3.95.20 \(\int \frac {24 e^{2 x}+(e^x (-120-24 e^5)+24 e^{2 x} x) \log (\frac {2}{x})+e^x (-120 x-24 e^5 x) \log ^2(\frac {2}{x})+(12 e^{3 x} x \log (\frac {2}{x})+e^{2 x} (-120 x-24 e^5 x) \log ^2(\frac {2}{x})+e^x (299 x+120 e^5 x+12 e^{10} x) \log ^3(\frac {2}{x})) \log (\frac {-12 e^{2 x}+e^x (120+24 e^5) \log (\frac {2}{x})+(-299-120 e^5-12 e^{10}) \log ^2(\frac {2}{x})}{\log ^2(\frac {2}{x})})}{(12 e^{2 x} x \log (\frac {2}{x})+e^x (-120 x-24 e^5 x) \log ^2(\frac {2}{x})+(299 x+120 e^5 x+12 e^{10} x) \log ^3(\frac {2}{x})) \log (\frac {-12 e^{2 x}+e^x (120+24 e^5) \log (\frac {2}{x})+(-299-120 e^5-12 e^{10}) \log ^2(\frac {2}{x})}{\log ^2(\frac {2}{x})})} \, dx\)

Optimal. Leaf size=31 \[ e^x+\log \left (\log \left (1-12 \left (-5-e^5+\frac {e^x}{\log \left (\frac {2}{x}\right )}\right )^2\right )\right ) \]

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Rubi [F]  time = 10.48, 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 {24 e^{2 x}+\left (e^x \left (-120-24 e^5\right )+24 e^{2 x} x\right ) \log \left (\frac {2}{x}\right )+e^x \left (-120 x-24 e^5 x\right ) \log ^2\left (\frac {2}{x}\right )+\left (12 e^{3 x} x \log \left (\frac {2}{x}\right )+e^{2 x} \left (-120 x-24 e^5 x\right ) \log ^2\left (\frac {2}{x}\right )+e^x \left (299 x+120 e^5 x+12 e^{10} x\right ) \log ^3\left (\frac {2}{x}\right )\right ) \log \left (\frac {-12 e^{2 x}+e^x \left (120+24 e^5\right ) \log \left (\frac {2}{x}\right )+\left (-299-120 e^5-12 e^{10}\right ) \log ^2\left (\frac {2}{x}\right )}{\log ^2\left (\frac {2}{x}\right )}\right )}{\left (12 e^{2 x} x \log \left (\frac {2}{x}\right )+e^x \left (-120 x-24 e^5 x\right ) \log ^2\left (\frac {2}{x}\right )+\left (299 x+120 e^5 x+12 e^{10} x\right ) \log ^3\left (\frac {2}{x}\right )\right ) \log \left (\frac {-12 e^{2 x}+e^x \left (120+24 e^5\right ) \log \left (\frac {2}{x}\right )+\left (-299-120 e^5-12 e^{10}\right ) \log ^2\left (\frac {2}{x}\right )}{\log ^2\left (\frac {2}{x}\right )}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(24*E^(2*x) + (E^x*(-120 - 24*E^5) + 24*E^(2*x)*x)*Log[2/x] + E^x*(-120*x - 24*E^5*x)*Log[2/x]^2 + (12*E^(
3*x)*x*Log[2/x] + E^(2*x)*(-120*x - 24*E^5*x)*Log[2/x]^2 + E^x*(299*x + 120*E^5*x + 12*E^10*x)*Log[2/x]^3)*Log
[(-12*E^(2*x) + E^x*(120 + 24*E^5)*Log[2/x] + (-299 - 120*E^5 - 12*E^10)*Log[2/x]^2)/Log[2/x]^2])/((12*E^(2*x)
*x*Log[2/x] + E^x*(-120*x - 24*E^5*x)*Log[2/x]^2 + (299*x + 120*E^5*x + 12*E^10*x)*Log[2/x]^3)*Log[(-12*E^(2*x
) + E^x*(120 + 24*E^5)*Log[2/x] + (-299 - 120*E^5 - 12*E^10)*Log[2/x]^2)/Log[2/x]^2]),x]

[Out]

E^x + 2*Defer[Int][Log[-299*(1 + (12*E^5*(10 + E^5))/299) - (12*E^(2*x))/Log[2/x]^2 + (24*E^x*(5 + E^5))/Log[2
/x]]^(-1), x] + 2*Defer[Int][1/(x*Log[2/x]*Log[-299*(1 + (12*E^5*(10 + E^5))/299) - (12*E^(2*x))/Log[2/x]^2 +
(24*E^x*(5 + E^5))/Log[2/x]]), x] + 24*(5 + E^5)*Defer[Int][E^x/(x*(12*E^(2*x) - 120*E^x*(1 + E^5/5)*Log[2/x]
+ 299*(1 + (12*E^5*(10 + E^5))/299)*Log[2/x]^2)*Log[-299*(1 + (12*E^5*(10 + E^5))/299) - (12*E^(2*x))/Log[2/x]
^2 + (24*E^x*(5 + E^5))/Log[2/x]]), x] + 24*(5 + E^5)*Defer[Int][(E^x*Log[2/x])/((12*E^(2*x) - 120*E^x*(1 + E^
5/5)*Log[2/x] + 299*(1 + (12*E^5*(10 + E^5))/299)*Log[2/x]^2)*Log[-299*(1 + (12*E^5*(10 + E^5))/299) - (12*E^(
2*x))/Log[2/x]^2 + (24*E^x*(5 + E^5))/Log[2/x]]), x] - 2*(299 + 120*E^5 + 12*E^10)*Defer[Int][Log[2/x]/(x*(12*
E^(2*x) - 120*E^x*(1 + E^5/5)*Log[2/x] + 299*(1 + (12*E^5*(10 + E^5))/299)*Log[2/x]^2)*Log[-299*(1 + (12*E^5*(
10 + E^5))/299) - (12*E^(2*x))/Log[2/x]^2 + (24*E^x*(5 + E^5))/Log[2/x]]), x] - 2*(299 + 120*E^5 + 12*E^10)*De
fer[Int][Log[2/x]^2/((12*E^(2*x) - 120*E^x*(1 + E^5/5)*Log[2/x] + 299*(1 + (12*E^5*(10 + E^5))/299)*Log[2/x]^2
)*Log[-299*(1 + (12*E^5*(10 + E^5))/299) - (12*E^(2*x))/Log[2/x]^2 + (24*E^x*(5 + E^5))/Log[2/x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^x+\frac {2 \left (1+x \log \left (\frac {2}{x}\right )\right )}{x \log \left (\frac {2}{x}\right ) \log \left (-299 \left (1+\frac {12}{299} e^5 \left (10+e^5\right )\right )-\frac {12 e^{2 x}}{\log ^2\left (\frac {2}{x}\right )}+\frac {24 e^x \left (5+e^5\right )}{\log \left (\frac {2}{x}\right )}\right )}+\frac {2 \left (60 e^x \left (1+\frac {e^5}{5}\right )-299 \left (1+\frac {12}{299} e^5 \left (10+e^5\right )\right ) \log \left (\frac {2}{x}\right )\right ) \left (1+x \log \left (\frac {2}{x}\right )\right )}{x \left (12 e^{2 x}-120 e^x \left (1+\frac {e^5}{5}\right ) \log \left (\frac {2}{x}\right )+299 \left (1+\frac {12}{299} e^5 \left (10+e^5\right )\right ) \log ^2\left (\frac {2}{x}\right )\right ) \log \left (-299 \left (1+\frac {12}{299} e^5 \left (10+e^5\right )\right )-\frac {12 e^{2 x}}{\log ^2\left (\frac {2}{x}\right )}+\frac {24 e^x \left (5+e^5\right )}{\log \left (\frac {2}{x}\right )}\right )}\right ) \, dx\\ &=2 \int \frac {1+x \log \left (\frac {2}{x}\right )}{x \log \left (\frac {2}{x}\right ) \log \left (-299 \left (1+\frac {12}{299} e^5 \left (10+e^5\right )\right )-\frac {12 e^{2 x}}{\log ^2\left (\frac {2}{x}\right )}+\frac {24 e^x \left (5+e^5\right )}{\log \left (\frac {2}{x}\right )}\right )} \, dx+2 \int \frac {\left (60 e^x \left (1+\frac {e^5}{5}\right )-299 \left (1+\frac {12}{299} e^5 \left (10+e^5\right )\right ) \log \left (\frac {2}{x}\right )\right ) \left (1+x \log \left (\frac {2}{x}\right )\right )}{x \left (12 e^{2 x}-120 e^x \left (1+\frac {e^5}{5}\right ) \log \left (\frac {2}{x}\right )+299 \left (1+\frac {12}{299} e^5 \left (10+e^5\right )\right ) \log ^2\left (\frac {2}{x}\right )\right ) \log \left (-299 \left (1+\frac {12}{299} e^5 \left (10+e^5\right )\right )-\frac {12 e^{2 x}}{\log ^2\left (\frac {2}{x}\right )}+\frac {24 e^x \left (5+e^5\right )}{\log \left (\frac {2}{x}\right )}\right )} \, dx+\int e^x \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 51, normalized size = 1.65 \begin {gather*} e^x+\log \left (\log \left (-299-120 e^5-12 e^{10}-\frac {12 e^{2 x}}{\log ^2\left (\frac {2}{x}\right )}+\frac {24 e^x \left (5+e^5\right )}{\log \left (\frac {2}{x}\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24*E^(2*x) + (E^x*(-120 - 24*E^5) + 24*E^(2*x)*x)*Log[2/x] + E^x*(-120*x - 24*E^5*x)*Log[2/x]^2 + (
12*E^(3*x)*x*Log[2/x] + E^(2*x)*(-120*x - 24*E^5*x)*Log[2/x]^2 + E^x*(299*x + 120*E^5*x + 12*E^10*x)*Log[2/x]^
3)*Log[(-12*E^(2*x) + E^x*(120 + 24*E^5)*Log[2/x] + (-299 - 120*E^5 - 12*E^10)*Log[2/x]^2)/Log[2/x]^2])/((12*E
^(2*x)*x*Log[2/x] + E^x*(-120*x - 24*E^5*x)*Log[2/x]^2 + (299*x + 120*E^5*x + 12*E^10*x)*Log[2/x]^3)*Log[(-12*
E^(2*x) + E^x*(120 + 24*E^5)*Log[2/x] + (-299 - 120*E^5 - 12*E^10)*Log[2/x]^2)/Log[2/x]^2]),x]

[Out]

E^x + Log[Log[-299 - 120*E^5 - 12*E^10 - (12*E^(2*x))/Log[2/x]^2 + (24*E^x*(5 + E^5))/Log[2/x]]]

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fricas [A]  time = 0.60, size = 55, normalized size = 1.77 \begin {gather*} e^{x} + \log \left (\log \left (\frac {24 \, {\left (e^{5} + 5\right )} e^{x} \log \left (\frac {2}{x}\right ) - {\left (12 \, e^{10} + 120 \, e^{5} + 299\right )} \log \left (\frac {2}{x}\right )^{2} - 12 \, e^{\left (2 \, x\right )}}{\log \left (\frac {2}{x}\right )^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x*exp(5)^2+120*x*exp(5)+299*x)*exp(x)*log(2/x)^3+(-24*x*exp(5)-120*x)*exp(x)^2*log(2/x)^2+12*x
*exp(x)^3*log(2/x))*log(((-12*exp(5)^2-120*exp(5)-299)*log(2/x)^2+(24*exp(5)+120)*exp(x)*log(2/x)-12*exp(x)^2)
/log(2/x)^2)+(-24*x*exp(5)-120*x)*exp(x)*log(2/x)^2+(24*x*exp(x)^2+(-24*exp(5)-120)*exp(x))*log(2/x)+24*exp(x)
^2)/((12*x*exp(5)^2+120*x*exp(5)+299*x)*log(2/x)^3+(-24*x*exp(5)-120*x)*exp(x)*log(2/x)^2+12*x*exp(x)^2*log(2/
x))/log(((-12*exp(5)^2-120*exp(5)-299)*log(2/x)^2+(24*exp(5)+120)*exp(x)*log(2/x)-12*exp(x)^2)/log(2/x)^2),x,
algorithm="fricas")

[Out]

e^x + log(log((24*(e^5 + 5)*e^x*log(2/x) - (12*e^10 + 120*e^5 + 299)*log(2/x)^2 - 12*e^(2*x))/log(2/x)^2))

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giac [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((((12*x*exp(5)^2+120*x*exp(5)+299*x)*exp(x)*log(2/x)^3+(-24*x*exp(5)-120*x)*exp(x)^2*log(2/x)^2+12*x
*exp(x)^3*log(2/x))*log(((-12*exp(5)^2-120*exp(5)-299)*log(2/x)^2+(24*exp(5)+120)*exp(x)*log(2/x)-12*exp(x)^2)
/log(2/x)^2)+(-24*x*exp(5)-120*x)*exp(x)*log(2/x)^2+(24*x*exp(x)^2+(-24*exp(5)-120)*exp(x))*log(2/x)+24*exp(x)
^2)/((12*x*exp(5)^2+120*x*exp(5)+299*x)*log(2/x)^3+(-24*x*exp(5)-120*x)*exp(x)*log(2/x)^2+12*x*exp(x)^2*log(2/
x))/log(((-12*exp(5)^2-120*exp(5)-299)*log(2/x)^2+(24*exp(5)+120)*exp(x)*log(2/x)-12*exp(x)^2)/log(2/x)^2),x,
algorithm="giac")

[Out]

Timed out

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maple [C]  time = 36.17, size = 1090, normalized size = 35.16

method	result	size
risch	\({\mathrm e}^{x}+\ln \left (\ln \left (-40 \,{\mathrm e}^{5} \ln \relax (x )^{2}-40 \,{\mathrm e}^{5} \ln \relax (2)^{2}+80 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )-\frac {299 \ln \relax (2)^{2}}{3}-\frac {299 \ln \relax (x )^{2}}{3}-4 \,{\mathrm e}^{2 x}+40 \,{\mathrm e}^{x} \ln \relax (2)+\frac {598 \ln \relax (2) \ln \relax (x )}{3}-40 \,{\mathrm e}^{x} \ln \relax (x )+8 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )-4 \,{\mathrm e}^{10} \ln \relax (x )^{2}-4 \,{\mathrm e}^{10} \ln \relax (2)^{2}+8 \ln \relax (2) {\mathrm e}^{5+x}-8 \ln \relax (x ) {\mathrm e}^{5+x}\right )+\frac {i \left (-2 \pi \mathrm {csgn}\left (\frac {i \left (40 \,{\mathrm e}^{5} \ln \relax (x )^{2}+40 \,{\mathrm e}^{5} \ln \relax (2)^{2}-80 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )+\frac {299 \ln \relax (2)^{2}}{3}+\frac {299 \ln \relax (x )^{2}}{3}+4 \,{\mathrm e}^{2 x}-40 \,{\mathrm e}^{x} \ln \relax (2)-\frac {598 \ln \relax (2) \ln \relax (x )}{3}+40 \,{\mathrm e}^{x} \ln \relax (x )-8 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+4 \,{\mathrm e}^{10} \ln \relax (x )^{2}+4 \,{\mathrm e}^{10} \ln \relax (2)^{2}-8 \ln \relax (2) {\mathrm e}^{5+x}+8 \ln \relax (x ) {\mathrm e}^{5+x}\right )}{\left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{\left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}}\right ) \mathrm {csgn}\left (i \left (40 \,{\mathrm e}^{5} \ln \relax (x )^{2}+40 \,{\mathrm e}^{5} \ln \relax (2)^{2}-80 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )+\frac {299 \ln \relax (2)^{2}}{3}+\frac {299 \ln \relax (x )^{2}}{3}+4 \,{\mathrm e}^{2 x}-40 \,{\mathrm e}^{x} \ln \relax (2)-\frac {598 \ln \relax (2) \ln \relax (x )}{3}+40 \,{\mathrm e}^{x} \ln \relax (x )-8 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+4 \,{\mathrm e}^{10} \ln \relax (x )^{2}+4 \,{\mathrm e}^{10} \ln \relax (2)^{2}-8 \ln \relax (2) {\mathrm e}^{5+x}+8 \ln \relax (x ) {\mathrm e}^{5+x}\right )\right ) \mathrm {csgn}\left (\frac {i \left (40 \,{\mathrm e}^{5} \ln \relax (x )^{2}+40 \,{\mathrm e}^{5} \ln \relax (2)^{2}-80 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )+\frac {299 \ln \relax (2)^{2}}{3}+\frac {299 \ln \relax (x )^{2}}{3}+4 \,{\mathrm e}^{2 x}-40 \,{\mathrm e}^{x} \ln \relax (2)-\frac {598 \ln \relax (2) \ln \relax (x )}{3}+40 \,{\mathrm e}^{x} \ln \relax (x )-8 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+4 \,{\mathrm e}^{10} \ln \relax (x )^{2}+4 \,{\mathrm e}^{10} \ln \relax (2)^{2}-8 \ln \relax (2) {\mathrm e}^{5+x}+8 \ln \relax (x ) {\mathrm e}^{5+x}\right )}{\left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{\left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (40 \,{\mathrm e}^{5} \ln \relax (x )^{2}+40 \,{\mathrm e}^{5} \ln \relax (2)^{2}-80 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )+\frac {299 \ln \relax (2)^{2}}{3}+\frac {299 \ln \relax (x )^{2}}{3}+4 \,{\mathrm e}^{2 x}-40 \,{\mathrm e}^{x} \ln \relax (2)-\frac {598 \ln \relax (2) \ln \relax (x )}{3}+40 \,{\mathrm e}^{x} \ln \relax (x )-8 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+4 \,{\mathrm e}^{10} \ln \relax (x )^{2}+4 \,{\mathrm e}^{10} \ln \relax (2)^{2}-8 \ln \relax (2) {\mathrm e}^{5+x}+8 \ln \relax (x ) {\mathrm e}^{5+x}\right )}{\left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}}\right )^{2}+\pi \mathrm {csgn}\left (i \left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )\right )^{2} \mathrm {csgn}\left (i \left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )\right ) \mathrm {csgn}\left (i \left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i \left (40 \,{\mathrm e}^{5} \ln \relax (x )^{2}+40 \,{\mathrm e}^{5} \ln \relax (2)^{2}-80 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )+\frac {299 \ln \relax (2)^{2}}{3}+\frac {299 \ln \relax (x )^{2}}{3}+4 \,{\mathrm e}^{2 x}-40 \,{\mathrm e}^{x} \ln \relax (2)-\frac {598 \ln \relax (2) \ln \relax (x )}{3}+40 \,{\mathrm e}^{x} \ln \relax (x )-8 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+4 \,{\mathrm e}^{10} \ln \relax (x )^{2}+4 \,{\mathrm e}^{10} \ln \relax (2)^{2}-8 \ln \relax (2) {\mathrm e}^{5+x}+8 \ln \relax (x ) {\mathrm e}^{5+x}\right )\right ) \mathrm {csgn}\left (\frac {i \left (40 \,{\mathrm e}^{5} \ln \relax (x )^{2}+40 \,{\mathrm e}^{5} \ln \relax (2)^{2}-80 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )+\frac {299 \ln \relax (2)^{2}}{3}+\frac {299 \ln \relax (x )^{2}}{3}+4 \,{\mathrm e}^{2 x}-40 \,{\mathrm e}^{x} \ln \relax (2)-\frac {598 \ln \relax (2) \ln \relax (x )}{3}+40 \,{\mathrm e}^{x} \ln \relax (x )-8 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+4 \,{\mathrm e}^{10} \ln \relax (x )^{2}+4 \,{\mathrm e}^{10} \ln \relax (2)^{2}-8 \ln \relax (2) {\mathrm e}^{5+x}+8 \ln \relax (x ) {\mathrm e}^{5+x}\right )}{\left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \left (40 \,{\mathrm e}^{5} \ln \relax (x )^{2}+40 \,{\mathrm e}^{5} \ln \relax (2)^{2}-80 \,{\mathrm e}^{5} \ln \relax (2) \ln \relax (x )+\frac {299 \ln \relax (2)^{2}}{3}+\frac {299 \ln \relax (x )^{2}}{3}+4 \,{\mathrm e}^{2 x}-40 \,{\mathrm e}^{x} \ln \relax (2)-\frac {598 \ln \relax (2) \ln \relax (x )}{3}+40 \,{\mathrm e}^{x} \ln \relax (x )-8 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+4 \,{\mathrm e}^{10} \ln \relax (x )^{2}+4 \,{\mathrm e}^{10} \ln \relax (2)^{2}-8 \ln \relax (2) {\mathrm e}^{5+x}+8 \ln \relax (x ) {\mathrm e}^{5+x}\right )}{\left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )^{2}}\right )^{3}-4 i \ln \relax (2)-2 i \ln \relax (3)+4 i \ln \left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )+2 \pi \right )}{2}\right )\)	\(1090\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((12*x*exp(5)^2+120*x*exp(5)+299*x)*exp(x)*ln(2/x)^3+(-24*x*exp(5)-120*x)*exp(x)^2*ln(2/x)^2+12*x*exp(x)^
3*ln(2/x))*ln(((-12*exp(5)^2-120*exp(5)-299)*ln(2/x)^2+(24*exp(5)+120)*exp(x)*ln(2/x)-12*exp(x)^2)/ln(2/x)^2)+
(-24*x*exp(5)-120*x)*exp(x)*ln(2/x)^2+(24*x*exp(x)^2+(-24*exp(5)-120)*exp(x))*ln(2/x)+24*exp(x)^2)/((12*x*exp(
5)^2+120*x*exp(5)+299*x)*ln(2/x)^3+(-24*x*exp(5)-120*x)*exp(x)*ln(2/x)^2+12*x*exp(x)^2*ln(2/x))/ln(((-12*exp(5
)^2-120*exp(5)-299)*ln(2/x)^2+(24*exp(5)+120)*exp(x)*ln(2/x)-12*exp(x)^2)/ln(2/x)^2),x,method=_RETURNVERBOSE)

[Out]

exp(x)+ln(ln(-40*exp(5)*ln(x)^2-40*exp(5)*ln(2)^2+80*exp(5)*ln(2)*ln(x)-299/3*ln(2)^2-299/3*ln(x)^2-4*exp(2*x)
+40*exp(x)*ln(2)+598/3*ln(2)*ln(x)-40*exp(x)*ln(x)+8*exp(10)*ln(2)*ln(x)-4*exp(10)*ln(x)^2-4*exp(10)*ln(2)^2+8
*ln(2)*exp(5+x)-8*ln(x)*exp(5+x))+1/2*I*(-2*Pi*csgn(I*(40*exp(5)*ln(x)^2+40*exp(5)*ln(2)^2-80*exp(5)*ln(2)*ln(
x)+299/3*ln(2)^2+299/3*ln(x)^2+4*exp(2*x)-40*exp(x)*ln(2)-598/3*ln(2)*ln(x)+40*exp(x)*ln(x)-8*exp(10)*ln(2)*ln
(x)+4*exp(10)*ln(x)^2+4*exp(10)*ln(2)^2-8*ln(2)*exp(5+x)+8*ln(x)*exp(5+x))/(2*I*ln(2)-2*I*ln(x))^2)^2-Pi*csgn(
I/(2*I*ln(2)-2*I*ln(x))^2)*csgn(I*(40*exp(5)*ln(x)^2+40*exp(5)*ln(2)^2-80*exp(5)*ln(2)*ln(x)+299/3*ln(2)^2+299
/3*ln(x)^2+4*exp(2*x)-40*exp(x)*ln(2)-598/3*ln(2)*ln(x)+40*exp(x)*ln(x)-8*exp(10)*ln(2)*ln(x)+4*exp(10)*ln(x)^
2+4*exp(10)*ln(2)^2-8*ln(2)*exp(5+x)+8*ln(x)*exp(5+x)))*csgn(I*(40*exp(5)*ln(x)^2+40*exp(5)*ln(2)^2-80*exp(5)*
ln(2)*ln(x)+299/3*ln(2)^2+299/3*ln(x)^2+4*exp(2*x)-40*exp(x)*ln(2)-598/3*ln(2)*ln(x)+40*exp(x)*ln(x)-8*exp(10)
*ln(2)*ln(x)+4*exp(10)*ln(x)^2+4*exp(10)*ln(2)^2-8*ln(2)*exp(5+x)+8*ln(x)*exp(5+x))/(2*I*ln(2)-2*I*ln(x))^2)+P
i*csgn(I/(2*I*ln(2)-2*I*ln(x))^2)*csgn(I*(40*exp(5)*ln(x)^2+40*exp(5)*ln(2)^2-80*exp(5)*ln(2)*ln(x)+299/3*ln(2
)^2+299/3*ln(x)^2+4*exp(2*x)-40*exp(x)*ln(2)-598/3*ln(2)*ln(x)+40*exp(x)*ln(x)-8*exp(10)*ln(2)*ln(x)+4*exp(10)
*ln(x)^2+4*exp(10)*ln(2)^2-8*ln(2)*exp(5+x)+8*ln(x)*exp(5+x))/(2*I*ln(2)-2*I*ln(x))^2)^2+Pi*csgn(I*(2*I*ln(2)-
2*I*ln(x)))^2*csgn(I*(2*I*ln(2)-2*I*ln(x))^2)-2*Pi*csgn(I*(2*I*ln(2)-2*I*ln(x)))*csgn(I*(2*I*ln(2)-2*I*ln(x))^
2)^2+Pi*csgn(I*(2*I*ln(2)-2*I*ln(x))^2)^3-Pi*csgn(I*(40*exp(5)*ln(x)^2+40*exp(5)*ln(2)^2-80*exp(5)*ln(2)*ln(x)
+299/3*ln(2)^2+299/3*ln(x)^2+4*exp(2*x)-40*exp(x)*ln(2)-598/3*ln(2)*ln(x)+40*exp(x)*ln(x)-8*exp(10)*ln(2)*ln(x
)+4*exp(10)*ln(x)^2+4*exp(10)*ln(2)^2-8*ln(2)*exp(5+x)+8*ln(x)*exp(5+x)))*csgn(I*(40*exp(5)*ln(x)^2+40*exp(5)*
ln(2)^2-80*exp(5)*ln(2)*ln(x)+299/3*ln(2)^2+299/3*ln(x)^2+4*exp(2*x)-40*exp(x)*ln(2)-598/3*ln(2)*ln(x)+40*exp(
x)*ln(x)-8*exp(10)*ln(2)*ln(x)+4*exp(10)*ln(x)^2+4*exp(10)*ln(2)^2-8*ln(2)*exp(5+x)+8*ln(x)*exp(5+x))/(2*I*ln(
2)-2*I*ln(x))^2)^2-Pi*csgn(I*(40*exp(5)*ln(x)^2+40*exp(5)*ln(2)^2-80*exp(5)*ln(2)*ln(x)+299/3*ln(2)^2+299/3*ln
(x)^2+4*exp(2*x)-40*exp(x)*ln(2)-598/3*ln(2)*ln(x)+40*exp(x)*ln(x)-8*exp(10)*ln(2)*ln(x)+4*exp(10)*ln(x)^2+4*e
xp(10)*ln(2)^2-8*ln(2)*exp(5+x)+8*ln(x)*exp(5+x))/(2*I*ln(2)-2*I*ln(x))^2)^3-4*I*ln(2)-2*I*ln(3)+4*I*ln(2*I*ln
(2)-2*I*ln(x))+2*Pi))

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maxima [B]  time = 0.60, size = 92, normalized size = 2.97 \begin {gather*} e^{x} + \log \left (\log \left (24 \, {\left (e^{5} + 5\right )} e^{x} \log \relax (2) - {\left (12 \, e^{10} + 120 \, e^{5} + 299\right )} \log \relax (2)^{2} - {\left (12 \, e^{10} + 120 \, e^{5} + 299\right )} \log \relax (x)^{2} - 2 \, {\left (12 \, {\left (e^{5} + 5\right )} e^{x} - {\left (12 \, e^{10} + 120 \, e^{5} + 299\right )} \log \relax (2)\right )} \log \relax (x) - 12 \, e^{\left (2 \, x\right )}\right ) - 2 \, \log \left (-\log \relax (2) + \log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x*exp(5)^2+120*x*exp(5)+299*x)*exp(x)*log(2/x)^3+(-24*x*exp(5)-120*x)*exp(x)^2*log(2/x)^2+12*x
*exp(x)^3*log(2/x))*log(((-12*exp(5)^2-120*exp(5)-299)*log(2/x)^2+(24*exp(5)+120)*exp(x)*log(2/x)-12*exp(x)^2)
/log(2/x)^2)+(-24*x*exp(5)-120*x)*exp(x)*log(2/x)^2+(24*x*exp(x)^2+(-24*exp(5)-120)*exp(x))*log(2/x)+24*exp(x)
^2)/((12*x*exp(5)^2+120*x*exp(5)+299*x)*log(2/x)^3+(-24*x*exp(5)-120*x)*exp(x)*log(2/x)^2+12*x*exp(x)^2*log(2/
x))/log(((-12*exp(5)^2-120*exp(5)-299)*log(2/x)^2+(24*exp(5)+120)*exp(x)*log(2/x)-12*exp(x)^2)/log(2/x)^2),x,
algorithm="maxima")

[Out]

e^x + log(log(24*(e^5 + 5)*e^x*log(2) - (12*e^10 + 120*e^5 + 299)*log(2)^2 - (12*e^10 + 120*e^5 + 299)*log(x)^
2 - 2*(12*(e^5 + 5)*e^x - (12*e^10 + 120*e^5 + 299)*log(2))*log(x) - 12*e^(2*x)) - 2*log(-log(2) + log(x)))

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mupad [B]  time = 9.18, size = 57, normalized size = 1.84 \begin {gather*} \ln \left (\ln \left (-\frac {\left (120\,{\mathrm {e}}^5+12\,{\mathrm {e}}^{10}+299\right )\,{\ln \left (\frac {2}{x}\right )}^2-{\mathrm {e}}^x\,\left (24\,{\mathrm {e}}^5+120\right )\,\ln \left (\frac {2}{x}\right )+12\,{\mathrm {e}}^{2\,x}}{{\ln \left (\frac {2}{x}\right )}^2}\right )\right )+{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((24*exp(2*x) + log(-(12*exp(2*x) + log(2/x)^2*(120*exp(5) + 12*exp(10) + 299) - exp(x)*log(2/x)*(24*exp(5)
+ 120))/log(2/x)^2)*(12*x*exp(3*x)*log(2/x) - exp(2*x)*log(2/x)^2*(120*x + 24*x*exp(5)) + exp(x)*log(2/x)^3*(
299*x + 120*x*exp(5) + 12*x*exp(10))) + log(2/x)*(24*x*exp(2*x) - exp(x)*(24*exp(5) + 120)) - exp(x)*log(2/x)^
2*(120*x + 24*x*exp(5)))/(log(-(12*exp(2*x) + log(2/x)^2*(120*exp(5) + 12*exp(10) + 299) - exp(x)*log(2/x)*(24
*exp(5) + 120))/log(2/x)^2)*(log(2/x)^3*(299*x + 120*x*exp(5) + 12*x*exp(10)) - exp(x)*log(2/x)^2*(120*x + 24*
x*exp(5)) + 12*x*exp(2*x)*log(2/x))),x)

[Out]

log(log(-(12*exp(2*x) + log(2/x)^2*(120*exp(5) + 12*exp(10) + 299) - exp(x)*log(2/x)*(24*exp(5) + 120))/log(2/
x)^2)) + exp(x)

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sympy [A]  time = 11.78, size = 54, normalized size = 1.74 \begin {gather*} e^{x} + \log {\left (\log {\left (\frac {- 12 e^{2 x} + \left (120 + 24 e^{5}\right ) e^{x} \log {\left (\frac {2}{x} \right )} + \left (- 12 e^{10} - 120 e^{5} - 299\right ) \log {\left (\frac {2}{x} \right )}^{2}}{\log {\left (\frac {2}{x} \right )}^{2}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((12*x*exp(5)**2+120*x*exp(5)+299*x)*exp(x)*ln(2/x)**3+(-24*x*exp(5)-120*x)*exp(x)**2*ln(2/x)**2+12
*x*exp(x)**3*ln(2/x))*ln(((-12*exp(5)**2-120*exp(5)-299)*ln(2/x)**2+(24*exp(5)+120)*exp(x)*ln(2/x)-12*exp(x)**
2)/ln(2/x)**2)+(-24*x*exp(5)-120*x)*exp(x)*ln(2/x)**2+(24*x*exp(x)**2+(-24*exp(5)-120)*exp(x))*ln(2/x)+24*exp(
x)**2)/((12*x*exp(5)**2+120*x*exp(5)+299*x)*ln(2/x)**3+(-24*x*exp(5)-120*x)*exp(x)*ln(2/x)**2+12*x*exp(x)**2*l
n(2/x))/ln(((-12*exp(5)**2-120*exp(5)-299)*ln(2/x)**2+(24*exp(5)+120)*exp(x)*ln(2/x)-12*exp(x)**2)/ln(2/x)**2)
,x)

[Out]

exp(x) + log(log((-12*exp(2*x) + (120 + 24*exp(5))*exp(x)*log(2/x) + (-12*exp(10) - 120*exp(5) - 299)*log(2/x)
**2)/log(2/x)**2))

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