3.11.68 \(\int \frac {9+3 \log (4)}{(144+24 x+x^2) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx\)

Optimal. Leaf size=25 \[ 1-\frac {15}{\log (2) \left (25-\frac {5 (3-x)}{3+\log (4)}\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 0.76, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 1981, 27, 32} \begin {gather*} -\frac {3 (3+\log (4))}{\log (2) (x+12+5 \log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9 + 3*Log[4])/((144 + 24*x + x^2)*Log[2] + (120 + 10*x)*Log[2]*Log[4] + 25*Log[2]*Log[4]^2),x]

[Out]

(-3*(3 + Log[4]))/(Log[2]*(12 + x + 5*Log[4]))

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]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=(9+3 \log (4)) \int \frac {1}{\left (144+24 x+x^2\right ) \log (2)+(120+10 x) \log (2) \log (4)+25 \log (2) \log ^2(4)} \, dx\\ &=(9+3 \log (4)) \int \frac {1}{x^2 \log (2)+2 x \log (2) (12+5 \log (4))+\log (2) (12+5 \log (4))^2} \, dx\\ &=(9+3 \log (4)) \int \frac {1}{\log (2) (12+x+5 \log (4))^2} \, dx\\ &=\frac {(9+3 \log (4)) \int \frac {1}{(12+x+5 \log (4))^2} \, dx}{\log (2)}\\ &=-\frac {3 (3+\log (4))}{\log (2) (12+x+5 \log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.76 \begin {gather*} -\frac {3 (3+\log (4))}{\log (2) (12+x+5 \log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 + 3*Log[4])/((144 + 24*x + x^2)*Log[2] + (120 + 10*x)*Log[2]*Log[4] + 25*Log[2]*Log[4]^2),x]

[Out]

(-3*(3 + Log[4]))/(Log[2]*(12 + x + 5*Log[4]))

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fricas [A]  time = 0.59, size = 23, normalized size = 0.92 \begin {gather*} -\frac {3 \, {\left (2 \, \log \relax (2) + 3\right )}}{{\left (x + 12\right )} \log \relax (2) + 10 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(2)+9)/(100*log(2)^3+2*(10*x+120)*log(2)^2+(x^2+24*x+144)*log(2)),x, algorithm="fricas")

[Out]

-3*(2*log(2) + 3)/((x + 12)*log(2) + 10*log(2)^2)

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giac [A]  time = 0.33, size = 21, normalized size = 0.84 \begin {gather*} -\frac {3 \, {\left (2 \, \log \relax (2) + 3\right )}}{{\left (x + 10 \, \log \relax (2) + 12\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(2)+9)/(100*log(2)^3+2*(10*x+120)*log(2)^2+(x^2+24*x+144)*log(2)),x, algorithm="giac")

[Out]

-3*(2*log(2) + 3)/((x + 10*log(2) + 12)*log(2))

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maple [A]  time = 0.19, size = 22, normalized size = 0.88

method	result	size
gosper	\(-\frac {3 \left (2 \ln \relax (2)+3\right )}{\ln \relax (2) \left (10 \ln \relax (2)+x +12\right )}\)	\(22\)
default	\(-\frac {6 \ln \relax (2)+9}{\ln \relax (2) \left (10 \ln \relax (2)+x +12\right )}\)	\(22\)
norman	\(-\frac {3 \left (2 \ln \relax (2)+3\right )}{\ln \relax (2) \left (10 \ln \relax (2)+x +12\right )}\)	\(22\)
risch	\(-\frac {6}{10 \ln \relax (2)+x +12}-\frac {9}{\ln \relax (2) \left (10 \ln \relax (2)+x +12\right )}\)	\(28\)
meijerg	\(\frac {3 x}{\left (10 \ln \relax (2)+12\right ) \left (5 \ln \relax (2)+6\right ) \left (1+\frac {x}{10 \ln \relax (2)+12}\right )}+\frac {9 x}{2 \ln \relax (2) \left (10 \ln \relax (2)+12\right ) \left (5 \ln \relax (2)+6\right ) \left (1+\frac {x}{10 \ln \relax (2)+12}\right )}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*ln(2)+9)/(100*ln(2)^3+2*(10*x+120)*ln(2)^2+(x^2+24*x+144)*ln(2)),x,method=_RETURNVERBOSE)

[Out]

-3*(2*ln(2)+3)/ln(2)/(10*ln(2)+x+12)

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maxima [A]  time = 0.42, size = 25, normalized size = 1.00 \begin {gather*} -\frac {3 \, {\left (2 \, \log \relax (2) + 3\right )}}{x \log \relax (2) + 10 \, \log \relax (2)^{2} + 12 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(2)+9)/(100*log(2)^3+2*(10*x+120)*log(2)^2+(x^2+24*x+144)*log(2)),x, algorithm="maxima")

[Out]

-3*(2*log(2) + 3)/(x*log(2) + 10*log(2)^2 + 12*log(2))

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mupad [B]  time = 0.09, size = 19, normalized size = 0.76 \begin {gather*} -\frac {\ln \left (64\right )+9}{\ln \relax (2)\,\left (x+10\,\ln \relax (2)+12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*log(2) + 9)/(2*log(2)^2*(10*x + 120) + 100*log(2)^3 + log(2)*(24*x + x^2 + 144)),x)

[Out]

-(log(64) + 9)/(log(2)*(x + 10*log(2) + 12))

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sympy [A]  time = 0.19, size = 24, normalized size = 0.96 \begin {gather*} - \frac {6 \log {\relax (2 )} + 9}{x \log {\relax (2 )} + 10 \log {\relax (2 )}^{2} + 12 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*ln(2)+9)/(100*ln(2)**3+2*(10*x+120)*ln(2)**2+(x**2+24*x+144)*ln(2)),x)

[Out]

-(6*log(2) + 9)/(x*log(2) + 10*log(2)**2 + 12*log(2))

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