3.17.12 \(\int \frac {e^{\log ^2(2)}-16 x}{e^{\log ^2(2)} x-8 x^2} \, dx\)

Optimal. Leaf size=18 \[ \log \left (\frac {1}{5} x \left (-\frac {1}{8} e^{\log ^2(2)}+x\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.83, number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {628} \begin {gather*} \log \left (x e^{\log ^2(2)}-8 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^Log[2]^2 - 16*x)/(E^Log[2]^2*x - 8*x^2),x]

[Out]

Log[E^Log[2]^2*x - 8*x^2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (e^{\log ^2(2)} x-8 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.78 \begin {gather*} \log \left (e^{\log ^2(2)}-8 x\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^Log[2]^2 - 16*x)/(E^Log[2]^2*x - 8*x^2),x]

[Out]

Log[E^Log[2]^2 - 8*x] + Log[x]

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fricas [A]  time = 0.66, size = 15, normalized size = 0.83 \begin {gather*} \log \left (8 \, x^{2} - x e^{\left (\log \relax (2)^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(8*x^2 - x*e^(log(2)^2))

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giac [A]  time = 0.22, size = 16, normalized size = 0.89 \begin {gather*} \log \left ({\left | 8 \, x^{2} - x e^{\left (\log \relax (2)^{2}\right )} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(8*x^2 - x*e^(log(2)^2)))

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maple [A]  time = 0.31, size = 13, normalized size = 0.72

method	result	size
default	\(\ln \left (x \left ({\mathrm e}^{\ln \relax (2)^{2}}-8 x \right )\right )\)	\(13\)
norman	\(\ln \relax (x )+\ln \left ({\mathrm e}^{\ln \relax (2)^{2}}-8 x \right )\)	\(14\)
derivativedivides	\(\ln \left (x \,{\mathrm e}^{\ln \relax (2)^{2}}-8 x^{2}\right )\)	\(15\)
risch	\(\ln \left (-x \,{\mathrm e}^{\ln \relax (2)^{2}}+8 x^{2}\right )\)	\(16\)
meijerg	\(\ln \left (1-8 x \,{\mathrm e}^{-\ln \relax (2)^{2}}\right )+\ln \relax (x )+3 \ln \relax (2)-\ln \relax (2)^{2}+i \pi \)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x*(exp(ln(2)^2)-8*x))

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maxima [A]  time = 0.36, size = 15, normalized size = 0.83 \begin {gather*} \log \left (8 \, x^{2} - x e^{\left (\log \relax (2)^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(8*x^2 - x*e^(log(2)^2))

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mupad [B]  time = 1.10, size = 14, normalized size = 0.78 \begin {gather*} \ln \left (x\,\left (8\,x-{\mathrm {e}}^{{\ln \relax (2)}^2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x - exp(log(2)^2))/(x*exp(log(2)^2) - 8*x^2),x)

[Out]

log(x*(8*x - exp(log(2)^2)))

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sympy [A]  time = 0.15, size = 14, normalized size = 0.78 \begin {gather*} \log {\left (8 x^{2} - x e^{\log {\relax (2 )}^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(ln(2)**2)-16*x)/(x*exp(ln(2)**2)-8*x**2),x)

[Out]

log(8*x**2 - x*exp(log(2)**2))

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