\(\int \frac {3+x+5 x^2+(30 x+10 x^2) \log (6+2 x)}{3 x+x^2+(15 x^2+5 x^3) \log (6+2 x)} \, dx\) [29]

Optimal result

Integrand size = 52, antiderivative size = 16 \[ \int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{3 x+x^2+\left (15 x^2+5 x^3\right ) \log (6+2 x)} \, dx=\log \left (-3 x-15 x^2 \log (2 (3+x))\right ) \]

[Out]

ln(-3*x-15*x^2*ln(2*x+6))

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6873, 6817} \[ \int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{3 x+x^2+\left (15 x^2+5 x^3\right ) \log (6+2 x)} \, dx=\log (x (5 x \log (2 (x+3))+1)) \]

[In]

Int[(3 + x + 5*x^2 + (30*x + 10*x^2)*Log[6 + 2*x])/(3*x + x^2 + (15*x^2 + 5*x^3)*Log[6 + 2*x]),x]

[Out]

Log[x*(1 + 5*x*Log[2*(3 + x)])]

Rule 6817

Int[(u_)/((w_)*(y_)), x_Symbol] :> With[{q = DerivativeDivides[y*w, u, x]}, Simp[q*Log[RemoveContent[y*w, x]],
x] /;  !FalseQ[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 {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{x (3+x) (1+5 x \log (2 (3+x)))} \, dx \\ & = \log (x (1+5 x \log (2 (3+x)))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{3 x+x^2+\left (15 x^2+5 x^3\right ) \log (6+2 x)} \, dx=\log (x)+\log (1+5 x \log (2 (3+x))) \]

[In]

Integrate[(3 + x + 5*x^2 + (30*x + 10*x^2)*Log[6 + 2*x])/(3*x + x^2 + (15*x^2 + 5*x^3)*Log[6 + 2*x]),x]

[Out]

Log[x] + Log[1 + 5*x*Log[2*(3 + x)]]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(\ln \left (x \right )+\ln \left (x \ln \left (2 x +6\right )+\frac {1}{5}\right )\)	\(15\)
norman	\(\ln \left (x \right )+\ln \left (5 x \ln \left (2 x +6\right )+1\right )\)	\(16\)
risch	\(2 \ln \left (x \right )+\ln \left (\ln \left (2 x +6\right )+\frac {1}{5 x}\right )\)	\(19\)
derivativedivides	\(\ln \left (5 \ln \left (2 x +6\right ) \left (2 x +6\right )^{2}-60 \ln \left (2 x +6\right ) \left (2 x +6\right )+180 \ln \left (2 x +6\right )+4 x \right )\)	\(42\)
default	\(\ln \left (5 \ln \left (2 x +6\right ) \left (2 x +6\right )^{2}-60 \ln \left (2 x +6\right ) \left (2 x +6\right )+180 \ln \left (2 x +6\right )+4 x \right )\)	\(42\)

[In]

int(((10*x^2+30*x)*ln(2*x+6)+5*x^2+x+3)/((5*x^3+15*x^2)*ln(2*x+6)+x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(x*ln(2*x+6)+1/5)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{3 x+x^2+\left (15 x^2+5 x^3\right ) \log (6+2 x)} \, dx=2 \, \log \left (x\right ) + \log \left (\frac {5 \, x \log \left (2 \, x + 6\right ) + 1}{x}\right ) \]

[In]

integrate(((10*x^2+30*x)*log(2*x+6)+5*x^2+x+3)/((5*x^3+15*x^2)*log(2*x+6)+x^2+3*x),x, algorithm="fricas")

[Out]

2*log(x) + log((5*x*log(2*x + 6) + 1)/x)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{3 x+x^2+\left (15 x^2+5 x^3\right ) \log (6+2 x)} \, dx=2 \log {\left (x \right )} + \log {\left (\log {\left (2 x + 6 \right )} + \frac {1}{5 x} \right )} \]

[In]

integrate(((10*x**2+30*x)*ln(2*x+6)+5*x**2+x+3)/((5*x**3+15*x**2)*ln(2*x+6)+x**2+3*x),x)

[Out]

2*log(x) + log(log(2*x + 6) + 1/(5*x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{3 x+x^2+\left (15 x^2+5 x^3\right ) \log (6+2 x)} \, dx=2 \, \log \left (x\right ) + \log \left (\frac {5 \, x \log \left (2\right ) + 5 \, x \log \left (x + 3\right ) + 1}{5 \, x}\right ) \]

[In]

integrate(((10*x^2+30*x)*log(2*x+6)+5*x^2+x+3)/((5*x^3+15*x^2)*log(2*x+6)+x^2+3*x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{3 x+x^2+\left (15 x^2+5 x^3\right ) \log (6+2 x)} \, dx=\log \left (5 \, x \log \left (2 \, x + 6\right ) + 1\right ) + \log \left (x\right ) \]

[In]

integrate(((10*x^2+30*x)*log(2*x+6)+5*x^2+x+3)/((5*x^3+15*x^2)*log(2*x+6)+x^2+3*x),x, algorithm="giac")

[Out]

log(5*x*log(2*x + 6) + 1) + log(x)

Mupad [B] (verification not implemented)

Time = 7.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {3+x+5 x^2+\left (30 x+10 x^2\right ) \log (6+2 x)}{3 x+x^2+\left (15 x^2+5 x^3\right ) \log (6+2 x)} \, dx=\ln \left (5\,x\,\ln \left (2\,x+6\right )+1\right )+\ln \left (x\right ) \]

[In]

int((x + log(2*x + 6)*(30*x + 10*x^2) + 5*x^2 + 3)/(3*x + log(2*x + 6)*(15*x^2 + 5*x^3) + x^2),x)

[Out]

log(5*x*log(2*x + 6) + 1) + log(x)