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

Optimal. Leaf size=17 \[ \left (x-\log \left (\log ^2\left (-e^{2+x}\right )\right )\right )^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6688, 12, 6686} \begin {gather*} \left (x-\log \left (\log ^2\left (-e^{x+2}\right )\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x - Log[Log[-E^(2 + x)]^2])^2

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 0.03, size = 74, normalized size = 4.35 \begin {gather*} -4 x+x^2+4 \log \left (-e^{2+x}\right )+4 \left (-x+\log \left (-e^{2+x}\right )\right ) \log \left (\log \left (-e^{2+x}\right )\right )-2 \log \left (-e^{2+x}\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )+\log ^2\left (\log ^2\left (-e^{2+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*x + x^2 + 4*Log[-E^(2 + x)] + 4*(-x + Log[-E^(2 + x)])*Log[Log[-E^(2 + x)]] - 2*Log[-E^(2 + x)]*Log[Log[-E^
(2 + x)]^2] + Log[Log[-E^(2 + x)]^2]^2

________________________________________________________________________________________

fricas [C]  time = 0.81, size = 49, normalized size = 2.88 \begin {gather*} x^{2} - 2 \, x \log \left (-\pi ^{2} + 2 i \, \pi {\left (x + 2\right )} + x^{2} + 4 \, x + 4\right ) + \log \left (-\pi ^{2} + 2 i \, \pi {\left (x + 2\right )} + x^{2} + 4 \, x + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(-exp(2+x))+4)*log(log(-exp(2+x))^2)+2*x*log(-exp(2+x))-4*x)/log(-exp(2+x)),x, algorithm="fr
icas")

[Out]

x^2 - 2*x*log(-pi^2 + 2*I*pi*(x + 2) + x^2 + 4*x + 4) + log(-pi^2 + 2*I*pi*(x + 2) + x^2 + 4*x + 4)^2

________________________________________________________________________________________

giac [C]  time = 0.21, size = 51, normalized size = 3.00 \begin {gather*} x^{2} - 2 \, x \log \left (4 i \, \pi - \pi ^{2} + 2 i \, \pi x + x^{2} + 4 \, x + 4\right ) + \log \left (4 i \, \pi - \pi ^{2} + 2 i \, \pi x + x^{2} + 4 \, x + 4\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(-exp(2+x))+4)*log(log(-exp(2+x))^2)+2*x*log(-exp(2+x))-4*x)/log(-exp(2+x)),x, algorithm="gi
ac")

[Out]

x^2 - 2*x*log(4*I*pi - pi^2 + 2*I*pi*x + x^2 + 4*x + 4) + log(4*I*pi - pi^2 + 2*I*pi*x + x^2 + 4*x + 4)^2

________________________________________________________________________________________

maple [B]  time = 0.37, size = 100, normalized size = 5.88

method	result	size
default	\(-2 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )^{2}\right ) \ln \left (-{\mathrm e}^{2+x}\right )+4 \ln \left (-{\mathrm e}^{2+x}\right )+4 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right ) \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )^{2}\right )-4 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right )^{2}+x^{2}-4 x +4 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right ) \left (\ln \left (-{\mathrm e}^{2+x}\right )-2-x \right )+8 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right )\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*ln(ln(-exp(2+x))^2)*ln(-exp(2+x))+4*ln(-exp(2+x))+4*ln(ln(-exp(2+x)))*ln(ln(-exp(2+x))^2)-4*ln(ln(-exp(2+x)
))^2+x^2-4*x+4*ln(ln(-exp(2+x)))*(ln(-exp(2+x))-2-x)+8*ln(ln(-exp(2+x)))

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 27, normalized size = 1.59 \begin {gather*} x^{2} - 4 \, x \log \left (\log \left (-e^{x}\right ) + 2\right ) + 4 \, \log \left (\log \left (-e^{x}\right ) + 2\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(-exp(2+x))+4)*log(log(-exp(2+x))^2)+2*x*log(-exp(2+x))-4*x)/log(-exp(2+x)),x, algorithm="ma
xima")

[Out]

x^2 - 4*x*log(log(-e^x) + 2) + 4*log(log(-e^x) + 2)^2

________________________________________________________________________________________

mupad [B]  time = 5.83, size = 16, normalized size = 0.94 \begin {gather*} {\left (x-\ln \left ({\left (x+2+\pi \,1{}\mathrm {i}\right )}^2\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - log((x + pi*1i + 2)^2))^2

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________