Integrand size = 144, antiderivative size = 29 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {3-x}{20 e^x-x-4 \log (2) \left (e^x-\log (x)\right )} \] Output:
(3-x)/(5*exp(x+2*ln(2))-4*(exp(x)-ln(x))*ln(2)-x)
Leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(29)=58\).
Time = 0.85 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.48 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x^2 \left (1+4 e^x (-5+\log (2))\right )-x \left (3+12 e^x (-5+\log (2))+\log (16)\right )+\log (4096)}{\left (x+4 e^x x (-5+\log (2))-4 \log (2)\right ) \left (x+4 e^x (-5+\log (2))-4 \log (2) \log (x)\right )} \] Input:
Integrate[(3*x + 4*E^x*(-20*x + 5*x^2) + (-12 + 4*x)*Log[2] + E^x*(16*x - 4*x^2)*Log[2] - 4*x*Log[2]*Log[x])/(400*E^(2*x)*x + x^3 + 8*E^x*x^2*Log[2] + 16*E^(2*x)*x*Log[2]^2 + 4*E^x*(-10*x^2 - 40*E^x*x*Log[2]) + (160*E^x*x* Log[2] - 8*x^2*Log[2] - 32*E^x*x*Log[2]^2)*Log[x] + 16*x*Log[2]^2*Log[x]^2 ),x]
Output:
(x^2*(1 + 4*E^x*(-5 + Log[2])) - x*(3 + 12*E^x*(-5 + Log[2]) + Log[16]) + Log[4096])/((x + 4*E^x*x*(-5 + Log[2]) - 4*Log[2])*(x + 4*E^x*(-5 + Log[2] ) - 4*Log[2]*Log[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.
| \(\displaystyle \int \frac {4 e^x \left (5 x^2-20 x\right )+e^x \left (16 x-4 x^2\right ) \log (2)+3 x-4 x \log (2) \log (x)+(4 x-12) \log (2)}{x^3+\left (-8 x^2 \log (2)-32 e^x x \log ^2(2)+160 e^x x \log (2)\right ) \log (x)+8 e^x x^2 \log (2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+400 e^{2 x} x+16 x \log ^2(2) \log ^2(x)+16 e^{2 x} x \log ^2(2)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {4 e^x \left (5 x^2-20 x\right )+e^x \left (16 x-4 x^2\right ) \log (2)+3 x-4 x \log (2) \log (x)+(4 x-12) \log (2)}{x^3+\left (-8 x^2 \log (2)-32 e^x x \log ^2(2)+160 e^x x \log (2)\right ) \log (x)+8 e^x x^2 \log (2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+16 x \log ^2(2) \log ^2(x)+e^{2 x} x \left (400+16 \log ^2(2)\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-4 e^x x^2 (\log (2)-5)+x \left (16 e^x (\log (2)-5)+3+\log (16)\right )-4 x \log (2) \log (x)-12 \log (2)}{x \left (x-4 \log (2) \log (x)+4 e^x (\log (2)-5)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {x^3-4 x^2-4 x^2 \log (2) \log (x)+12 x \log (2) \log (x)+3 x \left (1+\frac {\log (16)}{3}\right )-12 \log (2)}{x \left (x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )\right )^2}+\frac {4-x}{x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {x^2}{\left (x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )\right )^2}dx+(3+\log (16)) \int \frac {1}{\left (x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )\right )^2}dx-12 \log (2) \int \frac {1}{x \left (x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )\right )^2}dx-4 \int \frac {x}{\left (x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )\right )^2}dx+12 \log (2) \int \frac {\log (x)}{\left (x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )\right )^2}dx-4 \log (2) \int \frac {x \log (x)}{\left (x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )\right )^2}dx+4 \int \frac {1}{x-4 \log (2) \log (x)-20 e^x \left (1-\frac {\log (2)}{5}\right )}dx+\int \frac {x}{-x+4 \log (2) \log (x)+20 e^x \left (1-\frac {\log (2)}{5}\right )}dx\) |
Input:
Int[(3*x + 4*E^x*(-20*x + 5*x^2) + (-12 + 4*x)*Log[2] + E^x*(16*x - 4*x^2) *Log[2] - 4*x*Log[2]*Log[x])/(400*E^(2*x)*x + x^3 + 8*E^x*x^2*Log[2] + 16* E^(2*x)*x*Log[2]^2 + 4*E^x*(-10*x^2 - 40*E^x*x*Log[2]) + (160*E^x*x*Log[2] - 8*x^2*Log[2] - 32*E^x*x*Log[2]^2)*Log[x] + 16*x*Log[2]^2*Log[x]^2),x]
Output:
$Aborted
Time = 1.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
| method | result | size |
| risch | \(\frac {-3+x}{4 \,{\mathrm e}^{x} \ln \left (2\right )-4 \ln \left (2\right ) \ln \left (x \right )+x -20 \,{\mathrm e}^{x}}\) | \(25\) |
| parallelrisch | \(\frac {15-5 x}{20 \ln \left (2\right ) \ln \left (x \right )-20 \,{\mathrm e}^{x} \ln \left (2\right )+25 \,{\mathrm e}^{x +2 \ln \left (2\right )}-5 x}\) | \(35\) |
Input:
int((-4*x*ln(2)*ln(x)+(5*x^2-20*x)*exp(x+2*ln(2))+(-4*x^2+16*x)*ln(2)*exp( x)+(4*x-12)*ln(2)+3*x)/(16*x*ln(2)^2*ln(x)^2+(40*x*ln(2)*exp(x+2*ln(2))-32 *x*ln(2)^2*exp(x)-8*x^2*ln(2))*ln(x)+25*x*exp(x+2*ln(2))^2+(-40*x*ln(2)*ex p(x)-10*x^2)*exp(x+2*ln(2))+16*x*ln(2)^2*exp(x)^2+8*x^2*ln(2)*exp(x)+x^3), x,method=_RETURNVERBOSE)
Output:
(-3+x)/(4*exp(x)*ln(2)-4*ln(2)*ln(x)+x-20*exp(x))
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x - 3}{{\left (\log \left (2\right ) - 5\right )} e^{\left (x + 2 \, \log \left (2\right )\right )} - 4 \, \log \left (2\right ) \log \left (x\right ) + x} \] Input:
integrate((-4*x*log(2)*log(x)+(5*x^2-20*x)*exp(x+2*log(2))+(-4*x^2+16*x)*l og(2)*exp(x)+(4*x-12)*log(2)+3*x)/(16*x*log(2)^2*log(x)^2+(40*x*log(2)*exp (x+2*log(2))-32*x*log(2)^2*exp(x)-8*x^2*log(2))*log(x)+25*x*exp(x+2*log(2) )^2+(-40*x*log(2)*exp(x)-10*x^2)*exp(x+2*log(2))+16*x*log(2)^2*exp(x)^2+8* x^2*log(2)*exp(x)+x^3),x, algorithm="fricas")
Output:
(x - 3)/((log(2) - 5)*e^(x + 2*log(2)) - 4*log(2)*log(x) + x)
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x - 3}{x + \left (-20 + 4 \log {\left (2 \right )}\right ) e^{x} - 4 \log {\left (2 \right )} \log {\left (x \right )}} \] Input:
integrate((-4*x*ln(2)*ln(x)+(5*x**2-20*x)*exp(x+2*ln(2))+(-4*x**2+16*x)*ln (2)*exp(x)+(4*x-12)*ln(2)+3*x)/(16*x*ln(2)**2*ln(x)**2+(40*x*ln(2)*exp(x+2 *ln(2))-32*x*ln(2)**2*exp(x)-8*x**2*ln(2))*ln(x)+25*x*exp(x+2*ln(2))**2+(- 40*x*ln(2)*exp(x)-10*x**2)*exp(x+2*ln(2))+16*x*ln(2)**2*exp(x)**2+8*x**2*l n(2)*exp(x)+x**3),x)
Output:
(x - 3)/(x + (-20 + 4*log(2))*exp(x) - 4*log(2)*log(x))
Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x - 3}{4 \, {\left (\log \left (2\right ) - 5\right )} e^{x} - 4 \, \log \left (2\right ) \log \left (x\right ) + x} \] Input:
integrate((-4*x*log(2)*log(x)+(5*x^2-20*x)*exp(x+2*log(2))+(-4*x^2+16*x)*l og(2)*exp(x)+(4*x-12)*log(2)+3*x)/(16*x*log(2)^2*log(x)^2+(40*x*log(2)*exp (x+2*log(2))-32*x*log(2)^2*exp(x)-8*x^2*log(2))*log(x)+25*x*exp(x+2*log(2) )^2+(-40*x*log(2)*exp(x)-10*x^2)*exp(x+2*log(2))+16*x*log(2)^2*exp(x)^2+8* x^2*log(2)*exp(x)+x^3),x, algorithm="maxima")
Output:
(x - 3)/(4*(log(2) - 5)*e^x - 4*log(2)*log(x) + x)
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x - 3}{4 \, e^{x} \log \left (2\right ) - 4 \, \log \left (2\right ) \log \left (x\right ) + x - 20 \, e^{x}} \] Input:
integrate((-4*x*log(2)*log(x)+(5*x^2-20*x)*exp(x+2*log(2))+(-4*x^2+16*x)*l og(2)*exp(x)+(4*x-12)*log(2)+3*x)/(16*x*log(2)^2*log(x)^2+(40*x*log(2)*exp (x+2*log(2))-32*x*log(2)^2*exp(x)-8*x^2*log(2))*log(x)+25*x*exp(x+2*log(2) )^2+(-40*x*log(2)*exp(x)-10*x^2)*exp(x+2*log(2))+16*x*log(2)^2*exp(x)^2+8* x^2*log(2)*exp(x)+x^3),x, algorithm="giac")
Output:
(x - 3)/(4*e^x*log(2) - 4*log(2)*log(x) + x - 20*e^x)
Timed out. \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\int \frac {3\,x+\ln \left (2\right )\,\left (4\,x-12\right )-{\mathrm {e}}^{x+2\,\ln \left (2\right )}\,\left (20\,x-5\,x^2\right )+{\mathrm {e}}^x\,\ln \left (2\right )\,\left (16\,x-4\,x^2\right )-4\,x\,\ln \left (2\right )\,\ln \left (x\right )}{25\,x\,{\mathrm {e}}^{2\,x+4\,\ln \left (2\right )}-{\mathrm {e}}^{x+2\,\ln \left (2\right )}\,\left (10\,x^2+40\,x\,{\mathrm {e}}^x\,\ln \left (2\right )\right )-\ln \left (x\right )\,\left (8\,x^2\,\ln \left (2\right )-40\,x\,{\mathrm {e}}^{x+2\,\ln \left (2\right )}\,\ln \left (2\right )+32\,x\,{\mathrm {e}}^x\,{\ln \left (2\right )}^2\right )+x^3+8\,x^2\,{\mathrm {e}}^x\,\ln \left (2\right )+16\,x\,{\mathrm {e}}^{2\,x}\,{\ln \left (2\right )}^2+16\,x\,{\ln \left (2\right )}^2\,{\ln \left (x\right )}^2} \,d x \] Input:
int((3*x + log(2)*(4*x - 12) - exp(x + 2*log(2))*(20*x - 5*x^2) + exp(x)*l og(2)*(16*x - 4*x^2) - 4*x*log(2)*log(x))/(25*x*exp(2*x + 4*log(2)) - exp( x + 2*log(2))*(10*x^2 + 40*x*exp(x)*log(2)) - log(x)*(8*x^2*log(2) - 40*x* exp(x + 2*log(2))*log(2) + 32*x*exp(x)*log(2)^2) + x^3 + 8*x^2*exp(x)*log( 2) + 16*x*exp(2*x)*log(2)^2 + 16*x*log(2)^2*log(x)^2),x)
Output:
int((3*x + log(2)*(4*x - 12) - exp(x + 2*log(2))*(20*x - 5*x^2) + exp(x)*l og(2)*(16*x - 4*x^2) - 4*x*log(2)*log(x))/(25*x*exp(2*x + 4*log(2)) - exp( x + 2*log(2))*(10*x^2 + 40*x*exp(x)*log(2)) - log(x)*(8*x^2*log(2) - 40*x* exp(x + 2*log(2))*log(2) + 32*x*exp(x)*log(2)^2) + x^3 + 8*x^2*exp(x)*log( 2) + 16*x*exp(2*x)*log(2)^2 + 16*x*log(2)^2*log(x)^2), x)
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx=\frac {x -3}{4 e^{x} \mathrm {log}\left (2\right )-20 e^{x}-4 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )+x} \] Input:
int((-4*x*log(2)*log(x)+(5*x^2-20*x)*exp(x+2*log(2))+(-4*x^2+16*x)*log(2)* exp(x)+(4*x-12)*log(2)+3*x)/(16*x*log(2)^2*log(x)^2+(40*x*log(2)*exp(x+2*l og(2))-32*x*log(2)^2*exp(x)-8*x^2*log(2))*log(x)+25*x*exp(x+2*log(2))^2+(- 40*x*log(2)*exp(x)-10*x^2)*exp(x+2*log(2))+16*x*log(2)^2*exp(x)^2+8*x^2*lo g(2)*exp(x)+x^3),x)
Output:
(x - 3)/(4*e**x*log(2) - 20*e**x - 4*log(x)*log(2) + x)