\(\int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log (x^4)} \, dx\) [8478]

Optimal result

Integrand size = 40, antiderivative size = 17 \[ \int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx=\log \left (3-e^{e^x}-3 x+\log \left (x^4\right )\right ) \]

[Out]

ln(3-3*x-exp(exp(x))+ln(x^4))

Rubi [F]

\[ \int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx=\int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx \]

[In]

Int[(-4 + 3*x + E^(E^x + x)*x)/(-3*x + E^E^x*x + 3*x^2 - x*Log[x^4]),x]

[Out]

3*Defer[Int][(-3 + E^E^x + 3*x - Log[x^4])^(-1), x] + Defer[Int][E^(E^x + x)/(-3 + E^E^x + 3*x - Log[x^4]), x]
- 4*Defer[Int][1/(x*(-3 + E^E^x + 3*x - Log[x^4])), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^{e^x+x}}{-3+e^{e^x}+3 x-\log \left (x^4\right )}+\frac {-4+3 x}{x \left (-3+e^{e^x}+3 x-\log \left (x^4\right )\right )}\right ) \, dx \\ & = \int \frac {e^{e^x+x}}{-3+e^{e^x}+3 x-\log \left (x^4\right )} \, dx+\int \frac {-4+3 x}{x \left (-3+e^{e^x}+3 x-\log \left (x^4\right )\right )} \, dx \\ & = \int \left (\frac {3}{-3+e^{e^x}+3 x-\log \left (x^4\right )}-\frac {4}{x \left (-3+e^{e^x}+3 x-\log \left (x^4\right )\right )}\right ) \, dx+\int \frac {e^{e^x+x}}{-3+e^{e^x}+3 x-\log \left (x^4\right )} \, dx \\ & = 3 \int \frac {1}{-3+e^{e^x}+3 x-\log \left (x^4\right )} \, dx-4 \int \frac {1}{x \left (-3+e^{e^x}+3 x-\log \left (x^4\right )\right )} \, dx+\int \frac {e^{e^x+x}}{-3+e^{e^x}+3 x-\log \left (x^4\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx=\log \left (3-e^{e^x}-3 x+\log \left (x^4\right )\right ) \]

[In]

Integrate[(-4 + 3*x + E^(E^x + x)*x)/(-3*x + E^E^x*x + 3*x^2 - x*Log[x^4]),x]

[Out]

Log[3 - E^E^x - 3*x + Log[x^4]]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{x}}}{3}-\frac {\ln \left (x^{4}\right )}{3}+x -1\right )\)	\(16\)
risch	\(\ln \left ({\mathrm e}^{{\mathrm e}^{x}}+\frac {i \left (\pi \operatorname {csgn}\left (i x^{3}\right )^{3}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i x \right )+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )+\pi \operatorname {csgn}\left (i x^{4}\right )^{3}-\pi \operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x \right )+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-6 i x +8 i \ln \left (x \right )+6 i\right )}{2}\right )\)	\(201\)

[In]

int((x*exp(x)*exp(exp(x))+3*x-4)/(x*exp(exp(x))-x*ln(x^4)+3*x^2-3*x),x,method=_RETURNVERBOSE)

[Out]

ln(1/3*exp(exp(x))-1/3*ln(x^4)+x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx=-x + \log \left (3 \, {\left (x - 1\right )} e^{x} - e^{x} \log \left (x^{4}\right ) + e^{\left (x + e^{x}\right )}\right ) \]

[In]

integrate((x*exp(x)*exp(exp(x))+3*x-4)/(x*exp(exp(x))-x*log(x^4)+3*x^2-3*x),x, algorithm="fricas")

[Out]

-x + log(3*(x - 1)*e^x - e^x*log(x^4) + e^(x + e^x))

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx=\log {\left (3 x + e^{e^{x}} - \log {\left (x^{4} \right )} - 3 \right )} \]

[In]

integrate((x*exp(x)*exp(exp(x))+3*x-4)/(x*exp(exp(x))-x*ln(x**4)+3*x**2-3*x),x)

[Out]

log(3*x + exp(exp(x)) - log(x**4) - 3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx=\log \left (3 \, x + e^{\left (e^{x}\right )} - 4 \, \log \left (x\right ) - 3\right ) \]

[In]

integrate((x*exp(x)*exp(exp(x))+3*x-4)/(x*exp(exp(x))-x*log(x^4)+3*x^2-3*x),x, algorithm="maxima")

[Out]

log(3*x + e^(e^x) - 4*log(x) - 3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.65 \[ \int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx=-x + \log \left (3 \, x e^{x} - e^{x} \log \left (x^{4}\right ) + e^{\left (x + e^{x}\right )} - 3 \, e^{x}\right ) \]

[In]

integrate((x*exp(x)*exp(exp(x))+3*x-4)/(x*exp(exp(x))-x*log(x^4)+3*x^2-3*x),x, algorithm="giac")

[Out]

-x + log(3*x*e^x - e^x*log(x^4) + e^(x + e^x) - 3*e^x)

Mupad [B] (verification not implemented)

Time = 13.75 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-4+3 x+e^{e^x+x} x}{-3 x+e^{e^x} x+3 x^2-x \log \left (x^4\right )} \, dx=\ln \left (3\,x+{\mathrm {e}}^{{\mathrm {e}}^x}-\ln \left (x^4\right )-3\right ) \]

[In]

int(-(3*x + x*exp(exp(x))*exp(x) - 4)/(3*x - x*exp(exp(x)) + x*log(x^4) - 3*x^2),x)

[Out]

log(3*x + exp(exp(x)) - log(x^4) - 3)