3.47.88 \(\int \frac {e^{e^{\log ^2(\frac {-4 e^x+x}{16+12 x})}+\log ^2(\frac {-4 e^x+x}{16+12 x})} (-8+e^x (8+24 x)) \log (\frac {-4 e^x+x}{16+12 x})}{-4 x-3 x^2+e^x (16+12 x)} \, dx\)

Optimal. Leaf size=28 \[ e^{e^{\log ^2\left (\frac {-4 e^x+x}{2 (3+x+5 (1+x))}\right )}} \]

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Rubi [F]  time = 10.67, 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 {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \left (-8+e^x (8+24 x)\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{-4 x-3 x^2+e^x (16+12 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(E^Log[(-4*E^x + x)/(16 + 12*x)]^2 + Log[(-4*E^x + x)/(16 + 12*x)]^2)*(-8 + E^x*(8 + 24*x))*Log[(-4*E^x
+ x)/(16 + 12*x)])/(-4*x - 3*x^2 + E^x*(16 + 12*x)),x]

[Out]

2*Defer[Int][E^(E^Log[(-4*E^x + x)/(16 + 12*x)]^2 + Log[(-4*E^x + x)/(16 + 12*x)]^2)*Log[(-4*E^x + x)/(16 + 12
*x)], x] - 2*Defer[Int][(E^(E^Log[(-4*E^x + x)/(16 + 12*x)]^2 + Log[(-4*E^x + x)/(16 + 12*x)]^2)*Log[(-4*E^x +
x)/(16 + 12*x)])/(4*E^x - x), x] + 2*Defer[Int][(E^(E^Log[(-4*E^x + x)/(16 + 12*x)]^2 + Log[(-4*E^x + x)/(16
+ 12*x)]^2)*x*Log[(-4*E^x + x)/(16 + 12*x)])/(4*E^x - x), x] - 6*Defer[Int][(E^(E^Log[(-4*E^x + x)/(16 + 12*x)
]^2 + Log[(-4*E^x + x)/(16 + 12*x)]^2)*Log[(-4*E^x + x)/(16 + 12*x)])/(4 + 3*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 \exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \left (-1+e^x+3 e^x x\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{\left (4 e^x-x\right ) (4+3 x)} \, dx\\ &=8 \int \frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \left (-1+e^x+3 e^x x\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{\left (4 e^x-x\right ) (4+3 x)} \, dx\\ &=8 \int \left (\frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) (-1+x) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4 \left (4 e^x-x\right )}+\frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) (1+3 x) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4 (4+3 x)}\right ) \, dx\\ &=2 \int \frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) (-1+x) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4 e^x-x} \, dx+2 \int \frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) (1+3 x) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4+3 x} \, dx\\ &=2 \int \left (-\frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4 e^x-x}+\frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) x \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4 e^x-x}\right ) \, dx+2 \int \left (\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )-\frac {3 \exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4+3 x}\right ) \, dx\\ &=2 \int \exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right ) \, dx-2 \int \frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4 e^x-x} \, dx+2 \int \frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) x \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4 e^x-x} \, dx-6 \int \frac {\exp \left (e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{4+3 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 22, normalized size = 0.79 \begin {gather*} e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^Log[(-4*E^x + x)/(16 + 12*x)]^2 + Log[(-4*E^x + x)/(16 + 12*x)]^2)*(-8 + E^x*(8 + 24*x))*Log[(
-4*E^x + x)/(16 + 12*x)])/(-4*x - 3*x^2 + E^x*(16 + 12*x)),x]

[Out]

E^E^Log[(-4*E^x + x)/(16 + 12*x)]^2

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fricas [A]  time = 0.87, size = 20, normalized size = 0.71 \begin {gather*} e^{\left (e^{\left (\log \left (\frac {x - 4 \, e^{x}}{4 \, {\left (3 \, x + 4\right )}}\right )^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x+8)*exp(x)-8)*log((-4*exp(x)+x)/(12*x+16))*exp(log((-4*exp(x)+x)/(12*x+16))^2)*exp(exp(log((-4
*exp(x)+x)/(12*x+16))^2))/((12*x+16)*exp(x)-3*x^2-4*x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {8 \, {\left ({\left (3 \, x + 1\right )} e^{x} - 1\right )} e^{\left (\log \left (\frac {x - 4 \, e^{x}}{4 \, {\left (3 \, x + 4\right )}}\right )^{2} + e^{\left (\log \left (\frac {x - 4 \, e^{x}}{4 \, {\left (3 \, x + 4\right )}}\right )^{2}\right )}\right )} \log \left (\frac {x - 4 \, e^{x}}{4 \, {\left (3 \, x + 4\right )}}\right )}{3 \, x^{2} - 4 \, {\left (3 \, x + 4\right )} e^{x} + 4 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x+8)*exp(x)-8)*log((-4*exp(x)+x)/(12*x+16))*exp(log((-4*exp(x)+x)/(12*x+16))^2)*exp(exp(log((-4
*exp(x)+x)/(12*x+16))^2))/((12*x+16)*exp(x)-3*x^2-4*x),x, algorithm="giac")

[Out]

integrate(-8*((3*x + 1)*e^x - 1)*e^(log(1/4*(x - 4*e^x)/(3*x + 4))^2 + e^(log(1/4*(x - 4*e^x)/(3*x + 4))^2))*l
og(1/4*(x - 4*e^x)/(3*x + 4))/(3*x^2 - 4*(3*x + 4)*e^x + 4*x), x)

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maple [C]  time = 0.61, size = 163, normalized size = 5.82

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{\frac {\left (-i \pi \mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{x}-x \right )}{x +\frac {4}{3}}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{x}-x \right )}{x +\frac {4}{3}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x +\frac {4}{3}}\right )+i \pi \mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{x}-x \right )}{x +\frac {4}{3}}\right )^{2} \mathrm {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right )\right )+i \pi \,\mathrm {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{x}-x \right )}{x +\frac {4}{3}}\right ) \mathrm {csgn}\left (\frac {i}{x +\frac {4}{3}}\right ) \mathrm {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right )\right )+2 \ln \left (x +\frac {4}{3}\right )-2 \ln \left (-4 \,{\mathrm e}^{x}+x \right )+4 \ln \relax (2)+2 \ln \relax (3)\right )^{2}}{4}}}\)	\(163\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((24*x+8)*exp(x)-8)*ln((-4*exp(x)+x)/(12*x+16))*exp(ln((-4*exp(x)+x)/(12*x+16))^2)*exp(exp(ln((-4*exp(x)+x
)/(12*x+16))^2))/((12*x+16)*exp(x)-3*x^2-4*x),x,method=_RETURNVERBOSE)

[Out]

exp(exp(1/4*(-I*Pi*csgn(I/(x+4/3)*(4*exp(x)-x))^3-I*Pi*csgn(I/(x+4/3)*(4*exp(x)-x))^2*csgn(I/(x+4/3))+I*Pi*csg
n(I/(x+4/3)*(4*exp(x)-x))^2*csgn(I*(4*exp(x)-x))+I*Pi*csgn(I/(x+4/3)*(4*exp(x)-x))*csgn(I/(x+4/3))*csgn(I*(4*e
xp(x)-x))+2*ln(x+4/3)-2*ln(-4*exp(x)+x)+4*ln(2)+2*ln(3))^2))

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maxima [B]  time = 1.01, size = 62, normalized size = 2.21 \begin {gather*} e^{\left (e^{\left (4 \, \log \relax (2)^{2} + 4 \, \log \relax (2) \log \left (3 \, x + 4\right ) + \log \left (3 \, x + 4\right )^{2} - 4 \, \log \relax (2) \log \left (x - 4 \, e^{x}\right ) - 2 \, \log \left (3 \, x + 4\right ) \log \left (x - 4 \, e^{x}\right ) + \log \left (x - 4 \, e^{x}\right )^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x+8)*exp(x)-8)*log((-4*exp(x)+x)/(12*x+16))*exp(log((-4*exp(x)+x)/(12*x+16))^2)*exp(exp(log((-4
*exp(x)+x)/(12*x+16))^2))/((12*x+16)*exp(x)-3*x^2-4*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.62, size = 19, normalized size = 0.68 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{{\ln \left (\frac {x-4\,{\mathrm {e}}^x}{12\,x+16}\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp(log((x - 4*exp(x))/(12*x + 16))^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x+8)*exp(x)-8)*ln((-4*exp(x)+x)/(12*x+16))*exp(ln((-4*exp(x)+x)/(12*x+16))**2)*exp(exp(ln((-4*e
xp(x)+x)/(12*x+16))**2))/((12*x+16)*exp(x)-3*x**2-4*x),x)

[Out]

Timed out

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