Integrand size = 106, antiderivative size = 34 \[ \int \frac {e^{-2 x} \left (-4+14 x-8 x^2+e \left (-16+18 x-4 x^2\right )+e^{2 x} \left (-4 x+5 x^2+e \left (3-6 x+2 x^2\right )\right )+\left (-2+6 x-4 x^2+e (-4+4 x)+e^{2 x} \left (e (1-x)-2 x+2 x^2\right )\right ) \log (1-x)\right )}{-1+x} \, dx=\left (2 e^{-2 x}+x\right ) \left (-x+x^2+(e-x) (-3+x-\log (1-x))\right ) \] Output:
(2/exp(x)^2+x)*(x^2-x+(x-ln(1-x)-3)*(exp(1)-x))
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-2 x} \left (-4+14 x-8 x^2+e \left (-16+18 x-4 x^2\right )+e^{2 x} \left (-4 x+5 x^2+e \left (3-6 x+2 x^2\right )\right )+\left (-2+6 x-4 x^2+e (-4+4 x)+e^{2 x} \left (e (1-x)-2 x+2 x^2\right )\right ) \log (1-x)\right )}{-1+x} \, dx=e^{-2 x} \left (2+e^{2 x} x\right ) (e (-3+x)+2 x+(-e+x) \log (1-x)) \] Input:
Integrate[(-4 + 14*x - 8*x^2 + E*(-16 + 18*x - 4*x^2) + E^(2*x)*(-4*x + 5* x^2 + E*(3 - 6*x + 2*x^2)) + (-2 + 6*x - 4*x^2 + E*(-4 + 4*x) + E^(2*x)*(E *(1 - x) - 2*x + 2*x^2))*Log[1 - x])/(E^(2*x)*(-1 + x)),x]
Output:
((2 + E^(2*x)*x)*(E*(-3 + x) + 2*x + (-E + x)*Log[1 - x]))/E^(2*x)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.66 (sec) , antiderivative size = 167, normalized size of antiderivative = 4.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {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.
\(\displaystyle \int \frac {e^{-2 x} \left (-8 x^2+e \left (-4 x^2+18 x-16\right )+e^{2 x} \left (5 x^2+e \left (2 x^2-6 x+3\right )-4 x\right )+\left (-4 x^2+e^{2 x} \left (2 x^2-2 x+e (1-x)\right )+6 x+e (4 x-4)-2\right ) \log (1-x)+14 x-4\right )}{x-1} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {8 e^{-2 x} x^2}{x-1}-\frac {2 e^{1-2 x} \left (2 x^2-9 x+8\right )}{x-1}-\frac {4 e^{-2 x} x^2 \log (1-x)}{x-1}+\frac {-5 \left (1+\frac {2 e}{5}\right ) x^2-2 x^2 \log (1-x)+4 \left (1+\frac {3 e}{2}\right ) x+2 \left (1+\frac {e}{2}\right ) x \log (1-x)-e \log (1-x)-3 e}{1-x}+\frac {14 e^{-2 x} x}{x-1}-\frac {4 e^{-2 x}}{x-1}+\frac {6 e^{-2 x} x \log (1-x)}{x-1}+4 e^{1-2 x} \log (1-x)-\frac {2 e^{-2 x} \log (1-x)}{x-1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \operatorname {ExpIntegralEi}(2-2 x)}{e}-\frac {2 \operatorname {ExpIntegralEi}(2 (1-x))}{e}+\frac {1}{2} (5+2 e) x^2-\frac {1}{8} (e-2 x)^2-6 e^{1-2 x}+2 e^{1-2 x} x+4 e^{-2 x} x-\frac {1}{2} (2-e) x+(1-4 e) x+\frac {1}{4} (e-2 x)^2 \log (1-x)-2 e^{1-2 x} \log (1-x)+2 e^{-2 x} x \log (1-x)-\frac {1}{4} (2-e)^2 \log (1-x)+(1-e) \log (1-x)\) |
Input:
Int[(-4 + 14*x - 8*x^2 + E*(-16 + 18*x - 4*x^2) + E^(2*x)*(-4*x + 5*x^2 + E*(3 - 6*x + 2*x^2)) + (-2 + 6*x - 4*x^2 + E*(-4 + 4*x) + E^(2*x)*(E*(1 - x) - 2*x + 2*x^2))*Log[1 - x])/(E^(2*x)*(-1 + x)),x]
Output:
-6*E^(1 - 2*x) - (E - 2*x)^2/8 + (1 - 4*E)*x - ((2 - E)*x)/2 + 2*E^(1 - 2* x)*x + (4*x)/E^(2*x) + ((5 + 2*E)*x^2)/2 + (2*ExpIntegralEi[2 - 2*x])/E - (2*ExpIntegralEi[2*(1 - x)])/E + (1 - E)*Log[1 - x] - ((2 - E)^2*Log[1 - x ])/4 - 2*E^(1 - 2*x)*Log[1 - x] + ((E - 2*x)^2*Log[1 - x])/4 + (2*x*Log[1 - x])/E^(2*x)
Leaf count of result is larger than twice the leaf count of optimal. \(123\) vs. \(2(34)=68\).
Time = 0.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.65
\[\left (\left (4+2 \,{\mathrm e}\right ) x +2 \ln \left (1-x \right ) x -2 \,{\mathrm e} \ln \left (1-x \right )-6 \,{\mathrm e}\right ) {\mathrm e}^{-2 x}+{\mathrm e} \left (\ln \left (1-x \right ) \left (1-x \right )-1+x \right )-2 \ln \left (1-x \right ) \left (1-x \right )+2-x +\ln \left (1-x \right ) \left (1-x \right )^{2}-\frac {\left (1-x \right )^{2}}{2}+x^{2} {\mathrm e}-4 x \,{\mathrm e}+\frac {5 x^{2}}{2}+\left (1-{\mathrm e}\right ) \ln \left (-1+x \right )\]
Input:
int(((((1-x)*exp(1)+2*x^2-2*x)*exp(x)^2+(-4+4*x)*exp(1)-4*x^2+6*x-2)*ln(1- x)+((2*x^2-6*x+3)*exp(1)+5*x^2-4*x)*exp(x)^2+(-4*x^2+18*x-16)*exp(1)-8*x^2 +14*x-4)/(-1+x)/exp(x)^2,x)
Output:
((4+2*exp(1))*x+2*ln(1-x)*x-2*exp(1)*ln(1-x)-6*exp(1))/exp(x)^2+exp(1)*(ln (1-x)*(1-x)-1+x)-2*ln(1-x)*(1-x)+2-x+ln(1-x)*(1-x)^2-1/2*(1-x)^2+x^2*exp(1 )-4*x*exp(1)+5/2*x^2+(1-exp(1))*ln(-1+x)
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.94 \[ \int \frac {e^{-2 x} \left (-4+14 x-8 x^2+e \left (-16+18 x-4 x^2\right )+e^{2 x} \left (-4 x+5 x^2+e \left (3-6 x+2 x^2\right )\right )+\left (-2+6 x-4 x^2+e (-4+4 x)+e^{2 x} \left (e (1-x)-2 x+2 x^2\right )\right ) \log (1-x)\right )}{-1+x} \, dx={\left (2 \, {\left (x - 3\right )} e + {\left (2 \, x^{2} + {\left (x^{2} - 3 \, x\right )} e\right )} e^{\left (2 \, x\right )} + {\left ({\left (x^{2} - x e\right )} e^{\left (2 \, x\right )} + 2 \, x - 2 \, e\right )} \log \left (-x + 1\right ) + 4 \, x\right )} e^{\left (-2 \, x\right )} \] Input:
integrate(((((1-x)*exp(1)+2*x^2-2*x)*exp(x)^2+(-4+4*x)*exp(1)-4*x^2+6*x-2) *log(1-x)+((2*x^2-6*x+3)*exp(1)+5*x^2-4*x)*exp(x)^2+(-4*x^2+18*x-16)*exp(1 )-8*x^2+14*x-4)/(-1+x)/exp(x)^2,x, algorithm="fricas")
Output:
(2*(x - 3)*e + (2*x^2 + (x^2 - 3*x)*e)*e^(2*x) + ((x^2 - x*e)*e^(2*x) + 2* x - 2*e)*log(-x + 1) + 4*x)*e^(-2*x)
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (26) = 52\).
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.94 \[ \int \frac {e^{-2 x} \left (-4+14 x-8 x^2+e \left (-16+18 x-4 x^2\right )+e^{2 x} \left (-4 x+5 x^2+e \left (3-6 x+2 x^2\right )\right )+\left (-2+6 x-4 x^2+e (-4+4 x)+e^{2 x} \left (e (1-x)-2 x+2 x^2\right )\right ) \log (1-x)\right )}{-1+x} \, dx=x^{2} \cdot \left (2 + e\right ) - 3 e x + \left (x^{2} - e x\right ) \log {\left (1 - x \right )} + \left (2 x \log {\left (1 - x \right )} + 4 x + 2 e x - 2 e \log {\left (1 - x \right )} - 6 e\right ) e^{- 2 x} \] Input:
integrate(((((1-x)*exp(1)+2*x**2-2*x)*exp(x)**2+(-4+4*x)*exp(1)-4*x**2+6*x -2)*ln(1-x)+((2*x**2-6*x+3)*exp(1)+5*x**2-4*x)*exp(x)**2+(-4*x**2+18*x-16) *exp(1)-8*x**2+14*x-4)/(-1+x)/exp(x)**2,x)
Output:
x**2*(2 + E) - 3*E*x + (x**2 - E*x)*log(1 - x) + (2*x*log(1 - x) + 4*x + 2 *E*x - 2*E*log(1 - x) - 6*E)*exp(-2*x)
\[ \int \frac {e^{-2 x} \left (-4+14 x-8 x^2+e \left (-16+18 x-4 x^2\right )+e^{2 x} \left (-4 x+5 x^2+e \left (3-6 x+2 x^2\right )\right )+\left (-2+6 x-4 x^2+e (-4+4 x)+e^{2 x} \left (e (1-x)-2 x+2 x^2\right )\right ) \log (1-x)\right )}{-1+x} \, dx=\int { -\frac {{\left (8 \, x^{2} + 2 \, {\left (2 \, x^{2} - 9 \, x + 8\right )} e - {\left (5 \, x^{2} + {\left (2 \, x^{2} - 6 \, x + 3\right )} e - 4 \, x\right )} e^{\left (2 \, x\right )} + {\left (4 \, x^{2} - 4 \, {\left (x - 1\right )} e - {\left (2 \, x^{2} - {\left (x - 1\right )} e - 2 \, x\right )} e^{\left (2 \, x\right )} - 6 \, x + 2\right )} \log \left (-x + 1\right ) - 14 \, x + 4\right )} e^{\left (-2 \, x\right )}}{x - 1} \,d x } \] Input:
integrate(((((1-x)*exp(1)+2*x^2-2*x)*exp(x)^2+(-4+4*x)*exp(1)-4*x^2+6*x-2) *log(1-x)+((2*x^2-6*x+3)*exp(1)+5*x^2-4*x)*exp(x)^2+(-4*x^2+18*x-16)*exp(1 )-8*x^2+14*x-4)/(-1+x)/exp(x)^2,x, algorithm="maxima")
Output:
16*e^(-1)*exp_integral_e(1, 2*x - 2) + 4*e^(-2)*exp_integral_e(1, 2*x - 2) + (x^3*(e + 2) - 2*x^2*(2*e + 1) + 3*x*e + 2*(x^2*(e + 2) - 2*x*(2*e + 1) )*e^(-2*x) + (x^3 - x^2*(e + 1) + x*e + 2*(x^2 - x*(e + 1) + e)*e^(-2*x))* log(-x + 1))/(x - 1) + integrate(2*(2*x*(e + 1) - 5*e - 2)*e^(-2*x)/(x^2 - 2*x + 1), x)
Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (35) = 70\).
Time = 0.11 (sec) , antiderivative size = 177, normalized size of antiderivative = 5.21 \[ \int \frac {e^{-2 x} \left (-4+14 x-8 x^2+e \left (-16+18 x-4 x^2\right )+e^{2 x} \left (-4 x+5 x^2+e \left (3-6 x+2 x^2\right )\right )+\left (-2+6 x-4 x^2+e (-4+4 x)+e^{2 x} \left (e (1-x)-2 x+2 x^2\right )\right ) \log (1-x)\right )}{-1+x} \, dx={\left ({\left (x - 1\right )}^{2} e^{2} \log \left (-x + 1\right ) + {\left (x - 1\right )}^{2} e^{3} + 2 \, {\left (x - 1\right )}^{2} e^{2} - {\left (x - 1\right )} e^{3} \log \left (-x + 1\right ) + 2 \, {\left (x - 1\right )} e^{2} \log \left (-x + 1\right ) + 2 \, {\left (x - 1\right )} e^{\left (-2 \, x + 2\right )} \log \left (-x + 1\right ) - {\left (x - 1\right )} e^{3} + 4 \, {\left (x - 1\right )} e^{2} + 2 \, {\left (x - 1\right )} e^{\left (-2 \, x + 3\right )} + 4 \, {\left (x - 1\right )} e^{\left (-2 \, x + 2\right )} - e^{3} \log \left (-x + 1\right ) + e^{2} \log \left (-x + 1\right ) - 2 \, e^{\left (-2 \, x + 3\right )} \log \left (-x + 1\right ) + 2 \, e^{\left (-2 \, x + 2\right )} \log \left (-x + 1\right ) - 4 \, e^{\left (-2 \, x + 3\right )} + 4 \, e^{\left (-2 \, x + 2\right )}\right )} e^{\left (-2\right )} \] Input:
integrate(((((1-x)*exp(1)+2*x^2-2*x)*exp(x)^2+(-4+4*x)*exp(1)-4*x^2+6*x-2) *log(1-x)+((2*x^2-6*x+3)*exp(1)+5*x^2-4*x)*exp(x)^2+(-4*x^2+18*x-16)*exp(1 )-8*x^2+14*x-4)/(-1+x)/exp(x)^2,x, algorithm="giac")
Output:
((x - 1)^2*e^2*log(-x + 1) + (x - 1)^2*e^3 + 2*(x - 1)^2*e^2 - (x - 1)*e^3 *log(-x + 1) + 2*(x - 1)*e^2*log(-x + 1) + 2*(x - 1)*e^(-2*x + 2)*log(-x + 1) - (x - 1)*e^3 + 4*(x - 1)*e^2 + 2*(x - 1)*e^(-2*x + 3) + 4*(x - 1)*e^( -2*x + 2) - e^3*log(-x + 1) + e^2*log(-x + 1) - 2*e^(-2*x + 3)*log(-x + 1) + 2*e^(-2*x + 2)*log(-x + 1) - 4*e^(-2*x + 3) + 4*e^(-2*x + 2))*e^(-2)
Time = 4.40 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-2 x} \left (-4+14 x-8 x^2+e \left (-16+18 x-4 x^2\right )+e^{2 x} \left (-4 x+5 x^2+e \left (3-6 x+2 x^2\right )\right )+\left (-2+6 x-4 x^2+e (-4+4 x)+e^{2 x} \left (e (1-x)-2 x+2 x^2\right )\right ) \log (1-x)\right )}{-1+x} \, dx=\left (x+2\,{\mathrm {e}}^{-2\,x}\right )\,\left (2\,x-3\,\mathrm {e}+x\,\mathrm {e}-\mathrm {e}\,\ln \left (1-x\right )+x\,\ln \left (1-x\right )\right ) \] Input:
int(-(exp(-2*x)*(log(1 - x)*(exp(2*x)*(2*x + exp(1)*(x - 1) - 2*x^2) - 6*x + 4*x^2 - exp(1)*(4*x - 4) + 2) - 14*x + exp(1)*(4*x^2 - 18*x + 16) - exp (2*x)*(exp(1)*(2*x^2 - 6*x + 3) - 4*x + 5*x^2) + 8*x^2 + 4))/(x - 1),x)
Output:
(x + 2*exp(-2*x))*(2*x - 3*exp(1) + x*exp(1) - exp(1)*log(1 - x) + x*log(1 - x))
Time = 0.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.82 \[ \int \frac {e^{-2 x} \left (-4+14 x-8 x^2+e \left (-16+18 x-4 x^2\right )+e^{2 x} \left (-4 x+5 x^2+e \left (3-6 x+2 x^2\right )\right )+\left (-2+6 x-4 x^2+e (-4+4 x)+e^{2 x} \left (e (1-x)-2 x+2 x^2\right )\right ) \log (1-x)\right )}{-1+x} \, dx=\frac {-e^{2 x} \mathrm {log}\left (1-x \right ) e x +e^{2 x} \mathrm {log}\left (1-x \right ) x^{2}+e^{2 x} e \,x^{2}-3 e^{2 x} e x +2 e^{2 x} x^{2}-2 \,\mathrm {log}\left (1-x \right ) e +2 \,\mathrm {log}\left (1-x \right ) x +2 e x -6 e +4 x}{e^{2 x}} \] Input:
int(((((1-x)*exp(1)+2*x^2-2*x)*exp(x)^2+(-4+4*x)*exp(1)-4*x^2+6*x-2)*log(1 -x)+((2*x^2-6*x+3)*exp(1)+5*x^2-4*x)*exp(x)^2+(-4*x^2+18*x-16)*exp(1)-8*x^ 2+14*x-4)/(-1+x)/exp(x)^2,x)
Output:
( - e**(2*x)*log( - x + 1)*e*x + e**(2*x)*log( - x + 1)*x**2 + e**(2*x)*e* x**2 - 3*e**(2*x)*e*x + 2*e**(2*x)*x**2 - 2*log( - x + 1)*e + 2*log( - x + 1)*x + 2*e*x - 6*e + 4*x)/e**(2*x)