\(\int \frac {e^x (5+x)+(6+x+e^x (6+x)) \log (3+3 e^x)}{(5+x+e^x (5+x)) \log (3+3 e^x) \log (e^x (5+x) \log (3+3 e^x))} \, dx\) [6451]

Optimal result

Integrand size = 68, antiderivative size = 19 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (5 \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )\right ) \]

[Out]

ln(5*ln(exp(ln(5+x)+x)*ln(3*exp(x)+3)))

Rubi [F]

\[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx \]

[In]

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

[Out]

Defer[Int][Log[E^x*(5 + x)*Log[3*(1 + E^x)]]^(-1), x] + Defer[Int][1/((5 + x)*Log[E^x*(5 + x)*Log[3*(1 + E^x)]
]), x] + Defer[Int][1/(Log[3*(1 + E^x)]*Log[E^x*(5 + x)*Log[3*(1 + E^x)]]), x] - Defer[Int][1/((1 + E^x)*Log[3
*(1 + E^x)]*Log[E^x*(5 + x)*Log[3*(1 + E^x)]]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (1+e^x\right ) (5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx \\ & = \int \left (-\frac {1}{\left (1+e^x\right ) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}+\frac {5+x+6 \log \left (3 \left (1+e^x\right )\right )+x \log \left (3 \left (1+e^x\right )\right )}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}\right ) \, dx \\ & = -\int \frac {1}{\left (1+e^x\right ) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+\int \frac {5+x+6 \log \left (3 \left (1+e^x\right )\right )+x \log \left (3 \left (1+e^x\right )\right )}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx \\ & = -\int \frac {1}{\left (1+e^x\right ) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+\int \frac {5+x+(6+x) \log \left (3 \left (1+e^x\right )\right )}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx \\ & = \int \left (\frac {6}{(5+x) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}+\frac {x}{(5+x) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}+\frac {5}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}+\frac {x}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}\right ) \, dx-\int \frac {1}{\left (1+e^x\right ) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx \\ & = 5 \int \frac {1}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+6 \int \frac {1}{(5+x) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+\int \frac {x}{(5+x) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx-\int \frac {1}{\left (1+e^x\right ) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+\int \frac {x}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx \\ & = 5 \int \frac {1}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+6 \int \frac {1}{(5+x) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+\int \left (\frac {1}{\log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}-\frac {5}{(5+x) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}\right ) \, dx+\int \left (\frac {1}{\log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}-\frac {5}{(5+x) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )}\right ) \, dx-\int \frac {1}{\left (1+e^x\right ) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx \\ & = -\left (5 \int \frac {1}{(5+x) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx\right )+6 \int \frac {1}{(5+x) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+\int \frac {1}{\log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx+\int \frac {1}{\log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx-\int \frac {1}{\left (1+e^x\right ) \log \left (3 \left (1+e^x\right )\right ) \log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (\log \left (e^x (5+x) \log \left (3 \left (1+e^x\right )\right )\right )\right ) \]

[In]

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

[Out]

Log[Log[E^x*(5 + x)*Log[3*(1 + E^x)]]]

Maple [A] (verified)

Time = 49.67 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\ln \left (\ln \left ({\mathrm e}^{\ln \left (5+x \right )+x} \ln \left (3 \,{\mathrm e}^{x}+3\right )\right )\right )\)	\(18\)
risch	\(\ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (-\pi \,\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right )\right ) {\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )}^{2}+\pi \,\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right )\right ) \operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right ) \operatorname {csgn}\left (i \left (5+x \right )\right )+\pi {\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )}^{3}-\pi {\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )}^{2} \operatorname {csgn}\left (i \left (5+x \right )\right )-\pi \,\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )^{2}+\pi \,\operatorname {csgn}\left (i \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+\pi \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )^{3}-\pi \operatorname {csgn}\left (i {\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}+3\right ) \left (5+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+2 i \ln \left (5+x \right )+2 i \ln \left (\ln \left (3 \,{\mathrm e}^{x}+3\right )\right )\right )}{2}\right )\)	\(256\)

[In]

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

[Out]

ln(ln(exp(ln(5+x)+x)*ln(3*exp(x)+3)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (\log \left (e^{\left (x + \log \left (x + 5\right )\right )} \log \left (\frac {3 \, {\left (x + e^{\left (x + \log \left (x + 5\right )\right )} + 5\right )}}{x + 5}\right )\right )\right ) \]

[In]

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

[Out]

log(log(e^(x + log(x + 5))*log(3*(x + e^(x + log(x + 5)) + 5)/(x + 5))))

Sympy [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log {\left (\log {\left (\left (x + 5\right ) e^{x} \log {\left (3 e^{x} + 3 \right )} \right )} \right )} \]

[In]

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

[Out]

log(log((x + 5)*exp(x)*log(3*exp(x) + 3)))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (x + \log \left (x + 5\right ) + \log \left (\log \left (3\right ) + \log \left (e^{x} + 1\right )\right )\right ) \]

[In]

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

[Out]

log(x + log(x + 5) + log(log(3) + log(e^x + 1)))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\log \left (x + \log \left (x + 5\right ) + \log \left (\log \left (3\right ) + \log \left (e^{x} + 1\right )\right )\right ) \]

[In]

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

[Out]

log(x + log(x + 5) + log(log(3) + log(e^x + 1)))

Mupad [B] (verification not implemented)

Time = 11.77 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^x (5+x)+\left (6+x+e^x (6+x)\right ) \log \left (3+3 e^x\right )}{\left (5+x+e^x (5+x)\right ) \log \left (3+3 e^x\right ) \log \left (e^x (5+x) \log \left (3+3 e^x\right )\right )} \, dx=\ln \left (x+\ln \left (\ln \left (3\,{\mathrm {e}}^x+3\right )\,\left (x+5\right )\right )\right ) \]

[In]

int((exp(x)*(x + 5) + log(3*exp(x) + 3)*(x + exp(x)*(x + 6) + 6))/(log(3*exp(x) + 3)*log(log(3*exp(x) + 3)*exp
(x + log(x + 5)))*(x + exp(x)*(x + 5) + 5)),x)

[Out]

log(x + log(log(3*exp(x) + 3)*(x + 5)))