3.100.8 \(\int \frac {e^{e^{\frac {-5+6 x+(-2+x) \log (2)}{3+\log (2)}}+\frac {-5+6 x+(-2+x) \log (2)}{3+\log (2)}} (-6-\log (2))+(3+\log (2)) \log (6)}{(3+\log (2)) \log (6)} \, dx\)

Optimal. Leaf size=28 \[ -1+x-\frac {e^{e^{-2+x+\frac {1+3 x}{3+\log (2)}}}}{\log (6)} \]

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Rubi [A]  time = 0.18, antiderivative size = 41, normalized size of antiderivative = 1.46, number of steps used = 4, number of rules used = 3, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {12, 2282, 2194} \begin {gather*} x-\frac {e^{2^{-\frac {2-x}{3+\log (2)}} e^{-\frac {5-6 x}{3+\log (2)}}}}{\log (6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^((-5 + 6*x + (-2 + x)*Log[2])/(3 + Log[2])) + (-5 + 6*x + (-2 + x)*Log[2])/(3 + Log[2]))*(-6 - Log[2
]) + (3 + Log[2])*Log[6])/((3 + Log[2])*Log[6]),x]

[Out]

x - E^(1/(2^((2 - x)/(3 + Log[2]))*E^((5 - 6*x)/(3 + Log[2]))))/Log[6]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (\exp \left (e^{\frac {-5+6 x+(-2+x) \log (2)}{3+\log (2)}}+\frac {-5+6 x+(-2+x) \log (2)}{3+\log (2)}\right ) (-6-\log (2))+(3+\log (2)) \log (6)\right ) \, dx}{(3+\log (2)) \log (6)}\\ &=x-\frac {(6+\log (2)) \int \exp \left (e^{\frac {-5+6 x+(-2+x) \log (2)}{3+\log (2)}}+\frac {-5+6 x+(-2+x) \log (2)}{3+\log (2)}\right ) \, dx}{(3+\log (2)) \log (6)}\\ &=x-\frac {\operatorname {Subst}\left (\int e^x \, dx,x,e^{\frac {-5+6 x+(-2+x) \log (2)}{3+\log (2)}}\right )}{\log (6)}\\ &=x-\frac {e^{2^{-\frac {2-x}{3+\log (2)}} e^{-\frac {5-6 x}{3+\log (2)}}}}{\log (6)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 41, normalized size = 1.46 \begin {gather*} x+\frac {e^{e^{\frac {-5+6 x+(-2+x) \log (2)}{3+\log (2)}}} (-6-\log (2))}{(6+\log (2)) \log (6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^((-5 + 6*x + (-2 + x)*Log[2])/(3 + Log[2])) + (-5 + 6*x + (-2 + x)*Log[2])/(3 + Log[2]))*(-6 -
Log[2]) + (3 + Log[2])*Log[6])/((3 + Log[2])*Log[6]),x]

[Out]

x + (E^E^((-5 + 6*x + (-2 + x)*Log[2])/(3 + Log[2]))*(-6 - Log[2]))/((6 + Log[2])*Log[6])

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fricas [B]  time = 0.98, size = 94, normalized size = 3.36 \begin {gather*} \frac {{\left (x e^{\left (\frac {{\left (x - 2\right )} \log \relax (2) + 6 \, x - 5}{\log \relax (2) + 3}\right )} \log \relax (6) - e^{\left (\frac {{\left (\log \relax (2) + 3\right )} e^{\left (\frac {{\left (x - 2\right )} \log \relax (2) + 6 \, x - 5}{\log \relax (2) + 3}\right )} + {\left (x - 2\right )} \log \relax (2) + 6 \, x - 5}{\log \relax (2) + 3}\right )}\right )} e^{\left (-\frac {{\left (x - 2\right )} \log \relax (2) + 6 \, x - 5}{\log \relax (2) + 3}\right )}}{\log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(2)-6)*exp(((x-2)*log(2)+6*x-5)/(3+log(2)))*exp(exp(((x-2)*log(2)+6*x-5)/(3+log(2))))+(3+log(2
))*log(6))/(3+log(2))/log(6),x, algorithm="fricas")

[Out]

(x*e^(((x - 2)*log(2) + 6*x - 5)/(log(2) + 3))*log(6) - e^(((log(2) + 3)*e^(((x - 2)*log(2) + 6*x - 5)/(log(2)
+ 3)) + (x - 2)*log(2) + 6*x - 5)/(log(2) + 3)))*e^(-((x - 2)*log(2) + 6*x - 5)/(log(2) + 3))/log(6)

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giac [B]  time = 0.37, size = 254, normalized size = 9.07 \begin {gather*} \frac {2 \, x {\left (\log \relax (2) + 3\right )} \log \relax (6) - \frac {{\left (2^{\frac {8}{9}} e^{\left (\frac {x \log \relax (2) + e^{\left (\frac {x \log \relax (2) + 6 \, x - 2 \, \log \relax (2) - 5}{\log \relax (2) + 3}\right )} \log \relax (2) + 6 \, x + 3 \, e^{\left (\frac {x \log \relax (2) + 6 \, x - 2 \, \log \relax (2) - 5}{\log \relax (2) + 3}\right )}}{\log \relax (2) + 3} + \frac {\log \relax (2)^{2} - 15 \, \log \relax (2) - 45}{9 \, {\left (\log \relax (2) + 3\right )}}\right )} \log \relax (2) + 3 \cdot 2^{\frac {8}{9}} e^{\left (\frac {x \log \relax (2) + e^{\left (\frac {x \log \relax (2) + 6 \, x - 2 \, \log \relax (2) - 5}{\log \relax (2) + 3}\right )} \log \relax (2) + 6 \, x + 3 \, e^{\left (\frac {x \log \relax (2) + 6 \, x - 2 \, \log \relax (2) - 5}{\log \relax (2) + 3}\right )}}{\log \relax (2) + 3} + \frac {\log \relax (2)^{2} - 15 \, \log \relax (2) - 45}{9 \, {\left (\log \relax (2) + 3\right )}}\right )}\right )} {\left (\log \relax (2) + 6\right )}}{e^{\left (\frac {x \log \relax (2) + 6 \, x - 2 \, \log \relax (2) - 5}{\log \relax (2) + 3}\right )} \log \relax (2) + 6 \, e^{\left (\frac {x \log \relax (2) + 6 \, x - 2 \, \log \relax (2) - 5}{\log \relax (2) + 3}\right )}}}{2 \, {\left (\log \relax (2) + 3\right )} \log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(2)-6)*exp(((x-2)*log(2)+6*x-5)/(3+log(2)))*exp(exp(((x-2)*log(2)+6*x-5)/(3+log(2))))+(3+log(2
))*log(6))/(3+log(2))/log(6),x, algorithm="giac")

[Out]

1/2*(2*x*(log(2) + 3)*log(6) - (2^(8/9)*e^((x*log(2) + e^((x*log(2) + 6*x - 2*log(2) - 5)/(log(2) + 3))*log(2)
+ 6*x + 3*e^((x*log(2) + 6*x - 2*log(2) - 5)/(log(2) + 3)))/(log(2) + 3) + 1/9*(log(2)^2 - 15*log(2) - 45)/(l
og(2) + 3))*log(2) + 3*2^(8/9)*e^((x*log(2) + e^((x*log(2) + 6*x - 2*log(2) - 5)/(log(2) + 3))*log(2) + 6*x +
3*e^((x*log(2) + 6*x - 2*log(2) - 5)/(log(2) + 3)))/(log(2) + 3) + 1/9*(log(2)^2 - 15*log(2) - 45)/(log(2) + 3
)))*(log(2) + 6)/(e^((x*log(2) + 6*x - 2*log(2) - 5)/(log(2) + 3))*log(2) + 6*e^((x*log(2) + 6*x - 2*log(2) -
5)/(log(2) + 3))))/((log(2) + 3)*log(6))

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maple [A]  time = 0.12, size = 29, normalized size = 1.04

method	result	size
norman	\(x -\frac {{\mathrm e}^{{\mathrm e}^{\frac {\left (x -2\right ) \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}}}{\ln \relax (6)}\)	\(29\)
derivativedivides	\(\frac {\ln \relax (6) \ln \relax (2) \ln \left ({\mathrm e}^{\frac {\left (x -2\right ) \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}\right )+3 \ln \relax (6) \ln \left ({\mathrm e}^{\frac {\left (x -2\right ) \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}\right )-{\mathrm e}^{{\mathrm e}^{\frac {\left (x -2\right ) \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}} \ln \relax (2)-6 \,{\mathrm e}^{{\mathrm e}^{\frac {\left (x -2\right ) \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}}}{\ln \relax (6) \left (\ln \relax (2)+6\right )}\)	\(108\)
default	\(\frac {\ln \relax (6) x \ln \relax (2)+3 x \ln \relax (6)-\frac {{\mathrm e}^{{\mathrm e}^{\frac {\left (x -2\right ) \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}} \ln \relax (2)^{2}}{\ln \relax (2)+6}-\frac {9 \,{\mathrm e}^{{\mathrm e}^{\frac {\left (x -2\right ) \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}} \ln \relax (2)}{\ln \relax (2)+6}-\frac {18 \,{\mathrm e}^{{\mathrm e}^{\frac {\left (x -2\right ) \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}}}{\ln \relax (2)+6}}{\left (3+\ln \relax (2)\right ) \ln \relax (6)}\)	\(114\)
risch	\(\frac {x \ln \relax (2)^{2}}{\left (3+\ln \relax (2)\right ) \left (\ln \relax (2)+\ln \relax (3)\right )}+\frac {x \ln \relax (2) \ln \relax (3)}{\left (3+\ln \relax (2)\right ) \left (\ln \relax (2)+\ln \relax (3)\right )}+\frac {3 x \ln \relax (2)}{\left (3+\ln \relax (2)\right ) \left (\ln \relax (2)+\ln \relax (3)\right )}+\frac {3 x \ln \relax (3)}{\left (3+\ln \relax (2)\right ) \left (\ln \relax (2)+\ln \relax (3)\right )}-\frac {{\mathrm e}^{{\mathrm e}^{\frac {x \ln \relax (2)-2 \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}} \ln \relax (2)}{\left (3+\ln \relax (2)\right ) \left (\ln \relax (2)+\ln \relax (3)\right )}-\frac {3 \,{\mathrm e}^{{\mathrm e}^{\frac {x \ln \relax (2)-2 \ln \relax (2)+6 x -5}{3+\ln \relax (2)}}}}{\left (3+\ln \relax (2)\right ) \left (\ln \relax (2)+\ln \relax (3)\right )}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(2)-6)*exp(((x-2)*ln(2)+6*x-5)/(3+ln(2)))*exp(exp(((x-2)*ln(2)+6*x-5)/(3+ln(2))))+(3+ln(2))*ln(6))/(3
+ln(2))/ln(6),x,method=_RETURNVERBOSE)

[Out]

x-1/ln(6)*exp(exp(((x-2)*ln(2)+6*x-5)/(3+ln(2))))

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maxima [B]  time = 0.45, size = 69, normalized size = 2.46 \begin {gather*} \frac {x {\left (\log \relax (2) + 3\right )} \log \relax (6) - {\left (\log \relax (2) + 3\right )} e^{\left (\frac {e^{\left (\frac {x \log \relax (2)}{\log \relax (2) + 3} + \frac {6 \, x}{\log \relax (2) + 3} - \frac {5}{\log \relax (2) + 3}\right )}}{2^{\frac {2}{\log \relax (2) + 3}}}\right )}}{{\left (\log \relax (2) + 3\right )} \log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(2)-6)*exp(((x-2)*log(2)+6*x-5)/(3+log(2)))*exp(exp(((x-2)*log(2)+6*x-5)/(3+log(2))))+(3+log(2
))*log(6))/(3+log(2))/log(6),x, algorithm="maxima")

[Out]

(x*(log(2) + 3)*log(6) - (log(2) + 3)*e^(e^(x*log(2)/(log(2) + 3) + 6*x/(log(2) + 3) - 5/(log(2) + 3))/2^(2/(l
og(2) + 3))))/((log(2) + 3)*log(6))

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mupad [B]  time = 0.31, size = 61, normalized size = 2.18 \begin {gather*} x-\frac {{\mathrm {e}}^{\frac {2^{\frac {x}{\ln \relax (2)+3}}\,{\mathrm {e}}^{\frac {6\,x}{\ln \relax (2)+3}}\,{\mathrm {e}}^{-\frac {5}{\ln \relax (2)+3}}}{2^{\frac {2}{\ln \relax (2)+3}}}}\,\left (\ln \relax (2)+3\right )}{\ln \left (216\right )+\ln \relax (2)\,\ln \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(6)*(log(2) + 3) - exp(exp((6*x + log(2)*(x - 2) - 5)/(log(2) + 3)))*exp((6*x + log(2)*(x - 2) - 5)/(l
og(2) + 3))*(log(2) + 6))/(log(6)*(log(2) + 3)),x)

[Out]

x - (exp((2^(x/(log(2) + 3))*exp((6*x)/(log(2) + 3))*exp(-5/(log(2) + 3)))/2^(2/(log(2) + 3)))*(log(2) + 3))/(
log(216) + log(2)*log(6))

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sympy [A]  time = 0.27, size = 24, normalized size = 0.86 \begin {gather*} x - \frac {e^{e^{\frac {6 x + \left (x - 2\right ) \log {\relax (2 )} - 5}{\log {\relax (2 )} + 3}}}}{\log {\relax (6 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-ln(2)-6)*exp(((x-2)*ln(2)+6*x-5)/(3+ln(2)))*exp(exp(((x-2)*ln(2)+6*x-5)/(3+ln(2))))+(3+ln(2))*ln(
6))/(3+ln(2))/ln(6),x)

[Out]

x - exp(exp((6*x + (x - 2)*log(2) - 5)/(log(2) + 3)))/log(6)

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