3.34.77 \(\int -\frac {4 \log (2) \log (9)}{x^2+(-16 x-32 e^3 x) \log (2)+(64+256 e^3+256 e^6) \log ^2(2)} \, dx\)

Optimal. Leaf size=28 \[ \frac {x \log (9)}{x+\frac {x}{1+4 e^3-\frac {x}{4 \log (2)}}} \]

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 1981, 27, 32} \begin {gather*} \frac {4 \log (2) \log (9)}{x-8 \left (1+2 e^3\right ) \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*Log[2]*Log[9])/(x^2 + (-16*x - 32*E^3*x)*Log[2] + (64 + 256*E^3 + 256*E^6)*Log[2]^2),x]

[Out]

(4*Log[2]*Log[9])/(x - 8*(1 + 2*E^3)*Log[2])

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 ((4 \log (2) \log (9)) \int \frac {1}{x^2+\left (-16 x-32 e^3 x\right ) \log (2)+\left (64+256 e^3+256 e^6\right ) \log ^2(2)} \, dx\right )\\ &=-\left ((4 \log (2) \log (9)) \int \frac {1}{x^2-16 \left (1+2 e^3\right ) x \log (2)+64 \left (\log (2)+e^3 \log (4)\right )^2} \, dx\right )\\ &=-\left ((4 \log (2) \log (9)) \int \frac {1}{\left (x-8 \left (1+2 e^3\right ) \log (2)\right )^2} \, dx\right )\\ &=\frac {4 \log (2) \log (9)}{x-8 \left (1+2 e^3\right ) \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.75 \begin {gather*} \frac {4 \log (2) \log (9)}{x-8 \left (\log (2)+e^3 \log (4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*Log[2]*Log[9])/(x^2 + (-16*x - 32*E^3*x)*Log[2] + (64 + 256*E^3 + 256*E^6)*Log[2]^2),x]

[Out]

(4*Log[2]*Log[9])/(x - 8*(Log[2] + E^3*Log[4]))

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fricas [A]  time = 0.94, size = 22, normalized size = 0.79 \begin {gather*} -\frac {8 \, \log \relax (3) \log \relax (2)}{8 \, {\left (2 \, e^{3} + 1\right )} \log \relax (2) - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*log(2)*log(3)/((256*exp(3)^2+256*exp(3)+64)*log(2)^2+(-32*x*exp(3)-16*x)*log(2)+x^2),x, algorithm
="fricas")

[Out]

-8*log(3)*log(2)/(8*(2*e^3 + 1)*log(2) - x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*log(2)*log(3)/((256*exp(3)^2+256*exp(3)+64)*log(2)^2+(-32*x*exp(3)-16*x)*log(2)+x^2),x, algorithm
="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -ln(3)*ln(2)*8*2*1/32/sqrt(-exp(3)^2+exp
(6))/ln(2)*atan((sageVARx-16*exp(3)*ln(2)-8*ln(2))*1/16/sqrt(-exp(3)^2+exp(6))/ln(2))

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maple [A]  time = 0.52, size = 22, normalized size = 0.79

method	result	size
risch	\(-\frac {\ln \relax (2) \ln \relax (3)}{2 \left ({\mathrm e}^{3} \ln \relax (2)+\frac {\ln \relax (2)}{2}-\frac {x}{16}\right )}\)	\(22\)
gosper	\(-\frac {8 \ln \relax (3) \ln \relax (2)}{16 \,{\mathrm e}^{3} \ln \relax (2)+8 \ln \relax (2)-x}\)	\(23\)
norman	\(-\frac {8 \ln \relax (3) \ln \relax (2)}{16 \,{\mathrm e}^{3} \ln \relax (2)+8 \ln \relax (2)-x}\)	\(23\)
meijerg	\(\frac {\ln \relax (2) \ln \relax (3) x}{\left (-16 \,{\mathrm e}^{3} \ln \relax (2)-8 \ln \relax (2)\right ) \left (1-\frac {x}{8 \left (2 \,{\mathrm e}^{3} \ln \relax (2)+\ln \relax (2)\right )}\right ) \left (2 \,{\mathrm e}^{3} \ln \relax (2)+\ln \relax (2)\right )}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-8*ln(2)*ln(3)/((256*exp(3)^2+256*exp(3)+64)*ln(2)^2+(-32*x*exp(3)-16*x)*ln(2)+x^2),x,method=_RETURNVERBOS
E)

[Out]

-1/2*ln(2)*ln(3)/(exp(3)*ln(2)+1/2*ln(2)-1/16*x)

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maxima [A]  time = 0.46, size = 22, normalized size = 0.79 \begin {gather*} -\frac {8 \, \log \relax (3) \log \relax (2)}{8 \, {\left (2 \, e^{3} + 1\right )} \log \relax (2) - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*log(2)*log(3)/((256*exp(3)^2+256*exp(3)+64)*log(2)^2+(-32*x*exp(3)-16*x)*log(2)+x^2),x, algorithm
="maxima")

[Out]

-8*log(3)*log(2)/(8*(2*e^3 + 1)*log(2) - x)

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mupad [B]  time = 0.20, size = 22, normalized size = 0.79 \begin {gather*} -\frac {8\,\ln \relax (2)\,\ln \relax (3)}{8\,\ln \relax (2)-x+16\,{\mathrm {e}}^3\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*log(2)*log(3))/(log(2)^2*(256*exp(3) + 256*exp(6) + 64) - log(2)*(16*x + 32*x*exp(3)) + x^2),x)

[Out]

-(8*log(2)*log(3))/(8*log(2) - x + 16*exp(3)*log(2))

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sympy [A]  time = 0.18, size = 22, normalized size = 0.79 \begin {gather*} \frac {8 \log {\relax (2 )} \log {\relax (3 )}}{x - 16 e^{3} \log {\relax (2 )} - 8 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*ln(2)*ln(3)/((256*exp(3)**2+256*exp(3)+64)*ln(2)**2+(-32*x*exp(3)-16*x)*ln(2)+x**2),x)

[Out]

8*log(2)*log(3)/(x - 16*exp(3)*log(2) - 8*log(2))

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