\(\int \frac {-3-2 x+3 x^2+\log (\frac {3}{\log (5)})}{-3 x-x^2+x^3+x \log (\frac {3}{\log (5)})} \, dx\) [1732]

Optimal result

Integrand size = 41, antiderivative size = 20 \[ \int \frac {-3-2 x+3 x^2+\log \left (\frac {3}{\log (5)}\right )}{-3 x-x^2+x^3+x \log \left (\frac {3}{\log (5)}\right )} \, dx=\log \left (x \left (-3-(1-x) x+\log \left (\frac {3}{\log (5)}\right )\right )\right ) \]

[Out]

ln((ln(3/ln(5))-3-x*(1-x))*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6, 1601} \[ \int \frac {-3-2 x+3 x^2+\log \left (\frac {3}{\log (5)}\right )}{-3 x-x^2+x^3+x \log \left (\frac {3}{\log (5)}\right )} \, dx=\log \left (-x^3+x^2+x \left (3-\log \left (\frac {3}{\log (5)}\right )\right )\right ) \]

[In]

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

[Out]

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

Rule 6

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

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-3-2 x+3 x^2+\log \left (\frac {3}{\log (5)}\right )}{-3 x-x^2+x^3+x \log \left (\frac {3}{\log (5)}\right )} \, dx=\log (x)+\log \left (3+x-x^2-\log \left (\frac {3}{\log (5)}\right )\right ) \]

[In]

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

[Out]

Log[x] + Log[3 + x - x^2 - Log[3/Log[5]]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
default	\(\ln \left (x \left (x^{2}+\ln \left (\frac {3}{\ln \left (5\right )}\right )-x -3\right )\right )\)	\(19\)
norman	\(\ln \left (x \right )+\ln \left (x^{2}+\ln \left (\frac {3}{\ln \left (5\right )}\right )-x -3\right )\)	\(20\)
parallelrisch	\(\ln \left (x \right )+\ln \left (x^{2}+\ln \left (\frac {3}{\ln \left (5\right )}\right )-x -3\right )\)	\(20\)
risch	\(\ln \left (x^{3}-x^{2}+\left (\ln \left (3\right )-\ln \left (\ln \left (5\right )\right )-3\right ) x \right )\)	\(22\)
derivativedivides	\(\ln \left (x \ln \left (\frac {3}{\ln \left (5\right )}\right )+x^{3}-x^{2}-3 x \right )\)	\(23\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-3-2 x+3 x^2+\log \left (\frac {3}{\log (5)}\right )}{-3 x-x^2+x^3+x \log \left (\frac {3}{\log (5)}\right )} \, dx=\log \left (x^{3} - x^{2} + x \log \left (\frac {3}{\log \left (5\right )}\right ) - 3 \, x\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-3-2 x+3 x^2+\log \left (\frac {3}{\log (5)}\right )}{-3 x-x^2+x^3+x \log \left (\frac {3}{\log (5)}\right )} \, dx=\log {\left (x^{3} - x^{2} + x \left (-3 - \log {\left (\log {\left (5 \right )} \right )} + \log {\left (3 \right )}\right ) \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-3-2 x+3 x^2+\log \left (\frac {3}{\log (5)}\right )}{-3 x-x^2+x^3+x \log \left (\frac {3}{\log (5)}\right )} \, dx=\log \left (x^{3} - x^{2} + x \log \left (\frac {3}{\log \left (5\right )}\right ) - 3 \, x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-3-2 x+3 x^2+\log \left (\frac {3}{\log (5)}\right )}{-3 x-x^2+x^3+x \log \left (\frac {3}{\log (5)}\right )} \, dx=\log \left ({\left | x^{3} - x^{2} + x \log \left (\frac {3}{\log \left (5\right )}\right ) - 3 \, x \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-3-2 x+3 x^2+\log \left (\frac {3}{\log (5)}\right )}{-3 x-x^2+x^3+x \log \left (\frac {3}{\log (5)}\right )} \, dx=\ln \left (x^3-x^2+\left (\ln \left (3\right )-\ln \left (\ln \left (5\right )\right )-3\right )\,x\right ) \]

[In]

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

[Out]

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