\(\int \frac {e^x (6+6 x+3 x^2)}{32+32 x+8 x^2+e^x (6 x+3 x^2)} \, dx\) [7968]

Optimal result

Integrand size = 39, antiderivative size = 16 \[ \int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx=\log \left (4+\frac {3 e^x x}{2 (2+x)}\right ) \]

[Out]

ln(3/2*x*exp(x)/(2+x)+4)

Rubi [F]

\[ \int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx=\int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx \]

[In]

Int[(E^x*(6 + 6*x + 3*x^2))/(32 + 32*x + 8*x^2 + E^x*(6*x + 3*x^2)),x]

[Out]

3*Defer[Int][(E^x*x)/(16 + 8*x + 3*E^x*x), x] + 6*Defer[Int][E^x/((2 + x)*(16 + 8*x + 3*E^x*x)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (6+6 x+3 x^2\right )}{(2+x) \left (16+8 x+3 e^x x\right )} \, dx \\ & = \int \left (\frac {3 e^x x}{16+8 x+3 e^x x}+\frac {6 e^x}{(2+x) \left (16+8 x+3 e^x x\right )}\right ) \, dx \\ & = 3 \int \frac {e^x x}{16+8 x+3 e^x x} \, dx+6 \int \frac {e^x}{(2+x) \left (16+8 x+3 e^x x\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx=3 \left (-\frac {1}{3} \log (2+x)+\frac {1}{3} \log \left (16+8 x+3 e^x x\right )\right ) \]

[In]

Integrate[(E^x*(6 + 6*x + 3*x^2))/(32 + 32*x + 8*x^2 + E^x*(6*x + 3*x^2)),x]

[Out]

3*(-1/3*Log[2 + x] + Log[16 + 8*x + 3*E^x*x]/3)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(-\ln \left (2+x \right )+\ln \left ({\mathrm e}^{x} x +\frac {8 x}{3}+\frac {16}{3}\right )\)	\(18\)
norman	\(-\ln \left (2+x \right )+\ln \left (3 \,{\mathrm e}^{x} x +8 x +16\right )\)	\(19\)
risch	\(-\ln \left (2+x \right )+\ln \left (x \right )+\ln \left ({\mathrm e}^{x}+\frac {\frac {8 x}{3}+\frac {16}{3}}{x}\right )\)	\(22\)

[In]

int((3*x^2+6*x+6)*exp(x)/((3*x^2+6*x)*exp(x)+8*x^2+32*x+32),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx=-\log \left (x + 2\right ) + \log \left (x\right ) + \log \left (\frac {3 \, x e^{x} + 8 \, x + 16}{x}\right ) \]

[In]

integrate((3*x^2+6*x+6)*exp(x)/((3*x^2+6*x)*exp(x)+8*x^2+32*x+32),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx=\log {\left (x \right )} - \log {\left (x + 2 \right )} + \log {\left (e^{x} + \frac {8 x + 16}{3 x} \right )} \]

[In]

integrate((3*x**2+6*x+6)*exp(x)/((3*x**2+6*x)*exp(x)+8*x**2+32*x+32),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx=-\log \left (x + 2\right ) + \log \left (x\right ) + \log \left (\frac {3 \, x e^{x} + 8 \, x + 16}{3 \, x}\right ) \]

[In]

integrate((3*x^2+6*x+6)*exp(x)/((3*x^2+6*x)*exp(x)+8*x^2+32*x+32),x, algorithm="maxima")

[Out]

-log(x + 2) + log(x) + log(1/3*(3*x*e^x + 8*x + 16)/x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx=\log \left (3 \, x e^{x} + 8 \, x + 16\right ) - \log \left (x + 2\right ) \]

[In]

integrate((3*x^2+6*x+6)*exp(x)/((3*x^2+6*x)*exp(x)+8*x^2+32*x+32),x, algorithm="giac")

[Out]

log(3*x*e^x + 8*x + 16) - log(x + 2)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {e^x \left (6+6 x+3 x^2\right )}{32+32 x+8 x^2+e^x \left (6 x+3 x^2\right )} \, dx=\ln \left (8\,x+3\,x\,{\mathrm {e}}^x+16\right )-\ln \left (x+2\right ) \]

[In]

int((exp(x)*(6*x + 3*x^2 + 6))/(32*x + exp(x)*(6*x + 3*x^2) + 8*x^2 + 32),x)

[Out]

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