Integrand size = 183, antiderivative size = 30 \[ \int \frac {e^{-10-2 x} \left (9216-4608 x+\left (4608+6912 x-4608 x^2+e^{5+x} \left (576 x-288 x^2\right )+e^{5+x} (192-96 x) \log (2-x)\right ) \log (x)+\left (e^{5+x} \left (96 x+576 x^2-288 x^3\right )+e^{5+x} \left (192+96 x-96 x^2\right ) \log (2-x)\right ) \log ^2(x)+\left (e^{10+2 x} \left (-12 x^2+9 x^3\right )+2 e^{10+2 x} x \log (2-x)+e^{10+2 x} (2-x) \log ^2(2-x)\right ) \log ^3(x)\right )}{\left (-2 x^2+x^3\right ) \log ^3(x)} \, dx=\frac {\left (\log (2-x)+3 \left (x+\frac {16 e^{-5-x}}{\log (x)}\right )\right )^2}{x} \]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 3.87 \[ \int \frac {e^{-10-2 x} \left (9216-4608 x+\left (4608+6912 x-4608 x^2+e^{5+x} \left (576 x-288 x^2\right )+e^{5+x} (192-96 x) \log (2-x)\right ) \log (x)+\left (e^{5+x} \left (96 x+576 x^2-288 x^3\right )+e^{5+x} \left (192+96 x-96 x^2\right ) \log (2-x)\right ) \log ^2(x)+\left (e^{10+2 x} \left (-12 x^2+9 x^3\right )+2 e^{10+2 x} x \log (2-x)+e^{10+2 x} (2-x) \log ^2(2-x)\right ) \log ^3(x)\right )}{\left (-2 x^2+x^3\right ) \log ^3(x)} \, dx=9 x+6 \log (2-x)-\log (2) \log (2-x)+\frac {\log ^2(2-x)}{x}+\frac {2304 e^{-2 (5+x)}}{x \log ^2(x)}+\frac {288 e^{-5-x}}{\log (x)}+\frac {96 e^{-5-x} \log (2-x)}{x \log (x)}-\log (2) \log (x)+\log (2-x) \log (x)+\operatorname {PolyLog}\left (2,1-\frac {x}{2}\right )+\operatorname {PolyLog}\left (2,\frac {x}{2}\right ) \]
Integrate[(E^(-10 - 2*x)*(9216 - 4608*x + (4608 + 6912*x - 4608*x^2 + E^(5 + x)*(576*x - 288*x^2) + E^(5 + x)*(192 - 96*x)*Log[2 - x])*Log[x] + (E^( 5 + x)*(96*x + 576*x^2 - 288*x^3) + E^(5 + x)*(192 + 96*x - 96*x^2)*Log[2 - x])*Log[x]^2 + (E^(10 + 2*x)*(-12*x^2 + 9*x^3) + 2*E^(10 + 2*x)*x*Log[2 - x] + E^(10 + 2*x)*(2 - x)*Log[2 - x]^2)*Log[x]^3))/((-2*x^2 + x^3)*Log[x ]^3),x]
9*x + 6*Log[2 - x] - Log[2]*Log[2 - x] + Log[2 - x]^2/x + 2304/(E^(2*(5 + x))*x*Log[x]^2) + (288*E^(-5 - x))/Log[x] + (96*E^(-5 - x)*Log[2 - x])/(x* Log[x]) - Log[2]*Log[x] + Log[2 - x]*Log[x] + PolyLog[2, 1 - x/2] + PolyLo g[2, x/2]
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.
| \(\displaystyle \int \frac {e^{-2 x-10} \left (\left (-4608 x^2+e^{x+5} \left (576 x-288 x^2\right )+6912 x+e^{x+5} (192-96 x) \log (2-x)+4608\right ) \log (x)+\left (e^{x+5} \left (-96 x^2+96 x+192\right ) \log (2-x)+e^{x+5} \left (-288 x^3+576 x^2+96 x\right )\right ) \log ^2(x)+\left (e^{2 x+10} \left (9 x^3-12 x^2\right )+e^{2 x+10} (2-x) \log ^2(2-x)+2 e^{2 x+10} x \log (2-x)\right ) \log ^3(x)-4608 x+9216\right )}{\left (x^3-2 x^2\right ) \log ^3(x)} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{-2 x-10} \left (\left (-4608 x^2+e^{x+5} \left (576 x-288 x^2\right )+6912 x+e^{x+5} (192-96 x) \log (2-x)+4608\right ) \log (x)+\left (e^{x+5} \left (-96 x^2+96 x+192\right ) \log (2-x)+e^{x+5} \left (-288 x^3+576 x^2+96 x\right )\right ) \log ^2(x)+\left (e^{2 x+10} \left (9 x^3-12 x^2\right )+e^{2 x+10} (2-x) \log ^2(2-x)+2 e^{2 x+10} x \log (2-x)\right ) \log ^3(x)-4608 x+9216\right )}{(x-2) x^2 \log ^3(x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {2304 e^{-2 x-10} (2 x \log (x)+\log (x)+2)}{x^2 \log ^3(x)}+\frac {9 x^3-12 x^2-x \log ^2(2-x)+2 \log ^2(2-x)+2 x \log (2-x)}{(x-2) x^2}-\frac {96 e^{-x-5} \left (3 x^3 \log (x)+3 x^2+x^2 \log (2-x) \log (x)-6 x^2 \log (x)-6 x+x \log (2-x)-x \log (2-x) \log (x)-x \log (x)-2 \log (2-x)-2 \log (2-x) \log (x)\right )}{(x-2) x^2 \log ^2(x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -96 \int \frac {e^{-x-5} \log (2-x)}{x^2 \log ^2(x)}dx-96 \int \frac {e^{-x-5} \log (2-x)}{x^2 \log (x)}dx-288 \int \frac {e^{-x-5}}{x \log ^2(x)}dx-288 \int \frac {e^{-x-5}}{\log (x)}dx+48 \int \frac {e^{-x-5}}{(x-2) \log (x)}dx-48 \int \frac {e^{-x-5}}{x \log (x)}dx-96 \int \frac {e^{-x-5} \log (2-x)}{x \log (x)}dx+\operatorname {PolyLog}\left (2,\frac {2}{2-x}\right )-\operatorname {PolyLog}\left (2,\frac {x}{2}\right )+9 x+\frac {(2-x) \log ^2(2-x)}{2 x}+\frac {2304 e^{-2 x-10}}{x \log ^2(x)}-\log \left (1-\frac {2}{2-x}\right ) \log (2-x)+6 \log (2-x)+\log (2) \log (x)\) |
Int[(E^(-10 - 2*x)*(9216 - 4608*x + (4608 + 6912*x - 4608*x^2 + E^(5 + x)* (576*x - 288*x^2) + E^(5 + x)*(192 - 96*x)*Log[2 - x])*Log[x] + (E^(5 + x) *(96*x + 576*x^2 - 288*x^3) + E^(5 + x)*(192 + 96*x - 96*x^2)*Log[2 - x])* Log[x]^2 + (E^(10 + 2*x)*(-12*x^2 + 9*x^3) + 2*E^(10 + 2*x)*x*Log[2 - x] + E^(10 + 2*x)*(2 - x)*Log[2 - x]^2)*Log[x]^3))/((-2*x^2 + x^3)*Log[x]^3),x ]
3.25.76.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Leaf count of result is larger than twice the leaf count of optimal. \(77\) vs. \(2(28)=56\).
Time = 2.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.60
| method | result | size |
| risch | \(\frac {6 \ln \left (x \right )^{2} x \ln \left (-2+x \right )+9 x^{2} \ln \left (x \right )^{2}+\ln \left (2-x \right )^{2} \ln \left (x \right )^{2}+288 x \ln \left (x \right ) {\mathrm e}^{-x -5}+96 \,{\mathrm e}^{-x -5} \ln \left (2-x \right ) \ln \left (x \right )+2304 \,{\mathrm e}^{-2 x -10}}{\ln \left (x \right )^{2} x}\) | \(78\) |
| parallelrisch | \(\frac {\left (9216+24 \ln \left (-2+x \right ) x \,{\mathrm e}^{2 x +10} \ln \left (x \right )^{2}+36 \ln \left (x \right )^{2} {\mathrm e}^{2 x +10} x^{2}+4 \,{\mathrm e}^{2 x +10} \ln \left (2-x \right )^{2} \ln \left (x \right )^{2}+36 \,{\mathrm e}^{2 x +10} \ln \left (x \right )^{2} x +1152 \ln \left (x \right ) {\mathrm e}^{5+x} x +384 \ln \left (x \right ) {\mathrm e}^{5+x} \ln \left (2-x \right )\right ) {\mathrm e}^{-2 x -10}}{4 x \ln \left (x \right )^{2}}\) | \(106\) |
int((((2-x)*exp(5+x)^2*ln(2-x)^2+2*x*exp(5+x)^2*ln(2-x)+(9*x^3-12*x^2)*exp (5+x)^2)*ln(x)^3+((-96*x^2+96*x+192)*exp(5+x)*ln(2-x)+(-288*x^3+576*x^2+96 *x)*exp(5+x))*ln(x)^2+((-96*x+192)*exp(5+x)*ln(2-x)+(-288*x^2+576*x)*exp(5 +x)-4608*x^2+6912*x+4608)*ln(x)-4608*x+9216)/(x^3-2*x^2)/exp(5+x)^2/ln(x)^ 3,x,method=_RETURNVERBOSE)
(6*ln(x)^2*x*ln(-2+x)+9*x^2*ln(x)^2+ln(2-x)^2*ln(x)^2+288*x*ln(x)*exp(-x-5 )+96*exp(-x-5)*ln(2-x)*ln(x)+2304*exp(-2*x-10))/ln(x)^2/x
Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (28) = 56\).
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.87 \[ \int \frac {e^{-10-2 x} \left (9216-4608 x+\left (4608+6912 x-4608 x^2+e^{5+x} \left (576 x-288 x^2\right )+e^{5+x} (192-96 x) \log (2-x)\right ) \log (x)+\left (e^{5+x} \left (96 x+576 x^2-288 x^3\right )+e^{5+x} \left (192+96 x-96 x^2\right ) \log (2-x)\right ) \log ^2(x)+\left (e^{10+2 x} \left (-12 x^2+9 x^3\right )+2 e^{10+2 x} x \log (2-x)+e^{10+2 x} (2-x) \log ^2(2-x)\right ) \log ^3(x)\right )}{\left (-2 x^2+x^3\right ) \log ^3(x)} \, dx=\frac {{\left ({\left (9 \, x^{2} e^{\left (2 \, x + 10\right )} + 6 \, x e^{\left (2 \, x + 10\right )} \log \left (-x + 2\right ) + e^{\left (2 \, x + 10\right )} \log \left (-x + 2\right )^{2}\right )} \log \left (x\right )^{2} + 96 \, {\left (3 \, x e^{\left (x + 5\right )} + e^{\left (x + 5\right )} \log \left (-x + 2\right )\right )} \log \left (x\right ) + 2304\right )} e^{\left (-2 \, x - 10\right )}}{x \log \left (x\right )^{2}} \]
integrate((((2-x)*exp(5+x)^2*log(2-x)^2+2*x*exp(5+x)^2*log(2-x)+(9*x^3-12* x^2)*exp(5+x)^2)*log(x)^3+((-96*x^2+96*x+192)*exp(5+x)*log(2-x)+(-288*x^3+ 576*x^2+96*x)*exp(5+x))*log(x)^2+((-96*x+192)*exp(5+x)*log(2-x)+(-288*x^2+ 576*x)*exp(5+x)-4608*x^2+6912*x+4608)*log(x)-4608*x+9216)/(x^3-2*x^2)/exp( 5+x)^2/log(x)^3,x, algorithm=\
((9*x^2*e^(2*x + 10) + 6*x*e^(2*x + 10)*log(-x + 2) + e^(2*x + 10)*log(-x + 2)^2)*log(x)^2 + 96*(3*x*e^(x + 5) + e^(x + 5)*log(-x + 2))*log(x) + 230 4)*e^(-2*x - 10)/(x*log(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (22) = 44\).
Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.37 \[ \int \frac {e^{-10-2 x} \left (9216-4608 x+\left (4608+6912 x-4608 x^2+e^{5+x} \left (576 x-288 x^2\right )+e^{5+x} (192-96 x) \log (2-x)\right ) \log (x)+\left (e^{5+x} \left (96 x+576 x^2-288 x^3\right )+e^{5+x} \left (192+96 x-96 x^2\right ) \log (2-x)\right ) \log ^2(x)+\left (e^{10+2 x} \left (-12 x^2+9 x^3\right )+2 e^{10+2 x} x \log (2-x)+e^{10+2 x} (2-x) \log ^2(2-x)\right ) \log ^3(x)\right )}{\left (-2 x^2+x^3\right ) \log ^3(x)} \, dx=9 x + 6 \log {\left (x - 2 \right )} + \frac {\log {\left (2 - x \right )}^{2}}{x} + \frac {2304 x e^{- 2 x - 10} \log {\left (x \right )} + \left (288 x^{2} \log {\left (x \right )}^{2} + 96 x \log {\left (x \right )}^{2} \log {\left (2 - x \right )}\right ) e^{- x - 5}}{x^{2} \log {\left (x \right )}^{3}} \]
integrate((((2-x)*exp(5+x)**2*ln(2-x)**2+2*x*exp(5+x)**2*ln(2-x)+(9*x**3-1 2*x**2)*exp(5+x)**2)*ln(x)**3+((-96*x**2+96*x+192)*exp(5+x)*ln(2-x)+(-288* x**3+576*x**2+96*x)*exp(5+x))*ln(x)**2+((-96*x+192)*exp(5+x)*ln(2-x)+(-288 *x**2+576*x)*exp(5+x)-4608*x**2+6912*x+4608)*ln(x)-4608*x+9216)/(x**3-2*x* *2)/exp(5+x)**2/ln(x)**3,x)
9*x + 6*log(x - 2) + log(2 - x)**2/x + (2304*x*exp(-2*x - 10)*log(x) + (28 8*x**2*log(x)**2 + 96*x*log(x)**2*log(2 - x))*exp(-x - 5))/(x**2*log(x)**3 )
Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (28) = 56\).
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.70 \[ \int \frac {e^{-10-2 x} \left (9216-4608 x+\left (4608+6912 x-4608 x^2+e^{5+x} \left (576 x-288 x^2\right )+e^{5+x} (192-96 x) \log (2-x)\right ) \log (x)+\left (e^{5+x} \left (96 x+576 x^2-288 x^3\right )+e^{5+x} \left (192+96 x-96 x^2\right ) \log (2-x)\right ) \log ^2(x)+\left (e^{10+2 x} \left (-12 x^2+9 x^3\right )+2 e^{10+2 x} x \log (2-x)+e^{10+2 x} (2-x) \log ^2(2-x)\right ) \log ^3(x)\right )}{\left (-2 x^2+x^3\right ) \log ^3(x)} \, dx=\frac {{\left (9 \, x^{2} e^{10} \log \left (x\right )^{2} + e^{10} \log \left (x\right )^{2} \log \left (-x + 2\right )^{2} + 288 \, x e^{\left (-x + 5\right )} \log \left (x\right ) + 6 \, {\left (x e^{10} \log \left (x\right )^{2} + 16 \, e^{\left (-x + 5\right )} \log \left (x\right )\right )} \log \left (-x + 2\right ) + 2304 \, e^{\left (-2 \, x\right )}\right )} e^{\left (-10\right )}}{x \log \left (x\right )^{2}} \]
integrate((((2-x)*exp(5+x)^2*log(2-x)^2+2*x*exp(5+x)^2*log(2-x)+(9*x^3-12* x^2)*exp(5+x)^2)*log(x)^3+((-96*x^2+96*x+192)*exp(5+x)*log(2-x)+(-288*x^3+ 576*x^2+96*x)*exp(5+x))*log(x)^2+((-96*x+192)*exp(5+x)*log(2-x)+(-288*x^2+ 576*x)*exp(5+x)-4608*x^2+6912*x+4608)*log(x)-4608*x+9216)/(x^3-2*x^2)/exp( 5+x)^2/log(x)^3,x, algorithm=\
(9*x^2*e^10*log(x)^2 + e^10*log(x)^2*log(-x + 2)^2 + 288*x*e^(-x + 5)*log( x) + 6*(x*e^10*log(x)^2 + 16*e^(-x + 5)*log(x))*log(-x + 2) + 2304*e^(-2*x ))*e^(-10)/(x*log(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (28) = 56\).
Time = 0.34 (sec) , antiderivative size = 122, normalized size of antiderivative = 4.07 \[ \int \frac {e^{-10-2 x} \left (9216-4608 x+\left (4608+6912 x-4608 x^2+e^{5+x} \left (576 x-288 x^2\right )+e^{5+x} (192-96 x) \log (2-x)\right ) \log (x)+\left (e^{5+x} \left (96 x+576 x^2-288 x^3\right )+e^{5+x} \left (192+96 x-96 x^2\right ) \log (2-x)\right ) \log ^2(x)+\left (e^{10+2 x} \left (-12 x^2+9 x^3\right )+2 e^{10+2 x} x \log (2-x)+e^{10+2 x} (2-x) \log ^2(2-x)\right ) \log ^3(x)\right )}{\left (-2 x^2+x^3\right ) \log ^3(x)} \, dx=\frac {9 \, {\left (x + 5\right )}^{2} \log \left (x\right )^{2} + 6 \, {\left (x + 5\right )} \log \left (x - 2\right ) \log \left (x\right )^{2} + \log \left (x\right )^{2} \log \left (-x + 2\right )^{2} + 288 \, {\left (x + 5\right )} e^{\left (-x - 5\right )} \log \left (x\right ) - 45 \, {\left (x + 5\right )} \log \left (x\right )^{2} - 30 \, \log \left (x - 2\right ) \log \left (x\right )^{2} + 96 \, e^{\left (-x - 5\right )} \log \left (x\right ) \log \left (-x + 2\right ) - 1440 \, e^{\left (-x - 5\right )} \log \left (x\right ) + 2304 \, e^{\left (-2 \, x - 10\right )}}{{\left (x + 5\right )} \log \left (x\right )^{2} - 5 \, \log \left (x\right )^{2}} \]
integrate((((2-x)*exp(5+x)^2*log(2-x)^2+2*x*exp(5+x)^2*log(2-x)+(9*x^3-12* x^2)*exp(5+x)^2)*log(x)^3+((-96*x^2+96*x+192)*exp(5+x)*log(2-x)+(-288*x^3+ 576*x^2+96*x)*exp(5+x))*log(x)^2+((-96*x+192)*exp(5+x)*log(2-x)+(-288*x^2+ 576*x)*exp(5+x)-4608*x^2+6912*x+4608)*log(x)-4608*x+9216)/(x^3-2*x^2)/exp( 5+x)^2/log(x)^3,x, algorithm=\
(9*(x + 5)^2*log(x)^2 + 6*(x + 5)*log(x - 2)*log(x)^2 + log(x)^2*log(-x + 2)^2 + 288*(x + 5)*e^(-x - 5)*log(x) - 45*(x + 5)*log(x)^2 - 30*log(x - 2) *log(x)^2 + 96*e^(-x - 5)*log(x)*log(-x + 2) - 1440*e^(-x - 5)*log(x) + 23 04*e^(-2*x - 10))/((x + 5)*log(x)^2 - 5*log(x)^2)
Timed out. \[ \int \frac {e^{-10-2 x} \left (9216-4608 x+\left (4608+6912 x-4608 x^2+e^{5+x} \left (576 x-288 x^2\right )+e^{5+x} (192-96 x) \log (2-x)\right ) \log (x)+\left (e^{5+x} \left (96 x+576 x^2-288 x^3\right )+e^{5+x} \left (192+96 x-96 x^2\right ) \log (2-x)\right ) \log ^2(x)+\left (e^{10+2 x} \left (-12 x^2+9 x^3\right )+2 e^{10+2 x} x \log (2-x)+e^{10+2 x} (2-x) \log ^2(2-x)\right ) \log ^3(x)\right )}{\left (-2 x^2+x^3\right ) \log ^3(x)} \, dx=\int -\frac {{\mathrm {e}}^{-2\,x-10}\,\left (\left (-{\mathrm {e}}^{2\,x+10}\,\left (x-2\right )\,{\ln \left (2-x\right )}^2+2\,x\,{\mathrm {e}}^{2\,x+10}\,\ln \left (2-x\right )-{\mathrm {e}}^{2\,x+10}\,\left (12\,x^2-9\,x^3\right )\right )\,{\ln \left (x\right )}^3+\left ({\mathrm {e}}^{x+5}\,\left (-288\,x^3+576\,x^2+96\,x\right )+{\mathrm {e}}^{x+5}\,\ln \left (2-x\right )\,\left (-96\,x^2+96\,x+192\right )\right )\,{\ln \left (x\right )}^2+\left (6912\,x+{\mathrm {e}}^{x+5}\,\left (576\,x-288\,x^2\right )-4608\,x^2-{\mathrm {e}}^{x+5}\,\ln \left (2-x\right )\,\left (96\,x-192\right )+4608\right )\,\ln \left (x\right )-4608\,x+9216\right )}{{\ln \left (x\right )}^3\,\left (2\,x^2-x^3\right )} \,d x \]
int(-(exp(- 2*x - 10)*(log(x)*(6912*x + exp(x + 5)*(576*x - 288*x^2) - 460 8*x^2 - exp(x + 5)*log(2 - x)*(96*x - 192) + 4608) - 4608*x - log(x)^3*(ex p(2*x + 10)*(12*x^2 - 9*x^3) - 2*x*exp(2*x + 10)*log(2 - x) + exp(2*x + 10 )*log(2 - x)^2*(x - 2)) + log(x)^2*(exp(x + 5)*(96*x + 576*x^2 - 288*x^3) + exp(x + 5)*log(2 - x)*(96*x - 96*x^2 + 192)) + 9216))/(log(x)^3*(2*x^2 - x^3)),x)
int(-(exp(- 2*x - 10)*(log(x)*(6912*x + exp(x + 5)*(576*x - 288*x^2) - 460 8*x^2 - exp(x + 5)*log(2 - x)*(96*x - 192) + 4608) - 4608*x - log(x)^3*(ex p(2*x + 10)*(12*x^2 - 9*x^3) - 2*x*exp(2*x + 10)*log(2 - x) + exp(2*x + 10 )*log(2 - x)^2*(x - 2)) + log(x)^2*(exp(x + 5)*(96*x + 576*x^2 - 288*x^3) + exp(x + 5)*log(2 - x)*(96*x - 96*x^2 + 192)) + 9216))/(log(x)^3*(2*x^2 - x^3)), x)