Integrand size = 68, antiderivative size = 19 \[ \int \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx=e^{2 x} \left (-5+e^5\right ) (-5+x \log (x))^2 \] Output:
(x*ln(x)-5)^2*exp(x)^2*(exp(5)-5)
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx=e^{2 x} \left (-5+e^5\right ) (-5+x \log (x))^2 \] Input:
Integrate[E^(2*x)*(-200 + 40*E^5) + E^(2*x)*(50 + E^5*(-10 - 18*x) + 90*x) *Log[x] + E^(2*x)*(-10*x - 10*x^2 + E^5*(2*x + 2*x^2))*Log[x]^2,x]
Output:
E^(2*x)*(-5 + E^5)*(-5 + x*Log[x])^2
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.
\(\displaystyle \int \left (e^{2 x} \left (-10 x^2+e^5 \left (2 x^2+2 x\right )-10 x\right ) \log ^2(x)+\left (40 e^5-200\right ) e^{2 x}+e^{2 x} \left (e^5 (-18 x-10)+90 x+50\right ) \log (x)\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (5-e^5\right ) \int e^{2 x} x^2 \log ^2(x)dx-2 \left (5-e^5\right ) \int e^{2 x} x \log ^2(x)dx-\frac {1}{2} \left (5-e^5\right ) \operatorname {ExpIntegralEi}(2 x)-\frac {49}{2} \left (5-e^5\right ) e^{2 x}+\frac {5}{2} e^{2 x} \log (x)+\frac {9}{2} e^{2 x+5} \log (x)+45 e^{2 x} x \log (x)-e^{2 x+5} (9 x+5) \log (x)\) |
Input:
Int[E^(2*x)*(-200 + 40*E^5) + E^(2*x)*(50 + E^5*(-10 - 18*x) + 90*x)*Log[x ] + E^(2*x)*(-10*x - 10*x^2 + E^5*(2*x + 2*x^2))*Log[x]^2,x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(62\) vs. \(2(17)=34\).
Time = 0.75 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.32
method | result | size |
risch | \(x^{2} \ln \left (x \right )^{2} {\mathrm e}^{5+2 x}-5 \ln \left (x \right )^{2} {\mathrm e}^{2 x} x^{2}-10 x \ln \left (x \right ) {\mathrm e}^{5+2 x}+50 x \,{\mathrm e}^{2 x} \ln \left (x \right )+25 \,{\mathrm e}^{5+2 x}-125 \,{\mathrm e}^{2 x}\) | \(63\) |
parallelrisch | \({\mathrm e}^{5} \ln \left (x \right )^{2} {\mathrm e}^{2 x} x^{2}-5 \ln \left (x \right )^{2} {\mathrm e}^{2 x} x^{2}-10 \,{\mathrm e}^{5} \ln \left (x \right ) {\mathrm e}^{2 x} x +50 x \,{\mathrm e}^{2 x} \ln \left (x \right )+25 \,{\mathrm e}^{2 x} {\mathrm e}^{5}-125 \,{\mathrm e}^{2 x}\) | \(63\) |
orering | \(\frac {x \left (12 x^{3}+132 x^{2}+367 x +20\right ) \left (\left (\left (2 x^{2}+2 x \right ) {\mathrm e}^{5}-10 x^{2}-10 x \right ) {\mathrm e}^{2 x} \ln \left (x \right )^{2}+\left (\left (-18 x -10\right ) {\mathrm e}^{5}+90 x +50\right ) {\mathrm e}^{2 x} \ln \left (x \right )+\left (40 \,{\mathrm e}^{5}-200\right ) {\mathrm e}^{2 x}\right )}{8 x^{4}+92 x^{3}+274 x^{2}+48 x +10}-\frac {3 x^{2} \left (2 x^{2}+21 x +55\right ) \left (\left (\left (4 x +2\right ) {\mathrm e}^{5}-20 x -10\right ) {\mathrm e}^{2 x} \ln \left (x \right )^{2}+2 \left (\left (2 x^{2}+2 x \right ) {\mathrm e}^{5}-10 x^{2}-10 x \right ) {\mathrm e}^{2 x} \ln \left (x \right )^{2}+\frac {2 \left (\left (2 x^{2}+2 x \right ) {\mathrm e}^{5}-10 x^{2}-10 x \right ) {\mathrm e}^{2 x} \ln \left (x \right )}{x}+\left (-18 \,{\mathrm e}^{5}+90\right ) {\mathrm e}^{2 x} \ln \left (x \right )+2 \left (\left (-18 x -10\right ) {\mathrm e}^{5}+90 x +50\right ) {\mathrm e}^{2 x} \ln \left (x \right )+\frac {\left (\left (-18 x -10\right ) {\mathrm e}^{5}+90 x +50\right ) {\mathrm e}^{2 x}}{x}+2 \left (40 \,{\mathrm e}^{5}-200\right ) {\mathrm e}^{2 x}\right )}{2 \left (4 x^{4}+46 x^{3}+137 x^{2}+24 x +5\right )}+\frac {\left (x^{2}+10 x +25\right ) x^{2} \left (\left (4 \,{\mathrm e}^{5}-20\right ) {\mathrm e}^{2 x} \ln \left (x \right )^{2}+4 \left (\left (4 x +2\right ) {\mathrm e}^{5}-20 x -10\right ) {\mathrm e}^{2 x} \ln \left (x \right )^{2}+\frac {4 \left (\left (4 x +2\right ) {\mathrm e}^{5}-20 x -10\right ) {\mathrm e}^{2 x} \ln \left (x \right )}{x}+4 \left (\left (2 x^{2}+2 x \right ) {\mathrm e}^{5}-10 x^{2}-10 x \right ) {\mathrm e}^{2 x} \ln \left (x \right )^{2}+\frac {8 \left (\left (2 x^{2}+2 x \right ) {\mathrm e}^{5}-10 x^{2}-10 x \right ) {\mathrm e}^{2 x} \ln \left (x \right )}{x}+\frac {2 \left (\left (2 x^{2}+2 x \right ) {\mathrm e}^{5}-10 x^{2}-10 x \right ) {\mathrm e}^{2 x}}{x^{2}}-\frac {2 \left (\left (2 x^{2}+2 x \right ) {\mathrm e}^{5}-10 x^{2}-10 x \right ) {\mathrm e}^{2 x} \ln \left (x \right )}{x^{2}}+4 \left (-18 \,{\mathrm e}^{5}+90\right ) {\mathrm e}^{2 x} \ln \left (x \right )+\frac {2 \left (-18 \,{\mathrm e}^{5}+90\right ) {\mathrm e}^{2 x}}{x}+4 \left (\left (-18 x -10\right ) {\mathrm e}^{5}+90 x +50\right ) {\mathrm e}^{2 x} \ln \left (x \right )+\frac {4 \left (\left (-18 x -10\right ) {\mathrm e}^{5}+90 x +50\right ) {\mathrm e}^{2 x}}{x}-\frac {\left (\left (-18 x -10\right ) {\mathrm e}^{5}+90 x +50\right ) {\mathrm e}^{2 x}}{x^{2}}+4 \left (40 \,{\mathrm e}^{5}-200\right ) {\mathrm e}^{2 x}\right )}{8 x^{4}+92 x^{3}+274 x^{2}+48 x +10}\) | \(623\) |
Input:
int(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*ln(x)^2+((-18*x-10)*exp(5)+9 0*x+50)*exp(x)^2*ln(x)+(40*exp(5)-200)*exp(x)^2,x,method=_RETURNVERBOSE)
Output:
x^2*ln(x)^2*exp(5+2*x)-5*ln(x)^2*exp(2*x)*x^2-10*x*ln(x)*exp(5+2*x)+50*x*e xp(2*x)*ln(x)+25*exp(5+2*x)-125*exp(2*x)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (17) = 34\).
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.53 \[ \int \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx={\left (x^{2} e^{5} - 5 \, x^{2}\right )} e^{\left (2 \, x\right )} \log \left (x\right )^{2} - 10 \, {\left (x e^{5} - 5 \, x\right )} e^{\left (2 \, x\right )} \log \left (x\right ) + 25 \, {\left (e^{5} - 5\right )} e^{\left (2 \, x\right )} \] Input:
integrate(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*log(x)^2+((-18*x-10)*e xp(5)+90*x+50)*exp(x)^2*log(x)+(40*exp(5)-200)*exp(x)^2,x, algorithm="fric as")
Output:
(x^2*e^5 - 5*x^2)*e^(2*x)*log(x)^2 - 10*(x*e^5 - 5*x)*e^(2*x)*log(x) + 25* (e^5 - 5)*e^(2*x)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (17) = 34\).
Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.58 \[ \int \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx=\left (- 5 x^{2} \log {\left (x \right )}^{2} + x^{2} e^{5} \log {\left (x \right )}^{2} - 10 x e^{5} \log {\left (x \right )} + 50 x \log {\left (x \right )} - 125 + 25 e^{5}\right ) e^{2 x} \] Input:
integrate(((2*x**2+2*x)*exp(5)-10*x**2-10*x)*exp(x)**2*ln(x)**2+((-18*x-10 )*exp(5)+90*x+50)*exp(x)**2*ln(x)+(40*exp(5)-200)*exp(x)**2,x)
Output:
(-5*x**2*log(x)**2 + x**2*exp(5)*log(x)**2 - 10*x*exp(5)*log(x) + 50*x*log (x) - 125 + 25*exp(5))*exp(2*x)
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (17) = 34\).
Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.26 \[ \int \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx={\left (x^{2} {\left (e^{5} - 5\right )} \log \left (x\right )^{2} - 10 \, x {\left (e^{5} - 5\right )} \log \left (x\right ) + 5 \, e^{5} - 25\right )} e^{\left (2 \, x\right )} + 20 \, {\left (e^{5} - 5\right )} e^{\left (2 \, x\right )} \] Input:
integrate(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*log(x)^2+((-18*x-10)*e xp(5)+90*x+50)*exp(x)^2*log(x)+(40*exp(5)-200)*exp(x)^2,x, algorithm="maxi ma")
Output:
(x^2*(e^5 - 5)*log(x)^2 - 10*x*(e^5 - 5)*log(x) + 5*e^5 - 25)*e^(2*x) + 20 *(e^5 - 5)*e^(2*x)
Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (17) = 34\).
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 5.74 \[ \int \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx=\frac {5}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \left (x\right ) - \frac {1}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x + 5\right )} \log \left (x\right ) - {\left (5 \, x^{2} e^{\left (2 \, x\right )} - x^{2} e^{\left (2 \, x + 5\right )}\right )} \log \left (x\right )^{2} + 20 \, {\left (e^{5} - 5\right )} e^{\left (2 \, x\right )} + \frac {1}{2} \, {\left (5 \, {\left (18 \, x + 1\right )} e^{\left (2 \, x\right )} - {\left (18 \, x + 1\right )} e^{\left (2 \, x + 5\right )}\right )} \log \left (x\right ) - 25 \, e^{\left (2 \, x\right )} + 5 \, e^{\left (2 \, x + 5\right )} \] Input:
integrate(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*log(x)^2+((-18*x-10)*e xp(5)+90*x+50)*exp(x)^2*log(x)+(40*exp(5)-200)*exp(x)^2,x, algorithm="giac ")
Output:
5/2*(2*x - 1)*e^(2*x)*log(x) - 1/2*(2*x - 1)*e^(2*x + 5)*log(x) - (5*x^2*e ^(2*x) - x^2*e^(2*x + 5))*log(x)^2 + 20*(e^5 - 5)*e^(2*x) + 1/2*(5*(18*x + 1)*e^(2*x) - (18*x + 1)*e^(2*x + 5))*log(x) - 25*e^(2*x) + 5*e^(2*x + 5)
Time = 1.75 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx={\mathrm {e}}^{2\,x}\,{\left (x\,\ln \left (x\right )-5\right )}^2\,\left ({\mathrm {e}}^5-5\right ) \] Input:
int(exp(2*x)*(40*exp(5) - 200) - exp(2*x)*log(x)^2*(10*x - exp(5)*(2*x + 2 *x^2) + 10*x^2) + exp(2*x)*log(x)*(90*x - exp(5)*(18*x + 10) + 50),x)
Output:
exp(2*x)*(x*log(x) - 5)^2*(exp(5) - 5)
Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.42 \[ \int \left (e^{2 x} \left (-200+40 e^5\right )+e^{2 x} \left (50+e^5 (-10-18 x)+90 x\right ) \log (x)+e^{2 x} \left (-10 x-10 x^2+e^5 \left (2 x+2 x^2\right )\right ) \log ^2(x)\right ) \, dx=e^{2 x} \left (\mathrm {log}\left (x \right )^{2} e^{5} x^{2}-5 \mathrm {log}\left (x \right )^{2} x^{2}-10 \,\mathrm {log}\left (x \right ) e^{5} x +50 \,\mathrm {log}\left (x \right ) x +25 e^{5}-125\right ) \] Input:
int(((2*x^2+2*x)*exp(5)-10*x^2-10*x)*exp(x)^2*log(x)^2+((-18*x-10)*exp(5)+ 90*x+50)*exp(x)^2*log(x)+(40*exp(5)-200)*exp(x)^2,x)
Output:
e**(2*x)*(log(x)**2*e**5*x**2 - 5*log(x)**2*x**2 - 10*log(x)*e**5*x + 50*l og(x)*x + 25*e**5 - 125)