3.104.2 \(\int \frac {12+e^{1-x} (-12-4 x)+4 x+(-12-4 e^{1-x}-4 x+4 \log (2)) \log (3+e^{1-x}+x-\log (2))}{27+27 x+9 x^2+x^3+e^{1-x} (9+6 x+x^2)+(-9-6 x-x^2) \log (2)+(-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)) \log (3+e^{1-x}+x-\log (2))+(48+16 e^{1-x}+16 x-16 \log (2)) \log ^2(3+e^{1-x}+x-\log (2))} \, dx\)

Optimal. Leaf size=30 \[ \frac {2}{-2+\frac {3+x}{2 \log \left (3+e^{1-x}+x-\log (2)\right )}} \]

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Rubi [F]  time = 6.28, 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 {12+e^{1-x} (-12-4 x)+4 x+\left (-12-4 e^{1-x}-4 x+4 \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )}{27+27 x+9 x^2+x^3+e^{1-x} \left (9+6 x+x^2\right )+\left (-9-6 x-x^2\right ) \log (2)+\left (-72+e^{1-x} (-24-8 x)-48 x-8 x^2+(24+8 x) \log (2)\right ) \log \left (3+e^{1-x}+x-\log (2)\right )+\left (48+16 e^{1-x}+16 x-16 \log (2)\right ) \log ^2\left (3+e^{1-x}+x-\log (2)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(12 + E^(1 - x)*(-12 - 4*x) + 4*x + (-12 - 4*E^(1 - x) - 4*x + 4*Log[2])*Log[3 + E^(1 - x) + x - Log[2]])/
(27 + 27*x + 9*x^2 + x^3 + E^(1 - x)*(9 + 6*x + x^2) + (-9 - 6*x - x^2)*Log[2] + (-72 + E^(1 - x)*(-24 - 8*x)
- 48*x - 8*x^2 + (24 + 8*x)*Log[2])*Log[3 + E^(1 - x) + x - Log[2]] + (48 + 16*E^(1 - x) + 16*x - 16*Log[2])*L
og[3 + E^(1 - x) + x - Log[2]]^2),x]

[Out]

Defer[Int][(3 + x - 4*Log[3 + E^(1 - x) + x - Log[2]])^(-2), x] - Defer[Int][x/(3 + x - 4*Log[3 + E^(1 - x) +
x - Log[2]])^2, x] + 16*E*Defer[Int][1/((-E - E^x*(3 + x - Log[2]))*(3 + x - 4*Log[3 + E^(1 - x) + x - Log[2]]
)^2), x] + 4*E*Defer[Int][x/((-E - E^x*(3 + x - Log[2]))*(3 + x - 4*Log[3 + E^(1 - x) + x - Log[2]])^2), x] +
4*Log[2]*Defer[Int][1/((3 + x - Log[2])*(3 + x - 4*Log[3 + E^(1 - x) + x - Log[2]])^2), x] + 4*E*Log[2]*Defer[
Int][1/((-E - E^x*(3 + x - Log[2]))*(3 + x - Log[2])*(3 + x - 4*Log[3 + E^(1 - x) + x - Log[2]])^2), x] + Defe
r[Int][(3 + x - 4*Log[3 + E^(1 - x) + x - Log[2]])^(-1), x]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 26, normalized size = 0.87 \begin {gather*} \frac {3+x}{3+x-4 \log \left (3+e^{1-x}+x-\log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + E^(1 - x)*(-12 - 4*x) + 4*x + (-12 - 4*E^(1 - x) - 4*x + 4*Log[2])*Log[3 + E^(1 - x) + x - Log
[2]])/(27 + 27*x + 9*x^2 + x^3 + E^(1 - x)*(9 + 6*x + x^2) + (-9 - 6*x - x^2)*Log[2] + (-72 + E^(1 - x)*(-24 -
8*x) - 48*x - 8*x^2 + (24 + 8*x)*Log[2])*Log[3 + E^(1 - x) + x - Log[2]] + (48 + 16*E^(1 - x) + 16*x - 16*Log
[2])*Log[3 + E^(1 - x) + x - Log[2]]^2),x]

[Out]

(3 + x)/(3 + x - 4*Log[3 + E^(1 - x) + x - Log[2]])

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fricas [A]  time = 1.08, size = 25, normalized size = 0.83 \begin {gather*} \frac {x + 3}{x - 4 \, \log \left (x + e^{\left (-x + 1\right )} - \log \relax (2) + 3\right ) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(-x+1)+4*log(2)-4*x-12)*log(exp(-x+1)-log(2)+3+x)+(-4*x-12)*exp(-x+1)+4*x+12)/((16*exp(-x+1)
-16*log(2)+16*x+48)*log(exp(-x+1)-log(2)+3+x)^2+((-8*x-24)*exp(-x+1)+(8*x+24)*log(2)-8*x^2-48*x-72)*log(exp(-x
+1)-log(2)+3+x)+(x^2+6*x+9)*exp(-x+1)+(-x^2-6*x-9)*log(2)+x^3+9*x^2+27*x+27),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.72, size = 25, normalized size = 0.83 \begin {gather*} \frac {x + 3}{x - 4 \, \log \left (x + e^{\left (-x + 1\right )} - \log \relax (2) + 3\right ) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(-x+1)+4*log(2)-4*x-12)*log(exp(-x+1)-log(2)+3+x)+(-4*x-12)*exp(-x+1)+4*x+12)/((16*exp(-x+1)
-16*log(2)+16*x+48)*log(exp(-x+1)-log(2)+3+x)^2+((-8*x-24)*exp(-x+1)+(8*x+24)*log(2)-8*x^2-48*x-72)*log(exp(-x
+1)-log(2)+3+x)+(x^2+6*x+9)*exp(-x+1)+(-x^2-6*x-9)*log(2)+x^3+9*x^2+27*x+27),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 26, normalized size = 0.87

method	result	size
risch	\(\frac {3+x}{x -4 \ln \left ({\mathrm e}^{1-x}-\ln \relax (2)+3+x \right )+3}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*exp(1-x)+4*ln(2)-4*x-12)*ln(exp(1-x)-ln(2)+3+x)+(-4*x-12)*exp(1-x)+4*x+12)/((16*exp(1-x)-16*ln(2)+16*
x+48)*ln(exp(1-x)-ln(2)+3+x)^2+((-8*x-24)*exp(1-x)+(8*x+24)*ln(2)-8*x^2-48*x-72)*ln(exp(1-x)-ln(2)+3+x)+(x^2+6
*x+9)*exp(1-x)+(-x^2-6*x-9)*ln(2)+x^3+9*x^2+27*x+27),x,method=_RETURNVERBOSE)

[Out]

(3+x)/(x-4*ln(exp(1-x)-ln(2)+3+x)+3)

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maxima [A]  time = 0.61, size = 27, normalized size = 0.90 \begin {gather*} \frac {x + 3}{5 \, x - 4 \, \log \left ({\left (x - \log \relax (2) + 3\right )} e^{x} + e\right ) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(-x+1)+4*log(2)-4*x-12)*log(exp(-x+1)-log(2)+3+x)+(-4*x-12)*exp(-x+1)+4*x+12)/((16*exp(-x+1)
-16*log(2)+16*x+48)*log(exp(-x+1)-log(2)+3+x)^2+((-8*x-24)*exp(-x+1)+(8*x+24)*log(2)-8*x^2-48*x-72)*log(exp(-x
+1)-log(2)+3+x)+(x^2+6*x+9)*exp(-x+1)+(-x^2-6*x-9)*log(2)+x^3+9*x^2+27*x+27),x, algorithm="maxima")

[Out]

(x + 3)/(5*x - 4*log((x - log(2) + 3)*e^x + e) + 3)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - log(x - log(2) + exp(1 - x) + 3)*(4*x - 4*log(2) + 4*exp(1 - x) + 12) - exp(1 - x)*(4*x + 12) + 12)
/(27*x + exp(1 - x)*(6*x + x^2 + 9) + log(x - log(2) + exp(1 - x) + 3)^2*(16*x - 16*log(2) + 16*exp(1 - x) + 4
8) - log(x - log(2) + exp(1 - x) + 3)*(48*x - log(2)*(8*x + 24) + exp(1 - x)*(8*x + 24) + 8*x^2 + 72) + 9*x^2
+ x^3 - log(2)*(6*x + x^2 + 9) + 27),x)

[Out]

\text{Hanged}

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sympy [A]  time = 0.34, size = 22, normalized size = 0.73 \begin {gather*} \frac {- x - 3}{- x + 4 \log {\left (x + e^{1 - x} - \log {\relax (2 )} + 3 \right )} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(-x+1)+4*ln(2)-4*x-12)*ln(exp(-x+1)-ln(2)+3+x)+(-4*x-12)*exp(-x+1)+4*x+12)/((16*exp(-x+1)-16
*ln(2)+16*x+48)*ln(exp(-x+1)-ln(2)+3+x)**2+((-8*x-24)*exp(-x+1)+(8*x+24)*ln(2)-8*x**2-48*x-72)*ln(exp(-x+1)-ln
(2)+3+x)+(x**2+6*x+9)*exp(-x+1)+(-x**2-6*x-9)*ln(2)+x**3+9*x**2+27*x+27),x)

[Out]

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

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