) \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+19 \ln \left (5\right ) \left (x \ln \left (x \right )-x \right )-2 \,{\mathrm e}^{3} \left (x \ln \left (x \right )-x \right )-x^{2} \ln \left (x \right )+\frac {5 x^{2} \ln \left (5\right )}{2}-9 x \ln \left (x \right )+19 x \ln \left (5\right )-\ln \left (x \right ) {\mathrm e}^{6}-2 x \,{\mathrm e}^{3}+20 \ln \left (5\right ) \ln \left (x \right )-8 \ln \left (x \right ) {\mathrm e}^{3}-19 \ln \left (x \right )\) | \(96\) |
int((((10*x^2+19*x)*ln(5)-2*x*exp(3)-2*x^2-9*x)*ln(x)+(5*x^2+19*x+20)*ln(5 )-exp(3)^2+(-2*x-8)*exp(3)-x^2-9*x-19)/x,x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {-19-e^6+e^3 (-8-2 x)-9 x-x^2+\left (20+19 x+5 x^2\right ) \log (5)+\left (-9 x-2 e^3 x-2 x^2+\left (19 x+10 x^2\right ) \log (5)\right ) \log (x)}{x} \, dx=-{\left (x^{2} + 2 \, {\left (x + 4\right )} e^{3} - {\left (5 \, x^{2} + 19 \, x + 20\right )} \log \left (5\right ) + 9 \, x + e^{6} + 19\right )} \log \left (x\right ) \]
integrate((((10*x^2+19*x)*log(5)-2*x*exp(3)-2*x^2-9*x)*log(x)+(5*x^2+19*x+ 20)*log(5)-exp(3)^2+(-2*x-8)*exp(3)-x^2-9*x-19)/x,x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {-19-e^6+e^3 (-8-2 x)-9 x-x^2+\left (20+19 x+5 x^2\right ) \log (5)+\left (-9 x-2 e^3 x-2 x^2+\left (19 x+10 x^2\right ) \log (5)\right ) \log (x)}{x} \, dx=\left (- x^{2} + 5 x^{2} \log {\left (5 \right )} - 2 x e^{3} - 9 x + 19 x \log {\left (5 \right )}\right ) \log {\left (x \right )} + \left (- e^{6} - 8 e^{3} - 19 + 20 \log {\left (5 \right )}\right ) \log {\left (x \right )} \]
integrate((((10*x**2+19*x)*ln(5)-2*x*exp(3)-2*x**2-9*x)*ln(x)+(5*x**2+19*x +20)*ln(5)-exp(3)**2+(-2*x-8)*exp(3)-x**2-9*x-19)/x,x)
(-x**2 + 5*x**2*log(5) - 2*x*exp(3) - 9*x + 19*x*log(5))*log(x) + (-exp(6) - 8*exp(3) - 19 + 20*log(5))*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (29) = 58\).
Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.91 \[ \int \frac {-19-e^6+e^3 (-8-2 x)-9 x-x^2+\left (20+19 x+5 x^2\right ) \log (5)+\left (-9 x-2 e^3 x-2 x^2+\left (19 x+10 x^2\right ) \log (5)\right ) \log (x)}{x} \, dx=\frac {5}{2} \, x^{2} \log \left (5\right ) - x^{2} \log \left (x\right ) - 2 \, {\left (x \log \left (x\right ) - x\right )} e^{3} - 2 \, x e^{3} + \frac {5}{2} \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} \log \left (5\right ) + 19 \, {\left (x \log \left (x\right ) - x\right )} \log \left (5\right ) + 19 \, x \log \left (5\right ) - 9 \, x \log \left (x\right ) - e^{6} \log \left (x\right ) - 8 \, e^{3} \log \left (x\right ) + 20 \, \log \left (5\right ) \log \left (x\right ) - 19 \, \log \left (x\right ) \]
integrate((((10*x^2+19*x)*log(5)-2*x*exp(3)-2*x^2-9*x)*log(x)+(5*x^2+19*x+ 20)*log(5)-exp(3)^2+(-2*x-8)*exp(3)-x^2-9*x-19)/x,x, algorithm=\
5/2*x^2*log(5) - x^2*log(x) - 2*(x*log(x) - x)*e^3 - 2*x*e^3 + 5/2*(2*x^2* log(x) - x^2)*log(5) + 19*(x*log(x) - x)*log(5) + 19*x*log(5) - 9*x*log(x) - e^6*log(x) - 8*e^3*log(x) + 20*log(5)*log(x) - 19*log(x)
Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {-19-e^6+e^3 (-8-2 x)-9 x-x^2+\left (20+19 x+5 x^2\right ) \log (5)+\left (-9 x-2 e^3 x-2 x^2+\left (19 x+10 x^2\right ) \log (5)\right ) \log (x)}{x} \, dx=5 \, x^{2} \log \left (5\right ) \log \left (x\right ) - x^{2} \log \left (x\right ) - 2 \, x e^{3} \log \left (x\right ) + 19 \, x \log \left (5\right ) \log \left (x\right ) - 9 \, x \log \left (x\right ) - e^{6} \log \left (x\right ) - 8 \, e^{3} \log \left (x\right ) + 20 \, \log \left (5\right ) \log \left (x\right ) - 19 \, \log \left (x\right ) \]
integrate((((10*x^2+19*x)*log(5)-2*x*exp(3)-2*x^2-9*x)*log(x)+(5*x^2+19*x+ 20)*log(5)-exp(3)^2+(-2*x-8)*exp(3)-x^2-9*x-19)/x,x, algorithm=\
5*x^2*log(5)*log(x) - x^2*log(x) - 2*x*e^3*log(x) + 19*x*log(5)*log(x) - 9 *x*log(x) - e^6*log(x) - 8*e^3*log(x) + 20*log(5)*log(x) - 19*log(x)
Time = 12.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {-19-e^6+e^3 (-8-2 x)-9 x-x^2+\left (20+19 x+5 x^2\right ) \log (5)+\left (-9 x-2 e^3 x-2 x^2+\left (19 x+10 x^2\right ) \log (5)\right ) \log (x)}{x} \, dx=x^2\,\ln \left (x\right )\,\left (5\,\ln \left (5\right )-1\right )-\ln \left (x\right )\,\left (8\,{\mathrm {e}}^3+{\mathrm {e}}^6-20\,\ln \left (5\right )+19\right )-x\,\ln \left (x\right )\,\left (2\,{\mathrm {e}}^3-19\,\ln \left (5\right )+9\right ) \]
int(-(9*x + exp(6) + log(x)*(9*x - log(5)*(19*x + 10*x^2) + 2*x*exp(3) + 2 *x^2) - log(5)*(19*x + 5*x^2 + 20) + x^2 + exp(3)*(2*x + 8) + 19)/x,x)