3.103.66 \(\int \frac {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+(-3-e^2-5 x) \log (4)+e^x (-4 x+x \log (4))} \, dx\) [10266]

Optimal. Leaf size=27 \[ \log \left (3+e^2+x+\left (4-e^x\right ) x-\frac {4 x}{4-\log (4)}\right ) \]

[Out]

ln(exp(2)-4*x/(4-2*ln(2))+3+x+x*(-exp(x)+4))

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Rubi [F]
time = 2.88, 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 {16-5 \log (4)+e^x (-4-4 x+(1+x) \log (4))}{12+4 e^2+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16 - 5*Log[4] + E^x*(-4 - 4*x + (1 + x)*Log[4]))/(12 + 4*E^2 + 16*x + (-3 - E^2 - 5*x)*Log[4] + E^x*(-4*x
+ x*Log[4])),x]

[Out]

x + Log[x] - (12 + E^2*(4 - Log[4]) - Log[64])*Defer[Int][1/(x*(-3*(1 + E^2/3)*(-4 + Log[4]) + E^x*x*(-4 + Log
[4]) - 2*x*(-8 + Log[32]))), x] - (16 - 5*Log[4])*Defer[Int][x/(-3*(1 + E^2/3)*(-4 + Log[4]) + E^x*x*(-4 + Log
[4]) - 2*x*(-8 + Log[32])), x] + (12 + E^2*(4 - Log[4]) - Log[64])*Defer[Int][(-16*x*(1 - (5*Log[2])/8) + 4*E^
x*x*(1 - Log[2]/2) - 12*(1 + (-(E^2*(-4 + Log[4])) - Log[64])/12))^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16 \left (1-\frac {5 \log (2)}{8}\right )+e^x (-4-4 x+(1+x) \log (4))}{12 \left (1+\frac {e^2}{3}\right )+16 x+\left (-3-e^2-5 x\right ) \log (4)+e^x (-4 x+x \log (4))} \, dx\\ &=\int \left (\frac {1+x}{x}+\frac {-12-x^2 (16-5 \log (4))-e^2 (4-\log (4))-x \left (12+e^2 (4-\log (4))-\log (64)\right )+\log (64)}{x \left (16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )}\right ) \, dx\\ &=\int \frac {1+x}{x} \, dx+\int \frac {-12-x^2 (16-5 \log (4))-e^2 (4-\log (4))-x \left (12+e^2 (4-\log (4))-\log (64)\right )+\log (64)}{x \left (16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )} \, dx\\ &=\int \left (1+\frac {1}{x}\right ) \, dx+\int \left (\frac {x (-16+5 \log (4))}{16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )}+\frac {12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )}{-16 x \left (1-\frac {5 \log (2)}{8}\right )+4 e^x x \left (1-\frac {\log (2)}{2}\right )-12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )}+\frac {-12-e^2 (4-\log (4))+\log (64)}{x \left (16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )}\right ) \, dx\\ &=x+\log (x)+(-16+5 \log (4)) \int \frac {x}{16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )} \, dx+\left (12+e^2 (4-\log (4))-\log (64)\right ) \int \frac {1}{-16 x \left (1-\frac {5 \log (2)}{8}\right )+4 e^x x \left (1-\frac {\log (2)}{2}\right )-12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )} \, dx+\left (-12-e^2 (4-\log (4))+\log (64)\right ) \int \frac {1}{x \left (16 x \left (1-\frac {5 \log (2)}{8}\right )-4 e^x x \left (1-\frac {\log (2)}{2}\right )+12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-3 \log (4)\right )\right )\right )} \, dx\\ &=x+\log (x)+(-16+5 \log (4)) \int \frac {x}{-3 \left (1+\frac {e^2}{3}\right ) (-4+\log (4))+e^x x (-4+\log (4))-2 x (-8+\log (32))} \, dx+\left (12+e^2 (4-\log (4))-\log (64)\right ) \int \frac {1}{-16 x \left (1-\frac {5 \log (2)}{8}\right )+4 e^x x \left (1-\frac {\log (2)}{2}\right )-12 \left (1+\frac {1}{12} \left (-e^2 (-4+\log (4))-\log (64)\right )\right )} \, dx+\left (-12-e^2 (4-\log (4))+\log (64)\right ) \int \frac {1}{x \left (-3 \left (1+\frac {e^2}{3}\right ) (-4+\log (4))+e^x x (-4+\log (4))-2 x (-8+\log (32))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.17, size = 40, normalized size = 1.48 \begin {gather*} \log \left (12+4 e^2+16 x-4 e^x x-3 \log (4)-e^2 \log (4)-5 x \log (4)+e^x x \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 - 5*Log[4] + E^x*(-4 - 4*x + (1 + x)*Log[4]))/(12 + 4*E^2 + 16*x + (-3 - E^2 - 5*x)*Log[4] + E^x
*(-4*x + x*Log[4])),x]

[Out]

Log[12 + 4*E^2 + 16*x - 4*E^x*x - 3*Log[4] - E^2*Log[4] - 5*x*Log[4] + E^x*x*Log[4]]

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Maple [A]
time = 0.04, size = 37, normalized size = 1.37

method	result	size
norman	\(\ln \left (-x \ln \left (2\right ) {\mathrm e}^{x}+{\mathrm e}^{2} \ln \left (2\right )+2 \,{\mathrm e}^{x} x +5 x \ln \left (2\right )-2 \,{\mathrm e}^{2}+3 \ln \left (2\right )-8 x -6\right )\)	\(37\)
risch	\(\ln \left (x \right )+\ln \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2} \ln \left (2\right )+5 x \ln \left (2\right )-2 \,{\mathrm e}^{2}+3 \ln \left (2\right )-8 x -6}{x \left (\ln \left (2\right )-2\right )}\right )\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*ln(2)*(x+1)-4*x-4)*exp(x)-10*ln(2)+16)/((2*x*ln(2)-4*x)*exp(x)+2*(-exp(2)-5*x-3)*ln(2)+4*exp(2)+16*x+1
2),x,method=_RETURNVERBOSE)

[Out]

ln(-x*ln(2)*exp(x)+exp(2)*ln(2)+2*exp(x)*x+5*x*ln(2)-2*exp(2)+3*ln(2)-8*x-6)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
time = 0.50, size = 45, normalized size = 1.67 \begin {gather*} \log \left (x\right ) + \log \left (\frac {x {\left (\log \left (2\right ) - 2\right )} e^{x} - x {\left (5 \, \log \left (2\right ) - 8\right )} - {\left (\log \left (2\right ) - 2\right )} e^{2} - 3 \, \log \left (2\right ) + 6}{x {\left (\log \left (2\right ) - 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(2)*(1+x)-4*x-4)*exp(x)-10*log(2)+16)/((2*x*log(2)-4*x)*exp(x)+2*(-exp(2)-5*x-3)*log(2)+4*exp
(2)+16*x+12),x, algorithm="maxima")

[Out]

log(x) + log((x*(log(2) - 2)*e^x - x*(5*log(2) - 8) - (log(2) - 2)*e^2 - 3*log(2) + 6)/(x*(log(2) - 2)))

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Fricas [A]
time = 0.36, size = 39, normalized size = 1.44 \begin {gather*} \log \left (x\right ) + \log \left (\frac {{\left (x \log \left (2\right ) - 2 \, x\right )} e^{x} - {\left (5 \, x + e^{2} + 3\right )} \log \left (2\right ) + 8 \, x + 2 \, e^{2} + 6}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(2)*(1+x)-4*x-4)*exp(x)-10*log(2)+16)/((2*x*log(2)-4*x)*exp(x)+2*(-exp(2)-5*x-3)*log(2)+4*exp
(2)+16*x+12),x, algorithm="fricas")

[Out]

log(x) + log(((x*log(2) - 2*x)*e^x - (5*x + e^2 + 3)*log(2) + 8*x + 2*e^2 + 6)/x)

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Sympy [A]
time = 0.38, size = 44, normalized size = 1.63 \begin {gather*} \log {\left (x \right )} + \log {\left (e^{x} + \frac {- 5 x \log {\left (2 \right )} + 8 x - e^{2} \log {\left (2 \right )} - 3 \log {\left (2 \right )} + 6 + 2 e^{2}}{- 2 x + x \log {\left (2 \right )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*ln(2)*(1+x)-4*x-4)*exp(x)-10*ln(2)+16)/((2*x*ln(2)-4*x)*exp(x)+2*(-exp(2)-5*x-3)*ln(2)+4*exp(2)+
16*x+12),x)

[Out]

log(x) + log(exp(x) + (-5*x*log(2) + 8*x - exp(2)*log(2) - 3*log(2) + 6 + 2*exp(2))/(-2*x + x*log(2)))

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Giac [A]
time = 0.41, size = 36, normalized size = 1.33 \begin {gather*} \log \left (x e^{x} \log \left (2\right ) - 2 \, x e^{x} - 5 \, x \log \left (2\right ) - e^{2} \log \left (2\right ) + 8 \, x + 2 \, e^{2} - 3 \, \log \left (2\right ) + 6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*log(2)*(1+x)-4*x-4)*exp(x)-10*log(2)+16)/((2*x*log(2)-4*x)*exp(x)+2*(-exp(2)-5*x-3)*log(2)+4*exp
(2)+16*x+12),x, algorithm="giac")

[Out]

log(x*e^x*log(2) - 2*x*e^x - 5*x*log(2) - e^2*log(2) + 8*x + 2*e^2 - 3*log(2) + 6)

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Mupad [B]
time = 7.21, size = 34, normalized size = 1.26 \begin {gather*} \ln \left (16\,x+4\,{\mathrm {e}}^2-2\,\ln \left (2\right )\,\left (5\,x+{\mathrm {e}}^2+3\right )-{\mathrm {e}}^x\,\left (4\,x-2\,x\,\ln \left (2\right )\right )+12\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*log(2) + exp(x)*(4*x - 2*log(2)*(x + 1) + 4) - 16)/(16*x + 4*exp(2) - 2*log(2)*(5*x + exp(2) + 3) - e
xp(x)*(4*x - 2*x*log(2)) + 12),x)

[Out]

log(16*x + 4*exp(2) - 2*log(2)*(5*x + exp(2) + 3) - exp(x)*(4*x - 2*x*log(2)) + 12)

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