3.43.56 \(\int \frac {e^x (40+30 x-10 x^2)+(20-5 x+e^x (80 x+20 x^2-10 x^3)) \log (\log (2))+(-10 e^x x+(-5 x-10 e^x x^2) \log (\log (2))) \log (\frac {2 e^x x+(x+2 e^x x^2) \log (\log (2))}{\log (\log (2))})}{2 e^x x+(x+2 e^x x^2) \log (\log (2))} \, dx\)

Optimal. Leaf size=26 \[ 5 (4-x) \log \left (x+e^x x \left (2 x+\frac {2}{\log (\log (2))}\right )\right ) \]

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Rubi [F]  time = 2.78, 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 {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(40 + 30*x - 10*x^2) + (20 - 5*x + E^x*(80*x + 20*x^2 - 10*x^3))*Log[Log[2]] + (-10*E^x*x + (-5*x - 1
0*E^x*x^2)*Log[Log[2]])*Log[(2*E^x*x + (x + 2*E^x*x^2)*Log[Log[2]])/Log[Log[2]]])/(2*E^x*x + (x + 2*E^x*x^2)*L
og[Log[2]]),x]

[Out]

20*x + 20*Log[x] - 5*x*Log[x + 2*E^x*x*(x + Log[Log[2]]^(-1))] + 5*(4 + Log[Log[2]]^(-1))*Log[1 + x*Log[Log[2]
]] - (5*Log[1 + x*Log[Log[2]]])/Log[Log[2]] - 20*Log[Log[2]]*Defer[Int][(2*E^x + Log[Log[2]] + 2*E^x*x*Log[Log
[2]])^(-1), x] + 5*Log[Log[2]]*Defer[Int][1/((1 + x*Log[Log[2]])*(2*E^x + Log[Log[2]] + 2*E^x*x*Log[Log[2]])),
x] - 5*Log[Log[2]]*(1 + 4*Log[Log[2]])*Defer[Int][1/((1 + x*Log[Log[2]])*(2*E^x + Log[Log[2]] + 2*E^x*x*Log[L
og[2]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {5 (-4+x) \left (\log (\log (2))+2 e^x \left (1+x+2 x \log (\log (2))+x^2 \log (\log (2))\right )\right )}{x \left (\log (\log (2))+2 e^x (1+x \log (\log (2)))\right )}-5 \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )\right ) \, dx\\ &=-\left (5 \int \frac {(-4+x) \left (\log (\log (2))+2 e^x \left (1+x+2 x \log (\log (2))+x^2 \log (\log (2))\right )\right )}{x \left (\log (\log (2))+2 e^x (1+x \log (\log (2)))\right )} \, dx\right )-5 \int \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right ) \, dx\\ &=-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+5 \int \frac {\log (\log (2))+2 e^x \left (1+x+2 x \log (\log (2))+x^2 \log (\log (2))\right )}{\log (\log (2))+2 e^x (1+x \log (\log (2)))} \, dx-5 \int \left (-\frac {(-4+x) \log (\log (2)) (1+\log (\log (2))+x \log (\log (2)))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )}+\frac {(4-x) \left (-1-x^2 \log (\log (2))-x (1+2 \log (\log (2)))\right )}{x (1+x \log (\log (2)))}\right ) \, dx\\ &=-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )-5 \int \frac {(4-x) \left (-1-x^2 \log (\log (2))-x (1+2 \log (\log (2)))\right )}{x (1+x \log (\log (2)))} \, dx+5 \int \left (-\frac {x \log (\log (2)) (1+\log (\log (2))+x \log (\log (2)))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )}+\frac {1+x^2 \log (\log (2))+x (1+2 \log (\log (2)))}{1+x \log (\log (2))}\right ) \, dx+(5 \log (\log (2))) \int \frac {(-4+x) (1+\log (\log (2))+x \log (\log (2)))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx\\ &=-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+5 \int \frac {1+x^2 \log (\log (2))+x (1+2 \log (\log (2)))}{1+x \log (\log (2))} \, dx-5 \int \left (-2-\frac {4}{x}+x+\frac {-1-4 \log (\log (2))}{1+x \log (\log (2))}\right ) \, dx-(5 \log (\log (2))) \int \frac {x (1+\log (\log (2))+x \log (\log (2)))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx+(5 \log (\log (2))) \int \left (-\frac {3}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))}+\frac {x}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))}-\frac {1+4 \log (\log (2))}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )}\right ) \, dx\\ &=10 x-\frac {5 x^2}{2}+20 \log (x)-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+5 \left (4+\frac {1}{\log (\log (2))}\right ) \log (1+x \log (\log (2)))+5 \int \left (2+x+\frac {1}{-1-x \log (\log (2))}\right ) \, dx+(5 \log (\log (2))) \int \frac {x}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))} \, dx-(5 \log (\log (2))) \int \left (\frac {1}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))}+\frac {x}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))}-\frac {1}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )}\right ) \, dx-(15 \log (\log (2))) \int \frac {1}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))} \, dx-(5 \log (\log (2)) (1+4 \log (\log (2)))) \int \frac {1}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx\\ &=20 x+20 \log (x)-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+5 \left (4+\frac {1}{\log (\log (2))}\right ) \log (1+x \log (\log (2)))-\frac {5 \log (1+x \log (\log (2)))}{\log (\log (2))}-(5 \log (\log (2))) \int \frac {1}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))} \, dx+(5 \log (\log (2))) \int \frac {1}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx-(15 \log (\log (2))) \int \frac {1}{2 e^x+\log (\log (2))+2 e^x x \log (\log (2))} \, dx-(5 \log (\log (2)) (1+4 \log (\log (2)))) \int \frac {1}{(1+x \log (\log (2))) \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.62, size = 45, normalized size = 1.73 \begin {gather*} 20 \log (x)-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+20 \log \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(40 + 30*x - 10*x^2) + (20 - 5*x + E^x*(80*x + 20*x^2 - 10*x^3))*Log[Log[2]] + (-10*E^x*x + (-5
*x - 10*E^x*x^2)*Log[Log[2]])*Log[(2*E^x*x + (x + 2*E^x*x^2)*Log[Log[2]])/Log[Log[2]]])/(2*E^x*x + (x + 2*E^x*
x^2)*Log[Log[2]]),x]

[Out]

20*Log[x] - 5*x*Log[x + 2*E^x*x*(x + Log[Log[2]]^(-1))] + 20*Log[2*E^x + Log[Log[2]] + 2*E^x*x*Log[Log[2]]]

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fricas [A]  time = 0.64, size = 31, normalized size = 1.19 \begin {gather*} -5 \, {\left (x - 4\right )} \log \left (\frac {2 \, x e^{x} + {\left (2 \, x^{2} e^{x} + x\right )} \log \left (\log \relax (2)\right )}{\log \left (\log \relax (2)\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x^2+x)*log(log(2))+2*exp(x)*x)/log(lo
g(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5*x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2
))+2*exp(x)*x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.23, size = 56, normalized size = 2.15 \begin {gather*} -5 \, x \log \left (2 \, x^{2} e^{x} \log \left (\log \relax (2)\right ) + 2 \, x e^{x} + x \log \left (\log \relax (2)\right )\right ) + 5 \, x \log \left (\log \left (\log \relax (2)\right )\right ) + 20 \, \log \left (2 \, x e^{x} \log \left (\log \relax (2)\right ) + 2 \, e^{x} + \log \left (\log \relax (2)\right )\right ) + 20 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x^2+x)*log(log(2))+2*exp(x)*x)/log(lo
g(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5*x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2
))+2*exp(x)*x),x, algorithm="giac")

[Out]

-5*x*log(2*x^2*e^x*log(log(2)) + 2*x*e^x + x*log(log(2))) + 5*x*log(log(log(2))) + 20*log(2*x*e^x*log(log(2))
+ 2*e^x + log(log(2))) + 20*log(x)

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maple [B]  time = 0.38, size = 56, normalized size = 2.15

method	result	size
default	\(20 \ln \left (-x \left (2 \,{\mathrm e}^{x} \ln \left (\ln \relax (2)\right ) x +2 \,{\mathrm e}^{x}+\ln \left (\ln \relax (2)\right )\right )\right )+5 \ln \left (-\ln \left (\ln \relax (2)\right )\right ) x -5 x \ln \left (-x \left (2 \,{\mathrm e}^{x} \ln \left (\ln \relax (2)\right ) x +2 \,{\mathrm e}^{x}+\ln \left (\ln \relax (2)\right )\right )\right )\)	\(56\)
norman	\(20 \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \relax (2)\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \relax (2)\right )}\right )-5 x \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \relax (2)\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \relax (2)\right )}\right )\)	\(59\)
risch	\(-5 x \ln \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )-5 x \ln \relax (x )+\frac {5 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )\right ) \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )\right )}{2}-\frac {5 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )\right )^{2}}{2}-\frac {5 i \pi x \,\mathrm {csgn}\left (i \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )\right ) \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )\right )^{2}}{2}+5 i \pi x \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )\right )^{2}-\frac {5 i \pi x \mathrm {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )\right )^{3}}{2}-5 i \pi x -5 x \ln \relax (2)+5 \ln \left (-\ln \left (\ln \relax (2)\right )\right ) x +20 \ln \left (x^{2} \ln \left (\ln \relax (2)\right )+x \right )+20 \ln \left ({\mathrm e}^{x}+\frac {\ln \left (\ln \relax (2)\right )}{2 x \ln \left (\ln \relax (2)\right )+2}\right )\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-10*exp(x)*x^2-5*x)*ln(ln(2))-10*exp(x)*x)*ln(((2*exp(x)*x^2+x)*ln(ln(2))+2*exp(x)*x)/ln(ln(2)))+((-10*
x^3+20*x^2+80*x)*exp(x)-5*x+20)*ln(ln(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*ln(ln(2))+2*exp(x)*x),x,
method=_RETURNVERBOSE)

[Out]

20*ln(-x*(2*exp(x)*ln(ln(2))*x+2*exp(x)+ln(ln(2))))+5*ln(-ln(ln(2)))*x-5*x*ln(-x*(2*exp(x)*ln(ln(2))*x+2*exp(x
)+ln(ln(2))))

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maxima [B]  time = 0.48, size = 75, normalized size = 2.88 \begin {gather*} -5 \, x \log \left (2 \, {\left (x \log \left (\log \relax (2)\right ) + 1\right )} e^{x} + \log \left (\log \relax (2)\right )\right ) - 5 \, x \log \relax (x) + 5 \, x \log \left (\log \left (\log \relax (2)\right )\right ) + 20 \, \log \left (x \log \left (\log \relax (2)\right ) + 1\right ) + 20 \, \log \relax (x) + 20 \, \log \left (\frac {2 \, {\left (x \log \left (\log \relax (2)\right ) + 1\right )} e^{x} + \log \left (\log \relax (2)\right )}{2 \, {\left (x \log \left (\log \relax (2)\right ) + 1\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x^2+x)*log(log(2))+2*exp(x)*x)/log(lo
g(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5*x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2
))+2*exp(x)*x),x, algorithm="maxima")

[Out]

-5*x*log(2*(x*log(log(2)) + 1)*e^x + log(log(2))) - 5*x*log(x) + 5*x*log(log(log(2))) + 20*log(x*log(log(2)) +
1) + 20*log(x) + 20*log(1/2*(2*(x*log(log(2)) + 1)*e^x + log(log(2)))/(x*log(log(2)) + 1))

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mupad [B]  time = 4.19, size = 71, normalized size = 2.73 \begin {gather*} 20\,\ln \left (\frac {\ln \left (\ln \relax (2)\right )+2\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^x\,\ln \left (\ln \relax (2)\right )}{x\,\ln \left (\ln \relax (2)\right )+1}\right )+20\,\ln \left (\ln \left (\ln \relax (2)\right )\,x^2+x\right )-5\,x\,\ln \left (\frac {\ln \left (\ln \relax (2)\right )\,\left (x+2\,x^2\,{\mathrm {e}}^x\right )+2\,x\,{\mathrm {e}}^x}{\ln \left (\ln \relax (2)\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(2))*(exp(x)*(80*x + 20*x^2 - 10*x^3) - 5*x + 20) - log((log(log(2))*(x + 2*x^2*exp(x)) + 2*x*exp(
x))/log(log(2)))*(log(log(2))*(5*x + 10*x^2*exp(x)) + 10*x*exp(x)) + exp(x)*(30*x - 10*x^2 + 40))/(log(log(2))
*(x + 2*x^2*exp(x)) + 2*x*exp(x)),x)

[Out]

20*log((log(log(2)) + 2*exp(x) + 2*x*exp(x)*log(log(2)))/(x*log(log(2)) + 1)) + 20*log(x + x^2*log(log(2))) -
5*x*log((log(log(2))*(x + 2*x^2*exp(x)) + 2*x*exp(x))/log(log(2)))

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sympy [B]  time = 0.90, size = 66, normalized size = 2.54 \begin {gather*} - 5 x \log {\left (\frac {2 x e^{x} + \left (2 x^{2} e^{x} + x\right ) \log {\left (\log {\relax (2 )} \right )}}{\log {\left (\log {\relax (2 )} \right )}} \right )} + 20 \log {\left (x^{2} \log {\left (\log {\relax (2 )} \right )} + x \right )} + 20 \log {\left (e^{x} + \frac {\log {\left (\log {\relax (2 )} \right )}}{2 x \log {\left (\log {\relax (2 )} \right )} + 2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*exp(x)*x**2-5*x)*ln(ln(2))-10*exp(x)*x)*ln(((2*exp(x)*x**2+x)*ln(ln(2))+2*exp(x)*x)/ln(ln(2))
)+((-10*x**3+20*x**2+80*x)*exp(x)-5*x+20)*ln(ln(2))+(-10*x**2+30*x+40)*exp(x))/((2*exp(x)*x**2+x)*ln(ln(2))+2*
exp(x)*x),x)

[Out]

-5*x*log((2*x*exp(x) + (2*x**2*exp(x) + x)*log(log(2)))/log(log(2))) + 20*log(x**2*log(log(2)) + x) + 20*log(e
xp(x) + log(log(2))/(2*x*log(log(2)) + 2))

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