\(\int \frac {e^{e^{-2+2 x}} (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4))}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx\) [7644]

Optimal result

Integrand size = 95, antiderivative size = 25 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {\log (4)}{1+e^{20+e^{-2+2 x}+\frac {2 x}{3}} x} \]

[Out]

2*ln(2)/(x*exp(1/6*x+5)^4*exp(exp(-1+x)^2)+1)

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6820, 12, 6818} \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {\log (4)}{e^{\frac {2 x}{3}+e^{2 x-2}+20} x+1} \]

[In]

Int[(E^E^(-2 + 2*x)*(E^((2*(30 + x))/3)*(-3 - 2*x)*Log[4] - 6*E^(-2 + 2*x + (2*(30 + x))/3)*x*Log[4]))/(3 + 6*
E^(E^(-2 + 2*x) + (2*(30 + x))/3)*x + 3*E^(2*E^(-2 + 2*x) + (4*(30 + x))/3)*x^2),x]

[Out]

Log[4]/(1 + E^(20 + E^(-2 + 2*x) + (2*x)/3)*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 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 {e^{18+e^{-2+2 x}+\frac {2 x}{3}} \left (-6 e^{2 x} x-e^2 (3+2 x)\right ) \log (4)}{3 \left (1+e^{20+e^{-2+2 x}+\frac {2 x}{3}} x\right )^2} \, dx \\ & = \frac {1}{3} \log (4) \int \frac {e^{18+e^{-2+2 x}+\frac {2 x}{3}} \left (-6 e^{2 x} x-e^2 (3+2 x)\right )}{\left (1+e^{20+e^{-2+2 x}+\frac {2 x}{3}} x\right )^2} \, dx \\ & = \frac {\log (4)}{1+e^{20+e^{-2+2 x}+\frac {2 x}{3}} x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {\log (4)}{1+e^{20+e^{-2+2 x}+\frac {2 x}{3}} x} \]

[In]

Integrate[(E^E^(-2 + 2*x)*(E^((2*(30 + x))/3)*(-3 - 2*x)*Log[4] - 6*E^(-2 + 2*x + (2*(30 + x))/3)*x*Log[4]))/(
3 + 6*E^(E^(-2 + 2*x) + (2*(30 + x))/3)*x + 3*E^(2*E^(-2 + 2*x) + (4*(30 + x))/3)*x^2),x]

[Out]

Log[4]/(1 + E^(20 + E^(-2 + 2*x) + (2*x)/3)*x)

Maple [A] (verified)

Time = 485.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {2 \ln \left (2\right )}{x \,{\mathrm e}^{\frac {2 x}{3}+20+{\mathrm e}^{-2+2 x}}+1}\)	\(23\)
parallelrisch	\(\frac {2 \ln \left (2\right )}{x \,{\mathrm e}^{\frac {2 x}{3}+20} {\mathrm e}^{{\mathrm e}^{-2+2 x}}+1}\)	\(26\)

[In]

int((-12*x*ln(2)*exp(1/6*x+5)^4*exp(-1+x)^2+2*(-2*x-3)*ln(2)*exp(1/6*x+5)^4)*exp(exp(-1+x)^2)/(3*x^2*exp(1/6*x
+5)^8*exp(exp(-1+x)^2)^2+6*x*exp(1/6*x+5)^4*exp(exp(-1+x)^2)+3),x,method=_RETURNVERBOSE)

[Out]

2*ln(2)/(x*exp(2/3*x+20+exp(-2+2*x))+1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {2 \, \log \left (2\right )}{x e^{\left (\frac {1}{3} \, {\left (2 \, {\left (x + 30\right )} e^{62} + 3 \, e^{\left (2 \, x + 60\right )}\right )} e^{\left (-62\right )}\right )} + 1} \]

[In]

integrate((-12*x*log(2)*exp(1/6*x+5)^4*exp(-1+x)^2+2*(-2*x-3)*log(2)*exp(1/6*x+5)^4)*exp(exp(-1+x)^2)/(3*x^2*e
xp(1/6*x+5)^8*exp(exp(-1+x)^2)^2+6*x*exp(1/6*x+5)^4*exp(exp(-1+x)^2)+3),x, algorithm="fricas")

[Out]

2*log(2)/(x*e^(1/3*(2*(x + 30)*e^62 + 3*e^(2*x + 60))*e^(-62)) + 1)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {2 \log {\left (2 \right )}}{x e^{\frac {e^{2 x + 60}}{e^{62}}} e^{\frac {2 x}{3} + 20} + 1} \]

[In]

integrate((-12*x*ln(2)*exp(1/6*x+5)**4*exp(-1+x)**2+2*(-2*x-3)*ln(2)*exp(1/6*x+5)**4)*exp(exp(-1+x)**2)/(3*x**
2*exp(1/6*x+5)**8*exp(exp(-1+x)**2)**2+6*x*exp(1/6*x+5)**4*exp(exp(-1+x)**2)+3),x)

[Out]

2*log(2)/(x*exp(exp(-62)*exp(2*x + 60))*exp(2*x/3 + 20) + 1)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (22) = 44\).

Time = 0.47 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.20 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {2 \, {\left (2 \, x e^{2} \log \left (2\right ) + 6 \, x e^{\left (2 \, x\right )} \log \left (2\right ) + 3 \, e^{2} \log \left (2\right )\right )}}{2 \, x e^{2} + 6 \, x e^{\left (2 \, x\right )} + {\left (2 \, x^{2} e^{22} + 6 \, x^{2} e^{\left (2 \, x + 20\right )} + 3 \, x e^{22}\right )} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )}\right )} + 3 \, e^{2}} \]

[In]

integrate((-12*x*log(2)*exp(1/6*x+5)^4*exp(-1+x)^2+2*(-2*x-3)*log(2)*exp(1/6*x+5)^4)*exp(exp(-1+x)^2)/(3*x^2*e
xp(1/6*x+5)^8*exp(exp(-1+x)^2)^2+6*x*exp(1/6*x+5)^4*exp(exp(-1+x)^2)+3),x, algorithm="maxima")

[Out]

2*(2*x*e^2*log(2) + 6*x*e^(2*x)*log(2) + 3*e^2*log(2))/(2*x*e^2 + 6*x*e^(2*x) + (2*x^2*e^22 + 6*x^2*e^(2*x + 2
0) + 3*x*e^22)*e^(2/3*x + e^(2*x - 2)) + 3*e^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (22) = 44\).

Time = 0.29 (sec) , antiderivative size = 261, normalized size of antiderivative = 10.44 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=-\frac {2 \, {\left (36 \, x^{3} e^{\left (\frac {14}{3} \, x + e^{\left (2 \, x - 2\right )} + 18\right )} \log \left (2\right ) + 24 \, x^{3} e^{\left (\frac {8}{3} \, x + e^{\left (2 \, x - 2\right )} + 20\right )} \log \left (2\right ) + 4 \, x^{3} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} \log \left (2\right ) + 36 \, x^{2} e^{\left (\frac {8}{3} \, x + e^{\left (2 \, x - 2\right )} + 20\right )} \log \left (2\right ) + 12 \, x^{2} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} \log \left (2\right ) + 9 \, x e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} \log \left (2\right )\right )}}{36 \, x^{3} e^{\left (\frac {14}{3} \, x + e^{\left (2 \, x - 2\right )} + 18\right )} + 24 \, x^{3} e^{\left (\frac {8}{3} \, x + e^{\left (2 \, x - 2\right )} + 20\right )} + 4 \, x^{3} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} + 4 \, x^{2} e^{2} + 24 \, x^{2} e^{\left (2 \, x\right )} + 36 \, x^{2} e^{\left (4 \, x - 2\right )} + 36 \, x^{2} e^{\left (\frac {8}{3} \, x + e^{\left (2 \, x - 2\right )} + 20\right )} + 12 \, x^{2} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} + 12 \, x e^{2} + 36 \, x e^{\left (2 \, x\right )} + 9 \, x e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} + 9 \, e^{2}} \]

[In]

integrate((-12*x*log(2)*exp(1/6*x+5)^4*exp(-1+x)^2+2*(-2*x-3)*log(2)*exp(1/6*x+5)^4)*exp(exp(-1+x)^2)/(3*x^2*e
xp(1/6*x+5)^8*exp(exp(-1+x)^2)^2+6*x*exp(1/6*x+5)^4*exp(exp(-1+x)^2)+3),x, algorithm="giac")

[Out]

-2*(36*x^3*e^(14/3*x + e^(2*x - 2) + 18)*log(2) + 24*x^3*e^(8/3*x + e^(2*x - 2) + 20)*log(2) + 4*x^3*e^(2/3*x
+ e^(2*x - 2) + 22)*log(2) + 36*x^2*e^(8/3*x + e^(2*x - 2) + 20)*log(2) + 12*x^2*e^(2/3*x + e^(2*x - 2) + 22)*
log(2) + 9*x*e^(2/3*x + e^(2*x - 2) + 22)*log(2))/(36*x^3*e^(14/3*x + e^(2*x - 2) + 18) + 24*x^3*e^(8/3*x + e^
(2*x - 2) + 20) + 4*x^3*e^(2/3*x + e^(2*x - 2) + 22) + 4*x^2*e^2 + 24*x^2*e^(2*x) + 36*x^2*e^(4*x - 2) + 36*x^
2*e^(8/3*x + e^(2*x - 2) + 20) + 12*x^2*e^(2/3*x + e^(2*x - 2) + 22) + 12*x*e^2 + 36*x*e^(2*x) + 9*x*e^(2/3*x
+ e^(2*x - 2) + 22) + 9*e^2)

Mupad [B] (verification not implemented)

Time = 13.94 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.04 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {2\,x\,\ln \left (2\right )\,\left (2\,x+6\,x\,{\mathrm {e}}^{2\,x-2}+3\right )}{\left ({\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-2}}+\frac {{\mathrm {e}}^{-\frac {2\,x}{3}-20}}{x}\right )\,\left (3\,x^2\,{\mathrm {e}}^{\frac {2\,x}{3}+20}+2\,x^3\,{\mathrm {e}}^{\frac {2\,x}{3}+20}+6\,x^3\,{\mathrm {e}}^{\frac {8\,x}{3}+18}\right )} \]

[In]

int(-(exp(exp(2*x - 2))*(2*exp((2*x)/3 + 20)*log(2)*(2*x + 3) + 12*x*exp(2*x - 2)*exp((2*x)/3 + 20)*log(2)))/(
3*x^2*exp(2*exp(2*x - 2))*exp((4*x)/3 + 40) + 6*x*exp(exp(2*x - 2))*exp((2*x)/3 + 20) + 3),x)

[Out]

(2*x*log(2)*(2*x + 6*x*exp(2*x - 2) + 3))/((exp(exp(2*x)*exp(-2)) + exp(- (2*x)/3 - 20)/x)*(3*x^2*exp((2*x)/3
+ 20) + 2*x^3*exp((2*x)/3 + 20) + 6*x^3*exp((8*x)/3 + 18)))