\(\int \frac {e^x (x^4-x^5)+e^{\frac {e^{4 x}}{x^4}} (e^{5 x} (-4+4 x)+e^{4 x} (12 x-12 x^2))}{e^x x^5-3 x^6} \, dx\) [6300]

Optimal result

Integrand size = 69, antiderivative size = 26 \[ \int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx=3+e^{\frac {e^{4 x}}{x^4}}-\log \left (3-\frac {e^x}{x}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx=\int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx \]

[In]

Int[(E^x*(x^4 - x^5) + E^(E^(4*x)/x^4)*(E^(5*x)*(-4 + 4*x) + E^(4*x)*(12*x - 12*x^2)))/(E^x*x^5 - 3*x^6),x]

[Out]

-Log[3 - E^x/x] - 4*Defer[Int][E^(E^(4*x)/x^4 + 4*x)/x^5, x] + 4*Defer[Int][E^(E^(4*x)/x^4 + 4*x)/x^4, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x (1-x) \left (-4 e^{\frac {e^{4 x}}{x^4}+4 x}+12 e^{\frac {e^{4 x}}{x^4}+3 x} x+x^4\right )}{e^x x^5-3 x^6} \, dx \\ & = \int \left (\frac {4 e^{\frac {e^{4 x}}{x^4}+4 x} (-1+x)}{x^5}-\frac {e^x (-1+x)}{\left (e^x-3 x\right ) x}\right ) \, dx \\ & = 4 \int \frac {e^{\frac {e^{4 x}}{x^4}+4 x} (-1+x)}{x^5} \, dx-\int \frac {e^x (-1+x)}{\left (e^x-3 x\right ) x} \, dx \\ & = 4 \int \left (-\frac {e^{\frac {e^{4 x}}{x^4}+4 x}}{x^5}+\frac {e^{\frac {e^{4 x}}{x^4}+4 x}}{x^4}\right ) \, dx-\text {Subst}\left (\int \frac {1}{-3+x} \, dx,x,\frac {e^x}{x}\right ) \\ & = -\log \left (3-\frac {e^x}{x}\right )-4 \int \frac {e^{\frac {e^{4 x}}{x^4}+4 x}}{x^5} \, dx+4 \int \frac {e^{\frac {e^{4 x}}{x^4}+4 x}}{x^4} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx=e^{\frac {e^{4 x}}{x^4}}-\log \left (e^x-3 x\right )+\log (x) \]

[In]

Integrate[(E^x*(x^4 - x^5) + E^(E^(4*x)/x^4)*(E^(5*x)*(-4 + 4*x) + E^(4*x)*(12*x - 12*x^2)))/(E^x*x^5 - 3*x^6)
,x]

[Out]

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

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85

method	result	size
risch	\(\ln \left (x \right )-\ln \left ({\mathrm e}^{x}-3 x \right )+{\mathrm e}^{\frac {{\mathrm e}^{4 x}}{x^{4}}}\)	\(22\)
parallelrisch	\({\mathrm e}^{\frac {{\mathrm e}^{4 x}}{x^{4}}}+\ln \left (x \right )-\ln \left (-\frac {{\mathrm e}^{x}}{3}+x \right )\)	\(22\)

[In]

int((((-4+4*x)*exp(x)^5+(-12*x^2+12*x)*exp(x)^4)*exp(exp(x)^4/x^4)+(-x^5+x^4)*exp(x))/(x^5*exp(x)-3*x^6),x,met
hod=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx=e^{\left (\frac {e^{\left (4 \, x\right )}}{x^{4}}\right )} + \log \left (x\right ) - \log \left (-3 \, x + e^{x}\right ) \]

[In]

integrate((((-4+4*x)*exp(x)^5+(-12*x^2+12*x)*exp(x)^4)*exp(exp(x)^4/x^4)+(-x^5+x^4)*exp(x))/(x^5*exp(x)-3*x^6)
,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx=e^{\frac {e^{4 x}}{x^{4}}} + \log {\left (x \right )} - \log {\left (- 3 x + e^{x} \right )} \]

[In]

integrate((((-4+4*x)*exp(x)**5+(-12*x**2+12*x)*exp(x)**4)*exp(exp(x)**4/x**4)+(-x**5+x**4)*exp(x))/(x**5*exp(x
)-3*x**6),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx=e^{\left (\frac {e^{\left (4 \, x\right )}}{x^{4}}\right )} + \log \left (x\right ) - \log \left (-3 \, x + e^{x}\right ) \]

[In]

integrate((((-4+4*x)*exp(x)^5+(-12*x^2+12*x)*exp(x)^4)*exp(exp(x)^4/x^4)+(-x^5+x^4)*exp(x))/(x^5*exp(x)-3*x^6)
,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx=-{\left (e^{\left (4 \, x\right )} \log \left (3 \, x - e^{x}\right ) - e^{\left (4 \, x\right )} \log \left (x\right ) - e^{\left (\frac {4 \, x^{5} + e^{\left (4 \, x\right )}}{x^{4}}\right )}\right )} e^{\left (-4 \, x\right )} \]

[In]

integrate((((-4+4*x)*exp(x)^5+(-12*x^2+12*x)*exp(x)^4)*exp(exp(x)^4/x^4)+(-x^5+x^4)*exp(x))/(x^5*exp(x)-3*x^6)
,x, algorithm="giac")

[Out]

-(e^(4*x)*log(3*x - e^x) - e^(4*x)*log(x) - e^((4*x^5 + e^(4*x))/x^4))*e^(-4*x)

Mupad [B] (verification not implemented)

Time = 11.88 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^x \left (x^4-x^5\right )+e^{\frac {e^{4 x}}{x^4}} \left (e^{5 x} (-4+4 x)+e^{4 x} \left (12 x-12 x^2\right )\right )}{e^x x^5-3 x^6} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,x}}{x^4}}-\ln \left ({\mathrm {e}}^x-3\,x\right )+\ln \left (x\right ) \]

[In]

int((exp(exp(4*x)/x^4)*(exp(4*x)*(12*x - 12*x^2) + exp(5*x)*(4*x - 4)) + exp(x)*(x^4 - x^5))/(x^5*exp(x) - 3*x
^6),x)

[Out]

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