3.2.12 \(\int \frac {2 x+5 \log (4)}{x^2+5 x \log (4)} \, dx\) [112]

Optimal. Leaf size=17 \[ \log \left (5 x+\frac {x^2}{\log (4)}\right )+\log (\log (27)) \]

[Out]

ln(1/2*x^2/ln(2)+5*x)+ln(3*ln(3))

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Rubi [A]
time = 0.00, antiderivative size = 10, normalized size of antiderivative = 0.59, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {642} \begin {gather*} \log \left (x^2+5 x \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x + 5*Log[4])/(x^2 + 5*x*Log[4]),x]

[Out]

Log[x^2 + 5*x*Log[4]]

Rule 642

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 (x^2+5 x \log (4)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 10, normalized size = 0.59 \begin {gather*} \log (x)+\log (x+5 \log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x + 5*Log[4])/(x^2 + 5*x*Log[4]),x]

[Out]

Log[x] + Log[x + 5*Log[4]]

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Maple [A]
time = 0.05, size = 10, normalized size = 0.59

method	result	size
default	\(\ln \left (x \left (10 \ln \left (2\right )+x \right )\right )\)	\(10\)
derivativedivides	\(\ln \left (10 x \ln \left (2\right )+x^{2}\right )\)	\(11\)
norman	\(\ln \left (x \right )+\ln \left (10 \ln \left (2\right )+x \right )\)	\(11\)
risch	\(\ln \left (10 x \ln \left (2\right )+x^{2}\right )\)	\(11\)
meijerg	\(\ln \left (1+\frac {x}{10 \ln \left (2\right )}\right )+\ln \left (x \right )-\ln \left (2\right )-\ln \left (5\right )-\ln \left (\ln \left (2\right )\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*ln(2)+2*x)/(10*x*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x*(10*ln(2)+x))

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Maxima [A]
time = 0.30, size = 10, normalized size = 0.59 \begin {gather*} \log \left (x^{2} + 10 \, x \log \left (2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(2)+2*x)/(10*x*log(2)+x^2),x, algorithm="maxima")

[Out]

log(x^2 + 10*x*log(2))

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Fricas [A]
time = 0.29, size = 10, normalized size = 0.59 \begin {gather*} \log \left (x^{2} + 10 \, x \log \left (2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(2)+2*x)/(10*x*log(2)+x^2),x, algorithm="fricas")

[Out]

log(x^2 + 10*x*log(2))

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Sympy [A]
time = 0.05, size = 10, normalized size = 0.59 \begin {gather*} \log {\left (x^{2} + 10 x \log {\left (2 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*ln(2)+2*x)/(10*x*ln(2)+x**2),x)

[Out]

log(x**2 + 10*x*log(2))

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Giac [A]
time = 0.39, size = 15, normalized size = 0.88 \begin {gather*} \log \left (2 \, {\left | \frac {1}{2} \, x^{2} + 5 \, x \log \left (2\right ) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(2)+2*x)/(10*x*log(2)+x^2),x, algorithm="giac")

[Out]

log(2*abs(1/2*x^2 + 5*x*log(2)))

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Mupad [B]
time = 0.08, size = 9, normalized size = 0.53 \begin {gather*} \ln \left (x\,\left (x+10\,\ln \left (2\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 10*log(2))/(10*x*log(2) + x^2),x)

[Out]

log(x*(x + 10*log(2)))

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