3.103.48 \(\int \frac {1}{-2 e-2 x+(e+x) \log (5 e+5 x)} \, dx\)

Optimal. Leaf size=16 \[ \log \left (\frac {93}{7}\right )+\log (2-\log (5 (e+x))) \]

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Rubi [A]  time = 0.07, antiderivative size = 11, normalized size of antiderivative = 0.69, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6741, 2390, 2302, 29} \begin {gather*} \log (2-\log (5 (x+e))) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E - 2*x + (E + x)*Log[5*E + 5*x])^(-1),x]

[Out]

Log[2 - Log[5*(E + x)]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 9, normalized size = 0.56 \begin {gather*} \log (-2+\log (5 (e+x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E - 2*x + (E + x)*Log[5*E + 5*x])^(-1),x]

[Out]

Log[-2 + Log[5*(E + x)]]

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fricas [A]  time = 0.95, size = 12, normalized size = 0.75 \begin {gather*} \log \left (\log \left (5 \, x + 5 \, e\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((x+exp(1))*log(5*exp(1)+5*x)-2*exp(1)-2*x),x, algorithm="fricas")

[Out]

log(log(5*x + 5*e) - 2)

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giac [A]  time = 0.14, size = 12, normalized size = 0.75 \begin {gather*} \log \left (\log \left (5 \, x + 5 \, e\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((x+exp(1))*log(5*exp(1)+5*x)-2*exp(1)-2*x),x, algorithm="giac")

[Out]

log(log(5*x + 5*e) - 2)

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maple [A]  time = 0.05, size = 13, normalized size = 0.81

method	result	size
derivativedivides	\(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\)	\(13\)
default	\(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\)	\(13\)
norman	\(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\)	\(13\)
risch	\(\ln \left (\ln \left (5 \,{\mathrm e}+5 x \right )-2\right )\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x+exp(1))*ln(5*exp(1)+5*x)-2*exp(1)-2*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(5*exp(1)+5*x)-2)

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maxima [A]  time = 0.45, size = 10, normalized size = 0.62 \begin {gather*} \log \left (\log \relax (5) + \log \left (x + e\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((x+exp(1))*log(5*exp(1)+5*x)-2*exp(1)-2*x),x, algorithm="maxima")

[Out]

log(log(5) + log(x + e) - 2)

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mupad [B]  time = 0.51, size = 12, normalized size = 0.75 \begin {gather*} \ln \left (\ln \left (5\,x+5\,\mathrm {e}\right )-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(2*x + 2*exp(1) - log(5*x + 5*exp(1))*(x + exp(1))),x)

[Out]

log(log(5*x + 5*exp(1)) - 2)

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sympy [A]  time = 0.13, size = 12, normalized size = 0.75 \begin {gather*} \log {\left (\log {\left (5 x + 5 e \right )} - 2 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((x+exp(1))*ln(5*exp(1)+5*x)-2*exp(1)-2*x),x)

[Out]

log(log(5*x + 5*E) - 2)

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