\(\int 12 \log (\log (4+e^5)) \, dx\) [6352]

Optimal result

Integrand size = 9, antiderivative size = 19 \[ \int 12 \log \left (\log \left (4+e^5\right )\right ) \, dx=3 \left (3+4 e^5+4 x \log \left (\log \left (4+e^5\right )\right )\right ) \]

[Out]

9+12*ln(ln(4+exp(5)))*x+3*exp(2*ln(2)+5)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {8} \[ \int 12 \log \left (\log \left (4+e^5\right )\right ) \, dx=12 x \log \left (\log \left (4+e^5\right )\right ) \]

[In]

Int[12*Log[Log[4 + E^5]],x]

[Out]

12*x*Log[Log[4 + E^5]]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = 12 x \log \left (\log \left (4+e^5\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int 12 \log \left (\log \left (4+e^5\right )\right ) \, dx=12 x \log \left (\log \left (4+e^5\right )\right ) \]

[In]

Integrate[12*Log[Log[4 + E^5]],x]

[Out]

12*x*Log[Log[4 + E^5]]

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53

method	result	size
default	\(12 \ln \left (\ln \left (4+{\mathrm e}^{5}\right )\right ) x\)	\(10\)
norman	\(12 \ln \left (\ln \left (4+{\mathrm e}^{5}\right )\right ) x\)	\(10\)
risch	\(12 \ln \left (\ln \left (4+{\mathrm e}^{5}\right )\right ) x\)	\(10\)
parallelrisch	\(12 \ln \left (\ln \left (4+{\mathrm e}^{5}\right )\right ) x\)	\(10\)

[In]

int(12*ln(ln(4+exp(5))),x,method=_RETURNVERBOSE)

[Out]

12*ln(ln(4+exp(5)))*x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.47 \[ \int 12 \log \left (\log \left (4+e^5\right )\right ) \, dx=12 \, x \log \left (\log \left (e^{5} + 4\right )\right ) \]

[In]

integrate(12*log(log(4+exp(5))),x, algorithm="fricas")

[Out]

12*x*log(log(e^5 + 4))

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int 12 \log \left (\log \left (4+e^5\right )\right ) \, dx=12 x \log {\left (\log {\left (4 + e^{5} \right )} \right )} \]

[In]

integrate(12*ln(ln(4+exp(5))),x)

[Out]

12*x*log(log(4 + exp(5)))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.47 \[ \int 12 \log \left (\log \left (4+e^5\right )\right ) \, dx=12 \, x \log \left (\log \left (e^{5} + 4\right )\right ) \]

[In]

integrate(12*log(log(4+exp(5))),x, algorithm="maxima")

[Out]

12*x*log(log(e^5 + 4))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.47 \[ \int 12 \log \left (\log \left (4+e^5\right )\right ) \, dx=12 \, x \log \left (\log \left (e^{5} + 4\right )\right ) \]

[In]

integrate(12*log(log(4+exp(5))),x, algorithm="giac")

[Out]

12*x*log(log(e^5 + 4))

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.47 \[ \int 12 \log \left (\log \left (4+e^5\right )\right ) \, dx=12\,x\,\ln \left (\ln \left ({\mathrm {e}}^5+4\right )\right ) \]

[In]

int(12*log(log(exp(5) + 4)),x)

[Out]

12*x*log(log(exp(5) + 4))