3.77.6 \(\int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx\)

Optimal. Leaf size=21 \[ -1+\log (-2 (1+x)+4 x \log (e x)+\log (x \log (2))) \]

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Rubi [F]  time = 0.28, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1+2 x+4 x \log (e x)}{-2 x-2 x^2+4 x^2 \log (e x)+x \log (x \log (2))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 + 2*x + 4*x*Log[E*x])/(-2*x - 2*x^2 + 4*x^2*Log[E*x] + x*Log[x*Log[2]]),x]

[Out]

6*Defer[Int][(-2 + 2*x + 4*x*Log[x] + Log[x*Log[2]])^(-1), x] + Defer[Int][1/(x*(-2 + 2*x + 4*x*Log[x] + Log[x
*Log[2]])), x] + 4*Defer[Int][Log[x]/(-2 + 2*x + 4*x*Log[x] + Log[x*Log[2]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2}{-2+2 x+4 x \log (x)+\log (x \log (2))}+\frac {1}{x (-2+2 x+4 x \log (x)+\log (x \log (2)))}+\frac {4 (1+\log (x))}{-2+2 x+4 x \log (x)+\log (x \log (2))}\right ) \, dx\\ &=2 \int \frac {1}{-2+2 x+4 x \log (x)+\log (x \log (2))} \, dx+4 \int \frac {1+\log (x)}{-2+2 x+4 x \log (x)+\log (x \log (2))} \, dx+\int \frac {1}{x (-2+2 x+4 x \log (x)+\log (x \log (2)))} \, dx\\ &=2 \int \frac {1}{-2+2 x+4 x \log (x)+\log (x \log (2))} \, dx+4 \int \left (\frac {1}{-2+2 x+4 x \log (x)+\log (x \log (2))}+\frac {\log (x)}{-2+2 x+4 x \log (x)+\log (x \log (2))}\right ) \, dx+\int \frac {1}{x (-2+2 x+4 x \log (x)+\log (x \log (2)))} \, dx\\ &=2 \int \frac {1}{-2+2 x+4 x \log (x)+\log (x \log (2))} \, dx+4 \int \frac {1}{-2+2 x+4 x \log (x)+\log (x \log (2))} \, dx+4 \int \frac {\log (x)}{-2+2 x+4 x \log (x)+\log (x \log (2))} \, dx+\int \frac {1}{x (-2+2 x+4 x \log (x)+\log (x \log (2)))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 16, normalized size = 0.76 \begin {gather*} \log (-2+2 x+4 x \log (x)+\log (x \log (2))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x + 4*x*Log[E*x])/(-2*x - 2*x^2 + 4*x^2*Log[E*x] + x*Log[x*Log[2]]),x]

[Out]

Log[-2 + 2*x + 4*x*Log[x] + Log[x*Log[2]]]

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fricas [A]  time = 0.60, size = 38, normalized size = 1.81 \begin {gather*} \log \left (4 \, x + 1\right ) + \log \left (\frac {{\left (4 \, x + 1\right )} \log \left (x e\right ) - 2 \, x + \log \left (e^{\left (-1\right )} \log \relax (2)\right ) - 2}{4 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x*exp(1))+2*x+1)/(x*log(x*log(2))+4*x^2*log(x*exp(1))-2*x^2-2*x),x, algorithm="fricas")

[Out]

log(4*x + 1) + log(((4*x + 1)*log(x*e) - 2*x + log(e^(-1)*log(2)) - 2)/(4*x + 1))

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giac [A]  time = 0.13, size = 16, normalized size = 0.76 \begin {gather*} \log \left (4 \, x \log \relax (x) + 2 \, x + \log \relax (x) + \log \left (\log \relax (2)\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x*exp(1))+2*x+1)/(x*log(x*log(2))+4*x^2*log(x*exp(1))-2*x^2-2*x),x, algorithm="giac")

[Out]

log(4*x*log(x) + 2*x + log(x) + log(log(2)) - 2)

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maple [A]  time = 0.08, size = 20, normalized size = 0.95

method	result	size
norman	\(\ln \left (\ln \left (x \ln \relax (2)\right )+4 x \ln \left (x \,{\mathrm e}\right )-2 x -2\right )\)	\(20\)
risch	\(\ln \left (4 x +1\right )+\ln \left (\ln \relax (x )+\frac {i \left (-2 i \ln \left (\ln \relax (2)\right )-4 i x +4 i\right )}{8 x +2}\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(x*exp(1))+2*x+1)/(x*ln(x*ln(2))+4*x^2*ln(x*exp(1))-2*x^2-2*x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x*ln(2))+4*x*ln(x*exp(1))-2*x-2)

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maxima [A]  time = 0.48, size = 32, normalized size = 1.52 \begin {gather*} \log \left (4 \, x + 1\right ) + \log \left (\frac {{\left (4 \, x + 1\right )} \log \relax (x) + 2 \, x + \log \left (\log \relax (2)\right ) - 2}{4 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x*exp(1))+2*x+1)/(x*log(x*log(2))+4*x^2*log(x*exp(1))-2*x^2-2*x),x, algorithm="maxima")

[Out]

log(4*x + 1) + log(((4*x + 1)*log(x) + 2*x + log(log(2)) - 2)/(4*x + 1))

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mupad [B]  time = 5.18, size = 16, normalized size = 0.76 \begin {gather*} \ln \left (2\,x+\ln \left (\ln \relax (2)\right )+\ln \relax (x)+4\,x\,\ln \relax (x)-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + 4*x*log(x*exp(1)) + 1)/(2*x - x*log(x*log(2)) + 2*x^2 - 4*x^2*log(x*exp(1))),x)

[Out]

log(2*x + log(log(2)) + log(x) + 4*x*log(x) - 2)

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sympy [A]  time = 0.60, size = 29, normalized size = 1.38 \begin {gather*} \log {\left (4 x + 1 \right )} + \log {\left (\log {\left (e x \right )} + \frac {- 2 x - 3 + \log {\left (\log {\relax (2 )} \right )}}{4 x + 1} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(x*exp(1))+2*x+1)/(x*ln(x*ln(2))+4*x**2*ln(x*exp(1))-2*x**2-2*x),x)

[Out]

log(4*x + 1) + log(log(E*x) + (-2*x - 3 + log(log(2)))/(4*x + 1))

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