\(\int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x (-48-6 x+2 x^2+2 x \log (5))}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x (-16 x+4 x^2)} \, dx\) [6446]

Optimal result

Integrand size = 78, antiderivative size = 25 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {-8+x+\log (5)}{4-\frac {e^x}{2}-x}+3 \log (x) \]

[Out]

(x+ln(5)-8)/(4-x-1/2*exp(x))+3*ln(x)

Rubi [F]

\[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx \]

[In]

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]

[Out]

3*Log[x] - 20*(8 - Log[5])*Defer[Int][(-8 + E^x + 2*x)^(-2), x] + 4*(13 - Log[5])*Defer[Int][x/(-8 + E^x + 2*x
)^2, x] - 4*Defer[Int][x^2/(-8 + E^x + 2*x)^2, x] - 2*(9 - Log[5])*Defer[Int][(-8 + E^x + 2*x)^(-1), x] + 2*De
fer[Int][x/(-8 + E^x + 2*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {192+3 e^{2 x}+12 x^2+x (-112+4 \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 \\ & = \int \frac {3 e^{2 x}+4 \left (48+3 x^2+x (-28+\log (5))\right )+2 e^x \left (-24+x^2+x (-3+\log (5))\right )}{\left (8-e^x-2 x\right )^2 x} \, dx \\ & = \int \left (\frac {3}{x}+\frac {2 (-9+x+\log (5))}{-8+e^x+2 x}-\frac {4 (-5+x) (-8+x+\log (5))}{\left (-8+e^x+2 x\right )^2}\right ) \, dx \\ & = 3 \log (x)+2 \int \frac {-9+x+\log (5)}{-8+e^x+2 x} \, dx-4 \int \frac {(-5+x) (-8+x+\log (5))}{\left (-8+e^x+2 x\right )^2} \, dx \\ & = 3 \log (x)+2 \int \left (\frac {x}{-8+e^x+2 x}-\frac {9 \left (1-\frac {\log (5)}{9}\right )}{-8+e^x+2 x}\right ) \, dx-4 \int \left (\frac {x^2}{\left (-8+e^x+2 x\right )^2}+\frac {x (-13+\log (5))}{\left (-8+e^x+2 x\right )^2}-\frac {5 (-8+\log (5))}{\left (-8+e^x+2 x\right )^2}\right ) \, dx \\ & = 3 \log (x)+2 \int \frac {x}{-8+e^x+2 x} \, dx-4 \int \frac {x^2}{\left (-8+e^x+2 x\right )^2} \, dx-(20 (8-\log (5))) \int \frac {1}{\left (-8+e^x+2 x\right )^2} \, dx-(2 (9-\log (5))) \int \frac {1}{-8+e^x+2 x} \, dx-(4 (-13+\log (5))) \int \frac {x}{\left (-8+e^x+2 x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=-\frac {2 (-8+x+\log (5))}{-8+e^x+2 x}+3 \log (x) \]

[In]

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]

[Out]

(-2*(-8 + x + Log[5]))/(-8 + E^x + 2*x) + 3*Log[x]

Maple [A] (verified)

Time = 0.13 (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	\(-\frac {-3 \,{\mathrm e}^{x} \ln \left (x \right )-6 x \ln \left (x \right )+2 \ln \left (5\right )+24 \ln \left (x \right )+2 x -16}{-8+{\mathrm e}^{x}+2 x}\)	\(36\)

[In]

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=_RETURNVERBOSE)

[Out]

3*ln(x)-2*(x+ln(5)-8)/(-8+exp(x)+2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {3 \, {\left (2 \, x + e^{x} - 8\right )} \log \left (x\right ) - 2 \, x - 2 \, \log \left (5\right ) + 16}{2 \, x + e^{x} - 8} \]

[In]

integrate((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+192)/(x*exp(x)^2+(4*x^2-16*x)*e
xp(x)+4*x^3-32*x^2+64*x),x, algorithm="fricas")

[Out]

(3*(2*x + e^x - 8)*log(x) - 2*x - 2*log(5) + 16)/(2*x + e^x - 8)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {- 2 x - 2 \log {\left (5 \right )} + 16}{2 x + e^{x} - 8} + 3 \log {\left (x \right )} \]

[In]

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)

[Out]

(-2*x - 2*log(5) + 16)/(2*x + exp(x) - 8) + 3*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=-\frac {2 \, {\left (x + \log \left (5\right ) - 8\right )}}{2 \, x + e^{x} - 8} + 3 \, \log \left (x\right ) \]

[In]

integrate((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+192)/(x*exp(x)^2+(4*x^2-16*x)*e
xp(x)+4*x^3-32*x^2+64*x),x, algorithm="maxima")

[Out]

-2*(x + log(5) - 8)/(2*x + e^x - 8) + 3*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=\frac {6 \, x \log \left (x\right ) + 3 \, e^{x} \log \left (x\right ) - 2 \, x - 2 \, \log \left (5\right ) - 24 \, \log \left (x\right ) + 16}{2 \, x + e^{x} - 8} \]

[In]

integrate((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+192)/(x*exp(x)^2+(4*x^2-16*x)*e
xp(x)+4*x^3-32*x^2+64*x),x, algorithm="giac")

[Out]

(6*x*log(x) + 3*e^x*log(x) - 2*x - 2*log(5) - 24*log(x) + 16)/(2*x + e^x - 8)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {192+3 e^{2 x}-112 x+12 x^2+4 x \log (5)+e^x \left (-48-6 x+2 x^2+2 x \log (5)\right )}{64 x+e^{2 x} x-32 x^2+4 x^3+e^x \left (-16 x+4 x^2\right )} \, dx=3\,\ln \left (x\right )-\frac {x\,\left (\frac {\ln \left (5\right )}{2}-2\right )+{\mathrm {e}}^x\,\left (\frac {\ln \left (5\right )}{4}-2\right )}{2\,x+{\mathrm {e}}^x-8} \]

[In]

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)

[Out]

3*log(x) - (x*(log(5)/2 - 2) + exp(x)*(log(5)/4 - 2))/(2*x + exp(x) - 8)