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

Optimal. Leaf size=22 \[ -\frac {2}{(3+2 x) \log (5)}+\log \left (4+e^{e^4}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 0.59, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 27, 32} \begin {gather*} -\frac {2}{(2 x+3) \log (5)} \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.59 \begin {gather*} -\frac {2}{(3+2 x) \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2/((3 + 2*x)*Log[5])

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fricas [A]  time = 0.43, size = 13, normalized size = 0.59 \begin {gather*} -\frac {2}{{\left (2 \, x + 3\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/((2*x + 3)*log(5))

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giac [A]  time = 0.22, size = 13, normalized size = 0.59 \begin {gather*} -\frac {2}{{\left (2 \, x + 3\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/((2*x + 3)*log(5))

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maple [A]  time = 0.20, size = 12, normalized size = 0.55

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 13, normalized size = 0.59 \begin {gather*} -\frac {2}{{\left (2 \, x + 3\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/((2*x + 3)*log(5))

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mupad [B]  time = 0.04, size = 13, normalized size = 0.59 \begin {gather*} -\frac {2}{\ln \relax (5)\,\left (2\,x+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4/(log(5)*(12*x + 4*x^2 + 9)),x)

[Out]

-2/(log(5)*(2*x + 3))

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sympy [A]  time = 0.11, size = 14, normalized size = 0.64 \begin {gather*} - \frac {4}{4 x \log {\relax (5 )} + 6 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(4*x*log(5) + 6*log(5))

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