3.70.55 \(\int -\frac {2}{(-e^4+5 e^{16}-x) \log (e^4-5 e^{16}+x)} \, dx\)

Optimal. Leaf size=14 \[ \log \left (\log ^2\left (e^4-5 e^{16}+x\right )\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 17, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {12, 2390, 2302, 29} \begin {gather*} 2 \log \left (\log \left (x+e^4 \left (1-5 e^{12}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-2/((-E^4 + 5*E^16 - x)*Log[E^4 - 5*E^16 + x]),x]

[Out]

2*Log[Log[E^4*(1 - 5*E^12) + x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (2 \int \frac {1}{\left (-e^4+5 e^{16}-x\right ) \log \left (e^4-5 e^{16}+x\right )} \, dx\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {e^4-5 e^{16}}{\left (-e^4+5 e^{16}\right ) x \log (x)} \, dx,x,e^4-5 e^{16}+x\right )\right )\\ &=2 \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,e^4-5 e^{16}+x\right )\\ &=2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (e^4-5 e^{16}+x\right )\right )\\ &=2 \log \left (\log \left (e^4 \left (1-5 e^{12}\right )+x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 14, normalized size = 1.00 \begin {gather*} 2 \log \left (\log \left (e^4-5 e^{16}+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-2/((-E^4 + 5*E^16 - x)*Log[E^4 - 5*E^16 + x]),x]

[Out]

2*Log[Log[E^4 - 5*E^16 + x]]

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fricas [A]  time = 0.59, size = 12, normalized size = 0.86 \begin {gather*} 2 \, \log \left (\log \left (x - 5 \, e^{16} + e^{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/(5*exp(16)-exp(4)-x)/log(-5*exp(16)+x+exp(4)),x, algorithm="fricas")

[Out]

2*log(log(x - 5*e^16 + e^4))

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giac [A]  time = 0.12, size = 12, normalized size = 0.86 \begin {gather*} 2 \, \log \left (\log \left (x - 5 \, e^{16} + e^{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/(5*exp(16)-exp(4)-x)/log(-5*exp(16)+x+exp(4)),x, algorithm="giac")

[Out]

2*log(log(x - 5*e^16 + e^4))

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maple [A]  time = 0.11, size = 13, normalized size = 0.93

method	result	size
derivativedivides	\(2 \ln \left (\ln \left (-5 \,{\mathrm e}^{16}+x +{\mathrm e}^{4}\right )\right )\)	\(13\)
default	\(2 \ln \left (\ln \left (-5 \,{\mathrm e}^{16}+x +{\mathrm e}^{4}\right )\right )\)	\(13\)
norman	\(2 \ln \left (\ln \left (-5 \,{\mathrm e}^{16}+x +{\mathrm e}^{4}\right )\right )\)	\(13\)
risch	\(2 \ln \left (\ln \left (-5 \,{\mathrm e}^{16}+x +{\mathrm e}^{4}\right )\right )\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2/(5*exp(16)-exp(4)-x)/ln(-5*exp(16)+x+exp(4)),x,method=_RETURNVERBOSE)

[Out]

2*ln(ln(-5*exp(16)+x+exp(4)))

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maxima [A]  time = 0.41, size = 12, normalized size = 0.86 \begin {gather*} 2 \, \log \left (\log \left (x - 5 \, e^{16} + e^{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/(5*exp(16)-exp(4)-x)/log(-5*exp(16)+x+exp(4)),x, algorithm="maxima")

[Out]

2*log(log(x - 5*e^16 + e^4))

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mupad [B]  time = 4.35, size = 12, normalized size = 0.86 \begin {gather*} 2\,\ln \left (\ln \left (x+{\mathrm {e}}^4-5\,{\mathrm {e}}^{16}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/(log(x + exp(4) - 5*exp(16))*(x + exp(4) - 5*exp(16))),x)

[Out]

2*log(log(x + exp(4) - 5*exp(16)))

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sympy [A]  time = 0.11, size = 14, normalized size = 1.00 \begin {gather*} 2 \log {\left (\log {\left (x - 5 e^{16} + e^{4} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/(5*exp(16)-exp(4)-x)/ln(-5*exp(16)+x+exp(4)),x)

[Out]

2*log(log(x - 5*exp(16) + exp(4)))

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