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

Optimal. Leaf size=33 \[ e^{\frac {8}{i \pi +\log (-\log (\log (2)))}}+\log \left (\frac {e^{e^5}}{x}-x\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 0.52, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1593, 446, 72} \begin {gather*} \log \left (e^{e^5}-x^2\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^E^5 - x^2)/(E^E^5*x - x^3),x]

[Out]

-Log[x] + Log[E^E^5 - x^2]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

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 1593

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{e^5}-x^2}{x \left (e^{e^5}-x^2\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {-e^{e^5}-x}{\left (e^{e^5}-x\right ) x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.52 \begin {gather*} -\log (x)+\log \left (e^{e^5}-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[x] + Log[E^E^5 - x^2]

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fricas [A]  time = 0.95, size = 15, normalized size = 0.45 \begin {gather*} \log \left (x^{2} - e^{\left (e^{5}\right )}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.18, size = 18, normalized size = 0.55 \begin {gather*} -\frac {1}{2} \, \log \left (x^{2}\right ) + \log \left ({\left | x^{2} - e^{\left (e^{5}\right )} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [A]  time = 0.26, size = 16, normalized size = 0.48

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 15, normalized size = 0.45 \begin {gather*} \log \left (x^{2} - e^{\left (e^{5}\right )}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.03, size = 15, normalized size = 0.45 \begin {gather*} \ln \left (x^2-{\mathrm {e}}^{{\mathrm {e}}^5}\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^2 - exp(exp(5))) - log(x)

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sympy [A]  time = 0.17, size = 12, normalized size = 0.36 \begin {gather*} - \log {\relax (x )} + \log {\left (x^{2} - e^{e^{5}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(x**2 - exp(exp(5)))

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