Integrand size = 93, antiderivative size = 35 \[ \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (5 x^2+50 x^3-3 x^4\right )+e^x \left (5-x^2-50 x^3+28 x^4-x^5\right )+\left (x^2-2 x^3+e^x \left (-5-5 x-x^2+x^3\right )\right ) \log (x)\right )}{4 x^2} \, dx=\frac {1}{4} x \left (\frac {5}{x}-x+e^{-x} x\right ) \left (-25+x+\frac {e^{-x} \log (x)}{x}\right ) \] Output:
1/4*(5/x+x/exp(x)-x)*(x+ln(x)/exp(x)/x-25)*x
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \[ \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (5 x^2+50 x^3-3 x^4\right )+e^x \left (5-x^2-50 x^3+28 x^4-x^5\right )+\left (x^2-2 x^3+e^x \left (-5-5 x-x^2+x^3\right )\right ) \log (x)\right )}{4 x^2} \, dx=\frac {1}{4} \left (x \left (5+25 x+e^{-x} (-25+x) x-x^2\right )+\frac {e^{-2 x} \left (x^2-e^x \left (-5+x^2\right )\right ) \log (x)}{x}\right ) \] Input:
Integrate[(x^2 + E^(2*x)*(5*x^2 + 50*x^3 - 3*x^4) + E^x*(5 - x^2 - 50*x^3 + 28*x^4 - x^5) + (x^2 - 2*x^3 + E^x*(-5 - 5*x - x^2 + x^3))*Log[x])/(4*E^ (2*x)*x^2),x]
Output:
(x*(5 + 25*x + ((-25 + x)*x)/E^x - x^2) + ((x^2 - E^x*(-5 + x^2))*Log[x])/ (E^(2*x)*x))/4
Time = 2.67 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {27, 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.
\(\displaystyle \int \frac {e^{-2 x} \left (x^2+\left (-2 x^3+x^2+e^x \left (x^3-x^2-5 x-5\right )\right ) \log (x)+e^{2 x} \left (-3 x^4+50 x^3+5 x^2\right )+e^x \left (-x^5+28 x^4-50 x^3-x^2+5\right )\right )}{4 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (-3 x^4+50 x^3+5 x^2\right )+e^x \left (-x^5+28 x^4-50 x^3-x^2+5\right )+\left (-2 x^3+x^2-e^x \left (-x^3+x^2+5 x+5\right )\right ) \log (x)\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (-3 x^2-2 e^{-2 x} \log (x) x+50 x+e^{-2 x}+e^{-2 x} \log (x)+5-\frac {e^{-x} \left (x^5-28 x^4-\log (x) x^3+50 x^3+\log (x) x^2+x^2+5 \log (x) x+5 \log (x)-5\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (e^{-x} x^3-x^3-25 e^{-x} x^2+25 x^2+5 x+e^{-2 x} x \log (x)-e^{-x} x \log (x)+\frac {5 e^{-x} \log (x)}{x}\right )\) |
Input:
Int[(x^2 + E^(2*x)*(5*x^2 + 50*x^3 - 3*x^4) + E^x*(5 - x^2 - 50*x^3 + 28*x ^4 - x^5) + (x^2 - 2*x^3 + E^x*(-5 - 5*x - x^2 + x^3))*Log[x])/(4*E^(2*x)* x^2),x]
Output:
(5*x + 25*x^2 - (25*x^2)/E^x - x^3 + x^3/E^x + (5*Log[x])/(E^x*x) + (x*Log [x])/E^(2*x) - (x*Log[x])/E^x)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 51.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71
method | result | size |
risch | \(-\frac {\left ({\mathrm e}^{x} x^{2}-x^{2}-5 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-2 x} \ln \left (x \right )}{4 x}-\frac {x \left ({\mathrm e}^{x} x^{2}-x^{2}-25 \,{\mathrm e}^{x} x +25 x -5 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{4}\) | \(60\) |
parallelrisch | \(-\frac {\left ({\mathrm e}^{2 x} x^{4}-{\mathrm e}^{x} x^{4}-25 \,{\mathrm e}^{2 x} x^{3}-5 \ln \left ({\mathrm e}^{x}\right ) x \,{\mathrm e}^{2 x}+25 \,{\mathrm e}^{x} x^{3}+x^{2} {\mathrm e}^{x} \ln \left (x \right )-x^{2} \ln \left (x \right )-5 \,{\mathrm e}^{x} \ln \left (x \right )\right ) {\mathrm e}^{-2 x}}{4 x}\) | \(73\) |
orering | \(\text {Expression too large to display}\) | \(6876\) |
Input:
int(1/4*(((x^3-x^2-5*x-5)*exp(x)-2*x^3+x^2)*ln(x)+(-3*x^4+50*x^3+5*x^2)*ex p(x)^2+(-x^5+28*x^4-50*x^3-x^2+5)*exp(x)+x^2)/exp(x)^2/x^2,x,method=_RETUR NVERBOSE)
Output:
-1/4*(exp(x)*x^2-x^2-5*exp(x))/x/exp(x)^2*ln(x)-1/4*x*(exp(x)*x^2-x^2-25*e xp(x)*x+25*x-5*exp(x))/exp(x)
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (5 x^2+50 x^3-3 x^4\right )+e^x \left (5-x^2-50 x^3+28 x^4-x^5\right )+\left (x^2-2 x^3+e^x \left (-5-5 x-x^2+x^3\right )\right ) \log (x)\right )}{4 x^2} \, dx=-\frac {{\left ({\left (x^{4} - 25 \, x^{3} - 5 \, x^{2}\right )} e^{\left (2 \, x\right )} - {\left (x^{4} - 25 \, x^{3}\right )} e^{x} - {\left (x^{2} - {\left (x^{2} - 5\right )} e^{x}\right )} \log \left (x\right )\right )} e^{\left (-2 \, x\right )}}{4 \, x} \] Input:
integrate(1/4*(((x^3-x^2-5*x-5)*exp(x)-2*x^3+x^2)*log(x)+(-3*x^4+50*x^3+5* x^2)*exp(x)^2+(-x^5+28*x^4-50*x^3-x^2+5)*exp(x)+x^2)/exp(x)^2/x^2,x, algor ithm="fricas")
Output:
-1/4*((x^4 - 25*x^3 - 5*x^2)*e^(2*x) - (x^4 - 25*x^3)*e^x - (x^2 - (x^2 - 5)*e^x)*log(x))*e^(-2*x)/x
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71 \[ \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (5 x^2+50 x^3-3 x^4\right )+e^x \left (5-x^2-50 x^3+28 x^4-x^5\right )+\left (x^2-2 x^3+e^x \left (-5-5 x-x^2+x^3\right )\right ) \log (x)\right )}{4 x^2} \, dx=- \frac {x^{3}}{4} + \frac {25 x^{2}}{4} + \frac {5 x}{4} + \frac {4 x^{2} e^{- 2 x} \log {\left (x \right )} + \left (4 x^{4} - 100 x^{3} - 4 x^{2} \log {\left (x \right )} + 20 \log {\left (x \right )}\right ) e^{- x}}{16 x} \] Input:
integrate(1/4*(((x**3-x**2-5*x-5)*exp(x)-2*x**3+x**2)*ln(x)+(-3*x**4+50*x* *3+5*x**2)*exp(x)**2+(-x**5+28*x**4-50*x**3-x**2+5)*exp(x)+x**2)/exp(x)**2 /x**2,x)
Output:
-x**3/4 + 25*x**2/4 + 5*x/4 + (4*x**2*exp(-2*x)*log(x) + (4*x**4 - 100*x** 3 - 4*x**2*log(x) + 20*log(x))*exp(-x))/(16*x)
\[ \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (5 x^2+50 x^3-3 x^4\right )+e^x \left (5-x^2-50 x^3+28 x^4-x^5\right )+\left (x^2-2 x^3+e^x \left (-5-5 x-x^2+x^3\right )\right ) \log (x)\right )}{4 x^2} \, dx=\int { \frac {{\left (x^{2} - {\left (3 \, x^{4} - 50 \, x^{3} - 5 \, x^{2}\right )} e^{\left (2 \, x\right )} - {\left (x^{5} - 28 \, x^{4} + 50 \, x^{3} + x^{2} - 5\right )} e^{x} - {\left (2 \, x^{3} - x^{2} - {\left (x^{3} - x^{2} - 5 \, x - 5\right )} e^{x}\right )} \log \left (x\right )\right )} e^{\left (-2 \, x\right )}}{4 \, x^{2}} \,d x } \] Input:
integrate(1/4*(((x^3-x^2-5*x-5)*exp(x)-2*x^3+x^2)*log(x)+(-3*x^4+50*x^3+5* x^2)*exp(x)^2+(-x^5+28*x^4-50*x^3-x^2+5)*exp(x)+x^2)/exp(x)^2/x^2,x, algor ithm="maxima")
Output:
-1/4*x^3 + 25/4*x^2 + 1/4*(x^3 + 3*x^2 + 6*x + 6)*e^(-x) - 7*(x^2 + 2*x + 2)*e^(-x) + 25/2*(x + 1)*e^(-x) - 1/8*e^(-2*x)*log(x) + 5/4*x - 1/8*(2*(x^ 2 - 5)*e^(-x)*log(x) - (2*x^2 + x)*e^(-2*x)*log(x))/x + 1/8*Ei(-2*x) + 1/4 *e^(-x) - 1/8*e^(-2*x) - 5/4*gamma(-1, x) - 1/8*integrate((2*x + 1)*e^(-2* x)/x, x) + 1/4*integrate((x^2 - 5)*e^(-x)/x^2, x)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (31) = 62\).
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.03 \[ \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (5 x^2+50 x^3-3 x^4\right )+e^x \left (5-x^2-50 x^3+28 x^4-x^5\right )+\left (x^2-2 x^3+e^x \left (-5-5 x-x^2+x^3\right )\right ) \log (x)\right )}{4 x^2} \, dx=-\frac {{\left (x^{4} e^{\left (2 \, x\right )} - x^{4} e^{x} - 25 \, x^{3} e^{\left (2 \, x\right )} + 25 \, x^{3} e^{x} + x^{2} e^{x} \log \left (x\right ) - 5 \, x^{2} e^{\left (2 \, x\right )} - x^{2} \log \left (x\right ) - 5 \, e^{x} \log \left (x\right )\right )} e^{\left (-2 \, x\right )}}{4 \, x} \] Input:
integrate(1/4*(((x^3-x^2-5*x-5)*exp(x)-2*x^3+x^2)*log(x)+(-3*x^4+50*x^3+5* x^2)*exp(x)^2+(-x^5+28*x^4-50*x^3-x^2+5)*exp(x)+x^2)/exp(x)^2/x^2,x, algor ithm="giac")
Output:
-1/4*(x^4*e^(2*x) - x^4*e^x - 25*x^3*e^(2*x) + 25*x^3*e^x + x^2*e^x*log(x) - 5*x^2*e^(2*x) - x^2*log(x) - 5*e^x*log(x))*e^(-2*x)/x
Time = 3.70 (sec) , antiderivative size = 140, normalized size of antiderivative = 4.00 \[ \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (5 x^2+50 x^3-3 x^4\right )+e^x \left (5-x^2-50 x^3+28 x^4-x^5\right )+\left (x^2-2 x^3+e^x \left (-5-5 x-x^2+x^3\right )\right ) \log (x)\right )}{4 x^2} \, dx=\frac {x\,\left (-x^2+25\,x+5\right )}{4}-\frac {5\,{\mathrm {e}}^{-x}-x\,{\mathrm {e}}^{-x}+25\,x^3\,{\mathrm {e}}^{-x}-x^4\,{\mathrm {e}}^{-x}}{4\,x}-\frac {{\mathrm {e}}^{-2\,x}}{8}+\frac {5\,x\,{\mathrm {e}}^{-x}-x^2\,{\mathrm {e}}^{-x}-x^3\,{\mathrm {e}}^{-x}\,\ln \left (x\right )+5\,x\,{\mathrm {e}}^{-x}\,\ln \left (x\right )}{4\,x^2}-\frac {{\mathrm {e}}^{-2\,x}\,\ln \left (x\right )}{8}+\frac {\frac {x\,{\mathrm {e}}^{-2\,x}}{4}+\frac {x^2\,{\mathrm {e}}^{-2\,x}\,\ln \left (x\right )}{2}+\frac {x\,{\mathrm {e}}^{-2\,x}\,\ln \left (x\right )}{4}}{2\,x} \] Input:
int(-(exp(-2*x)*((log(x)*(exp(x)*(5*x + x^2 - x^3 + 5) - x^2 + 2*x^3))/4 - (exp(2*x)*(5*x^2 + 50*x^3 - 3*x^4))/4 + (exp(x)*(x^2 + 50*x^3 - 28*x^4 + x^5 - 5))/4 - x^2/4))/x^2,x)
Output:
(x*(25*x - x^2 + 5))/4 - (5*exp(-x) - x*exp(-x) + 25*x^3*exp(-x) - x^4*exp (-x))/(4*x) - exp(-2*x)/8 + (5*x*exp(-x) - x^2*exp(-x) - x^3*exp(-x)*log(x ) + 5*x*exp(-x)*log(x))/(4*x^2) - (exp(-2*x)*log(x))/8 + ((x*exp(-2*x))/4 + (x^2*exp(-2*x)*log(x))/2 + (x*exp(-2*x)*log(x))/4)/(2*x)
Time = 0.15 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.31 \[ \int \frac {e^{-2 x} \left (x^2+e^{2 x} \left (5 x^2+50 x^3-3 x^4\right )+e^x \left (5-x^2-50 x^3+28 x^4-x^5\right )+\left (x^2-2 x^3+e^x \left (-5-5 x-x^2+x^3\right )\right ) \log (x)\right )}{4 x^2} \, dx=\frac {-e^{2 x} x^{4}+25 e^{2 x} x^{3}+5 e^{2 x} x^{2}-e^{x} \mathrm {log}\left (x \right ) x^{2}+5 e^{x} \mathrm {log}\left (x \right )+e^{x} x^{4}-25 e^{x} x^{3}+\mathrm {log}\left (x \right ) x^{2}}{4 e^{2 x} x} \] Input:
int(1/4*(((x^3-x^2-5*x-5)*exp(x)-2*x^3+x^2)*log(x)+(-3*x^4+50*x^3+5*x^2)*e xp(x)^2+(-x^5+28*x^4-50*x^3-x^2+5)*exp(x)+x^2)/exp(x)^2/x^2,x)
Output:
( - e**(2*x)*x**4 + 25*e**(2*x)*x**3 + 5*e**(2*x)*x**2 - e**x*log(x)*x**2 + 5*e**x*log(x) + e**x*x**4 - 25*e**x*x**3 + log(x)*x**2)/(4*e**(2*x)*x)