3.12.54 \(\int \frac {3 e^{x-\frac {e^{e^x}}{4+x}} ((-80-120 x-45 x^2-5 x^3) \log (2)+e^{e^x} (-5 x \log (2)+e^x (20 x+5 x^2) \log (2)))}{16+8 x+x^2} \, dx\)

Optimal. Leaf size=21 \[ -15 e^{x-\frac {e^{e^x}}{4+x}} x \log (2) \]

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Rubi [F]  time = 10.76, 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 {3 e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3*E^(x - E^E^x/(4 + x))*((-80 - 120*x - 45*x^2 - 5*x^3)*Log[2] + E^E^x*(-5*x*Log[2] + E^x*(20*x + 5*x^2)*
Log[2])))/(16 + 8*x + x^2),x]

[Out]

-15*Log[2]*Defer[Int][E^(x - E^E^x/(4 + x)), x] + 15*Log[2]*Defer[Int][E^(E^x + 2*x - E^E^x/(4 + x)), x] - 15*
Log[2]*Defer[Int][E^(x - E^E^x/(4 + x))*x, x] + 60*Log[2]*Defer[Int][E^(E^x + x - E^E^x/(4 + x))/(4 + x)^2, x]
- 15*Log[2]*Defer[Int][E^(E^x + x - E^E^x/(4 + x))/(4 + x), x] - 60*Log[2]*Defer[Int][E^(E^x + 2*x - E^E^x/(4
+ x))/(4 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=3 \int \frac {e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx\\ &=3 \int \frac {e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{(4+x)^2} \, dx\\ &=3 \int \frac {5 e^{x-\frac {e^{e^x}}{4+x}} \left (-16-24 x-e^{e^x} x+4 e^{e^x+x} x-9 x^2+e^{e^x+x} x^2-x^3\right ) \log (2)}{(4+x)^2} \, dx\\ &=(15 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}} \left (-16-24 x-e^{e^x} x+4 e^{e^x+x} x-9 x^2+e^{e^x+x} x^2-x^3\right )}{(4+x)^2} \, dx\\ &=(15 \log (2)) \int \left (-\frac {16 e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2}-\frac {24 e^{x-\frac {e^{e^x}}{4+x}} x}{(4+x)^2}-\frac {e^{e^x+x-\frac {e^{e^x}}{4+x}} x}{(4+x)^2}-\frac {9 e^{x-\frac {e^{e^x}}{4+x}} x^2}{(4+x)^2}-\frac {e^{x-\frac {e^{e^x}}{4+x}} x^3}{(4+x)^2}+\frac {e^{e^x+2 x-\frac {e^{e^x}}{4+x}} x}{4+x}\right ) \, dx\\ &=-\left ((15 \log (2)) \int \frac {e^{e^x+x-\frac {e^{e^x}}{4+x}} x}{(4+x)^2} \, dx\right )-(15 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}} x^3}{(4+x)^2} \, dx+(15 \log (2)) \int \frac {e^{e^x+2 x-\frac {e^{e^x}}{4+x}} x}{4+x} \, dx-(135 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}} x^2}{(4+x)^2} \, dx-(240 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2} \, dx-(360 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}} x}{(4+x)^2} \, dx\\ &=-\left ((15 \log (2)) \int \left (-8 e^{x-\frac {e^{e^x}}{4+x}}+e^{x-\frac {e^{e^x}}{4+x}} x-\frac {64 e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2}+\frac {48 e^{x-\frac {e^{e^x}}{4+x}}}{4+x}\right ) \, dx\right )-(15 \log (2)) \int \left (-\frac {4 e^{e^x+x-\frac {e^{e^x}}{4+x}}}{(4+x)^2}+\frac {e^{e^x+x-\frac {e^{e^x}}{4+x}}}{4+x}\right ) \, dx+(15 \log (2)) \int \left (e^{e^x+2 x-\frac {e^{e^x}}{4+x}}-\frac {4 e^{e^x+2 x-\frac {e^{e^x}}{4+x}}}{4+x}\right ) \, dx-(135 \log (2)) \int \left (e^{x-\frac {e^{e^x}}{4+x}}+\frac {16 e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2}-\frac {8 e^{x-\frac {e^{e^x}}{4+x}}}{4+x}\right ) \, dx-(240 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2} \, dx-(360 \log (2)) \int \left (-\frac {4 e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2}+\frac {e^{x-\frac {e^{e^x}}{4+x}}}{4+x}\right ) \, dx\\ &=(15 \log (2)) \int e^{e^x+2 x-\frac {e^{e^x}}{4+x}} \, dx-(15 \log (2)) \int e^{x-\frac {e^{e^x}}{4+x}} x \, dx-(15 \log (2)) \int \frac {e^{e^x+x-\frac {e^{e^x}}{4+x}}}{4+x} \, dx+(60 \log (2)) \int \frac {e^{e^x+x-\frac {e^{e^x}}{4+x}}}{(4+x)^2} \, dx-(60 \log (2)) \int \frac {e^{e^x+2 x-\frac {e^{e^x}}{4+x}}}{4+x} \, dx+(120 \log (2)) \int e^{x-\frac {e^{e^x}}{4+x}} \, dx-(135 \log (2)) \int e^{x-\frac {e^{e^x}}{4+x}} \, dx-(240 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2} \, dx-(360 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{4+x} \, dx-(720 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{4+x} \, dx+(960 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2} \, dx+(1080 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{4+x} \, dx+(1440 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2} \, dx-(2160 \log (2)) \int \frac {e^{x-\frac {e^{e^x}}{4+x}}}{(4+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.41, size = 21, normalized size = 1.00 \begin {gather*} -15 e^{x-\frac {e^{e^x}}{4+x}} x \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*E^(x - E^E^x/(4 + x))*((-80 - 120*x - 45*x^2 - 5*x^3)*Log[2] + E^E^x*(-5*x*Log[2] + E^x*(20*x + 5
*x^2)*Log[2])))/(16 + 8*x + x^2),x]

[Out]

-15*E^(x - E^E^x/(4 + x))*x*Log[2]

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fricas [A]  time = 0.69, size = 30, normalized size = 1.43 \begin {gather*} -5 \, x e^{\left (\frac {x^{2} + {\left (x + 4\right )} \log \relax (3) + 4 \, x - e^{\left (e^{x}\right )}}{x + 4}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2+20*x)*log(2)*exp(x)-5*x*log(2))*exp(exp(x))+(-5*x^3-45*x^2-120*x-80)*log(2))*exp(log(3/exp(
exp(exp(x))/(4+x)))+x)/(x^2+8*x+16),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5 \, {\left ({\left ({\left (x^{2} + 4 \, x\right )} e^{x} \log \relax (2) - x \log \relax (2)\right )} e^{\left (e^{x}\right )} - {\left (x^{3} + 9 \, x^{2} + 24 \, x + 16\right )} \log \relax (2)\right )} e^{\left (x + \log \left (3 \, e^{\left (-\frac {e^{\left (e^{x}\right )}}{x + 4}\right )}\right )\right )}}{x^{2} + 8 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2+20*x)*log(2)*exp(x)-5*x*log(2))*exp(exp(x))+(-5*x^3-45*x^2-120*x-80)*log(2))*exp(log(3/exp(
exp(exp(x))/(4+x)))+x)/(x^2+8*x+16),x, algorithm="giac")

[Out]

integrate(5*(((x^2 + 4*x)*e^x*log(2) - x*log(2))*e^(e^x) - (x^3 + 9*x^2 + 24*x + 16)*log(2))*e^(x + log(3*e^(-
e^(e^x)/(x + 4))))/(x^2 + 8*x + 16), x)

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (5 x^{2}+20 x \right ) \ln \relax (2) {\mathrm e}^{x}-5 x \ln \relax (2)\right ) {\mathrm e}^{{\mathrm e}^{x}}+\left (-5 x^{3}-45 x^{2}-120 x -80\right ) \ln \relax (2)\right ) {\mathrm e}^{\ln \left (3 \,{\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{4+x}}\right )+x}}{x^{2}+8 x +16}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*x^2+20*x)*ln(2)*exp(x)-5*x*ln(2))*exp(exp(x))+(-5*x^3-45*x^2-120*x-80)*ln(2))*exp(ln(3/exp(exp(exp(x)
)/(4+x)))+x)/(x^2+8*x+16),x)

[Out]

int((((5*x^2+20*x)*ln(2)*exp(x)-5*x*ln(2))*exp(exp(x))+(-5*x^3-45*x^2-120*x-80)*ln(2))*exp(ln(3/exp(exp(exp(x)
)/(4+x)))+x)/(x^2+8*x+16),x)

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maxima [A]  time = 0.88, size = 18, normalized size = 0.86 \begin {gather*} -15 \, x e^{\left (x - \frac {e^{\left (e^{x}\right )}}{x + 4}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2+20*x)*log(2)*exp(x)-5*x*log(2))*exp(exp(x))+(-5*x^3-45*x^2-120*x-80)*log(2))*exp(log(3/exp(
exp(exp(x))/(4+x)))+x)/(x^2+8*x+16),x, algorithm="maxima")

[Out]

-15*x*e^(x - e^(e^x)/(x + 4))*log(2)

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mupad [B]  time = 0.58, size = 18, normalized size = 0.86 \begin {gather*} -15\,x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{x+4}}\,{\mathrm {e}}^x\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + log(3*exp(-exp(exp(x))/(x + 4))))*(exp(exp(x))*(5*x*log(2) - exp(x)*log(2)*(20*x + 5*x^2)) + log
(2)*(120*x + 45*x^2 + 5*x^3 + 80)))/(8*x + x^2 + 16),x)

[Out]

-15*x*exp(-exp(exp(x))/(x + 4))*exp(x)*log(2)

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sympy [A]  time = 12.75, size = 20, normalized size = 0.95 \begin {gather*} - 15 x e^{x} e^{- \frac {e^{e^{x}}}{x + 4}} \log {\relax (2 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x**2+20*x)*ln(2)*exp(x)-5*x*ln(2))*exp(exp(x))+(-5*x**3-45*x**2-120*x-80)*ln(2))*exp(ln(3/exp(e
xp(exp(x))/(4+x)))+x)/(x**2+8*x+16),x)

[Out]

-15*x*exp(x)*exp(-exp(exp(x))/(x + 4))*log(2)

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