Integrand size = 95, antiderivative size = 25 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {\log (4)}{1+e^{20+e^{-2+2 x}+\frac {2 x}{3}} x} \]
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {\log (4)}{1+e^{20+e^{-2+2 x}+\frac {2 x}{3}} x} \]
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]
Time = 0.97 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {7292, 27, 25, 7237}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{e^{2 x-2}} \left (e^{\frac {2 (x+30)}{3}} (-2 x-3) \log (4)-6 e^{2 x+\frac {2 (x+30)}{3}-2} x \log (4)\right )}{3 e^{\frac {4 (x+30)}{3}+2 e^{2 x-2}} x^2+6 e^{\frac {2 (x+30)}{3}+e^{2 x-2}} x+3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {2 x}{3}+e^{2 x-2}+18} \left (-6 e^{2 x} x-2 e^2 x-3 e^2\right ) \log (4)}{3 \left (e^{\frac {2 x}{3}+e^{2 x-2}+20} x+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \log (4) \int -\frac {e^{\frac {2 x}{3}+e^{2 x-2}+18} \left (6 e^{2 x} x+2 e^2 x+3 e^2\right )}{\left (e^{\frac {2 x}{3}+e^{2 x-2}+20} x+1\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \log (4) \int \frac {e^{\frac {2 x}{3}+e^{2 x-2}+18} \left (6 e^{2 x} x+2 e^2 x+3 e^2\right )}{\left (e^{\frac {2 x}{3}+e^{2 x-2}+20} x+1\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {\log (4)}{e^{\frac {2 x}{3}+e^{2 x-2}+20} x+1}\) |
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]
3.17.44.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
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 | \(\frac {2 \ln \left (2\right )}{x \,{\mathrm e}^{\frac {2 x}{3}+20} {\mathrm e}^{{\mathrm e}^{-2+2 x}}+1}\) | \(26\) |
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)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {2 \, \log \left (2\right )}{x e^{\left (\frac {1}{3} \, {\left (2 \, {\left (x + 30\right )} e^{62} + 3 \, e^{\left (2 \, x + 60\right )}\right )} e^{\left (-62\right )}\right )} + 1} \]
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*exp(1/6*x+5)^8*exp(exp(-1+x)^2)^2+6*x*e xp(1/6*x+5)^4*exp(exp(-1+x)^2)+3),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {2 \log {\left (2 \right )}}{x e^{\frac {e^{2 x + 60}}{e^{62}}} e^{\frac {2 x}{3} + 20} + 1} \]
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)
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 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {2 \, {\left (2 \, x e^{2} \log \left (2\right ) + 6 \, x e^{\left (2 \, x\right )} \log \left (2\right ) + 3 \, e^{2} \log \left (2\right )\right )}}{2 \, x e^{2} + 6 \, x e^{\left (2 \, x\right )} + {\left (2 \, x^{2} e^{22} + 6 \, x^{2} e^{\left (2 \, x + 20\right )} + 3 \, x e^{22}\right )} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )}\right )} + 3 \, e^{2}} \]
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*exp(1/6*x+5)^8*exp(exp(-1+x)^2)^2+6*x*e xp(1/6*x+5)^4*exp(exp(-1+x)^2)+3),x, algorithm=\
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 + 20) + 3*x*e^22)*e^(2/3*x + e^(2*x - 2)) + 3*e^2)
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 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=-\frac {2 \, {\left (36 \, x^{3} e^{\left (\frac {14}{3} \, x + e^{\left (2 \, x - 2\right )} + 18\right )} \log \left (2\right ) + 24 \, x^{3} e^{\left (\frac {8}{3} \, x + e^{\left (2 \, x - 2\right )} + 20\right )} \log \left (2\right ) + 4 \, x^{3} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} \log \left (2\right ) + 36 \, x^{2} e^{\left (\frac {8}{3} \, x + e^{\left (2 \, x - 2\right )} + 20\right )} \log \left (2\right ) + 12 \, x^{2} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} \log \left (2\right ) + 9 \, x e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} \log \left (2\right )\right )}}{36 \, x^{3} e^{\left (\frac {14}{3} \, x + e^{\left (2 \, x - 2\right )} + 18\right )} + 24 \, x^{3} e^{\left (\frac {8}{3} \, x + e^{\left (2 \, x - 2\right )} + 20\right )} + 4 \, x^{3} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} + 4 \, x^{2} e^{2} + 24 \, x^{2} e^{\left (2 \, x\right )} + 36 \, x^{2} e^{\left (4 \, x - 2\right )} + 36 \, x^{2} e^{\left (\frac {8}{3} \, x + e^{\left (2 \, x - 2\right )} + 20\right )} + 12 \, x^{2} e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} + 12 \, x e^{2} + 36 \, x e^{\left (2 \, x\right )} + 9 \, x e^{\left (\frac {2}{3} \, x + e^{\left (2 \, x - 2\right )} + 22\right )} + 9 \, e^{2}} \]
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*exp(1/6*x+5)^8*exp(exp(-1+x)^2)^2+6*x*e xp(1/6*x+5)^4*exp(exp(-1+x)^2)+3),x, algorithm=\
-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)*lo g(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)
Time = 13.94 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.04 \[ \int \frac {e^{e^{-2+2 x}} \left (e^{\frac {2 (30+x)}{3}} (-3-2 x) \log (4)-6 e^{-2+2 x+\frac {2 (30+x)}{3}} x \log (4)\right )}{3+6 e^{e^{-2+2 x}+\frac {2 (30+x)}{3}} x+3 e^{2 e^{-2+2 x}+\frac {4 (30+x)}{3}} x^2} \, dx=\frac {2\,x\,\ln \left (2\right )\,\left (2\,x+6\,x\,{\mathrm {e}}^{2\,x-2}+3\right )}{\left ({\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-2}}+\frac {{\mathrm {e}}^{-\frac {2\,x}{3}-20}}{x}\right )\,\left (3\,x^2\,{\mathrm {e}}^{\frac {2\,x}{3}+20}+2\,x^3\,{\mathrm {e}}^{\frac {2\,x}{3}+20}+6\,x^3\,{\mathrm {e}}^{\frac {8\,x}{3}+18}\right )} \]
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)