Integrand size = 133, antiderivative size = 23 \[ \int \frac {10 e^x x-25 x^2+e^{2 x} \left (2 x+x^2\right )+\left (50 x+30 e^x x-25 x^2+e^{2 x} \left (6 x+3 x^2\right )\right ) \log (x)+\left (20 e^x x+e^{2 x} \left (6 x+3 x^2\right )\right ) \log ^2(x)+e^{2 x} \left (2 x+x^2\right ) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx=e^{-x} \left (e^x x+\frac {5 x}{1+\log (x)}\right )^2 \] Output:
(5*x/(1+ln(x))+exp(x)*x)^2/exp(x)
Time = 5.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {10 e^x x-25 x^2+e^{2 x} \left (2 x+x^2\right )+\left (50 x+30 e^x x-25 x^2+e^{2 x} \left (6 x+3 x^2\right )\right ) \log (x)+\left (20 e^x x+e^{2 x} \left (6 x+3 x^2\right )\right ) \log ^2(x)+e^{2 x} \left (2 x+x^2\right ) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx=\frac {e^{-x} x^2 \left (5+e^x+e^x \log (x)\right )^2}{(1+\log (x))^2} \] Input:
Integrate[(10*E^x*x - 25*x^2 + E^(2*x)*(2*x + x^2) + (50*x + 30*E^x*x - 25 *x^2 + E^(2*x)*(6*x + 3*x^2))*Log[x] + (20*E^x*x + E^(2*x)*(6*x + 3*x^2))* Log[x]^2 + E^(2*x)*(2*x + x^2)*Log[x]^3)/(E^x + 3*E^x*Log[x] + 3*E^x*Log[x ]^2 + E^x*Log[x]^3),x]
Output:
(x^2*(5 + E^x + E^x*Log[x])^2)/(E^x*(1 + Log[x])^2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.62 (sec) , antiderivative size = 96, normalized size of antiderivative = 4.17, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {7239, 7293, 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.
\(\displaystyle \int \frac {-25 x^2+e^{2 x} \left (x^2+2 x\right )+e^{2 x} \left (x^2+2 x\right ) \log ^3(x)+\left (e^{2 x} \left (3 x^2+6 x\right )+20 e^x x\right ) \log ^2(x)+\left (-25 x^2+e^{2 x} \left (3 x^2+6 x\right )+30 e^x x+50 x\right ) \log (x)+10 e^x x}{e^x+e^x \log ^3(x)+3 e^x \log ^2(x)+3 e^x \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{-x} x \left (e^x+e^x \log (x)+5\right ) \left (-5 x+e^x (x+2)+e^x (x+2) \log ^2(x)+\left (2 e^x (x+2)-5 (x-2)\right ) \log (x)\right )}{(\log (x)+1)^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^x x (x+2)+\frac {10 x (2 \log (x)+1)}{(\log (x)+1)^2}-\frac {25 e^{-x} x (x+x \log (x)-2 \log (x))}{(\log (x)+1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {20 \operatorname {ExpIntegralEi}(2 (\log (x)+1))}{e^2}-\frac {40 (\log (x)+1) \operatorname {ExpIntegralEi}(2 (\log (x)+1))}{e^2}+\frac {20 (2 \log (x)+1) \operatorname {ExpIntegralEi}(2 (\log (x)+1))}{e^2}+e^x x^2+20 x^2-\frac {10 x^2 (2 \log (x)+1)}{\log (x)+1}+\frac {25 e^{-x} x (x+x \log (x))}{(\log (x)+1)^3}\) |
Input:
Int[(10*E^x*x - 25*x^2 + E^(2*x)*(2*x + x^2) + (50*x + 30*E^x*x - 25*x^2 + E^(2*x)*(6*x + 3*x^2))*Log[x] + (20*E^x*x + E^(2*x)*(6*x + 3*x^2))*Log[x] ^2 + E^(2*x)*(2*x + x^2)*Log[x]^3)/(E^x + 3*E^x*Log[x] + 3*E^x*Log[x]^2 + E^x*Log[x]^3),x]
Output:
20*x^2 + E^x*x^2 + (20*ExpIntegralEi[2*(1 + Log[x])])/E^2 - (40*ExpIntegra lEi[2*(1 + Log[x])]*(1 + Log[x]))/E^2 + (20*ExpIntegralEi[2*(1 + Log[x])]* (1 + 2*Log[x]))/E^2 - (10*x^2*(1 + 2*Log[x]))/(1 + Log[x]) + (25*x*(x + x* Log[x]))/(E^x*(1 + Log[x])^3)
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 40.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52
method | result | size |
risch | \({\mathrm e}^{x} x^{2}+\frac {5 x^{2} \left (2 \,{\mathrm e}^{x} \ln \left (x \right )+2 \,{\mathrm e}^{x}+5\right ) {\mathrm e}^{-x}}{\left (\ln \left (x \right )+1\right )^{2}}\) | \(35\) |
parallelrisch | \(-\frac {\left (-2 \,{\mathrm e}^{2 x} \ln \left (x \right )^{2} x^{2}-4 \,{\mathrm e}^{2 x} \ln \left (x \right ) x^{2}-20 x^{2} {\mathrm e}^{x} \ln \left (x \right )-2 \,{\mathrm e}^{2 x} x^{2}-20 \,{\mathrm e}^{x} x^{2}-50 x^{2}\right ) {\mathrm e}^{-x}}{2 \left (\ln \left (x \right )^{2}+2 \ln \left (x \right )+1\right )}\) | \(74\) |
Input:
int(((x^2+2*x)*exp(x)^2*ln(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)*ln(x)^2 +((3*x^2+6*x)*exp(x)^2+30*exp(x)*x-25*x^2+50*x)*ln(x)+(x^2+2*x)*exp(x)^2+1 0*exp(x)*x-25*x^2)/(exp(x)*ln(x)^3+3*exp(x)*ln(x)^2+3*exp(x)*ln(x)+exp(x)) ,x,method=_RETURNVERBOSE)
Output:
exp(x)*x^2+5*x^2*(2*exp(x)*ln(x)+2*exp(x)+5)/(ln(x)+1)^2/exp(x)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (21) = 42\).
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.13 \[ \int \frac {10 e^x x-25 x^2+e^{2 x} \left (2 x+x^2\right )+\left (50 x+30 e^x x-25 x^2+e^{2 x} \left (6 x+3 x^2\right )\right ) \log (x)+\left (20 e^x x+e^{2 x} \left (6 x+3 x^2\right )\right ) \log ^2(x)+e^{2 x} \left (2 x+x^2\right ) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx=\frac {x^{2} e^{\left (2 \, x\right )} \log \left (x\right )^{2} + x^{2} e^{\left (2 \, x\right )} + 10 \, x^{2} e^{x} + 25 \, x^{2} + 2 \, {\left (x^{2} e^{\left (2 \, x\right )} + 5 \, x^{2} e^{x}\right )} \log \left (x\right )}{e^{x} \log \left (x\right )^{2} + 2 \, e^{x} \log \left (x\right ) + e^{x}} \] Input:
integrate(((x^2+2*x)*exp(x)^2*log(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)* log(x)^2+((3*x^2+6*x)*exp(x)^2+30*exp(x)*x-25*x^2+50*x)*log(x)+(x^2+2*x)*e xp(x)^2+10*exp(x)*x-25*x^2)/(exp(x)*log(x)^3+3*exp(x)*log(x)^2+3*exp(x)*lo g(x)+exp(x)),x, algorithm="fricas")
Output:
(x^2*e^(2*x)*log(x)^2 + x^2*e^(2*x) + 10*x^2*e^x + 25*x^2 + 2*(x^2*e^(2*x) + 5*x^2*e^x)*log(x))/(e^x*log(x)^2 + 2*e^x*log(x) + e^x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (17) = 34\).
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {10 e^x x-25 x^2+e^{2 x} \left (2 x+x^2\right )+\left (50 x+30 e^x x-25 x^2+e^{2 x} \left (6 x+3 x^2\right )\right ) \log (x)+\left (20 e^x x+e^{2 x} \left (6 x+3 x^2\right )\right ) \log ^2(x)+e^{2 x} \left (2 x+x^2\right ) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx=\frac {10 x^{2}}{\log {\left (x \right )} + 1} + \frac {25 x^{2} e^{- x} + \left (x^{2} \log {\left (x \right )}^{2} + 2 x^{2} \log {\left (x \right )} + x^{2}\right ) e^{x}}{\log {\left (x \right )}^{2} + 2 \log {\left (x \right )} + 1} \] Input:
integrate(((x**2+2*x)*exp(x)**2*ln(x)**3+((3*x**2+6*x)*exp(x)**2+20*exp(x) *x)*ln(x)**2+((3*x**2+6*x)*exp(x)**2+30*exp(x)*x-25*x**2+50*x)*ln(x)+(x**2 +2*x)*exp(x)**2+10*exp(x)*x-25*x**2)/(exp(x)*ln(x)**3+3*exp(x)*ln(x)**2+3* exp(x)*ln(x)+exp(x)),x)
Output:
10*x**2/(log(x) + 1) + (25*x**2*exp(-x) + (x**2*log(x)**2 + 2*x**2*log(x) + x**2)*exp(x))/(log(x)**2 + 2*log(x) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (21) = 42\).
Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.48 \[ \int \frac {10 e^x x-25 x^2+e^{2 x} \left (2 x+x^2\right )+\left (50 x+30 e^x x-25 x^2+e^{2 x} \left (6 x+3 x^2\right )\right ) \log (x)+\left (20 e^x x+e^{2 x} \left (6 x+3 x^2\right )\right ) \log ^2(x)+e^{2 x} \left (2 x+x^2\right ) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx=\frac {25 \, x^{2} e^{\left (-x\right )} + 10 \, x^{2} \log \left (x\right ) + 10 \, x^{2} + {\left (x^{2} \log \left (x\right )^{2} + 2 \, x^{2} \log \left (x\right ) + x^{2}\right )} e^{x}}{\log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 1} \] Input:
integrate(((x^2+2*x)*exp(x)^2*log(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)* log(x)^2+((3*x^2+6*x)*exp(x)^2+30*exp(x)*x-25*x^2+50*x)*log(x)+(x^2+2*x)*e xp(x)^2+10*exp(x)*x-25*x^2)/(exp(x)*log(x)^3+3*exp(x)*log(x)^2+3*exp(x)*lo g(x)+exp(x)),x, algorithm="maxima")
Output:
(25*x^2*e^(-x) + 10*x^2*log(x) + 10*x^2 + (x^2*log(x)^2 + 2*x^2*log(x) + x ^2)*e^x)/(log(x)^2 + 2*log(x) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (21) = 42\).
Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.13 \[ \int \frac {10 e^x x-25 x^2+e^{2 x} \left (2 x+x^2\right )+\left (50 x+30 e^x x-25 x^2+e^{2 x} \left (6 x+3 x^2\right )\right ) \log (x)+\left (20 e^x x+e^{2 x} \left (6 x+3 x^2\right )\right ) \log ^2(x)+e^{2 x} \left (2 x+x^2\right ) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx=\frac {x^{2} e^{\left (2 \, x\right )} \log \left (x\right )^{2} + 2 \, x^{2} e^{\left (2 \, x\right )} \log \left (x\right ) + 10 \, x^{2} e^{x} \log \left (x\right ) + x^{2} e^{\left (2 \, x\right )} + 10 \, x^{2} e^{x} + 25 \, x^{2}}{e^{x} \log \left (x\right )^{2} + 2 \, e^{x} \log \left (x\right ) + e^{x}} \] Input:
integrate(((x^2+2*x)*exp(x)^2*log(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)* log(x)^2+((3*x^2+6*x)*exp(x)^2+30*exp(x)*x-25*x^2+50*x)*log(x)+(x^2+2*x)*e xp(x)^2+10*exp(x)*x-25*x^2)/(exp(x)*log(x)^3+3*exp(x)*log(x)^2+3*exp(x)*lo g(x)+exp(x)),x, algorithm="giac")
Output:
(x^2*e^(2*x)*log(x)^2 + 2*x^2*e^(2*x)*log(x) + 10*x^2*e^x*log(x) + x^2*e^( 2*x) + 10*x^2*e^x + 25*x^2)/(e^x*log(x)^2 + 2*e^x*log(x) + e^x)
Time = 2.69 (sec) , antiderivative size = 169, normalized size of antiderivative = 7.35 \[ \int \frac {10 e^x x-25 x^2+e^{2 x} \left (2 x+x^2\right )+\left (50 x+30 e^x x-25 x^2+e^{2 x} \left (6 x+3 x^2\right )\right ) \log (x)+\left (20 e^x x+e^{2 x} \left (6 x+3 x^2\right )\right ) \log ^2(x)+e^{2 x} \left (2 x+x^2\right ) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx=x^2\,{\mathrm {e}}^x+20\,x^2\,\ln \left (x\right )-\frac {10\,x^2\,{\ln \left (x\right )}^2-\frac {5\,x^2\,{\mathrm {e}}^{-x}\,\left (5\,x-2\,{\mathrm {e}}^x\right )}{2}+\frac {5\,x^2\,{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left (6\,{\mathrm {e}}^x-5\,x+10\right )}{2}}{{\ln \left (x\right )}^2+2\,\ln \left (x\right )+1}+{\mathrm {e}}^{-x}\,\left (\frac {25\,x^4}{2}-\frac {125\,x^3}{2}+50\,x^2\right )+40\,x^2-\frac {20\,x^2\,{\ln \left (x\right )}^2+\frac {25\,x\,{\mathrm {e}}^{-x}\,\left (2\,x+2\,x\,{\mathrm {e}}^x-4\,x^2+x^3\right )}{2}+\frac {25\,x\,{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left (4\,x+4\,x\,{\mathrm {e}}^x-5\,x^2+x^3\right )}{2}}{\ln \left (x\right )+1} \] Input:
int((log(x)*(50*x + exp(2*x)*(6*x + 3*x^2) + 30*x*exp(x) - 25*x^2) + exp(2 *x)*(2*x + x^2) + 10*x*exp(x) + log(x)^2*(exp(2*x)*(6*x + 3*x^2) + 20*x*ex p(x)) - 25*x^2 + exp(2*x)*log(x)^3*(2*x + x^2))/(exp(x) + 3*exp(x)*log(x) + 3*exp(x)*log(x)^2 + exp(x)*log(x)^3),x)
Output:
x^2*exp(x) + 20*x^2*log(x) - (10*x^2*log(x)^2 - (5*x^2*exp(-x)*(5*x - 2*ex p(x)))/2 + (5*x^2*exp(-x)*log(x)*(6*exp(x) - 5*x + 10))/2)/(2*log(x) + log (x)^2 + 1) + exp(-x)*(50*x^2 - (125*x^3)/2 + (25*x^4)/2) + 40*x^2 - (20*x^ 2*log(x)^2 + (25*x*exp(-x)*(2*x + 2*x*exp(x) - 4*x^2 + x^3))/2 + (25*x*exp (-x)*log(x)*(4*x + 4*x*exp(x) - 5*x^2 + x^3))/2)/(log(x) + 1)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.57 \[ \int \frac {10 e^x x-25 x^2+e^{2 x} \left (2 x+x^2\right )+\left (50 x+30 e^x x-25 x^2+e^{2 x} \left (6 x+3 x^2\right )\right ) \log (x)+\left (20 e^x x+e^{2 x} \left (6 x+3 x^2\right )\right ) \log ^2(x)+e^{2 x} \left (2 x+x^2\right ) \log ^3(x)}{e^x+3 e^x \log (x)+3 e^x \log ^2(x)+e^x \log ^3(x)} \, dx=\frac {x^{2} \left (e^{2 x} \mathrm {log}\left (x \right )^{2}+2 e^{2 x} \mathrm {log}\left (x \right )+e^{2 x}+10 e^{x} \mathrm {log}\left (x \right )+10 e^{x}+25\right )}{e^{x} \left (\mathrm {log}\left (x \right )^{2}+2 \,\mathrm {log}\left (x \right )+1\right )} \] Input:
int(((x^2+2*x)*exp(x)^2*log(x)^3+((3*x^2+6*x)*exp(x)^2+20*exp(x)*x)*log(x) ^2+((3*x^2+6*x)*exp(x)^2+30*exp(x)*x-25*x^2+50*x)*log(x)+(x^2+2*x)*exp(x)^ 2+10*exp(x)*x-25*x^2)/(exp(x)*log(x)^3+3*exp(x)*log(x)^2+3*exp(x)*log(x)+e xp(x)),x)
Output:
(x**2*(e**(2*x)*log(x)**2 + 2*e**(2*x)*log(x) + e**(2*x) + 10*e**x*log(x) + 10*e**x + 25))/(e**x*(log(x)**2 + 2*log(x) + 1))