3.40.75 \(\int \frac {-4-e^x+4 x+8 x \log (2 x)}{-e^x-4 x+4 x^2 \log (2 x)} \, dx\)

Optimal. Leaf size=17 \[ \log \left (e^x+4 x (1-x \log (2 x))\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6684} \begin {gather*} \log \left (-4 x^2 \log (2 x)+4 x+e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - E^x + 4*x + 8*x*Log[2*x])/(-E^x - 4*x + 4*x^2*Log[2*x]),x]

[Out]

Log[E^x + 4*x - 4*x^2*Log[2*x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (e^x+4 x-4 x^2 \log (2 x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 17, normalized size = 1.00 \begin {gather*} \log \left (e^x+4 x-4 x^2 \log (2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - E^x + 4*x + 8*x*Log[2*x])/(-E^x - 4*x + 4*x^2*Log[2*x]),x]

[Out]

Log[E^x + 4*x - 4*x^2*Log[2*x]]

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fricas [A]  time = 0.81, size = 29, normalized size = 1.71 \begin {gather*} 2 \, \log \left (2 \, x\right ) + \log \left (\frac {4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(2*x)-exp(x)+4*x-4)/(4*x^2*log(2*x)-exp(x)-4*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 18, normalized size = 1.06 \begin {gather*} \log \left (4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(2*x)-exp(x)+4*x-4)/(4*x^2*log(2*x)-exp(x)-4*x),x, algorithm="giac")

[Out]

log(4*x^2*log(2*x) - 4*x - e^x)

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maple [A]  time = 0.03, size = 19, normalized size = 1.12

method	result	size
derivativedivides	\(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\)	\(19\)
default	\(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\)	\(19\)
norman	\(\ln \left (4 x^{2} \ln \left (2 x \right )-{\mathrm e}^{x}-4 x \right )\)	\(19\)
risch	\(2 \ln \relax (x )+\ln \left (\ln \left (2 x \right )-\frac {4 x +{\mathrm e}^{x}}{4 x^{2}}\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x*ln(2*x)-exp(x)+4*x-4)/(4*x^2*ln(2*x)-exp(x)-4*x),x,method=_RETURNVERBOSE)

[Out]

ln(4*x^2*ln(2*x)-exp(x)-4*x)

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maxima [A]  time = 0.35, size = 18, normalized size = 1.06 \begin {gather*} \log \left (4 \, x^{2} \log \left (2 \, x\right ) - 4 \, x - e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(2*x)-exp(x)+4*x-4)/(4*x^2*log(2*x)-exp(x)-4*x),x, algorithm="maxima")

[Out]

log(4*x^2*log(2*x) - 4*x - e^x)

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mupad [B]  time = 2.63, size = 18, normalized size = 1.06 \begin {gather*} \ln \left (4\,x^2\,\ln \left (2\,x\right )-{\mathrm {e}}^x-4\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - exp(x) + 8*x*log(2*x) - 4)/(4*x + exp(x) - 4*x^2*log(2*x)),x)

[Out]

log(4*x^2*log(2*x) - exp(x) - 4*x)

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sympy [A]  time = 0.28, size = 17, normalized size = 1.00 \begin {gather*} \log {\left (- 4 x^{2} \log {\left (2 x \right )} + 4 x + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*ln(2*x)-exp(x)+4*x-4)/(4*x**2*ln(2*x)-exp(x)-4*x),x)

[Out]

log(-4*x**2*log(2*x) + 4*x + exp(x))

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