∫ 9+e4+2x+log(3)___________________________________________ −27x−3e4x−3x2+(−27−3e4−3x) log(3)+(9x+e4x+x2+(9+e4+x) log(3)) log(9+e4+x)+(−9x−e4x−x2+(−9−e4−x) log(3)) log( 5____

Optimal. Leaf size=21 \[ \log \left (3-\log \left (9+e^4+x\right )+\log \left (\frac {5}{x+\log (3)}\right )\right ) \]

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Rubi [A] time = 0.32, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6, 6688, 6684} \begin {gather*} \log \left (-\log \left (x+e^4+9\right )+\log \left (\frac {5}{x+\log (3)}\right )+3\right ) \end {gather*}

Int[(9 + E^4 + 2*x + Log[3])/(-27*x - 3*E^4*x - 3*x^2 + (-27 - 3*E^4 - 3*x)*Log[3] + (9*x + E^4*x + x^2 + (9 +

E^4 + x)*Log[3])*Log[9 + E^4 + x] + (-9*x - E^4*x - x^2 + (-9 - E^4 - x)*Log[3])*Log[5/(x + Log[3])]),x]

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /; !Fa

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9+e^4+2 x+\log (3)}{\left (-27-3 e^4\right ) x-3 x^2+\left (-27-3 e^4-3 x\right ) \log (3)+\left (9 x+e^4 x+x^2+\left (9+e^4+x\right ) \log (3)\right ) \log \left (9+e^4+x\right )+\left (-9 x-e^4 x-x^2+\left (-9-e^4-x\right ) \log (3)\right ) \log \left (\frac {5}{x+\log (3)}\right )} \, dx\\ &=\int \frac {-9-e^4-2 x-\log (3)}{\left (9+e^4+x\right ) (x+\log (3)) \left (3-\log \left (9+e^4+x\right )+\log \left (\frac {5}{x+\log (3)}\right )\right )} \, dx\\ &=\log \left (3-\log \left (9+e^4+x\right )+\log \left (\frac {5}{x+\log (3)}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.08, size = 21, normalized size = 1.00 \begin {gather*} \log \left (-3+\log \left (9+e^4+x\right )-\log \left (\frac {5}{x+\log (3)}\right )\right ) \end {gather*}

Integrate[(9 + E^4 + 2*x + Log[3])/(-27*x - 3*E^4*x - 3*x^2 + (-27 - 3*E^4 - 3*x)*Log[3] + (9*x + E^4*x + x^2

+ (9 + E^4 + x)*Log[3])*Log[9 + E^4 + x] + (-9*x - E^4*x - x^2 + (-9 - E^4 - x)*Log[3])*Log[5/(x + Log[3])]),x

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fricas [A] time = 0.51, size = 20, normalized size = 0.95 \begin {gather*} \log \left (\log \left (x + e^{4} + 9\right ) - \log \left (\frac {5}{x + \log \relax (3)}\right ) - 3\right ) \end {gather*}

integrate((log(3)+exp(4)+2*x+9)/(((exp(4)+x+9)*log(3)+x*exp(4)+x^2+9*x)*log(exp(4)+x+9)+((-exp(4)-x-9)*log(3)-

x*exp(4)-x^2-9*x)*log(5/(log(3)+x))+(-3*exp(4)-3*x-27)*log(3)-3*x*exp(4)-3*x^2-27*x),x, algorithm="fricas")

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giac [A] time = 0.26, size = 20, normalized size = 0.95 \begin {gather*} \log \left (\log \relax (5) - \log \left (x + e^{4} + 9\right ) - \log \left (x + \log \relax (3)\right ) + 3\right ) \end {gather*}

integrate((log(3)+exp(4)+2*x+9)/(((exp(4)+x+9)*log(3)+x*exp(4)+x^2+9*x)*log(exp(4)+x+9)+((-exp(4)-x-9)*log(3)-

x*exp(4)-x^2-9*x)*log(5/(log(3)+x))+(-3*exp(4)-3*x-27)*log(3)-3*x*exp(4)-3*x^2-27*x),x, algorithm="giac")

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method	result	size

norman	\(\ln \left (\ln \left ({\mathrm e}^{4}+x +9\right )-\ln \left (\frac {5}{\ln \relax (3)+x}\right )-3\right )\)	\(21\)
risch	\(\ln \left (\ln \left ({\mathrm e}^{4}+x +9\right )+\frac {i \left (2 i \ln \relax (5)-2 i \ln \left (\ln \relax (3)+x \right )+6 i\right )}{2}\right )\)	\(28\)

int((ln(3)+exp(4)+2*x+9)/(((exp(4)+x+9)*ln(3)+x*exp(4)+x^2+9*x)*ln(exp(4)+x+9)+((-exp(4)-x-9)*ln(3)-x*exp(4)-x

^2-9*x)*ln(5/(ln(3)+x))+(-3*exp(4)-3*x-27)*ln(3)-3*x*exp(4)-3*x^2-27*x),x,method=_RETURNVERBOSE)

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maxima [A] time = 0.81, size = 18, normalized size = 0.86 \begin {gather*} \log \left (-\log \relax (5) + \log \left (x + e^{4} + 9\right ) + \log \left (x + \log \relax (3)\right ) - 3\right ) \end {gather*}

integrate((log(3)+exp(4)+2*x+9)/(((exp(4)+x+9)*log(3)+x*exp(4)+x^2+9*x)*log(exp(4)+x+9)+((-exp(4)-x-9)*log(3)-

x*exp(4)-x^2-9*x)*log(5/(log(3)+x))+(-3*exp(4)-3*x-27)*log(3)-3*x*exp(4)-3*x^2-27*x),x, algorithm="maxima")

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mupad [B] time = 118.60, size = 20, normalized size = 0.95 \begin {gather*} \ln \left (\ln \left (\frac {5}{x+\ln \relax (3)}\right )-\ln \left (x+{\mathrm {e}}^4+9\right )+3\right ) \end {gather*}

int(-(2*x + exp(4) + log(3) + 9)/(27*x + log(5/(x + log(3)))*(9*x + log(3)*(x + exp(4) + 9) + x*exp(4) + x^2)

+ 3*x*exp(4) - log(x + exp(4) + 9)*(9*x + log(3)*(x + exp(4) + 9) + x*exp(4) + x^2) + log(3)*(3*x + 3*exp(4) +

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sympy [A] time = 0.64, size = 19, normalized size = 0.90 \begin {gather*} \log {\left (\log {\left (\frac {5}{x + \log {\relax (3 )}} \right )} - \log {\left (x + 9 + e^{4} \right )} + 3 \right )} \end {gather*}

integrate((ln(3)+exp(4)+2*x+9)/(((exp(4)+x+9)*ln(3)+x*exp(4)+x**2+9*x)*ln(exp(4)+x+9)+((-exp(4)-x-9)*ln(3)-x*e

xp(4)-x**2-9*x)*ln(5/(ln(3)+x))+(-3*exp(4)-3*x-27)*ln(3)-3*x*exp(4)-3*x**2-27*x),x)

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3.22.60 \(\int \frac {9+e^4+2 x+\log (3)}{-27 x-3 e^4 x-3 x^2+(-27-3 e^4-3 x) \log (3)+(9 x+e^4 x+x^2+(9+e^4+x) \log (3)) \log (9+e^4+x)+(-9 x-e^4 x-x^2+(-9-e^4-x) \log (3)) \log (\frac {5}{x+\log (3)})} \, dx\)