\(\int \frac {-580+600 x-150 x^2+(100-100 x+25 x^2) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+(100-100 x+25 x^2) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx\) [4242]

Optimal result

Integrand size = 107, antiderivative size = 27 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=3+x+\frac {x}{\left (5-x-\frac {3}{2} \left (e^x+x\right )\right ) (-6+\log (2))} \]

[Out]

x/(ln(2)-6)/(5-5/2*x-3/2*exp(x))+3+x

Rubi [F]

\[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=\int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx \]

[In]

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 - 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]

[Out]

x - (30*Defer[Int][x/(-10 + 3*E^x + 5*x)^2, x])/(6 - Log[2]) + (10*Defer[Int][x^2/(-10 + 3*E^x + 5*x)^2, x])/(
6 - Log[2]) + (2*Defer[Int][(-10 + 3*E^x + 5*x)^(-1), x])/(6 - Log[2]) - (2*Defer[Int][x/(-10 + 3*E^x + 5*x),
x])/(6 - Log[2])

Rubi steps \begin{align*} \text {integral}& = \int \frac {-9 e^{2 x} (-6+\log (2))-5 \left (-20 x (-6+\log (2))+5 x^2 (-6+\log (2))+4 (-29+\log (32))\right )-6 e^x (59-10 \log (2)+x (-29+\log (32)))}{\left (3 e^x+5 (-2+x)\right )^2 (6-\log (2))} \, dx \\ & = \frac {\int \frac {-9 e^{2 x} (-6+\log (2))-5 \left (-20 x (-6+\log (2))+5 x^2 (-6+\log (2))+4 (-29+\log (32))\right )-6 e^x (59-10 \log (2)+x (-29+\log (32)))}{\left (3 e^x+5 (-2+x)\right )^2} \, dx}{6-\log (2)} \\ & = \frac {\int \left (\frac {10 (-3+x) x}{\left (-10+3 e^x+5 x\right )^2}-\frac {2 (-1+x)}{-10+3 e^x+5 x}+6 \left (1-\frac {\log (2)}{6}\right )\right ) \, dx}{6-\log (2)} \\ & = x-\frac {2 \int \frac {-1+x}{-10+3 e^x+5 x} \, dx}{6-\log (2)}+\frac {10 \int \frac {(-3+x) x}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)} \\ & = x-\frac {2 \int \left (-\frac {1}{-10+3 e^x+5 x}+\frac {x}{-10+3 e^x+5 x}\right ) \, dx}{6-\log (2)}+\frac {10 \int \left (-\frac {3 x}{\left (-10+3 e^x+5 x\right )^2}+\frac {x^2}{\left (-10+3 e^x+5 x\right )^2}\right ) \, dx}{6-\log (2)} \\ & = x+\frac {2 \int \frac {1}{-10+3 e^x+5 x} \, dx}{6-\log (2)}-\frac {2 \int \frac {x}{-10+3 e^x+5 x} \, dx}{6-\log (2)}+\frac {10 \int \frac {x^2}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)}-\frac {30 \int \frac {x}{\left (-10+3 e^x+5 x\right )^2} \, dx}{6-\log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=\frac {x \left (-6+\frac {2}{10-3 e^x-5 x}+\log (2)\right )}{-6+\log (2)} \]

[In]

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]

[Out]

(x*(-6 + 2/(10 - 3*E^x - 5*x) + Log[2]))/(-6 + Log[2])

Maple [A] (verified)

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	\(\frac {\frac {6 \left (5 \ln \left (2\right )-29\right ) {\mathrm e}^{x}}{5 \left (\ln \left (2\right )-6\right )}+5 x^{2}+3 \,{\mathrm e}^{x} x -\frac {4 \left (5 \ln \left (2\right )-29\right )}{\ln \left (2\right )-6}}{-10+5 x +3 \,{\mathrm e}^{x}}\)	\(54\)

[In]

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)-180*x+360)*exp(x)+(25*x^2-100*x+100)*ln(2)-150*x^2+600*x-600),x,method=_RE
TURNVERBOSE)

[Out]

x-2*x/(ln(2)-6)/(-10+5*x+3*exp(x))

Fricas [B] (verification not implemented)

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 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=-\frac {30 \, x^{2} - 3 \, {\left (x \log \left (2\right ) - 6 \, x\right )} e^{x} - 5 \, {\left (x^{2} - 2 \, x\right )} \log \left (2\right ) - 58 \, x}{3 \, {\left (\log \left (2\right ) - 6\right )} e^{x} + 5 \, {\left (x - 2\right )} \log \left (2\right ) - 30 \, x + 60} \]

[In]

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="fricas")

[Out]

-(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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=x - \frac {2 x}{- 30 x + 5 x \log {\left (2 \right )} + \left (-18 + 3 \log {\left (2 \right )}\right ) e^{x} - 10 \log {\left (2 \right )} + 60} \]

[In]

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)

[Out]

x - 2*x/(-30*x + 5*x*log(2) + (-18 + 3*log(2))*exp(x) - 10*log(2) + 60)

Maxima [B] (verification not implemented)

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 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=\frac {5 \, x^{2} {\left (\log \left (2\right ) - 6\right )} + 3 \, x {\left (\log \left (2\right ) - 6\right )} e^{x} - 2 \, x {\left (5 \, \log \left (2\right ) - 29\right )}}{5 \, x {\left (\log \left (2\right ) - 6\right )} + 3 \, {\left (\log \left (2\right ) - 6\right )} e^{x} - 10 \, \log \left (2\right ) + 60} \]

[In]

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="maxima")

[Out]

(5*x^2*(log(2) - 6) + 3*x*(log(2) - 6)*e^x - 2*x*(5*log(2) - 29))/(5*x*(log(2) - 6) + 3*(log(2) - 6)*e^x - 10*
log(2) + 60)

Giac [B] (verification not implemented)

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 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=\frac {5 \, x^{2} \log \left (2\right ) + 3 \, x e^{x} \log \left (2\right ) - 30 \, x^{2} - 18 \, x e^{x} - 10 \, x \log \left (2\right ) + 58 \, x}{5 \, x \log \left (2\right ) + 3 \, e^{x} \log \left (2\right ) - 30 \, x - 18 \, e^{x} - 10 \, \log \left (2\right ) + 60} \]

[In]

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="giac")

[Out]

(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 - 1
8*e^x - 10*log(2) + 60)

Mupad [B] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {-580+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (354-174 x+(-60+30 x) \log (2))}{-600+600 x-150 x^2+\left (100-100 x+25 x^2\right ) \log (2)+e^{2 x} (-54+9 \log (2))+e^x (360-180 x+(-60+30 x) \log (2))} \, dx=-\frac {x^2\,\left (\ln \left (32\right )-30\right )-x\,\left (10\,\ln \left (2\right )-58\right )+x\,{\mathrm {e}}^x\,\left (\ln \left (8\right )-18\right )}{30\,x+10\,\ln \left (2\right )+18\,{\mathrm {e}}^x-5\,x\,\ln \left (2\right )-3\,{\mathrm {e}}^x\,\ln \left (2\right )-60} \]

[In]

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)

[Out]

-(x^2*(log(32) - 30) - x*(10*log(2) - 58) + x*exp(x)*(log(8) - 18))/(30*x + 10*log(2) + 18*exp(x) - 5*x*log(2)
- 3*exp(x)*log(2) - 60)