\(\int \frac {2 x+3 x^2+3 \log (5)+(-x-3 x^2-\log (5)) \log (\frac {x+3 x^2+\log (5)}{3 x})}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+(x^2+3 x^3+x \log (5)) \log (\frac {x+3 x^2+\log (5)}{3 x})} \, dx\) [4372]

Optimal result

Integrand size = 100, antiderivative size = 37 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=4+\log \left (\frac {x}{2 \left (-x+x^2 \left (-2+\log \left (\frac {x^2+\frac {1}{3} (x+\log (5))}{x}\right )\right )\right )}\right ) \]

[Out]

ln(1/2*x/((ln((x^2+1/3*ln(5)+1/3*x)/x)-2)*x^2-x))+4

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6873, 6816} \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\log \left (2 x+x \left (-\log \left (x+\frac {\log (5)}{3 x}+\frac {1}{3}\right )\right )+1\right ) \]

[In]

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

[Out]

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

Rule 6816

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

Rule 6873

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

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

Mathematica [F]

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

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70

method	result	size
norman	\(-\ln \left (\ln \left (\frac {\ln \left (5\right )+3 x^{2}+x}{3 x}\right ) x -2 x -1\right )\)	\(26\)
parallelrisch	\(-\ln \left (\ln \left (\frac {\ln \left (5\right )+3 x^{2}+x}{3 x}\right ) x -2 x -1\right )\)	\(26\)
risch	\(-\ln \left (x \right )-\ln \left (\ln \left (\frac {\ln \left (5\right )+3 x^{2}+x}{3 x}\right )-\frac {1+2 x}{x}\right )\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\log \left (x\right ) - \log \left (\frac {x \log \left (\frac {3 \, x^{2} + x + \log \left (5\right )}{3 \, x}\right ) - 2 \, x - 1}{x}\right ) \]

[In]

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

[Out]

-log(x) - log((x*log(1/3*(3*x^2 + x + log(5))/x) - 2*x - 1)/x)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=- \log {\left (x \right )} - \log {\left (\log {\left (\frac {x^{2} + \frac {x}{3} + \frac {\log {\left (5 \right )}}{3}}{x} \right )} + \frac {- 2 x - 1}{x} \right )} \]

[In]

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

[Out]

-log(x) - log(log((x**2 + x/3 + log(5)/3)/x) + (-2*x - 1)/x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\log \left (x\right ) - \log \left (-\frac {x {\left (\log \left (3\right ) + 2\right )} - x \log \left (3 \, x^{2} + x + \log \left (5\right )\right ) + x \log \left (x\right ) + 1}{x}\right ) \]

[In]

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

[Out]

-log(x) - log(-(x*(log(3) + 2) - x*log(3*x^2 + x + log(5)) + x*log(x) + 1)/x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\log \left (-x \log \left (3 \, x^{2} + x + \log \left (5\right )\right ) + x \log \left (3 \, x\right ) + 2 \, x + 1\right ) \]

[In]

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

[Out]

-log(-x*log(3*x^2 + x + log(5)) + x*log(3*x) + 2*x + 1)

Mupad [B] (verification not implemented)

Time = 15.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {2 x+3 x^2+3 \log (5)+\left (-x-3 x^2-\log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )}{-x-5 x^2-6 x^3+(-1-2 x) \log (5)+\left (x^2+3 x^3+x \log (5)\right ) \log \left (\frac {x+3 x^2+\log (5)}{3 x}\right )} \, dx=-\ln \left (\frac {2\,x-x\,\ln \left (\frac {3\,x^2+x+\ln \left (5\right )}{3\,x}\right )+1}{x}\right )-\ln \left (x\right ) \]

[In]

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

[Out]

- log((2*x - x*log((x + log(5) + 3*x^2)/(3*x)) + 1)/x) - log(x)