\(\int \frac {8+16 x+e^3 (6 x-3 x^2-6 x^3)}{-8+3 e^3 x^2} \, dx\) [4049]

Optimal result

Integrand size = 36, antiderivative size = 22 \[ \int \frac {8+16 x+e^3 \left (6 x-3 x^2-6 x^3\right )}{-8+3 e^3 x^2} \, dx=-x-x^2+\log \left (\frac {8}{3}-e^3 x^2\right ) \]

[Out]

ln(8/3-x^2*exp(3))-x^2-x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1824, 266} \[ \int \frac {8+16 x+e^3 \left (6 x-3 x^2-6 x^3\right )}{-8+3 e^3 x^2} \, dx=-x^2+\log \left (8-3 e^3 x^2\right )-x \]

[In]

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

[Out]

-x - x^2 + Log[8 - 3*E^3*x^2]

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1824

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \left (-1-2 x+\frac {6 e^3 x}{-8+3 e^3 x^2}\right ) \, dx \\ & = -x-x^2+\left (6 e^3\right ) \int \frac {x}{-8+3 e^3 x^2} \, dx \\ & = -x-x^2+\log \left (8-3 e^3 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {8+16 x+e^3 \left (6 x-3 x^2-6 x^3\right )}{-8+3 e^3 x^2} \, dx=-x-x^2+\log \left (8-3 e^3 x^2\right ) \]

[In]

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

[Out]

-x - x^2 + Log[8 - 3*E^3*x^2]

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
default	\(-x -x^{2}+\ln \left (3 x^{2} {\mathrm e}^{3}-8\right )\)	\(20\)
norman	\(-x -x^{2}+\ln \left (3 x^{2} {\mathrm e}^{3}-8\right )\)	\(20\)
risch	\(-x -x^{2}+\ln \left (3 x^{2} {\mathrm e}^{3}-8\right )\)	\(20\)
parallelrisch	\(-x^{2}-x +\ln \left (\frac {\left (3 x^{2} {\mathrm e}^{3}-8\right ) {\mathrm e}^{-3}}{3}\right )\)	\(26\)
meijerg	\(-\frac {2 \sqrt {3}\, \sqrt {2}\, {\mathrm e}^{-\frac {3}{2}} \operatorname {arctanh}\left (\frac {x \sqrt {3}\, \sqrt {2}\, {\mathrm e}^{\frac {3}{2}}}{4}\right )}{3}+\frac {8 \,{\mathrm e}^{-3} \left (-\frac {3 x^{2} {\mathrm e}^{3}}{8}-\ln \left (1-\frac {3 x^{2} {\mathrm e}^{3}}{8}\right )\right )}{3}+\frac {i \sqrt {6}\, {\mathrm e}^{-\frac {3}{2}} \left (\frac {i \sqrt {6}\, x \,{\mathrm e}^{\frac {3}{2}}}{2}-2 i \operatorname {arctanh}\left (\frac {x \sqrt {3}\, \sqrt {2}\, {\mathrm e}^{\frac {3}{2}}}{4}\right )\right )}{3}-\frac {4 \,{\mathrm e}^{-3} \left (-\frac {3 \,{\mathrm e}^{3}}{4}-2\right ) \ln \left (1-\frac {3 x^{2} {\mathrm e}^{3}}{8}\right )}{3}\)	\(101\)

[In]

int(((-6*x^3-3*x^2+6*x)*exp(3)+16*x+8)/(3*x^2*exp(3)-8),x,method=_RETURNVERBOSE)

[Out]

-x-x^2+ln(3*x^2*exp(3)-8)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {8+16 x+e^3 \left (6 x-3 x^2-6 x^3\right )}{-8+3 e^3 x^2} \, dx=-x^{2} - x + \log \left (3 \, x^{2} e^{3} - 8\right ) \]

[In]

integrate(((-6*x^3-3*x^2+6*x)*exp(3)+16*x+8)/(3*x^2*exp(3)-8),x, algorithm="fricas")

[Out]

-x^2 - x + log(3*x^2*e^3 - 8)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {8+16 x+e^3 \left (6 x-3 x^2-6 x^3\right )}{-8+3 e^3 x^2} \, dx=- x^{2} - x + \log {\left (3 x^{2} e^{3} - 8 \right )} \]

[In]

integrate(((-6*x**3-3*x**2+6*x)*exp(3)+16*x+8)/(3*x**2*exp(3)-8),x)

[Out]

-x**2 - x + log(3*x**2*exp(3) - 8)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {8+16 x+e^3 \left (6 x-3 x^2-6 x^3\right )}{-8+3 e^3 x^2} \, dx=-x^{2} - x + \log \left (3 \, x^{2} e^{3} - 8\right ) \]

[In]

integrate(((-6*x^3-3*x^2+6*x)*exp(3)+16*x+8)/(3*x^2*exp(3)-8),x, algorithm="maxima")

[Out]

-x^2 - x + log(3*x^2*e^3 - 8)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {8+16 x+e^3 \left (6 x-3 x^2-6 x^3\right )}{-8+3 e^3 x^2} \, dx=-{\left (x^{2} e^{6} + x e^{6}\right )} e^{\left (-6\right )} + \log \left ({\left | 3 \, x^{2} e^{3} - 8 \right |}\right ) \]

[In]

integrate(((-6*x^3-3*x^2+6*x)*exp(3)+16*x+8)/(3*x^2*exp(3)-8),x, algorithm="giac")

[Out]

-(x^2*e^6 + x*e^6)*e^(-6) + log(abs(3*x^2*e^3 - 8))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {8+16 x+e^3 \left (6 x-3 x^2-6 x^3\right )}{-8+3 e^3 x^2} \, dx=\ln \left (x^2-\frac {8\,{\mathrm {e}}^{-3}}{3}\right )-x-x^2 \]

[In]

int((16*x - exp(3)*(3*x^2 - 6*x + 6*x^3) + 8)/(3*x^2*exp(3) - 8),x)

[Out]

log(x^2 - (8*exp(-3))/3) - x - x^2