\(\int \frac {-3+e^x (3 x+6 x^2+3 x^3)}{(2 e^{e^x (1+x^2)}-x) \log (e^{-e^x (1+x^2)} (-12 e^{e^x (1+x^2)}+6 x))} \, dx\) [2338]

Optimal result

Integrand size = 73, antiderivative size = 22 \[ \int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx=3 \log \left (\log \left (6 \left (-2+e^{-e^x \left (1+x^2\right )} x\right )\right )\right ) \]

[Out]

3*ln(ln(6*x/exp((x^2+1)*exp(x))-12))

Rubi [F]

\[ \int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx=\int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx \]

[In]

Int[(-3 + E^x*(3*x + 6*x^2 + 3*x^3))/((2*E^(E^x*(1 + x^2)) - x)*Log[(-12*E^(E^x*(1 + x^2)) + 6*x)/E^(E^x*(1 +
x^2))]),x]

[Out]

-3*Defer[Int][1/((2*E^(E^x*(1 + x^2)) - x)*Log[-12 + (6*x)/E^(E^x*(1 + x^2))]), x] + 3*Defer[Int][(E^x*x)/((2*
E^(E^x*(1 + x^2)) - x)*Log[-12 + (6*x)/E^(E^x*(1 + x^2))]), x] + 6*Defer[Int][(E^x*x^2)/((2*E^(E^x*(1 + x^2))
- x)*Log[-12 + (6*x)/E^(E^x*(1 + x^2))]), x] + 3*Defer[Int][(E^x*x^3)/((2*E^(E^x*(1 + x^2)) - x)*Log[-12 + (6*
x)/E^(E^x*(1 + x^2))]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {3 \left (-1+e^x x (1+x)^2\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx \\ & = 3 \int \frac {-1+e^x x (1+x)^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx \\ & = 3 \int \left (-\frac {1}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {e^x x}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {2 e^x x^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}+\frac {e^x x^3}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )}\right ) \, dx \\ & = -\left (3 \int \frac {1}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx\right )+3 \int \frac {e^x x}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx+3 \int \frac {e^x x^3}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx+6 \int \frac {e^x x^2}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx=3 \log \left (\log \left (-12+6 e^{-e^x \left (1+x^2\right )} x\right )\right ) \]

[In]

Integrate[(-3 + E^x*(3*x + 6*x^2 + 3*x^3))/((2*E^(E^x*(1 + x^2)) - x)*Log[(-12*E^(E^x*(1 + x^2)) + 6*x)/E^(E^x
*(1 + x^2))]),x]

[Out]

3*Log[Log[-12 + (6*x)/E^(E^x*(1 + x^2))]]

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50

method	result	size
parallelrisch	\(3 \ln \left (\ln \left (-6 \left (2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}-x \right ) {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}}\right )\right )\)	\(33\)
risch	\(3 \ln \left (\ln \left ({\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}\right )+\frac {i \pi \,\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{2}}{2}+\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{-\left (x^{2}+1\right ) {\mathrm e}^{x}} \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )^{3}}{2}-\ln \left (2\right )-\ln \left (3\right )-\ln \left (-2 \,{\mathrm e}^{\left (x^{2}+1\right ) {\mathrm e}^{x}}+x \right )\right )\)	\(231\)

[In]

int(((3*x^3+6*x^2+3*x)*exp(x)-3)/(2*exp((x^2+1)*exp(x))-x)/ln((-12*exp((x^2+1)*exp(x))+6*x)/exp((x^2+1)*exp(x)
)),x,method=_RETURNVERBOSE)

[Out]

3*ln(ln(-6*(2*exp((x^2+1)*exp(x))-x)/exp((x^2+1)*exp(x))))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx=3 \, \log \left (\log \left (6 \, {\left (x - 2 \, e^{\left ({\left (x^{2} + 1\right )} e^{x}\right )}\right )} e^{\left (-{\left (x^{2} + 1\right )} e^{x}\right )}\right )\right ) \]

[In]

integrate(((3*x^3+6*x^2+3*x)*exp(x)-3)/(2*exp((x^2+1)*exp(x))-x)/log((-12*exp((x^2+1)*exp(x))+6*x)/exp((x^2+1)
*exp(x))),x, algorithm="fricas")

[Out]

3*log(log(6*(x - 2*e^((x^2 + 1)*e^x))*e^(-(x^2 + 1)*e^x)))

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx=3 \log {\left (\log {\left (\left (6 x - 12 e^{\left (x^{2} + 1\right ) e^{x}}\right ) e^{- \left (x^{2} + 1\right ) e^{x}} \right )} \right )} \]

[In]

integrate(((3*x**3+6*x**2+3*x)*exp(x)-3)/(2*exp((x**2+1)*exp(x))-x)/ln((-12*exp((x**2+1)*exp(x))+6*x)/exp((x**
2+1)*exp(x))),x)

[Out]

3*log(log((6*x - 12*exp((x**2 + 1)*exp(x)))*exp(-(x**2 + 1)*exp(x))))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx=3 \, \log \left (i \, \pi - {\left (x^{2} + 1\right )} e^{x} + \log \left (3\right ) + \log \left (2\right ) + \log \left (-x + 2 \, e^{\left (x^{2} e^{x} + e^{x}\right )}\right )\right ) \]

[In]

integrate(((3*x^3+6*x^2+3*x)*exp(x)-3)/(2*exp((x^2+1)*exp(x))-x)/log((-12*exp((x^2+1)*exp(x))+6*x)/exp((x^2+1)
*exp(x))),x, algorithm="maxima")

[Out]

3*log(I*pi - (x^2 + 1)*e^x + log(3) + log(2) + log(-x + 2*e^(x^2*e^x + e^x)))

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx=3 \, \log \left (\log \left (6 \, {\left (x - 2 \, e^{\left (x^{2} e^{x} + e^{x}\right )}\right )} e^{\left (-x^{2} e^{x} - e^{x}\right )}\right )\right ) \]

[In]

integrate(((3*x^3+6*x^2+3*x)*exp(x)-3)/(2*exp((x^2+1)*exp(x))-x)/log((-12*exp((x^2+1)*exp(x))+6*x)/exp((x^2+1)
*exp(x))),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-3+e^x \left (3 x+6 x^2+3 x^3\right )}{\left (2 e^{e^x \left (1+x^2\right )}-x\right ) \log \left (e^{-e^x \left (1+x^2\right )} \left (-12 e^{e^x \left (1+x^2\right )}+6 x\right )\right )} \, dx=3\,\ln \left (\ln \left (6\,x\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}-12\right )\right ) \]

[In]

int(-(exp(x)*(3*x + 6*x^2 + 3*x^3) - 3)/(log(exp(-exp(x)*(x^2 + 1))*(6*x - 12*exp(exp(x)*(x^2 + 1))))*(x - 2*e
xp(exp(x)*(x^2 + 1)))),x)

[Out]

3*log(log(6*x*exp(-x^2*exp(x))*exp(-exp(x)) - 12))