Integrand size = 78, antiderivative size = 25 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {-8+x+\log (5)}{4-\frac {e^x}{2}-x}+3 \log (x) \] Output:
(x+ln(5)-8)/(4-x-1/2*exp(x))+3*ln(x)
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=-\frac {2 (-8+x+\log (5))}{-8+e^x+2 x}+3 \log (x) \] Input:
Integrate[(192 + 3*E^(2*x) - 112*x + 12*x^2 + 4*x*Log[5] + E^x*(-48 - 6*x + 2*x^2 + 2*x*Log[5]))/(64*x + E^(2*x)*x - 32*x^2 + 4*x^3 + E^x*(-16*x + 4 *x^2)),x]
Output:
(-2*(-8 + x + Log[5]))/(-8 + E^x + 2*x) + 3*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.
\(\displaystyle \int \frac {12 x^2+e^x \left (2 x^2-6 x+2 x \log (5)-48\right )-112 x+3 e^{2 x}+4 x \log (5)+192}{4 x^3-32 x^2+e^x \left (4 x^2-16 x\right )+e^{2 x} x+64 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {12 x^2+e^x \left (2 x^2-6 x+2 x \log (5)-48\right )+3 e^{2 x}+x (4 \log (5)-112)+192}{4 x^3-32 x^2+e^x \left (4 x^2-16 x\right )+e^{2 x} x+64 x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {12 x^2+e^x \left (2 x^2-6 x+2 x \log (5)-48\right )+3 e^{2 x}+x (4 \log (5)-112)+192}{\left (-2 x-e^x+8\right )^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3}{x}+\frac {2 (x-9+\log (5))}{2 x+e^x-8}-\frac {4 (x-5) (x-8+\log (5))}{\left (2 x+e^x-8\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {x^2}{\left (2 x+e^x-8\right )^2}dx+2 \int \frac {x}{2 x+e^x-8}dx-20 (8-\log (5)) \int \frac {1}{\left (2 x+e^x-8\right )^2}dx+4 (13-\log (5)) \int \frac {x}{\left (2 x+e^x-8\right )^2}dx-2 (9-\log (5)) \int \frac {1}{2 x+e^x-8}dx+3 \log (x)\) |
Input:
Int[(192 + 3*E^(2*x) - 112*x + 12*x^2 + 4*x*Log[5] + E^x*(-48 - 6*x + 2*x^ 2 + 2*x*Log[5]))/(64*x + E^(2*x)*x - 32*x^2 + 4*x^3 + E^x*(-16*x + 4*x^2)) ,x]
Output:
$Aborted
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(3 \ln \left (x \right )-\frac {2 \left (x +\ln \left (5\right )-8\right )}{-8+{\mathrm e}^{x}+2 x}\) | \(22\) |
norman | \(\frac {{\mathrm e}^{x}+8-2 \ln \left (5\right )}{-8+{\mathrm e}^{x}+2 x}+3 \ln \left (x \right )\) | \(24\) |
parallelrisch | \(-\frac {-6 x \ln \left (x \right )-3 \,{\mathrm e}^{x} \ln \left (x \right )-16+2 \ln \left (5\right )+24 \ln \left (x \right )+2 x}{-8+{\mathrm e}^{x}+2 x}\) | \(36\) |
Input:
int((3*exp(x)^2+(2*x*ln(5)+2*x^2-6*x-48)*exp(x)+4*x*ln(5)+12*x^2-112*x+192 )/(x*exp(x)^2+(4*x^2-16*x)*exp(x)+4*x^3-32*x^2+64*x),x,method=_RETURNVERBO SE)
Output:
3*ln(x)-2*(x+ln(5)-8)/(-8+exp(x)+2*x)
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {3 \, {\left (2 \, x + e^{x} - 8\right )} \log \left (x\right ) - 2 \, x - 2 \, \log \left (5\right ) + 16}{2 \, x + e^{x} - 8} \] Input:
integrate((3*exp(x)^2+(2*x*log(5)+2*x^2-6*x-48)*exp(x)+4*x*log(5)+12*x^2-1 12*x+192)/(x*exp(x)^2+(4*x^2-16*x)*exp(x)+4*x^3-32*x^2+64*x),x, algorithm= "fricas")
Output:
(3*(2*x + e^x - 8)*log(x) - 2*x - 2*log(5) + 16)/(2*x + e^x - 8)
Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {- 2 x - 2 \log {\left (5 \right )} + 16}{2 x + e^{x} - 8} + 3 \log {\left (x \right )} \] Input:
integrate((3*exp(x)**2+(2*x*ln(5)+2*x**2-6*x-48)*exp(x)+4*x*ln(5)+12*x**2- 112*x+192)/(x*exp(x)**2+(4*x**2-16*x)*exp(x)+4*x**3-32*x**2+64*x),x)
Output:
(-2*x - 2*log(5) + 16)/(2*x + exp(x) - 8) + 3*log(x)
Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=-\frac {2 \, {\left (x + \log \left (5\right ) - 8\right )}}{2 \, x + e^{x} - 8} + 3 \, \log \left (x\right ) \] Input:
integrate((3*exp(x)^2+(2*x*log(5)+2*x^2-6*x-48)*exp(x)+4*x*log(5)+12*x^2-1 12*x+192)/(x*exp(x)^2+(4*x^2-16*x)*exp(x)+4*x^3-32*x^2+64*x),x, algorithm= "maxima")
Output:
-2*(x + log(5) - 8)/(2*x + e^x - 8) + 3*log(x)
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {6 \, x \log \left (x\right ) + 3 \, e^{x} \log \left (x\right ) - 2 \, x - 2 \, \log \left (5\right ) - 24 \, \log \left (x\right ) + 16}{2 \, x + e^{x} - 8} \] Input:
integrate((3*exp(x)^2+(2*x*log(5)+2*x^2-6*x-48)*exp(x)+4*x*log(5)+12*x^2-1 12*x+192)/(x*exp(x)^2+(4*x^2-16*x)*exp(x)+4*x^3-32*x^2+64*x),x, algorithm= "giac")
Output:
(6*x*log(x) + 3*e^x*log(x) - 2*x - 2*log(5) - 24*log(x) + 16)/(2*x + e^x - 8)
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=3\,\ln \left (x\right )-\frac {x\,\left (\frac {\ln \left (5\right )}{2}-2\right )+{\mathrm {e}}^x\,\left (\frac {\ln \left (5\right )}{4}-2\right )}{2\,x+{\mathrm {e}}^x-8} \] Input:
int((3*exp(2*x) - 112*x + 4*x*log(5) + 12*x^2 - exp(x)*(6*x - 2*x*log(5) - 2*x^2 + 48) + 192)/(64*x + x*exp(2*x) - exp(x)*(16*x - 4*x^2) - 32*x^2 + 4*x^3),x)
Output:
3*log(x) - (x*(log(5)/2 - 2) + exp(x)*(log(5)/4 - 2))/(2*x + exp(x) - 8)
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {3 e^{x} \mathrm {log}\left (x \right )+e^{x}+6 \,\mathrm {log}\left (x \right ) x -24 \,\mathrm {log}\left (x \right )-2 \,\mathrm {log}\left (5\right )+8}{e^{x}+2 x -8} \] Input:
int((3*exp(x)^2+(2*x*log(5)+2*x^2-6*x-48)*exp(x)+4*x*log(5)+12*x^2-112*x+1 92)/(x*exp(x)^2+(4*x^2-16*x)*exp(x)+4*x^3-32*x^2+64*x),x)
Output:
(3*e**x*log(x) + e**x + 6*log(x)*x - 24*log(x) - 2*log(5) + 8)/(e**x + 2*x - 8)