3.99.58 \(\int -\frac {8}{144+192 x+64 x^2+(-24-16 x) \log (16)+\log ^2(16)} \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{2 x+6 (2+x)-\log (16)} \]

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 0.73, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 1981, 27, 32} \begin {gather*} \frac {1}{8 x+12-\log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-8/(144 + 192*x + 64*x^2 + (-24 - 16*x)*Log[16] + Log[16]^2),x]

[Out]

(12 + 8*x - Log[16])^(-1)

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} &=-\left (8 \int \frac {1}{144+192 x+64 x^2+(-24-16 x) \log (16)+\log ^2(16)} \, dx\right )\\ &=-\left (8 \int \frac {1}{64 x^2+16 x (12-\log (16))+(-12+\log (16))^2} \, dx\right )\\ &=-\left (8 \int \frac {1}{(12+8 x-\log (16))^2} \, dx\right )\\ &=\frac {1}{12+8 x-\log (16)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 11, normalized size = 0.73 \begin {gather*} -\frac {1}{-12-8 x+\log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-8/(144 + 192*x + 64*x^2 + (-24 - 16*x)*Log[16] + Log[16]^2),x]

[Out]

-(-12 - 8*x + Log[16])^(-1)

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fricas [A]  time = 0.80, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{4 \, {\left (2 \, x - \log \relax (2) + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8/(16*log(2)^2+4*(-16*x-24)*log(2)+64*x^2+192*x+144),x, algorithm="fricas")

[Out]

1/4/(2*x - log(2) + 3)

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giac [A]  time = 0.16, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{4 \, {\left (2 \, x - \log \relax (2) + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8/(16*log(2)^2+4*(-16*x-24)*log(2)+64*x^2+192*x+144),x, algorithm="giac")

[Out]

1/4/(2*x - log(2) + 3)

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maple [A]  time = 0.12, size = 12, normalized size = 0.80

method	result	size
gosper	\(-\frac {1}{4 \left (\ln \relax (2)-2 x -3\right )}\)	\(12\)
norman	\(-\frac {1}{4 \left (\ln \relax (2)-2 x -3\right )}\)	\(12\)
risch	\(-\frac {1}{4 \left (\ln \relax (2)-2 x -3\right )}\)	\(12\)
default	\(\frac {1}{8 x +12-4 \ln \relax (2)}\)	\(14\)
meijerg	\(-\frac {x}{4 \left (-\frac {\ln \relax (2)}{2}+\frac {3}{2}\right ) \left (1+\frac {2 x}{-\ln \relax (2)+3}\right ) \left (-\ln \relax (2)+3\right )}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-8/(16*ln(2)^2+4*(-16*x-24)*ln(2)+64*x^2+192*x+144),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.34, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{4 \, {\left (2 \, x - \log \relax (2) + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8/(16*log(2)^2+4*(-16*x-24)*log(2)+64*x^2+192*x+144),x, algorithm="maxima")

[Out]

1/4/(2*x - log(2) + 3)

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mupad [B]  time = 5.58, size = 51, normalized size = 3.40 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {16\,x-\ln \left (256\right )+24}{\sqrt {\ln \left (256\right )-8\,\ln \relax (2)}\,\sqrt {8\,\ln \relax (2)+\ln \left (256\right )-48}}\right )}{\sqrt {\ln \left (256\right )-8\,\ln \relax (2)}\,\sqrt {8\,\ln \relax (2)+\ln \left (256\right )-48}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-8/(192*x - 4*log(2)*(16*x + 24) + 16*log(2)^2 + 64*x^2 + 144),x)

[Out]

(2*atanh((16*x - log(256) + 24)/((log(256) - 8*log(2))^(1/2)*(8*log(2) + log(256) - 48)^(1/2))))/((log(256) -
8*log(2))^(1/2)*(8*log(2) + log(256) - 48)^(1/2))

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sympy [A]  time = 0.17, size = 10, normalized size = 0.67 \begin {gather*} \frac {1}{8 x - 4 \log {\relax (2 )} + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8/(16*ln(2)**2+4*(-16*x-24)*ln(2)+64*x**2+192*x+144),x)

[Out]

1/(8*x - 4*log(2) + 12)

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