Integrand size = 72, antiderivative size = 30 \[ \int \frac {2+2 x+2 x^2+(1+x) \log \left (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 \left (-40-80 x-40 x^2\right )+e^6 \left (4+8 x+4 x^2\right )}\right )}{2+2 x} \, dx=\frac {1}{2} x \log \left (\frac {e^{2 x} x^2}{4 \left (-5+e^3\right )^2 (1+x)^2}\right ) \] Output:
1/2*ln(1/4*x^2/(exp(3)-5)^2/(1+x)^2*exp(x)^2)*x
Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {2+2 x+2 x^2+(1+x) \log \left (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 \left (-40-80 x-40 x^2\right )+e^6 \left (4+8 x+4 x^2\right )}\right )}{2+2 x} \, dx=\frac {1}{2} \left (1+x \log \left (\frac {e^{2 x} x^2}{4 \left (-5+e^3\right )^2 (1+x)^2}\right )\right ) \] Input:
Integrate[(2 + 2*x + 2*x^2 + (1 + x)*Log[(E^(2*x)*x^2)/(100 + 200*x + 100* x^2 + E^3*(-40 - 80*x - 40*x^2) + E^6*(4 + 8*x + 4*x^2))])/(2 + 2*x),x]
Output:
(1 + x*Log[(E^(2*x)*x^2)/(4*(-5 + E^3)^2*(1 + x)^2)])/2
Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {7239, 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.
| \(\displaystyle \int \frac {2 x^2+(x+1) \log \left (\frac {e^{2 x} x^2}{100 x^2+e^3 \left (-40 x^2-80 x-40\right )+e^6 \left (4 x^2+8 x+4\right )+200 x+100}\right )+2 x+2}{2 x+2} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (\frac {x^2+x+1}{x+1}+\frac {1}{2} \log \left (\frac {e^{2 x} x^2}{4 \left (e^3-5\right )^2 (x+1)^2}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} x \log \left (\frac {e^{2 x} x^2}{4 \left (5-e^3\right )^2 (x+1)^2}\right )\) |
Input:
Int[(2 + 2*x + 2*x^2 + (1 + x)*Log[(E^(2*x)*x^2)/(100 + 200*x + 100*x^2 + E^3*(-40 - 80*x - 40*x^2) + E^6*(4 + 8*x + 4*x^2))])/(2 + 2*x),x]
Output:
(x*Log[(E^(2*x)*x^2)/(4*(5 - E^3)^2*(1 + x)^2)])/2
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(52\) vs. \(2(24)=48\).
Time = 0.44 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77
| method | result | size |
| norman | \(\frac {\ln \left (\frac {x^{2} {\mathrm e}^{2 x}}{\left (4 x^{2}+8 x +4\right ) {\mathrm e}^{6}+\left (-40 x^{2}-80 x -40\right ) {\mathrm e}^{3}+100 x^{2}+200 x +100}\right ) x}{2}\) | \(53\) |
| parallelrisch | \(\frac {\ln \left (\frac {x^{2} {\mathrm e}^{2 x}}{4 x^{2} {\mathrm e}^{6}+8 x \,{\mathrm e}^{6}-40 x^{2} {\mathrm e}^{3}+4 \,{\mathrm e}^{6}-80 x \,{\mathrm e}^{3}+100 x^{2}-40 \,{\mathrm e}^{3}+200 x +100}\right ) x}{2}\) | \(61\) |
| default | \(\frac {x^{2}}{2}+\ln \left (1+x \right )-\frac {20 \,{\mathrm e}^{9} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {5 \,{\mathrm e}^{3} x^{2}}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}+\frac {\ln \left (\frac {x^{2} {\mathrm e}^{2 x}}{\left (4 x^{2}+8 x +4\right ) {\mathrm e}^{6}+\left (-40 x^{2}-80 x -40\right ) {\mathrm e}^{3}+100 x^{2}+200 x +100}\right ) x}{2}+\frac {10 \,{\mathrm e}^{3} x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}-\frac {{\mathrm e}^{12} \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {625 x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {25 x^{2}}{2 \left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )}-\frac {625 \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {25 x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}-\frac {150 \,{\mathrm e}^{6} \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {{\mathrm e}^{6} x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}+\frac {150 \,{\mathrm e}^{6} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {500 \,{\mathrm e}^{3} \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {{\mathrm e}^{12} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {500 \,{\mathrm e}^{3} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {{\mathrm e}^{6} x^{2}}{2 \left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )}+\frac {20 \,{\mathrm e}^{9} \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}\) | \(363\) |
| parts | \(\frac {x^{2}}{2}+\ln \left (1+x \right )-\frac {20 \,{\mathrm e}^{9} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {5 \,{\mathrm e}^{3} x^{2}}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}+\frac {\ln \left (\frac {x^{2} {\mathrm e}^{2 x}}{\left (4 x^{2}+8 x +4\right ) {\mathrm e}^{6}+\left (-40 x^{2}-80 x -40\right ) {\mathrm e}^{3}+100 x^{2}+200 x +100}\right ) x}{2}+\frac {10 \,{\mathrm e}^{3} x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}-\frac {{\mathrm e}^{12} \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {625 x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {25 x^{2}}{2 \left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )}-\frac {625 \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {25 x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}-\frac {150 \,{\mathrm e}^{6} \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {{\mathrm e}^{6} x}{{\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25}+\frac {150 \,{\mathrm e}^{6} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {500 \,{\mathrm e}^{3} \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}+\frac {{\mathrm e}^{12} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {500 \,{\mathrm e}^{3} x}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}-\frac {{\mathrm e}^{6} x^{2}}{2 \left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )}+\frac {20 \,{\mathrm e}^{9} \ln \left (1+x \right )}{\left ({\mathrm e}^{6}-10 \,{\mathrm e}^{3}+25\right )^{2}}\) | \(363\) |
| risch | \(-x \ln \left (2\right )+x \ln \left (x \right )-\ln \left (1+x \right ) x +\frac {i x \pi \operatorname {csgn}\left (i \left (1+x \right )^{2}\right )^{3}}{4}+\frac {i x \pi \,\operatorname {csgn}\left (\frac {i}{\left (1+x \right )^{2}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right )^{2}}{4}-\frac {i x \pi \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )}{4}+\frac {i x \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}}{2}+\frac {i x \pi \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{4}-\frac {i x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{4}+\frac {i x \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right ) \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right )^{2}}{4}-\frac {i x \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}}{4}-\frac {i x \pi \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right )^{3}}{4}+\frac {i x \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )}{4}-\frac {i x \pi \operatorname {csgn}\left (i \left (1+x \right )^{2}\right )^{2} \operatorname {csgn}\left (i \left (1+x \right )\right )}{2}-\frac {i x \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{4}-x \ln \left ({\mathrm e}^{3}-5\right )+x \ln \left ({\mathrm e}^{x}\right )+\frac {i x \pi \,\operatorname {csgn}\left (i \left (1+x \right )^{2}\right ) \operatorname {csgn}\left (i \left (1+x \right )\right )^{2}}{4}-\frac {i x \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right )^{3}}{4}-\frac {i x \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right ) \operatorname {csgn}\left (\frac {i x^{2} {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right ) \operatorname {csgn}\left (i x^{2}\right )}{4}+\frac {i x \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}}{2}-\frac {i x \pi \,\operatorname {csgn}\left (\frac {i}{\left (1+x \right )^{2}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 x}}{\left (1+x \right )^{2}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )}{4}\) | \(444\) |
Input:
int(((1+x)*ln(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40)*exp(3 )+100*x^2+200*x+100))+2*x^2+2*x+2)/(2+2*x),x,method=_RETURNVERBOSE)
Output:
1/2*ln(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40)*exp(3)+100*x ^2+200*x+100))*x
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {2+2 x+2 x^2+(1+x) \log \left (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 \left (-40-80 x-40 x^2\right )+e^6 \left (4+8 x+4 x^2\right )}\right )}{2+2 x} \, dx=\frac {1}{2} \, x \log \left (\frac {x^{2} e^{\left (2 \, x\right )}}{4 \, {\left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{6} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{3} + 50 \, x + 25\right )}}\right ) \] Input:
integrate(((1+x)*log(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40 )*exp(3)+100*x^2+200*x+100))+2*x^2+2*x+2)/(2+2*x),x, algorithm="fricas")
Output:
1/2*x*log(1/4*x^2*e^(2*x)/(25*x^2 + (x^2 + 2*x + 1)*e^6 - 10*(x^2 + 2*x + 1)*e^3 + 50*x + 25))
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (26) = 52\).
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.33 \[ \int \frac {2+2 x+2 x^2+(1+x) \log \left (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 \left (-40-80 x-40 x^2\right )+e^6 \left (4+8 x+4 x^2\right )}\right )}{2+2 x} \, dx=- \frac {x}{6} + \left (\frac {x}{2} + \frac {1}{12}\right ) \log {\left (\frac {x^{2} e^{2 x}}{100 x^{2} + 200 x + \left (- 40 x^{2} - 80 x - 40\right ) e^{3} + \left (4 x^{2} + 8 x + 4\right ) e^{6} + 100} \right )} - \frac {\log {\left (x \right )}}{6} + \frac {\log {\left (x + 1 \right )}}{6} \] Input:
integrate(((1+x)*ln(x**2*exp(x)**2/((4*x**2+8*x+4)*exp(3)**2+(-40*x**2-80* x-40)*exp(3)+100*x**2+200*x+100))+2*x**2+2*x+2)/(2+2*x),x)
Output:
-x/6 + (x/2 + 1/12)*log(x**2*exp(2*x)/(100*x**2 + 200*x + (-40*x**2 - 80*x - 40)*exp(3) + (4*x**2 + 8*x + 4)*exp(6) + 100)) - log(x)/6 + log(x + 1)/ 6
Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {2+2 x+2 x^2+(1+x) \log \left (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 \left (-40-80 x-40 x^2\right )+e^6 \left (4+8 x+4 x^2\right )}\right )}{2+2 x} \, dx=x^{2} - x {\left (\log \left (2\right ) + \log \left (e^{3} - 5\right )\right )} - {\left (x + 1\right )} \log \left (x + 1\right ) + x \log \left (x\right ) + \log \left (x + 1\right ) \] Input:
integrate(((1+x)*log(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40 )*exp(3)+100*x^2+200*x+100))+2*x^2+2*x+2)/(2+2*x),x, algorithm="maxima")
Output:
x^2 - x*(log(2) + log(e^3 - 5)) - (x + 1)*log(x + 1) + x*log(x) + log(x + 1)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (24) = 48\).
Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {2+2 x+2 x^2+(1+x) \log \left (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 \left (-40-80 x-40 x^2\right )+e^6 \left (4+8 x+4 x^2\right )}\right )}{2+2 x} \, dx=x^{2} + \frac {1}{2} \, x \log \left (\frac {x^{2}}{4 \, {\left (x^{2} e^{6} - 10 \, x^{2} e^{3} + 25 \, x^{2} + 2 \, x e^{6} - 20 \, x e^{3} + 50 \, x + e^{6} - 10 \, e^{3} + 25\right )}}\right ) \] Input:
integrate(((1+x)*log(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40 )*exp(3)+100*x^2+200*x+100))+2*x^2+2*x+2)/(2+2*x),x, algorithm="giac")
Output:
x^2 + 1/2*x*log(1/4*x^2/(x^2*e^6 - 10*x^2*e^3 + 25*x^2 + 2*x*e^6 - 20*x*e^ 3 + 50*x + e^6 - 10*e^3 + 25))
Time = 2.91 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.87 \[ \int \frac {2+2 x+2 x^2+(1+x) \log \left (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 \left (-40-80 x-40 x^2\right )+e^6 \left (4+8 x+4 x^2\right )}\right )}{2+2 x} \, dx=\frac {x\,\left (2\,x-\ln \left (200\,x-40\,{\mathrm {e}}^3+4\,{\mathrm {e}}^6-80\,x\,{\mathrm {e}}^3+8\,x\,{\mathrm {e}}^6-40\,x^2\,{\mathrm {e}}^3+4\,x^2\,{\mathrm {e}}^6+100\,x^2+100\right )+\ln \left (x^2\right )\right )}{2} \] Input:
int((2*x + log((x^2*exp(2*x))/(200*x + exp(6)*(8*x + 4*x^2 + 4) - exp(3)*( 80*x + 40*x^2 + 40) + 100*x^2 + 100))*(x + 1) + 2*x^2 + 2)/(2*x + 2),x)
Output:
(x*(2*x - log(200*x - 40*exp(3) + 4*exp(6) - 80*x*exp(3) + 8*x*exp(6) - 40 *x^2*exp(3) + 4*x^2*exp(6) + 100*x^2 + 100) + log(x^2)))/2
Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10 \[ \int \frac {2+2 x+2 x^2+(1+x) \log \left (\frac {e^{2 x} x^2}{100+200 x+100 x^2+e^3 \left (-40-80 x-40 x^2\right )+e^6 \left (4+8 x+4 x^2\right )}\right )}{2+2 x} \, dx=\frac {\mathrm {log}\left (\frac {e^{2 x} x^{2}}{4 e^{6} x^{2}+8 e^{6} x +4 e^{6}-40 e^{3} x^{2}-80 e^{3} x -40 e^{3}+100 x^{2}+200 x +100}\right ) x}{2} \] Input:
int(((1+x)*log(x^2*exp(x)^2/((4*x^2+8*x+4)*exp(3)^2+(-40*x^2-80*x-40)*exp( 3)+100*x^2+200*x+100))+2*x^2+2*x+2)/(2+2*x),x)
Output:
(log((e**(2*x)*x**2)/(4*e**6*x**2 + 8*e**6*x + 4*e**6 - 40*e**3*x**2 - 80* e**3*x - 40*e**3 + 100*x**2 + 200*x + 100))*x)/2