\(\int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx\) [5665]

Optimal result

Integrand size = 17, antiderivative size = 34 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=5-4 (2-x)+2 \left (\log (3)-x \left (-x+\frac {x^2}{x-x \log (3)}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12} \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {4 x}{1-\log (3)}-\frac {2 (x+1)^2 \log (3)}{1-\log (3)} \]

[In]

Int[(-4 + (4 + 4*x)*Log[3])/(-1 + Log[3]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (-4+(4+4 x) \log (3)) \, dx}{-1+\log (3)} \\ & = \frac {4 x}{1-\log (3)}-\frac {2 (1+x)^2 \log (3)}{1-\log (3)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {4 \left (-x+x \log (3)+\frac {1}{2} x^2 \log (3)\right )}{-1+\log (3)} \]

[In]

Integrate[(-4 + (4 + 4*x)*Log[3])/(-1 + Log[3]),x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53

method	result	size
norman	\(4 x +\frac {2 \ln \left (3\right ) x^{2}}{\ln \left (3\right )-1}\)	\(18\)
gosper	\(\frac {2 x \left (x \ln \left (3\right )+2 \ln \left (3\right )-2\right )}{\ln \left (3\right )-1}\)	\(20\)
default	\(\frac {2 x^{2} \ln \left (3\right )+4 x \ln \left (3\right )-4 x}{\ln \left (3\right )-1}\)	\(24\)
parallelrisch	\(\frac {\ln \left (3\right ) \left (2 x^{2}+4 x \right )-4 x}{\ln \left (3\right )-1}\)	\(24\)
risch	\(\frac {2 \ln \left (3\right ) x^{2}}{\ln \left (3\right )-1}+\frac {4 x \ln \left (3\right )}{\ln \left (3\right )-1}-\frac {4 x}{\ln \left (3\right )-1}\)	\(35\)

[In]

int(((4+4*x)*ln(3)-4)/(ln(3)-1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \left (3\right ) - 2 \, x\right )}}{\log \left (3\right ) - 1} \]

[In]

integrate(((4+4*x)*log(3)-4)/(log(3)-1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.44 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {2 x^{2} \log {\left (3 \right )}}{-1 + \log {\left (3 \right )}} + 4 x \]

[In]

integrate(((4+4*x)*ln(3)-4)/(ln(3)-1),x)

[Out]

2*x**2*log(3)/(-1 + log(3)) + 4*x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \left (3\right ) - 2 \, x\right )}}{\log \left (3\right ) - 1} \]

[In]

integrate(((4+4*x)*log(3)-4)/(log(3)-1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \left (3\right ) - 2 \, x\right )}}{\log \left (3\right ) - 1} \]

[In]

integrate(((4+4*x)*log(3)-4)/(log(3)-1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {{\left (\ln \left (3\right )\,\left (4\,x+4\right )-4\right )}^2}{4\,\ln \left (3\right )\,\left (\ln \left (9\right )-2\right )} \]

[In]

int((log(3)*(4*x + 4) - 4)/(log(3) - 1),x)

[Out]

(log(3)*(4*x + 4) - 4)^2/(4*log(3)*(log(9) - 2))