\(\int -\frac {64 e^{4-16 \log ^2(\frac {2+e^4 (-4-2 x)}{3 e^4})} \log (\frac {2+e^4 (-4-2 x)}{3 e^4})}{-1+e^4 (2+x)} \, dx\) [9913]

Optimal result

Integrand size = 59, antiderivative size = 26 \[ \int -\frac {64 e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx=2 e^{-16 \log ^2\left (-2-x+\frac {1}{3} \left (2+\frac {2}{e^4}+x\right )\right )} \]

[Out]

2/exp(16*ln(2/3/exp(4)-4/3-2/3*x)^2)

Rubi [A] (verified)

Time = 19.72 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 2494, 1, 2308, 2236, 6838} \[ \int -\frac {64 e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx=2 e^{-16 \log ^2\left (\frac {2 \left (1-e^4 (x+2)\right )}{3 e^4}\right )} \]

[In]

Int[(-64*E^(4 - 16*Log[(2 + E^4*(-4 - 2*x))/(3*E^4)]^2)*Log[(2 + E^4*(-4 - 2*x))/(3*E^4)])/(-1 + E^4*(2 + x)),
x]

[Out]

2/E^(16*Log[(2*(1 - E^4*(2 + x)))/(3*E^4)]^2)

Rule 1

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n)^p, x] /; FreeQ[{a, b, n, p}, x] && EqQ[a
, 0]

Rule 12

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

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 2308

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(g + h*x)^(m + 1)/(h*n*(c*(d + e*x)^n)^((m + 1)/n)), Subst[Int[E^(a*f*Log[F] + ((m + 1)*x)/n + b*f*Lo
g[F]*x^2), x], x, Log[c*(d + e*x)^n]], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*g - d*h, 0]

Rule 2494

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = -\left (64 \int \frac {e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx\right ) \\ & = -\left (64 \int \frac {e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2 \left (1-2 e^4\right )-2 e^4 x}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx\right ) \\ & = 2 e^{-16 \log ^2\left (\frac {2 \left (1-e^4 (2+x)\right )}{3 e^4}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 8.94 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int -\frac {64 e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx=2 e^{-16 \log ^2\left (\frac {2}{3} \left (-2+\frac {1}{e^4}-x\right )\right )} \]

[In]

Integrate[(-64*E^(4 - 16*Log[(2 + E^4*(-4 - 2*x))/(3*E^4)]^2)*Log[(2 + E^4*(-4 - 2*x))/(3*E^4)])/(-1 + E^4*(2
+ x)),x]

[Out]

2/E^(16*Log[(2*(-2 + E^(-4) - x))/3]^2)

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
risch	\(2 \,{\mathrm e}^{-16 {\ln \left (\frac {\left (\left (-2 x -4\right ) {\mathrm e}^{4}+2\right ) {\mathrm e}^{-4}}{3}\right )}^{2}}\)	\(23\)
norman	\(2 \,{\mathrm e}^{-16 {\ln \left (\frac {\left (\left (-2 x -4\right ) {\mathrm e}^{4}+2\right ) {\mathrm e}^{-4}}{3}\right )}^{2}}\)	\(27\)
parallelrisch	\(2 \,{\mathrm e}^{-16 {\ln \left (\frac {\left (\left (-2 x -4\right ) {\mathrm e}^{4}+2\right ) {\mathrm e}^{-4}}{3}\right )}^{2}}\)	\(27\)

[In]

int(-64*exp(4)*ln(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*ln(1/3*((-2*x-4)*exp(4)+2)/exp(4))^2
),x,method=_RETURNVERBOSE)

[Out]

2*exp(-16*ln(1/3*((-2*x-4)*exp(4)+2)*exp(-4))^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int -\frac {64 e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx=2 \, e^{\left (-16 \, \log \left (-\frac {2}{3} \, {\left ({\left (x + 2\right )} e^{4} - 1\right )} e^{\left (-4\right )}\right )^{2}\right )} \]

[In]

integrate(-64*exp(4)*log(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*log(1/3*((-2*x-4)*exp(4)+2)/e
xp(4))^2),x, algorithm="fricas")

[Out]

2*e^(-16*log(-2/3*((x + 2)*e^4 - 1)*e^(-4))^2)

Sympy [A] (verification not implemented)

Time = 34.72 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int -\frac {64 e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx=2 e^{- 16 \log {\left (\frac {\frac {\left (- 2 x - 4\right ) e^{4}}{3} + \frac {2}{3}}{e^{4}} \right )}^{2}} \]

[In]

integrate(-64*exp(4)*ln(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*ln(1/3*((-2*x-4)*exp(4)+2)/exp
(4))**2),x)

[Out]

2*exp(-16*log(((-2*x - 4)*exp(4)/3 + 2/3)*exp(-4))**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int -\frac {64 e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx=2 \, e^{\left (-16 \, \log \left (-\frac {2}{3} \, {\left ({\left (x + 2\right )} e^{4} - 1\right )} e^{\left (-4\right )}\right )^{2}\right )} \]

[In]

integrate(-64*exp(4)*log(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*log(1/3*((-2*x-4)*exp(4)+2)/e
xp(4))^2),x, algorithm="maxima")

[Out]

2*e^(-16*log(-2/3*((x + 2)*e^4 - 1)*e^(-4))^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (17) = 34\).

Time = 0.97 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.35 \[ \int -\frac {64 e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx=2 \, e^{\left (-16 \, \log \left (3\right )^{2} + 32 \, \log \left (3\right ) \log \left (-2 \, x e^{4} - 4 \, e^{4} + 2\right ) - 16 \, \log \left (-2 \, x e^{4} - 4 \, e^{4} + 2\right )^{2} - 128 \, \log \left (3\right ) + 128 \, \log \left (-2 \, x e^{4} - 4 \, e^{4} + 2\right ) - 256\right )} \]

[In]

integrate(-64*exp(4)*log(1/3*((-2*x-4)*exp(4)+2)/exp(4))/((2+x)*exp(4)-1)/exp(16*log(1/3*((-2*x-4)*exp(4)+2)/e
xp(4))^2),x, algorithm="giac")

[Out]

2*e^(-16*log(3)^2 + 32*log(3)*log(-2*x*e^4 - 4*e^4 + 2) - 16*log(-2*x*e^4 - 4*e^4 + 2)^2 - 128*log(3) + 128*lo
g(-2*x*e^4 - 4*e^4 + 2) - 256)

Mupad [B] (verification not implemented)

Time = 14.43 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int -\frac {64 e^{4-16 \log ^2\left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )} \log \left (\frac {2+e^4 (-4-2 x)}{3 e^4}\right )}{-1+e^4 (2+x)} \, dx=2\,{\mathrm {e}}^{-16\,{\ln \left (\frac {2\,{\mathrm {e}}^{-4}}{3}-\frac {2\,x}{3}-\frac {4}{3}\right )}^2} \]

[In]

int(-(64*exp(4)*exp(-16*log(-exp(-4)*((exp(4)*(2*x + 4))/3 - 2/3))^2)*log(-exp(-4)*((exp(4)*(2*x + 4))/3 - 2/3
)))/(exp(4)*(x + 2) - 1),x)

[Out]

2*exp(-16*log((2*exp(-4))/3 - (2*x)/3 - 4/3)^2)