Integrand size = 64, antiderivative size = 31 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=2 \left (2+e^{-3-e^{e^{-4+x}}} \left (-\frac {3}{5}+e^{e^{1+x}}\right ) \log (4)\right ) \] Output:
4*(exp(exp(1+x))-3/5)/exp(exp(exp(-4+x))+3)*ln(2)+4
Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\frac {2}{5} e^{-3-e^{e^{-4+x}}} \left (-3+5 e^{e^{1+x}}\right ) \log (4) \] Input:
Integrate[(E^(-3 - E^E^(-4 + x))*(6*E^(-4 + E^(-4 + x) + x)*Log[4] + E^E^( 1 + x)*(10*E^(1 + x)*Log[4] - 10*E^(-4 + E^(-4 + x) + x)*Log[4])))/5,x]
Output:
(2*E^(-3 - E^E^(-4 + x))*(-3 + 5*E^E^(1 + x))*Log[4])/5
Time = 0.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {27, 27, 2720, 27, 2726}
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.
\(\displaystyle \int \frac {1}{5} e^{-e^{e^{x-4}}-3} \left (e^{e^{x+1}} \left (10 e^{x+1} \log (4)-10 e^{x+e^{x-4}-4} \log (4)\right )+6 e^{x+e^{x-4}-4} \log (4)\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int 2 e^{-3-e^{e^{x-4}}} \left (5 e^{e^{x+1}} \left (e^{x+1} \log (4)-e^{x+e^{x-4}-4} \log (4)\right )+3 e^{x+e^{x-4}-4} \log (4)\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} \int e^{-3-e^{e^{x-4}}} \left (5 e^{e^{x+1}} \left (e^{x+1} \log (4)-e^{x+e^{x-4}-4} \log (4)\right )+3 e^{x+e^{x-4}-4} \log (4)\right )dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {2}{5} \int e^{-7-e^{e^{x-4}}} \left (3 e^{e^{x-4}}-5 e^{e^x \left (\frac {1}{e^4}+e\right )}+5 e^{5+e^{x+1}}\right ) \log (4)de^x\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} \log (4) \int e^{-7-e^{e^{x-4}}} \left (3 e^{e^{x-4}}-5 e^{e^x \left (\frac {1}{e^4}+e\right )}+5 e^{5+e^{x+1}}\right )de^x\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {2}{5} e^{-e^{e^{x-4}}-e^{x-4}-3} \left (3 e^{e^{x-4}}-5 e^{\left (\frac {1}{e^4}+e\right ) e^x}\right ) \log (4)\) |
Input:
Int[(E^(-3 - E^E^(-4 + x))*(6*E^(-4 + E^(-4 + x) + x)*Log[4] + E^E^(1 + x) *(10*E^(1 + x)*Log[4] - 10*E^(-4 + E^(-4 + x) + x)*Log[4])))/5,x]
Output:
(-2*E^(-3 - E^E^(-4 + x) - E^(-4 + x))*(3*E^E^(-4 + x) - 5*E^(E^x*(E^(-4) + E)))*Log[4])/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 0.73 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {4 \ln \left (2\right ) \left (5 \,{\mathrm e}^{{\mathrm e}^{1+x}}-3\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x -4}}-3}}{5}\) | \(24\) |
parallelrisch | \(\frac {\left (20 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{1+x}}-12 \ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x -4}}-3}}{5}\) | \(27\) |
norman | \(\left (-\frac {12 \,{\mathrm e}^{-5} {\mathrm e}^{5} \ln \left (2\right )}{5}+4 \,{\mathrm e}^{-5} {\mathrm e}^{5} \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{1+x}}\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x -4}}-3}\) | \(45\) |
Input:
int(1/5*((-20*ln(2)*exp(x-4)*exp(exp(x-4))+20*ln(2)*exp(1+x))*exp(exp(1+x) )+12*ln(2)*exp(x-4)*exp(exp(x-4)))/exp(exp(exp(x-4))+3),x,method=_RETURNVE RBOSE)
Output:
4/5*ln(2)*(5*exp(exp(1+x))-3)*exp(-exp(exp(x-4))-3)
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\frac {4}{5} \, {\left (5 \, e^{\left (e^{\left (x + 1\right )}\right )} \log \left (2\right ) - 3 \, \log \left (2\right )\right )} e^{\left (-{\left (e^{\left ({\left ({\left (x - 4\right )} e^{5} + e^{\left (x + 1\right )}\right )} e^{\left (-5\right )} + 5\right )} + 3 \, e^{\left (x + 1\right )}\right )} e^{\left (-x - 1\right )}\right )} \] Input:
integrate(1/5*((-20*log(2)*exp(-4+x)*exp(exp(-4+x))+20*log(2)*exp(1+x))*ex p(exp(1+x))+12*log(2)*exp(-4+x)*exp(exp(-4+x)))/exp(exp(exp(-4+x))+3),x, a lgorithm="fricas")
Output:
4/5*(5*e^(e^(x + 1))*log(2) - 3*log(2))*e^(-(e^(((x - 4)*e^5 + e^(x + 1))* e^(-5) + 5) + 3*e^(x + 1))*e^(-x - 1))
Timed out. \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(1/5*((-20*ln(2)*exp(-4+x)*exp(exp(-4+x))+20*ln(2)*exp(1+x))*exp( exp(1+x))+12*ln(2)*exp(-4+x)*exp(exp(-4+x)))/exp(exp(exp(-4+x))+3),x)
Output:
Timed out
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=4 \, e^{\left (e^{\left (x + 1\right )} - e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )} \log \left (2\right ) - \frac {12}{5} \, e^{\left (-e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )} \log \left (2\right ) \] Input:
integrate(1/5*((-20*log(2)*exp(-4+x)*exp(exp(-4+x))+20*log(2)*exp(1+x))*ex p(exp(1+x))+12*log(2)*exp(-4+x)*exp(exp(-4+x)))/exp(exp(exp(-4+x))+3),x, a lgorithm="maxima")
Output:
4*e^(e^(x + 1) - e^(e^(x - 4)) - 3)*log(2) - 12/5*e^(-e^(e^(x - 4)) - 3)*l og(2)
\[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\int { -\frac {4}{5} \, {\left (5 \, {\left (e^{\left (x + e^{\left (x - 4\right )} - 4\right )} \log \left (2\right ) - e^{\left (x + 1\right )} \log \left (2\right )\right )} e^{\left (e^{\left (x + 1\right )}\right )} - 3 \, e^{\left (x + e^{\left (x - 4\right )} - 4\right )} \log \left (2\right )\right )} e^{\left (-e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )} \,d x } \] Input:
integrate(1/5*((-20*log(2)*exp(-4+x)*exp(exp(-4+x))+20*log(2)*exp(1+x))*ex p(exp(1+x))+12*log(2)*exp(-4+x)*exp(exp(-4+x)))/exp(exp(exp(-4+x))+3),x, a lgorithm="giac")
Output:
integrate(-4/5*(5*(e^(x + e^(x - 4) - 4)*log(2) - e^(x + 1)*log(2))*e^(e^( x + 1)) - 3*e^(x + e^(x - 4) - 4)*log(2))*e^(-e^(e^(x - 4)) - 3), x)
Time = 3.43 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\frac {4\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-{\mathrm {e}}^{{\mathrm {e}}^{-4}\,{\mathrm {e}}^x}}\,\ln \left (2\right )\,\left (5\,{\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^x}-3\right )}{5} \] Input:
int(exp(- exp(exp(x - 4)) - 3)*((exp(exp(x + 1))*(20*exp(x + 1)*log(2) - 2 0*exp(x - 4)*exp(exp(x - 4))*log(2)))/5 + (12*exp(x - 4)*exp(exp(x - 4))*l og(2))/5),x)
Output:
(4*exp(-3)*exp(-exp(exp(-4)*exp(x)))*log(2)*(5*exp(exp(1)*exp(x)) - 3))/5
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\frac {4 \,\mathrm {log}\left (2\right ) \left (5 e^{e^{x} e}-3\right )}{5 e^{e^{\frac {e^{x}}{e^{4}}}} e^{3}} \] Input:
int(1/5*((-20*log(2)*exp(-4+x)*exp(exp(-4+x))+20*log(2)*exp(1+x))*exp(exp( 1+x))+12*log(2)*exp(-4+x)*exp(exp(-4+x)))/exp(exp(exp(-4+x))+3),x)
Output:
(4*log(2)*(5*e**(e**x*e) - 3))/(5*e**(e**(e**x/e**4))*e**3)