\(\int \frac {-4 x+2 x \log (-e^{2+x})+(4-2 \log (-e^{2+x})) \log (\log ^2(-e^{2+x}))}{\log (-e^{2+x})} \, dx\) [7520]

Optimal result

Integrand size = 50, antiderivative size = 17 \[ \int \frac {-4 x+2 x \log \left (-e^{2+x}\right )+\left (4-2 \log \left (-e^{2+x}\right )\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )}{\log \left (-e^{2+x}\right )} \, dx=\left (x-\log \left (\log ^2\left (-e^{2+x}\right )\right )\right )^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6820, 12, 6818} \[ \int \frac {-4 x+2 x \log \left (-e^{2+x}\right )+\left (4-2 \log \left (-e^{2+x}\right )\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )}{\log \left (-e^{2+x}\right )} \, dx=\left (x-\log \left (\log ^2\left (-e^{x+2}\right )\right )\right )^2 \]

[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 6818

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 6820

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

Rubi steps \begin{align*} \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{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(17)=34\).

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 4.35 \[ \int \frac {-4 x+2 x \log \left (-e^{2+x}\right )+\left (4-2 \log \left (-e^{2+x}\right )\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )}{\log \left (-e^{2+x}\right )} \, dx=-4 x+x^2+4 \log \left (-e^{2+x}\right )+4 \left (-x+\log \left (-e^{2+x}\right )\right ) \log \left (\log \left (-e^{2+x}\right )\right )-2 \log \left (-e^{2+x}\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )+\log ^2\left (\log ^2\left (-e^{2+x}\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.76

method	result	size
parallelrisch	\(x^{2}-2 x \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )^{2}\right )+{\ln \left (\ln \left (-{\mathrm e}^{2+x}\right )^{2}\right )}^{2}\)	\(30\)
default	\(-2 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )^{2}\right ) \ln \left (-{\mathrm e}^{2+x}\right )+4 \ln \left (-{\mathrm e}^{2+x}\right )+4 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right ) \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )^{2}\right )-4 {\ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right )}^{2}+x^{2}-4 x -4 \left (-\ln \left (-{\mathrm e}^{2+x}\right )+x \right ) \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right )\)	\(89\)
parts	\(-2 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )^{2}\right ) \ln \left (-{\mathrm e}^{2+x}\right )+4 \ln \left (-{\mathrm e}^{2+x}\right )+4 \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right ) \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )^{2}\right )-4 {\ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right )}^{2}+x^{2}-4 x -4 \left (-\ln \left (-{\mathrm e}^{2+x}\right )+x \right ) \ln \left (\ln \left (-{\mathrm e}^{2+x}\right )\right )\)	\(89\)

[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]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.88 \[ \int \frac {-4 x+2 x \log \left (-e^{2+x}\right )+\left (4-2 \log \left (-e^{2+x}\right )\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )}{\log \left (-e^{2+x}\right )} \, dx=x^{2} - 2 \, x \log \left (-\pi ^{2} + 2 i \, \pi {\left (x + 2\right )} + x^{2} + 4 \, x + 4\right ) + \log \left (-\pi ^{2} + 2 i \, \pi {\left (x + 2\right )} + x^{2} + 4 \, x + 4\right )^{2} \]

[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

Sympy [F(-1)]

Timed out. \[ \int \frac {-4 x+2 x \log \left (-e^{2+x}\right )+\left (4-2 \log \left (-e^{2+x}\right )\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )}{\log \left (-e^{2+x}\right )} \, dx=\text {Timed out} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int \frac {-4 x+2 x \log \left (-e^{2+x}\right )+\left (4-2 \log \left (-e^{2+x}\right )\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )}{\log \left (-e^{2+x}\right )} \, dx=x^{2} - 4 \, x \log \left (\log \left (-e^{x}\right ) + 2\right ) + 4 \, \log \left (\log \left (-e^{x}\right ) + 2\right )^{2} \]

[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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 3.00 \[ \int \frac {-4 x+2 x \log \left (-e^{2+x}\right )+\left (4-2 \log \left (-e^{2+x}\right )\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )}{\log \left (-e^{2+x}\right )} \, dx=x^{2} - 2 \, x \log \left (4 i \, \pi - \pi ^{2} + 2 i \, \pi x + x^{2} + 4 \, x + 4\right ) + \log \left (4 i \, \pi - \pi ^{2} + 2 i \, \pi x + x^{2} + 4 \, x + 4\right )^{2} \]

[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

Mupad [B] (verification not implemented)

Time = 12.93 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {-4 x+2 x \log \left (-e^{2+x}\right )+\left (4-2 \log \left (-e^{2+x}\right )\right ) \log \left (\log ^2\left (-e^{2+x}\right )\right )}{\log \left (-e^{2+x}\right )} \, dx={\left (x-\ln \left ({\left (x+2+\pi \,1{}\mathrm {i}\right )}^2\right )\right )}^2 \]

[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