\(\int \frac {4}{(9+12 x+4 x^2) \log (5)} \, dx\) [5162]

Optimal result

Integrand size = 18, antiderivative size = 22 \[ \int \frac {4}{\left (9+12 x+4 x^2\right ) \log (5)} \, dx=-\frac {2}{(3+2 x) \log (5)}+\log \left (4+e^{e^4}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 27, 32} \[ \int \frac {4}{\left (9+12 x+4 x^2\right ) \log (5)} \, dx=-\frac {2}{(2 x+3) \log (5)} \]

[In]

Int[4/((9 + 12*x + 4*x^2)*Log[5]),x]

[Out]

-2/((3 + 2*x)*Log[5])

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {4 \int \frac {1}{9+12 x+4 x^2} \, dx}{\log (5)} \\ & = \frac {4 \int \frac {1}{(3+2 x)^2} \, dx}{\log (5)} \\ & = -\frac {2}{(3+2 x) \log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {4}{\left (9+12 x+4 x^2\right ) \log (5)} \, dx=-\frac {2}{(3+2 x) \log (5)} \]

[In]

Integrate[4/((9 + 12*x + 4*x^2)*Log[5]),x]

[Out]

-2/((3 + 2*x)*Log[5])

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.55

method	result	size
risch	\(-\frac {1}{\ln \left (5\right ) \left (x +\frac {3}{2}\right )}\)	\(12\)
gosper	\(-\frac {2}{\left (3+2 x \right ) \ln \left (5\right )}\)	\(14\)
default	\(-\frac {2}{\left (3+2 x \right ) \ln \left (5\right )}\)	\(14\)
norman	\(-\frac {2}{\left (3+2 x \right ) \ln \left (5\right )}\)	\(14\)
parallelrisch	\(-\frac {2}{\left (3+2 x \right ) \ln \left (5\right )}\)	\(14\)
meijerg	\(\frac {4 x}{9 \ln \left (5\right ) \left (1+\frac {2 x}{3}\right )}\)	\(15\)

[In]

int(4/(4*x^2+12*x+9)/ln(5),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {4}{\left (9+12 x+4 x^2\right ) \log (5)} \, dx=-\frac {2}{{\left (2 \, x + 3\right )} \log \left (5\right )} \]

[In]

integrate(4/(4*x^2+12*x+9)/log(5),x, algorithm="fricas")

[Out]

-2/((2*x + 3)*log(5))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {4}{\left (9+12 x+4 x^2\right ) \log (5)} \, dx=- \frac {4}{4 x \log {\left (5 \right )} + 6 \log {\left (5 \right )}} \]

[In]

integrate(4/(4*x**2+12*x+9)/ln(5),x)

[Out]

-4/(4*x*log(5) + 6*log(5))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {4}{\left (9+12 x+4 x^2\right ) \log (5)} \, dx=-\frac {2}{{\left (2 \, x + 3\right )} \log \left (5\right )} \]

[In]

integrate(4/(4*x^2+12*x+9)/log(5),x, algorithm="maxima")

[Out]

-2/((2*x + 3)*log(5))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {4}{\left (9+12 x+4 x^2\right ) \log (5)} \, dx=-\frac {2}{{\left (2 \, x + 3\right )} \log \left (5\right )} \]

[In]

integrate(4/(4*x^2+12*x+9)/log(5),x, algorithm="giac")

[Out]

-2/((2*x + 3)*log(5))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {4}{\left (9+12 x+4 x^2\right ) \log (5)} \, dx=-\frac {2}{\ln \left (5\right )\,\left (2\,x+3\right )} \]

[In]

int(4/(log(5)*(12*x + 4*x^2 + 9)),x)

[Out]

-2/(log(5)*(2*x + 3))