\(\int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx\) [5489]

Optimal result

Integrand size = 16, antiderivative size = 14 \[ \int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx=5 \log \left (-x+(1+x)^2+\log (3)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {642} \[ \int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx=5 \log \left (x^2+x+1+\log (3)\right ) \]

[In]

Int[(5 + 10*x)/(1 + x + x^2 + Log[3]),x]

[Out]

5*Log[1 + x + x^2 + Log[3]]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx=5 \log \left (1+x+x^2+\log (3)\right ) \]

[In]

Integrate[(5 + 10*x)/(1 + x + x^2 + Log[3]),x]

[Out]

5*Log[1 + x + x^2 + Log[3]]

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86

method	result	size
default	\(5 \ln \left (x^{2}+\ln \left (3\right )+x +1\right )\)	\(12\)
norman	\(5 \ln \left (x^{2}+\ln \left (3\right )+x +1\right )\)	\(12\)
risch	\(5 \ln \left (x^{2}+\ln \left (3\right )+x +1\right )\)	\(12\)
parallelrisch	\(5 \ln \left (x^{2}+\ln \left (3\right )+x +1\right )\)	\(12\)

[In]

int((10*x+5)/(x^2+ln(3)+x+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx=5 \, \log \left (x^{2} + x + \log \left (3\right ) + 1\right ) \]

[In]

integrate((10*x+5)/(x^2+log(3)+x+1),x, algorithm="fricas")

[Out]

5*log(x^2 + x + log(3) + 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx=5 \log {\left (x^{2} + x + 1 + \log {\left (3 \right )} \right )} \]

[In]

integrate((10*x+5)/(x**2+ln(3)+x+1),x)

[Out]

5*log(x**2 + x + 1 + log(3))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx=5 \, \log \left (x^{2} + x + \log \left (3\right ) + 1\right ) \]

[In]

integrate((10*x+5)/(x^2+log(3)+x+1),x, algorithm="maxima")

[Out]

5*log(x^2 + x + log(3) + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx=5 \, \log \left (x^{2} + x + \log \left (3\right ) + 1\right ) \]

[In]

integrate((10*x+5)/(x^2+log(3)+x+1),x, algorithm="giac")

[Out]

5*log(x^2 + x + log(3) + 1)

Mupad [B] (verification not implemented)

Time = 11.60 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {5+10 x}{1+x+x^2+\log (3)} \, dx=5\,\ln \left (x^2+x+\ln \left (3\right )+1\right ) \]

[In]

int((10*x + 5)/(x + log(3) + x^2 + 1),x)

[Out]

5*log(x + log(3) + x^2 + 1)