3.28.1 \(\int \frac {8 e^3 \log (2)}{64 x^2+16 x \log (2)+\log ^2(2)} \, dx\)

Optimal. Leaf size=17 \[ \frac {e^3}{1+\frac {\log (2)}{8 x}} \]

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Rubi [A]  time = 0.00, antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 27, 32} \begin {gather*} -\frac {e^3 \log (2)}{8 x+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*E^3*Log[2])/(64*x^2 + 16*x*Log[2] + Log[2]^2),x]

[Out]

-((E^3*Log[2])/(8*x + 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (8 e^3 \log (2)\right ) \int \frac {1}{64 x^2+16 x \log (2)+\log ^2(2)} \, dx\\ &=\left (8 e^3 \log (2)\right ) \int \frac {1}{(8 x+\log (2))^2} \, dx\\ &=-\frac {e^3 \log (2)}{8 x+\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.88 \begin {gather*} -\frac {e^3 \log (2)}{8 x+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^3*Log[2])/(64*x^2 + 16*x*Log[2] + Log[2]^2),x]

[Out]

-((E^3*Log[2])/(8*x + Log[2]))

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fricas [A]  time = 0.47, size = 14, normalized size = 0.82 \begin {gather*} -\frac {e^{3} \log \relax (2)}{8 \, x + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^3*log(2)/(8*x + log(2))

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giac [A]  time = 0.15, size = 14, normalized size = 0.82 \begin {gather*} -\frac {e^{3} \log \relax (2)}{8 \, x + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^3*log(2)/(8*x + log(2))

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maple [A]  time = 0.43, size = 15, normalized size = 0.88

method	result	size
gosper	\(-\frac {\ln \relax (2) {\mathrm e}^{3}}{\ln \relax (2)+8 x}\)	\(15\)
default	\(-\frac {\ln \relax (2) {\mathrm e}^{3}}{\ln \relax (2)+8 x}\)	\(15\)
norman	\(-\frac {\ln \relax (2) {\mathrm e}^{3}}{\ln \relax (2)+8 x}\)	\(15\)
risch	\(-\frac {\ln \relax (2) {\mathrm e}^{3}}{\ln \relax (2)+8 x}\)	\(15\)
meijerg	\(\frac {8 \,{\mathrm e}^{3} x}{\ln \relax (2) \left (1+\frac {8 x}{\ln \relax (2)}\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*exp(3)*ln(2)/(ln(2)^2+16*x*ln(2)+64*x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(2)*exp(3)/(ln(2)+8*x)

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maxima [A]  time = 0.71, size = 14, normalized size = 0.82 \begin {gather*} -\frac {e^{3} \log \relax (2)}{8 \, x + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^3*log(2)/(8*x + log(2))

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mupad [B]  time = 1.54, size = 14, normalized size = 0.82 \begin {gather*} -\frac {{\mathrm {e}}^3\,\ln \relax (2)}{8\,x+\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(3)*log(2))/(8*x + log(2))

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sympy [A]  time = 0.10, size = 17, normalized size = 1.00 \begin {gather*} - \frac {8 e^{3} \log {\relax (2 )}}{64 x + 8 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*exp(3)*log(2)/(64*x + 8*log(2))

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