Integrand size = 107, antiderivative size = 27 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=3+x+\frac {x}{\left (5-x-\frac {3}{2} \left (e^x+x\right )\right ) (-6+\log (2))} \]
Time = 0.55 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=\frac {x \left (-6+\frac {2}{10-3 e^x-5 x}+\log (2)\right )}{-6+\log (2)} \]
Integrate[(-580 + 600*x - 150*x^2 + (100 - 100*x + 25*x^2)*Log[2] + E^(2*x )*(-54 + 9*Log[2]) + E^x*(354 - 174*x + (-60 + 30*x)*Log[2]))/(-600 + 600* x - 150*x^2 + (100 - 100*x + 25*x^2)*Log[2] + E^(2*x)*(-54 + 9*Log[2]) + E ^x*(360 - 180*x + (-60 + 30*x)*Log[2])),x]
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.
\(\displaystyle \int \frac {-150 x^2+\left (25 x^2-100 x+100\right ) \log (2)+600 x+e^x (-174 x+(30 x-60) \log (2)+354)+e^{2 x} (9 \log (2)-54)-580}{-150 x^2+\left (25 x^2-100 x+100\right ) \log (2)+600 x+e^x (-180 x+(30 x-60) \log (2)+360)+e^{2 x} (9 \log (2)-54)-600} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {150 x^2-\left (25 x^2-100 x+100\right ) \log (2)-600 x-e^x (-174 x+(30 x-60) \log (2)+354)-e^{2 x} (9 \log (2)-54)+580}{\left (-5 x-3 e^x+10\right )^2 (6-\log (2))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {150 x^2-600 x-6 e^x (-5 \log (2) (2-x)-29 x+59)-25 \left (x^2-4 x+4\right ) \log (2)+9 e^{2 x} (6-\log (2))+580}{\left (-5 x-3 e^x+10\right )^2}dx}{6-\log (2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (-\frac {2 (x-1)}{5 x+3 e^x-10}+\frac {10 (x-3) x}{\left (5 x+3 e^x-10\right )^2}+6 \left (1-\frac {\log (2)}{6}\right )\right )dx}{6-\log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {10 \int \frac {x^2}{\left (5 x+3 e^x-10\right )^2}dx-30 \int \frac {x}{\left (5 x+3 e^x-10\right )^2}dx+2 \int \frac {1}{5 x+3 e^x-10}dx-2 \int \frac {x}{5 x+3 e^x-10}dx+x (6-\log (2))}{6-\log (2)}\) |
Int[(-580 + 600*x - 150*x^2 + (100 - 100*x + 25*x^2)*Log[2] + E^(2*x)*(-54 + 9*Log[2]) + E^x*(354 - 174*x + (-60 + 30*x)*Log[2]))/(-600 + 600*x - 15 0*x^2 + (100 - 100*x + 25*x^2)*Log[2] + E^(2*x)*(-54 + 9*Log[2]) + E^x*(36 0 - 180*x + (-60 + 30*x)*Log[2])),x]
3.13.42.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
risch | \(x -\frac {2 x}{\left (\ln \left (2\right )-6\right ) \left (-10+5 x +3 \,{\mathrm e}^{x}\right )}\) | \(23\) |
parallelrisch | \(\frac {15 x^{2} \ln \left (2\right )+9 x \ln \left (2\right ) {\mathrm e}^{x}-30 x \ln \left (2\right )-90 x^{2}-54 \,{\mathrm e}^{x} x +174 x}{3 \left (\ln \left (2\right )-6\right ) \left (-10+5 x +3 \,{\mathrm e}^{x}\right )}\) | \(53\) |
norman | \(\frac {\frac {6 \left (5 \ln \left (2\right )-29\right ) {\mathrm e}^{x}}{5 \left (\ln \left (2\right )-6\right )}+5 x^{2}+3 \,{\mathrm e}^{x} x -\frac {4 \left (5 \ln \left (2\right )-29\right )}{\ln \left (2\right )-6}}{-10+5 x +3 \,{\mathrm e}^{x}}\) | \(54\) |
int(((9*ln(2)-54)*exp(x)^2+((30*x-60)*ln(2)-174*x+354)*exp(x)+(25*x^2-100* x+100)*ln(2)-150*x^2+600*x-580)/((9*ln(2)-54)*exp(x)^2+((30*x-60)*ln(2)-18 0*x+360)*exp(x)+(25*x^2-100*x+100)*ln(2)-150*x^2+600*x-600),x,method=_RETU RNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (23) = 46\).
Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=-\frac {30 \, x^{2} - 3 \, {\left (x \log \left (2\right ) - 6 \, x\right )} e^{x} - 5 \, {\left (x^{2} - 2 \, x\right )} \log \left (2\right ) - 58 \, x}{3 \, {\left (\log \left (2\right ) - 6\right )} e^{x} + 5 \, {\left (x - 2\right )} \log \left (2\right ) - 30 \, x + 60} \]
integrate(((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-174*x+354)*exp(x)+(25* x^2-100*x+100)*log(2)-150*x^2+600*x-580)/((9*log(2)-54)*exp(x)^2+((30*x-60 )*log(2)-180*x+360)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-600),x, algorithm=\
-(30*x^2 - 3*(x*log(2) - 6*x)*e^x - 5*(x^2 - 2*x)*log(2) - 58*x)/(3*(log(2 ) - 6)*e^x + 5*(x - 2)*log(2) - 30*x + 60)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=x - \frac {2 x}{- 30 x + 5 x \log {\left (2 \right )} + \left (-18 + 3 \log {\left (2 \right )}\right ) e^{x} - 10 \log {\left (2 \right )} + 60} \]
integrate(((9*ln(2)-54)*exp(x)**2+((30*x-60)*ln(2)-174*x+354)*exp(x)+(25*x **2-100*x+100)*ln(2)-150*x**2+600*x-580)/((9*ln(2)-54)*exp(x)**2+((30*x-60 )*ln(2)-180*x+360)*exp(x)+(25*x**2-100*x+100)*ln(2)-150*x**2+600*x-600),x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=\frac {5 \, x^{2} {\left (\log \left (2\right ) - 6\right )} + 3 \, x {\left (\log \left (2\right ) - 6\right )} e^{x} - 2 \, x {\left (5 \, \log \left (2\right ) - 29\right )}}{5 \, x {\left (\log \left (2\right ) - 6\right )} + 3 \, {\left (\log \left (2\right ) - 6\right )} e^{x} - 10 \, \log \left (2\right ) + 60} \]
integrate(((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-174*x+354)*exp(x)+(25* x^2-100*x+100)*log(2)-150*x^2+600*x-580)/((9*log(2)-54)*exp(x)^2+((30*x-60 )*log(2)-180*x+360)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-600),x, algorithm=\
(5*x^2*(log(2) - 6) + 3*x*(log(2) - 6)*e^x - 2*x*(5*log(2) - 29))/(5*x*(lo g(2) - 6) + 3*(log(2) - 6)*e^x - 10*log(2) + 60)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (23) = 46\).
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=\frac {5 \, x^{2} \log \left (2\right ) + 3 \, x e^{x} \log \left (2\right ) - 30 \, x^{2} - 18 \, x e^{x} - 10 \, x \log \left (2\right ) + 58 \, x}{5 \, x \log \left (2\right ) + 3 \, e^{x} \log \left (2\right ) - 30 \, x - 18 \, e^{x} - 10 \, \log \left (2\right ) + 60} \]
integrate(((9*log(2)-54)*exp(x)^2+((30*x-60)*log(2)-174*x+354)*exp(x)+(25* x^2-100*x+100)*log(2)-150*x^2+600*x-580)/((9*log(2)-54)*exp(x)^2+((30*x-60 )*log(2)-180*x+360)*exp(x)+(25*x^2-100*x+100)*log(2)-150*x^2+600*x-600),x, algorithm=\
(5*x^2*log(2) + 3*x*e^x*log(2) - 30*x^2 - 18*x*e^x - 10*x*log(2) + 58*x)/( 5*x*log(2) + 3*e^x*log(2) - 30*x - 18*e^x - 10*log(2) + 60)
Time = 0.59 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=-\frac {x^2\,\left (\ln \left (32\right )-30\right )-x\,\left (10\,\ln \left (2\right )-58\right )+x\,{\mathrm {e}}^x\,\left (\ln \left (8\right )-18\right )}{30\,x+10\,\ln \left (2\right )+18\,{\mathrm {e}}^x-5\,x\,\ln \left (2\right )-3\,{\mathrm {e}}^x\,\ln \left (2\right )-60} \]
int((600*x + log(2)*(25*x^2 - 100*x + 100) + exp(x)*(log(2)*(30*x - 60) - 174*x + 354) + exp(2*x)*(9*log(2) - 54) - 150*x^2 - 580)/(600*x + log(2)*( 25*x^2 - 100*x + 100) + exp(x)*(log(2)*(30*x - 60) - 180*x + 360) + exp(2* x)*(9*log(2) - 54) - 150*x^2 - 600),x)