\(\int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx\) [1341]

Optimal result

Integrand size = 29, antiderivative size = 33 \[ \int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx=e^{\frac {8}{i \pi +\log (-\log (\log (2)))}}+\log \left (\frac {e^{e^5}}{x}-x\right ) \]

[Out]

exp(4/ln(ln(ln(2))))^2+ln(1/x*exp(exp(5))-x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.52, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1607, 457, 78} \[ \int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx=\log \left (e^{e^5}-x^2\right )-\log (x) \]

[In]

Int[(-E^E^5 - x^2)/(E^E^5*x - x^3),x]

[Out]

-Log[x] + Log[E^E^5 - x^2]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-e^{e^5}-x^2}{x \left (e^{e^5}-x^2\right )} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {-e^{e^5}-x}{\left (e^{e^5}-x\right ) x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {2}{e^{e^5}-x}-\frac {1}{x}\right ) \, dx,x,x^2\right ) \\ & = -\log (x)+\log \left (e^{e^5}-x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.52 \[ \int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx=-\log (x)+\log \left (e^{e^5}-x^2\right ) \]

[In]

Integrate[(-E^E^5 - x^2)/(E^E^5*x - x^3),x]

[Out]

-Log[x] + Log[E^E^5 - x^2]

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.48

method	result	size
default	\(\ln \left (x^{2}-{\mathrm e}^{{\mathrm e}^{5}}\right )-\ln \left (x \right )\)	\(16\)
norman	\(-\ln \left (x \right )+\ln \left (-x^{2}+{\mathrm e}^{{\mathrm e}^{5}}\right )\)	\(16\)
risch	\(\ln \left (x^{2}-{\mathrm e}^{{\mathrm e}^{5}}\right )-\ln \left (x \right )\)	\(16\)
parallelrisch	\(\ln \left (x^{2}-{\mathrm e}^{{\mathrm e}^{5}}\right )-\ln \left (x \right )\)	\(16\)
meijerg	\(-\ln \left (x \right )+\frac {{\mathrm e}^{5}}{2}-\frac {i \pi }{2}+\ln \left (1-x^{2} {\mathrm e}^{-{\mathrm e}^{5}}\right )\)	\(27\)

[In]

int((-exp(exp(5))-x^2)/(x*exp(exp(5))-x^3),x,method=_RETURNVERBOSE)

[Out]

ln(x^2-exp(exp(5)))-ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.45 \[ \int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx=\log \left (x^{2} - e^{\left (e^{5}\right )}\right ) - \log \left (x\right ) \]

[In]

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

[Out]

log(x^2 - e^(e^5)) - log(x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.36 \[ \int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx=- \log {\left (x \right )} + \log {\left (x^{2} - e^{e^{5}} \right )} \]

[In]

integrate((-exp(exp(5))-x**2)/(x*exp(exp(5))-x**3),x)

[Out]

-log(x) + log(x**2 - exp(exp(5)))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.45 \[ \int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx=\log \left (x^{2} - e^{\left (e^{5}\right )}\right ) - \log \left (x\right ) \]

[In]

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

[Out]

log(x^2 - e^(e^5)) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.55 \[ \int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx=-\frac {1}{2} \, \log \left (x^{2}\right ) + \log \left ({\left | x^{2} - e^{\left (e^{5}\right )} \right |}\right ) \]

[In]

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

[Out]

-1/2*log(x^2) + log(abs(x^2 - e^(e^5)))

Mupad [B] (verification not implemented)

Time = 9.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.45 \[ \int \frac {-e^{e^5}-x^2}{e^{e^5} x-x^3} \, dx=\ln \left (x^2-{\mathrm {e}}^{{\mathrm {e}}^5}\right )-\ln \left (x\right ) \]

[In]

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

[Out]

log(x^2 - exp(exp(5))) - log(x)