Integrand size = 76, antiderivative size = 32 \[ \int \frac {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx=\frac {1}{4} x \left (1+x+x \log ^2\left (e^{\frac {1}{25} e^{10-2 x}-x} x\right )\right ) \] Output:
1/4*(x+1+x*ln(x*exp(1/25*exp(5-x)^2-x))^2)*x
Time = 0.56 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx=\frac {1}{4} x \left (1+x+x \log ^2\left (e^{\frac {1}{25} e^{10-2 x}-x} x\right )\right ) \] Input:
Integrate[(25 + 50*x + (50*x - 50*x^2 - 4*E^(10 - 2*x)*x^2)*Log[E^((E^(10 - 2*x) - 25*x)/25)*x] + 50*x*Log[E^((E^(10 - 2*x) - 25*x)/25)*x]^2)/100,x]
Output:
(x*(1 + x + x*Log[E^(E^(10 - 2*x)/25 - x)*x]^2))/4
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.
\(\displaystyle \int \frac {1}{100} \left (\left (-4 e^{10-2 x} x^2-50 x^2+50 x\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+25\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{100} \int \left (50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+2 \left (-2 e^{10-2 x} x^2-25 x^2+25 x\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x+25\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{100} \left (50 \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )dx-e^{10} \operatorname {ExpIntegralEi}(-2 x)-\frac {25 x^4}{6}+\frac {2}{3} e^{10-2 x} x^3+\frac {125 x^3}{9}-\frac {50}{3} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{25} e^{20-4 x} x^2-e^{10-2 x} x^2+\frac {25 x^2}{2}+2 e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+25 x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {3}{50} e^{20-4 x} x-e^{10-2 x} x+25 x-\frac {7}{200} e^{20-4 x}+2 e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right )\) |
Input:
Int[(25 + 50*x + (50*x - 50*x^2 - 4*E^(10 - 2*x)*x^2)*Log[E^((E^(10 - 2*x) - 25*x)/25)*x] + 50*x*Log[E^((E^(10 - 2*x) - 25*x)/25)*x]^2)/100,x]
Output:
$Aborted
Time = 3.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\frac {\ln \left (x \,{\mathrm e}^{\frac {{\mathrm e}^{10-2 x}}{25}-x}\right )^{2} x^{2}}{4}+\frac {x^{2}}{4}+\frac {x}{4}\) | \(35\) |
risch | \(\text {Expression too large to display}\) | \(768\) |
Input:
int(1/2*x*ln(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-50*x^2+5 0*x)*ln(x*exp(1/25*exp(5-x)^2-x))+1/2*x+1/4,x,method=_RETURNVERBOSE)
Output:
1/4*ln(x*exp(1/25*exp(5-x)^2-x))^2*x^2+1/4*x^2+1/4*x
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx=\frac {1}{4} \, x^{2} \log \left (x e^{\left (-x + \frac {1}{25} \, e^{\left (-2 \, x + 10\right )}\right )}\right )^{2} + \frac {1}{4} \, x^{2} + \frac {1}{4} \, x \] Input:
integrate(1/2*x*log(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-5 0*x^2+50*x)*log(x*exp(1/25*exp(5-x)^2-x))+1/2*x+1/4,x, algorithm="fricas")
Output:
1/4*x^2*log(x*e^(-x + 1/25*e^(-2*x + 10)))^2 + 1/4*x^2 + 1/4*x
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx=\frac {x^{2} \log {\left (x e^{- x + \frac {e^{10 - 2 x}}{25}} \right )}^{2}}{4} + \frac {x^{2}}{4} + \frac {x}{4} \] Input:
integrate(1/2*x*ln(x*exp(1/25*exp(5-x)**2-x))**2+1/100*(-4*x**2*exp(5-x)** 2-50*x**2+50*x)*ln(x*exp(1/25*exp(5-x)**2-x))+1/2*x+1/4,x)
Output:
x**2*log(x*exp(-x + exp(10 - 2*x)/25))**2/4 + x**2/4 + x/4
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (26) = 52\).
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.97 \[ \int \frac {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx=\frac {1}{4} \, x^{4} - \frac {1}{2} \, x^{3} \log \left (x\right ) + \frac {1}{4} \, x^{2} \log \left (x\right )^{2} + \frac {1}{2500} \, x^{2} e^{\left (-4 \, x + 20\right )} + \frac {1}{4} \, x^{2} - \frac {1}{50} \, {\left (x^{3} e^{10} - x^{2} e^{10} \log \left (x\right )\right )} e^{\left (-2 \, x\right )} + \frac {1}{4} \, x \] Input:
integrate(1/2*x*log(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-5 0*x^2+50*x)*log(x*exp(1/25*exp(5-x)^2-x))+1/2*x+1/4,x, algorithm="maxima")
Output:
1/4*x^4 - 1/2*x^3*log(x) + 1/4*x^2*log(x)^2 + 1/2500*x^2*e^(-4*x + 20) + 1 /4*x^2 - 1/50*(x^3*e^10 - x^2*e^10*log(x))*e^(-2*x) + 1/4*x
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (26) = 52\).
Time = 0.12 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.03 \[ \int \frac {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx=\frac {1}{4} \, x^{4} - \frac {1}{50} \, x^{3} e^{\left (-2 \, x + 10\right )} - \frac {1}{2} \, x^{3} \log \left (x\right ) + \frac {1}{50} \, x^{2} e^{\left (-2 \, x + 10\right )} \log \left (x\right ) + \frac {1}{4} \, x^{2} \log \left (x\right )^{2} + \frac {1}{2500} \, x^{2} e^{\left (-4 \, x + 20\right )} + \frac {1}{4} \, x^{2} + \frac {1}{4} \, x \] Input:
integrate(1/2*x*log(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-5 0*x^2+50*x)*log(x*exp(1/25*exp(5-x)^2-x))+1/2*x+1/4,x, algorithm="giac")
Output:
1/4*x^4 - 1/50*x^3*e^(-2*x + 10) - 1/2*x^3*log(x) + 1/50*x^2*e^(-2*x + 10) *log(x) + 1/4*x^2*log(x)^2 + 1/2500*x^2*e^(-4*x + 20) + 1/4*x^2 + 1/4*x
Time = 2.61 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.03 \[ \int \frac {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx=\frac {x}{4}-\frac {x^3\,\ln \left (x\right )}{2}+\frac {x^2\,{\ln \left (x\right )}^2}{4}-\frac {x^3\,{\mathrm {e}}^{10-2\,x}}{50}+\frac {x^2\,{\mathrm {e}}^{20-4\,x}}{2500}+\frac {x^2}{4}+\frac {x^4}{4}+\frac {x^2\,{\mathrm {e}}^{10-2\,x}\,\ln \left (x\right )}{50} \] Input:
int(x/2 + (x*log(x*exp(exp(10 - 2*x)/25 - x))^2)/2 - (log(x*exp(exp(10 - 2 *x)/25 - x))*(4*x^2*exp(10 - 2*x) - 50*x + 50*x^2))/100 + 1/4,x)
Output:
x/4 - (x^3*log(x))/2 + (x^2*log(x)^2)/4 - (x^3*exp(10 - 2*x))/50 + (x^2*ex p(20 - 4*x))/2500 + x^2/4 + x^4/4 + (x^2*exp(10 - 2*x)*log(x))/50
\[ \int \frac {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx=-\frac {\left (\int \frac {\mathrm {log}\left (\frac {e^{\frac {e^{10}}{25 e^{2 x}}} x}{e^{x}}\right ) x^{2}}{e^{2 x}}d x \right ) e^{10}}{25}+\frac {\left (\int \mathrm {log}\left (\frac {e^{\frac {e^{10}}{25 e^{2 x}}} x}{e^{x}}\right )^{2} x d x \right )}{2}-\frac {\left (\int \mathrm {log}\left (\frac {e^{\frac {e^{10}}{25 e^{2 x}}} x}{e^{x}}\right ) x^{2}d x \right )}{2}+\frac {\left (\int \mathrm {log}\left (\frac {e^{\frac {e^{10}}{25 e^{2 x}}} x}{e^{x}}\right ) x d x \right )}{2}+\frac {x^{2}}{4}+\frac {x}{4} \] Input:
int(1/2*x*log(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-50*x^2+ 50*x)*log(x*exp(1/25*exp(5-x)^2-x))+1/2*x+1/4,x)
Output:
( - 4*int((log((e**(e**10/(25*e**(2*x)))*x)/e**x)*x**2)/e**(2*x),x)*e**10 + 50*int(log((e**(e**10/(25*e**(2*x)))*x)/e**x)**2*x,x) - 50*int(log((e**( e**10/(25*e**(2*x)))*x)/e**x)*x**2,x) + 50*int(log((e**(e**10/(25*e**(2*x) ))*x)/e**x)*x,x) + 25*x**2 + 25*x)/100