Integrand size = 92, antiderivative size = 21 \[ \int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx=e^{-1+e^{3+\frac {4}{\log (\log (3+x+2 e x))}}} \] Output:
exp(exp(4/ln(ln(2*x*exp(1)+3+x))+3)-1)
Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx=e^{-1+e^{3+\frac {4}{\log (\log (3+x+2 e x))}}} \] Input:
Integrate[((-4 - 8*E)*E^(-1 + E^((4 + 3*Log[Log[3 + x + 2*E*x]])/Log[Log[3 + x + 2*E*x]]) + (4 + 3*Log[Log[3 + x + 2*E*x]])/Log[Log[3 + x + 2*E*x]]) )/((3 + x + 2*E*x)*Log[3 + x + 2*E*x]*Log[Log[3 + x + 2*E*x]]^2),x]
Output:
E^(-1 + E^(3 + 4/Log[Log[3 + x + 2*E*x]]))
Time = 1.56 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {6, 27, 2894, 7281, 3039, 3039, 7266, 2720, 2624}
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.
\(\displaystyle \int \frac {(-4-8 e) \exp \left (\exp \left (\frac {3 \log (\log (2 e x+x+3))+4}{\log (\log (2 e x+x+3))}\right )+\frac {3 \log (\log (2 e x+x+3))+4}{\log (\log (2 e x+x+3))}-1\right )}{(2 e x+x+3) \log (2 e x+x+3) \log ^2(\log (2 e x+x+3))} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {(-4-8 e) \exp \left (\exp \left (\frac {3 \log (\log (2 e x+x+3))+4}{\log (\log (2 e x+x+3))}\right )+\frac {3 \log (\log (2 e x+x+3))+4}{\log (\log (2 e x+x+3))}-1\right )}{((1+2 e) x+3) \log (2 e x+x+3) \log ^2(\log (2 e x+x+3))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -4 (1+2 e) \int \frac {\exp \left (e^{\frac {4}{\log (\log (2 e x+x+3))}} \log ^{\frac {3}{\log (\log (2 e x+x+3))}}(2 e x+x+3)+\frac {3 \log (\log (2 e x+x+3))+4}{\log (\log (2 e x+x+3))}-1\right )}{((1+2 e) x+3) \log (2 e x+x+3) \log ^2(\log (2 e x+x+3))}dx\) |
\(\Big \downarrow \) 2894 |
\(\displaystyle -4 (1+2 e) \int \frac {\exp \left (e^{\frac {4}{\log (\log (2 e x+x+3))}} \log ^{\frac {3}{\log (\log (2 e x+x+3))}}(2 e x+x+3)+\frac {3 \log (\log (2 e x+x+3))+4}{\log (\log (2 e x+x+3))}-1\right )}{((1+2 e) x+3) \log ((1+2 e) x+3) \log ^2(\log (2 e x+x+3))}dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle -4 \int \frac {\exp \left (e^{3+\frac {4}{\log (\log ((1+2 e) x+3))}}+2+\frac {4}{\log (\log ((1+2 e) x+3))}\right )}{((1+2 e) x+3) \log ((1+2 e) x+3) \log ^2(\log ((1+2 e) x+3))}d((1+2 e) x+3)\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle -4 \int \frac {\exp \left (e^{3+\frac {4}{\log (\log ((1+2 e) x+3))}}+2+\frac {4}{\log (\log ((1+2 e) x+3))}\right )}{\log ((1+2 e) x+3) \log ^2(\log ((1+2 e) x+3))}d\log ((1+2 e) x+3)\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle -4 \int \frac {\exp \left (e^{3+\frac {4}{\log (\log ((1+2 e) x+3))}}+2+\frac {4}{\log (\log ((1+2 e) x+3))}\right )}{\log ^2(\log ((1+2 e) x+3))}d\log (\log ((1+2 e) x+3))\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle 4 \int e^{e^3 ((1+2 e) x+3)^4+2} ((1+2 e) x+3)^4d\frac {1}{\log (\log ((1+2 e) x+3))}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int e^{e^3 ((1+2 e) x+3)^4+2}d((1+2 e) x+3)^4\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle e^{e^3 ((1+2 e) x+3)^4-1}\) |
Input:
Int[((-4 - 8*E)*E^(-1 + E^((4 + 3*Log[Log[3 + x + 2*E*x]])/Log[Log[3 + x + 2*E*x]]) + (4 + 3*Log[Log[3 + x + 2*E*x]])/Log[Log[3 + x + 2*E*x]]))/((3 + x + 2*E*x)*Log[3 + x + 2*E*x]*Log[Log[3 + x + 2*E*x]]^2),x]
Output:
E^(-1 + E^3*(3 + (1 + 2*E)*x)^4)
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, 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[((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] && Lin earQ[v, x] && !LinearMatchQ[v, x] && !(EqQ[n, 1] && MatchQ[c*v, (e_.)*((f _) + (g_.)*x) /; FreeQ[{e, f, g}, x]])
Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst [[3]] Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /; !FalseQ[lst]] /; NonsumQ[u]
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Leaf count of result is larger than twice the leaf count of optimal. \(47\) vs. \(2(20)=40\).
Time = 72.95 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29
method | result | size |
derivativedivides | \(-\frac {\left (-2 \,{\mathrm e}-1\right ) {\mathrm e}^{{\mathrm e}^{\frac {3 \ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )+4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}}-1}}{2 \,{\mathrm e}+1}\) | \(48\) |
default | \(-\frac {\left (-8 \,{\mathrm e}-4\right ) {\mathrm e}^{{\mathrm e}^{\frac {3 \ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )+4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}}-1}}{4 \left (2 \,{\mathrm e}+1\right )}\) | \(48\) |
parallelrisch | \(-\frac {\left (-8 \,{\mathrm e}-4\right ) {\mathrm e}^{{\mathrm e}^{\frac {3 \ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )+4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}}-1}}{4 \left (2 \,{\mathrm e}+1\right )}\) | \(48\) |
risch | \(\frac {2 \,{\mathrm e}^{{\mathrm e}^{\frac {3 \ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )+4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}}-1} {\mathrm e}}{2 \,{\mathrm e}+1}+\frac {{\mathrm e}^{{\mathrm e}^{\frac {3 \ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )+4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}}-1}}{2 \,{\mathrm e}+1}\) | \(85\) |
Input:
int((-8*exp(1)-4)*exp((3*ln(ln(2*x*exp(1)+3+x))+4)/ln(ln(2*x*exp(1)+3+x))) *exp(exp((3*ln(ln(2*x*exp(1)+3+x))+4)/ln(ln(2*x*exp(1)+3+x)))-1)/(2*x*exp( 1)+3+x)/ln(2*x*exp(1)+3+x)/ln(ln(2*x*exp(1)+3+x))^2,x,method=_RETURNVERBOS E)
Output:
-(-2*exp(1)-1)/(2*exp(1)+1)*exp(exp((3*ln(ln(2*x*exp(1)+3+x))+4)/ln(ln(2*x *exp(1)+3+x)))-1)
Leaf count of result is larger than twice the leaf count of optimal. 96 vs. \(2 (20) = 40\).
Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 4.57 \[ \int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx=e^{\left (\frac {e^{\left (\frac {3 \, \log \left (\log \left (2 \, x e + x + 3\right )\right ) + 4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )}\right )} \log \left (\log \left (2 \, x e + x + 3\right )\right ) + 2 \, \log \left (\log \left (2 \, x e + x + 3\right )\right ) + 4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )} - \frac {3 \, \log \left (\log \left (2 \, x e + x + 3\right )\right ) + 4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )}\right )} \] Input:
integrate((-8*exp(1)-4)*exp((3*log(log(2*exp(1)*x+3+x))+4)/log(log(2*exp(1 )*x+3+x)))*exp(exp((3*log(log(2*exp(1)*x+3+x))+4)/log(log(2*exp(1)*x+3+x)) )-1)/(2*exp(1)*x+3+x)/log(2*exp(1)*x+3+x)/log(log(2*exp(1)*x+3+x))^2,x, al gorithm="fricas")
Output:
e^((e^((3*log(log(2*x*e + x + 3)) + 4)/log(log(2*x*e + x + 3)))*log(log(2* x*e + x + 3)) + 2*log(log(2*x*e + x + 3)) + 4)/log(log(2*x*e + x + 3)) - ( 3*log(log(2*x*e + x + 3)) + 4)/log(log(2*x*e + x + 3)))
Time = 0.65 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx=e^{e^{\frac {3 \log {\left (\log {\left (x + 2 e x + 3 \right )} \right )} + 4}{\log {\left (\log {\left (x + 2 e x + 3 \right )} \right )}}} - 1} \] Input:
integrate((-8*exp(1)-4)*exp((3*ln(ln(2*exp(1)*x+3+x))+4)/ln(ln(2*exp(1)*x+ 3+x)))*exp(exp((3*ln(ln(2*exp(1)*x+3+x))+4)/ln(ln(2*exp(1)*x+3+x)))-1)/(2* exp(1)*x+3+x)/ln(2*exp(1)*x+3+x)/ln(ln(2*exp(1)*x+3+x))**2,x)
Output:
exp(exp((3*log(log(x + 2*E*x + 3)) + 4)/log(log(x + 2*E*x + 3))) - 1)
Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx=e^{\left (e^{\left (\frac {4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )} + 3\right )} - 1\right )} \] Input:
integrate((-8*exp(1)-4)*exp((3*log(log(2*exp(1)*x+3+x))+4)/log(log(2*exp(1 )*x+3+x)))*exp(exp((3*log(log(2*exp(1)*x+3+x))+4)/log(log(2*exp(1)*x+3+x)) )-1)/(2*exp(1)*x+3+x)/log(2*exp(1)*x+3+x)/log(log(2*exp(1)*x+3+x))^2,x, al gorithm="maxima")
Output:
e^(e^(4/log(log(2*x*e + x + 3)) + 3) - 1)
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx=\frac {{\left (2 \, e + 1\right )} e^{\left (e^{\left (\frac {4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )} + 3\right )}\right )}}{2 \, e^{2} + e} \] Input:
integrate((-8*exp(1)-4)*exp((3*log(log(2*exp(1)*x+3+x))+4)/log(log(2*exp(1 )*x+3+x)))*exp(exp((3*log(log(2*exp(1)*x+3+x))+4)/log(log(2*exp(1)*x+3+x)) )-1)/(2*exp(1)*x+3+x)/log(2*exp(1)*x+3+x)/log(log(2*exp(1)*x+3+x))^2,x, al gorithm="giac")
Output:
(2*e + 1)*e^(e^(4/log(log(2*x*e + x + 3)) + 3))/(2*e^2 + e)
Time = 3.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx={\mathrm {e}}^{{\mathrm {e}}^{\frac {4}{\ln \left (\ln \left (x+2\,x\,\mathrm {e}+3\right )\right )}}\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-1} \] Input:
int(-(exp(exp((3*log(log(x + 2*x*exp(1) + 3)) + 4)/log(log(x + 2*x*exp(1) + 3))) - 1)*exp((3*log(log(x + 2*x*exp(1) + 3)) + 4)/log(log(x + 2*x*exp(1 ) + 3)))*(8*exp(1) + 4))/(log(x + 2*x*exp(1) + 3)*log(log(x + 2*x*exp(1) + 3))^2*(x + 2*x*exp(1) + 3)),x)
Output:
exp(exp(4/log(log(x + 2*x*exp(1) + 3)))*exp(3))*exp(-1)
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx=\frac {e^{e^{\frac {4}{\mathrm {log}\left (\mathrm {log}\left (2 e x +x +3\right )\right )}} e^{3}}}{e} \] Input:
int((-8*exp(1)-4)*exp((3*log(log(2*exp(1)*x+3+x))+4)/log(log(2*exp(1)*x+3+ x)))*exp(exp((3*log(log(2*exp(1)*x+3+x))+4)/log(log(2*exp(1)*x+3+x)))-1)/( 2*exp(1)*x+3+x)/log(2*exp(1)*x+3+x)/log(log(2*exp(1)*x+3+x))^2,x)
Output:
e**(e**(4/log(log(2*e*x + x + 3)))*e**3)/e