Integrand size = 86, antiderivative size = 29 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7}{10 \left (e^{3-e^2-e^{x+\log ^2(3)} x}+\log (3)\right )} \] Output:
7/10/(ln(3)+exp(-x*exp(ln(3)^2+x)-exp(2)+3))
Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=-\frac {7 e^3}{10 \log (3) \left (e^3+e^{e^2+e^{x+\log ^2(3)} x} \log (3)\right )} \] Input:
Integrate[(E^(3 - E^2 + x - E^(x + Log[3]^2)*x + Log[3]^2)*(7 + 7*x))/(10* E^(6 - 2*E^2 - 2*E^(x + Log[3]^2)*x) + 20*E^(3 - E^2 - E^(x + Log[3]^2)*x) *Log[3] + 10*Log[3]^2),x]
Output:
(-7*E^3)/(10*Log[3]*(E^3 + E^(E^2 + E^(x + Log[3]^2)*x)*Log[3]))
Time = 1.67 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {7292, 27, 27, 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.
\(\displaystyle \int \frac {(7 x+7) e^{x+x \left (-e^{x+\log ^2(3)}\right )-e^2+3+\log ^2(3)}}{10 e^{-2 x e^{x+\log ^2(3)}-2 e^2+6}+20 \log (3) e^{x \left (-e^{x+\log ^2(3)}\right )-e^2+3}+10 \log ^2(3)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(7 x+7) \exp \left (x+x \left (-e^{x+\log ^2(3)}\right )+2 \left (x e^{x+\log ^2(3)}+e^2\right )+3 \left (1+\frac {1}{3} \left (\log ^2(3)-e^2\right )\right )\right )}{10 \left (\log (3) e^{x e^{x+\log ^2(3)}+e^2}+e^3\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \int \frac {7 \exp \left (-e^{x+\log ^2(3)} x+x+2 \left (e^{x+\log ^2(3)} x+e^2\right )+\log ^2(3)-e^2+3\right ) (x+1)}{\left (e^3+e^{e^{x+\log ^2(3)} x+e^2} \log (3)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7}{10} \int \frac {\exp \left (-e^{x+\log ^2(3)} x+x+2 \left (e^{x+\log ^2(3)} x+e^2\right )+\log ^2(3)-e^2+3\right ) (x+1)}{\left (e^3+e^{e^{x+\log ^2(3)} x+e^2} \log (3)\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {7 e^3}{10 \log (3) \left (\log (3) e^{x e^{x+\log ^2(3)}+e^2}+e^3\right )}\) |
Input:
Int[(E^(3 - E^2 + x - E^(x + Log[3]^2)*x + Log[3]^2)*(7 + 7*x))/(10*E^(6 - 2*E^2 - 2*E^(x + Log[3]^2)*x) + 20*E^(3 - E^2 - E^(x + Log[3]^2)*x)*Log[3 ] + 10*Log[3]^2),x]
Output:
(-7*E^3)/(10*Log[3]*(E^3 + E^(E^2 + E^(x + Log[3]^2)*x)*Log[3]))
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 = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {7}{10 \left (\ln \left (3\right )+{\mathrm e}^{-x \,{\mathrm e}^{\ln \left (3\right )^{2}+x}-{\mathrm e}^{2}+3}\right )}\) | \(25\) |
norman | \(\frac {7}{10 \left (\ln \left (3\right )+{\mathrm e}^{-x \,{\mathrm e}^{\ln \left (3\right )^{2}+x}-{\mathrm e}^{2}+3}\right )}\) | \(25\) |
risch | \(\frac {7}{10 \left (\ln \left (3\right )+{\mathrm e}^{-x \,{\mathrm e}^{\ln \left (3\right )^{2}+x}-{\mathrm e}^{2}+3}\right )}\) | \(25\) |
parallelrisch | \(\frac {7}{10 \left (\ln \left (3\right )+{\mathrm e}^{-x \,{\mathrm e}^{\ln \left (3\right )^{2}+x}-{\mathrm e}^{2}+3}\right )}\) | \(25\) |
Input:
int((7*x+7)*exp(ln(3)^2+x)*exp(-x*exp(ln(3)^2+x)-exp(2)+3)/(10*exp(-x*exp( ln(3)^2+x)-exp(2)+3)^2+20*ln(3)*exp(-x*exp(ln(3)^2+x)-exp(2)+3)+10*ln(3)^2 ),x,method=_RETURNVERBOSE)
Output:
7/10/(ln(3)+exp(-x*exp(ln(3)^2+x)-exp(2)+3))
Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7 \, e^{\left (\log \left (3\right )^{2} + x\right )}}{10 \, {\left (e^{\left (\log \left (3\right )^{2} + x\right )} \log \left (3\right ) + e^{\left (-x e^{\left (\log \left (3\right )^{2} + x\right )} + \log \left (3\right )^{2} + x - e^{2} + 3\right )}\right )}} \] Input:
integrate((7*x+7)*exp(log(3)^2+x)*exp(-x*exp(log(3)^2+x)-exp(2)+3)/(10*exp (-x*exp(log(3)^2+x)-exp(2)+3)^2+20*log(3)*exp(-x*exp(log(3)^2+x)-exp(2)+3) +10*log(3)^2),x, algorithm="fricas")
Output:
7/10*e^(log(3)^2 + x)/(e^(log(3)^2 + x)*log(3) + e^(-x*e^(log(3)^2 + x) + log(3)^2 + x - e^2 + 3))
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7}{10 e^{- x e^{x + \log {\left (3 \right )}^{2}} - e^{2} + 3} + 10 \log {\left (3 \right )}} \] Input:
integrate((7*x+7)*exp(ln(3)**2+x)*exp(-x*exp(ln(3)**2+x)-exp(2)+3)/(10*exp (-x*exp(ln(3)**2+x)-exp(2)+3)**2+20*ln(3)*exp(-x*exp(ln(3)**2+x)-exp(2)+3) +10*ln(3)**2),x)
Output:
7/(10*exp(-x*exp(x + log(3)**2) - exp(2) + 3) + 10*log(3))
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=-\frac {7 \, e^{3}}{10 \, {\left (e^{\left (x e^{\left (\log \left (3\right )^{2} + x\right )} + e^{2}\right )} \log \left (3\right )^{2} + e^{3} \log \left (3\right )\right )}} \] Input:
integrate((7*x+7)*exp(log(3)^2+x)*exp(-x*exp(log(3)^2+x)-exp(2)+3)/(10*exp (-x*exp(log(3)^2+x)-exp(2)+3)^2+20*log(3)*exp(-x*exp(log(3)^2+x)-exp(2)+3) +10*log(3)^2),x, algorithm="maxima")
Output:
-7/10*e^3/(e^(x*e^(log(3)^2 + x) + e^2)*log(3)^2 + e^3*log(3))
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=-\frac {7 \, e^{3}}{10 \, {\left (e^{\left (x e^{\left (\log \left (3\right )^{2} + x\right )} + e^{2}\right )} \log \left (3\right ) + e^{3}\right )} \log \left (3\right )} \] Input:
integrate((7*x+7)*exp(log(3)^2+x)*exp(-x*exp(log(3)^2+x)-exp(2)+3)/(10*exp (-x*exp(log(3)^2+x)-exp(2)+3)^2+20*log(3)*exp(-x*exp(log(3)^2+x)-exp(2)+3) +10*log(3)^2),x, algorithm="giac")
Output:
-7/10*e^3/((e^(x*e^(log(3)^2 + x) + e^2)*log(3) + e^3)*log(3))
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7}{10\,\left (\ln \left (3\right )+{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{{\ln \left (3\right )}^2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^3\right )} \] Input:
int((exp(x + log(3)^2)*exp(3 - x*exp(x + log(3)^2) - exp(2))*(7*x + 7))/(1 0*exp(6 - 2*x*exp(x + log(3)^2) - 2*exp(2)) + 20*exp(3 - x*exp(x + log(3)^ 2) - exp(2))*log(3) + 10*log(3)^2),x)
Output:
7/(10*(log(3) + exp(-exp(2))*exp(-x*exp(log(3)^2)*exp(x))*exp(3)))
Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.59 \[ \int \frac {e^{3-e^2+x-e^{x+\log ^2(3)} x+\log ^2(3)} (7+7 x)}{10 e^{6-2 e^2-2 e^{x+\log ^2(3)} x}+20 e^{3-e^2-e^{x+\log ^2(3)} x} \log (3)+10 \log ^2(3)} \, dx=\frac {7 e^{e^{\mathrm {log}\left (3\right )^{2}+x} x +e^{2}}}{10 e^{e^{\mathrm {log}\left (3\right )^{2}+x} x +e^{2}} \mathrm {log}\left (3\right )+10 e^{3}} \] Input:
int((7*x+7)*exp(log(3)^2+x)*exp(-x*exp(log(3)^2+x)-exp(2)+3)/(10*exp(-x*ex p(log(3)^2+x)-exp(2)+3)^2+20*log(3)*exp(-x*exp(log(3)^2+x)-exp(2)+3)+10*lo g(3)^2),x)
Output:
(7*e**(e**(log(3)**2 + x)*x + e**2))/(10*(e**(e**(log(3)**2 + x)*x + e**2) *log(3) + e**3))