\(\int \frac {18+5 x+(6+2 x) \log (\frac {1}{3+x})+\log (x) (-9-3 x+(-3-x) \log (\frac {1}{3+x}))+(-9-3 x+(-3-x) \log (\frac {1}{3+x})) \log (3 x+x \log (\frac {1}{3+x}))}{9 x^2+3 x^3+(3 x^2+x^3) \log (\frac {1}{3+x})} \, dx\) [8379]

Optimal result

Integrand size = 98, antiderivative size = 19 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {x+\log (x)+\log \left (x \left (3+\log \left (\frac {1}{3+x}\right )\right )\right )}{x} \]

[Out]

(ln((ln(1/(3+x))+3)*x)+x+ln(x))/x

Rubi [F]

\[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx \]

[In]

Int[(18 + 5*x + (6 + 2*x)*Log[(3 + x)^(-1)] + Log[x]*(-9 - 3*x + (-3 - x)*Log[(3 + x)^(-1)]) + (-9 - 3*x + (-3
- x)*Log[(3 + x)^(-1)])*Log[3*x + x*Log[(3 + x)^(-1)]])/(9*x^2 + 3*x^3 + (3*x^2 + x^3)*Log[(3 + x)^(-1)]),x]

[Out]

x^(-1) - (2 - Log[x])/x - Log[3 + Log[(3 + x)^(-1)]]/3 - Defer[Int][1/(x*(3 + Log[(3 + x)^(-1)])), x]/3 - Defe
r[Int][Log[3*x + x*Log[(3 + x)^(-1)]]/x^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2 (3+x) \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx \\ & = \int \left (\frac {18+5 x-9 \log (x)-3 x \log (x)+6 \log \left (\frac {1}{3+x}\right )+2 x \log \left (\frac {1}{3+x}\right )-3 \log (x) \log \left (\frac {1}{3+x}\right )-x \log (x) \log \left (\frac {1}{3+x}\right )}{x^2 (3+x) \left (3+\log \left (\frac {1}{3+x}\right )\right )}-\frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2}\right ) \, dx \\ & = \int \frac {18+5 x-9 \log (x)-3 x \log (x)+6 \log \left (\frac {1}{3+x}\right )+2 x \log \left (\frac {1}{3+x}\right )-3 \log (x) \log \left (\frac {1}{3+x}\right )-x \log (x) \log \left (\frac {1}{3+x}\right )}{x^2 (3+x) \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ & = \int \frac {18+5 x+2 (3+x) \log \left (\frac {1}{3+x}\right )-(3+x) \log (x) \left (3+\log \left (\frac {1}{3+x}\right )\right )}{x^2 (3+x) \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ & = \int \left (\frac {2-\log (x)}{x^2}-\frac {1}{x (3+x) \left (3+\log \left (\frac {1}{3+x}\right )\right )}\right ) \, dx-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ & = \int \frac {2-\log (x)}{x^2} \, dx-\int \frac {1}{x (3+x) \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ & = \frac {1}{x}-\frac {2-\log (x)}{x}-\int \left (\frac {1}{3 x \left (3+\log \left (\frac {1}{3+x}\right )\right )}-\frac {1}{3 (3+x) \left (3+\log \left (\frac {1}{3+x}\right )\right )}\right ) \, dx-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ & = \frac {1}{x}-\frac {2-\log (x)}{x}-\frac {1}{3} \int \frac {1}{x \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx+\frac {1}{3} \int \frac {1}{(3+x) \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ & = \frac {1}{x}-\frac {2-\log (x)}{x}-\frac {1}{3} \int \frac {1}{x \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx+\frac {1}{3} \text {Subst}\left (\int \frac {1}{x \left (3+\log \left (\frac {1}{x}\right )\right )} \, dx,x,3+x\right )-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ & = \frac {1}{x}-\frac {2-\log (x)}{x}-\frac {1}{3} \int \frac {1}{x \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx-\frac {1}{3} \text {Subst}\left (\int \frac {1}{x} \, dx,x,3+\log \left (\frac {1}{3+x}\right )\right )-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ & = \frac {1}{x}-\frac {2-\log (x)}{x}-\frac {1}{3} \log \left (3+\log \left (\frac {1}{3+x}\right )\right )-\frac {1}{3} \int \frac {1}{x \left (3+\log \left (\frac {1}{3+x}\right )\right )} \, dx-\int \frac {\log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\log (x)}{x}+\frac {\log \left (x \left (3+\log \left (\frac {1}{3+x}\right )\right )\right )}{x} \]

[In]

Integrate[(18 + 5*x + (6 + 2*x)*Log[(3 + x)^(-1)] + Log[x]*(-9 - 3*x + (-3 - x)*Log[(3 + x)^(-1)]) + (-9 - 3*x
+ (-3 - x)*Log[(3 + x)^(-1)])*Log[3*x + x*Log[(3 + x)^(-1)]])/(9*x^2 + 3*x^3 + (3*x^2 + x^3)*Log[(3 + x)^(-1)
]),x]

[Out]

Log[x]/x + Log[x*(3 + Log[(3 + x)^(-1)])]/x

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(54\) vs. \(2(19)=38\).

Time = 2.80 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.89

method	result	size
parallelrisch	\(\frac {27 \ln \left (\ln \left (\frac {1}{3+x}\right )+3\right ) x +27 x \ln \left (x \right )-27 \ln \left (\left (\ln \left (\frac {1}{3+x}\right )+3\right ) x \right ) x +162 \ln \left (x \right )+162 \ln \left (\left (\ln \left (\frac {1}{3+x}\right )+3\right ) x \right )}{162 x}\)	\(55\)
risch	\(\frac {\ln \left (\ln \left (3+x \right )-3\right )}{x}+\frac {-2 i \pi \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\ln \left (3+x \right )-3\right )\right ) \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )^{2}+i \pi \,\operatorname {csgn}\left (i \left (\ln \left (3+x \right )-3\right )\right ) \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )^{2}+i \pi \operatorname {csgn}\left (i x \left (\ln \left (3+x \right )-3\right )\right )^{3}+2 i \pi +4 \ln \left (x \right )}{2 x}\)	\(140\)

[In]

int((((-3-x)*ln(1/(3+x))-3*x-9)*ln(x*ln(1/(3+x))+3*x)+((-3-x)*ln(1/(3+x))-3*x-9)*ln(x)+(2*x+6)*ln(1/(3+x))+5*x
+18)/((x^3+3*x^2)*ln(1/(3+x))+3*x^3+9*x^2),x,method=_RETURNVERBOSE)

[Out]

1/162/x*(27*ln(ln(1/(3+x))+3)*x+27*x*ln(x)-27*ln((ln(1/(3+x))+3)*x)*x+162*ln(x)+162*ln((ln(1/(3+x))+3)*x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\log \left (x \log \left (\frac {1}{x + 3}\right ) + 3 \, x\right ) + \log \left (x\right )}{x} \]

[In]

integrate((((-3-x)*log(1/(3+x))-3*x-9)*log(x*log(1/(3+x))+3*x)+((-3-x)*log(1/(3+x))-3*x-9)*log(x)+(2*x+6)*log(
1/(3+x))+5*x+18)/((x^3+3*x^2)*log(1/(3+x))+3*x^3+9*x^2),x, algorithm="fricas")

[Out]

(log(x*log(1/(x + 3)) + 3*x) + log(x))/x

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\log {\left (x \right )}}{x} + \frac {\log {\left (x \log {\left (\frac {1}{x + 3} \right )} + 3 x \right )}}{x} \]

[In]

integrate((((-3-x)*ln(1/(3+x))-3*x-9)*ln(x*ln(1/(3+x))+3*x)+((-3-x)*ln(1/(3+x))-3*x-9)*ln(x)+(2*x+6)*ln(1/(3+x
))+5*x+18)/((x**3+3*x**2)*ln(1/(3+x))+3*x**3+9*x**2),x)

[Out]

log(x)/x + log(x*log(1/(x + 3)) + 3*x)/x

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {2 \, \log \left (x\right ) + \log \left (-\log \left (x + 3\right ) + 3\right )}{x} \]

[In]

integrate((((-3-x)*log(1/(3+x))-3*x-9)*log(x*log(1/(3+x))+3*x)+((-3-x)*log(1/(3+x))-3*x-9)*log(x)+(2*x+6)*log(
1/(3+x))+5*x+18)/((x^3+3*x^2)*log(1/(3+x))+3*x^3+9*x^2),x, algorithm="maxima")

[Out]

(2*log(x) + log(-log(x + 3) + 3))/x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {2 \, \log \left (x\right )}{x} + \frac {\log \left (-\log \left (x + 3\right ) + 3\right )}{x} \]

[In]

integrate((((-3-x)*log(1/(3+x))-3*x-9)*log(x*log(1/(3+x))+3*x)+((-3-x)*log(1/(3+x))-3*x-9)*log(x)+(2*x+6)*log(
1/(3+x))+5*x+18)/((x^3+3*x^2)*log(1/(3+x))+3*x^3+9*x^2),x, algorithm="giac")

[Out]

2*log(x)/x + log(-log(x + 3) + 3)/x

Mupad [B] (verification not implemented)

Time = 13.72 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {18+5 x+(6+2 x) \log \left (\frac {1}{3+x}\right )+\log (x) \left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right )+\left (-9-3 x+(-3-x) \log \left (\frac {1}{3+x}\right )\right ) \log \left (3 x+x \log \left (\frac {1}{3+x}\right )\right )}{9 x^2+3 x^3+\left (3 x^2+x^3\right ) \log \left (\frac {1}{3+x}\right )} \, dx=\frac {\ln \left (3\,x+x\,\ln \left (\frac {1}{x+3}\right )\right )+\ln \left (x\right )}{x} \]

[In]

int((5*x + log(1/(x + 3))*(2*x + 6) - log(x)*(3*x + log(1/(x + 3))*(x + 3) + 9) - log(3*x + x*log(1/(x + 3)))*
(3*x + log(1/(x + 3))*(x + 3) + 9) + 18)/(log(1/(x + 3))*(3*x^2 + x^3) + 9*x^2 + 3*x^3),x)

[Out]

(log(3*x + x*log(1/(x + 3))) + log(x))/x