3.78.73 \(\int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {e^{-80+8 x-\frac {32}{3+\log (4)}}}{\log ^8(x)} \]

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Rubi [A]  time = 0.86, antiderivative size = 35, normalized size of antiderivative = 1.75, number of steps used = 2, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2274, 2288} \begin {gather*} \frac {4^{-\frac {8 (10-x)}{3+\log (4)}} e^{-\frac {8 (34-3 x)}{3+\log (4)}}}{\log ^8(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((8*(-34 + 3*x + (-10 + x)*Log[4] + (-3 - Log[4])*Log[Log[x]]))/(3 + Log[4]))*(-8 + 8*x*Log[x]))/(x*Log
[x]),x]

[Out]

1/(4^((8*(10 - x))/(3 + Log[4]))*E^((8*(34 - 3*x))/(3 + Log[4]))*Log[x]^8)

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {8 (-34+3 x+(-10+x) \log (4))}{3+\log (4)}\right ) \log ^{-1+\frac {8 (-3-\log (4))}{3+\log (4)}}(x) (-8+8 x \log (x))}{x} \, dx\\ &=\frac {4^{-\frac {8 (10-x)}{3+\log (4)}} e^{-\frac {8 (34-3 x)}{3+\log (4)}}}{\log ^8(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 25, normalized size = 1.25 \begin {gather*} \frac {e^{8 x-\frac {16 (17+5 \log (4))}{3+\log (4)}}}{\log ^8(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((8*(-34 + 3*x + (-10 + x)*Log[4] + (-3 - Log[4])*Log[Log[x]]))/(3 + Log[4]))*(-8 + 8*x*Log[x]))/
(x*Log[x]),x]

[Out]

E^(8*x - (16*(17 + 5*Log[4]))/(3 + Log[4]))/Log[x]^8

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fricas [A]  time = 1.07, size = 34, normalized size = 1.70 \begin {gather*} e^{\left (\frac {8 \, {\left (2 \, {\left (x - 10\right )} \log \relax (2) - {\left (2 \, \log \relax (2) + 3\right )} \log \left (\log \relax (x)\right ) + 3 \, x - 34\right )}}{2 \, \log \relax (2) + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(x)-8)*exp(((-2*log(2)-3)*log(log(x))+2*(x-10)*log(2)+3*x-34)/(2*log(2)+3))^8/x/log(x),x, al
gorithm="fricas")

[Out]

e^(8*(2*(x - 10)*log(2) - (2*log(2) + 3)*log(log(x)) + 3*x - 34)/(2*log(2) + 3))

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giac [B]  time = 0.37, size = 76, normalized size = 3.80 \begin {gather*} e^{\left (\frac {16 \, x \log \relax (2)}{2 \, \log \relax (2) + 3} - \frac {16 \, \log \relax (2) \log \left (\log \relax (x)\right )}{2 \, \log \relax (2) + 3} + \frac {24 \, x}{2 \, \log \relax (2) + 3} - \frac {160 \, \log \relax (2)}{2 \, \log \relax (2) + 3} - \frac {24 \, \log \left (\log \relax (x)\right )}{2 \, \log \relax (2) + 3} - \frac {272}{2 \, \log \relax (2) + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(x)-8)*exp(((-2*log(2)-3)*log(log(x))+2*(x-10)*log(2)+3*x-34)/(2*log(2)+3))^8/x/log(x),x, al
gorithm="giac")

[Out]

e^(16*x*log(2)/(2*log(2) + 3) - 16*log(2)*log(log(x))/(2*log(2) + 3) + 24*x/(2*log(2) + 3) - 160*log(2)/(2*log
(2) + 3) - 24*log(log(x))/(2*log(2) + 3) - 272/(2*log(2) + 3))

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maple [A]  time = 0.04, size = 38, normalized size = 1.90

method	result	size
risch	\({\mathrm e}^{\frac {-16 \ln \relax (2) \ln \left (\ln \relax (x )\right )+16 x \ln \relax (2)-24 \ln \left (\ln \relax (x )\right )-160 \ln \relax (2)+24 x -272}{2 \ln \relax (2)+3}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x*ln(x)-8)*exp(((-2*ln(2)-3)*ln(ln(x))+2*(x-10)*ln(2)+3*x-34)/(2*ln(2)+3))^8/x/ln(x),x,method=_RETURNVE
RBOSE)

[Out]

exp(8*(-2*ln(2)*ln(ln(x))+2*x*ln(2)-3*ln(ln(x))-20*ln(2)+3*x-34)/(2*ln(2)+3))

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maxima [B]  time = 0.60, size = 79, normalized size = 3.95 \begin {gather*} \frac {e^{\left (\frac {16 \, x \log \relax (2)}{2 \, \log \relax (2) + 3} - \frac {16 \, \log \relax (2) \log \left (\log \relax (x)\right )}{2 \, \log \relax (2) + 3} + \frac {24 \, x}{2 \, \log \relax (2) + 3} - \frac {24 \, \log \left (\log \relax (x)\right )}{2 \, \log \relax (2) + 3} - \frac {272}{2 \, \log \relax (2) + 3}\right )}}{2^{\frac {160}{2 \, \log \relax (2) + 3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*log(x)-8)*exp(((-2*log(2)-3)*log(log(x))+2*(x-10)*log(2)+3*x-34)/(2*log(2)+3))^8/x/log(x),x, al
gorithm="maxima")

[Out]

e^(16*x*log(2)/(2*log(2) + 3) - 16*log(2)*log(log(x))/(2*log(2) + 3) + 24*x/(2*log(2) + 3) - 24*log(log(x))/(2
*log(2) + 3) - 272/(2*log(2) + 3))/2^(160/(2*log(2) + 3))

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mupad [B]  time = 7.22, size = 83, normalized size = 4.15 \begin {gather*} \frac {2^{\frac {16\,x}{2\,\ln \relax (2)+3}}\,{\mathrm {e}}^{-\frac {272}{2\,\ln \relax (2)+3}}\,{\mathrm {e}}^{\frac {24\,x}{2\,\ln \relax (2)+3}}}{2^{\frac {160}{2\,\ln \relax (2)+3}}\,{\ln \relax (x)}^{\frac {24}{2\,\ln \relax (2)+3}}\,{\ln \relax (x)}^{\frac {16\,\ln \relax (2)}{2\,\ln \relax (2)+3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((8*(3*x + 2*log(2)*(x - 10) - log(log(x))*(2*log(2) + 3) - 34))/(2*log(2) + 3))*(8*x*log(x) - 8))/(x*
log(x)),x)

[Out]

(2^((16*x)/(2*log(2) + 3))*exp(-272/(2*log(2) + 3))*exp((24*x)/(2*log(2) + 3)))/(2^(160/(2*log(2) + 3))*log(x)
^(24/(2*log(2) + 3))*log(x)^((16*log(2))/(2*log(2) + 3)))

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sympy [A]  time = 0.51, size = 37, normalized size = 1.85 \begin {gather*} e^{\frac {8 \left (3 x + \left (2 x - 20\right ) \log {\relax (2 )} + \left (-3 - 2 \log {\relax (2 )}\right ) \log {\left (\log {\relax (x )} \right )} - 34\right )}{2 \log {\relax (2 )} + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x*ln(x)-8)*exp(((-2*ln(2)-3)*ln(ln(x))+2*(x-10)*ln(2)+3*x-34)/(2*ln(2)+3))**8/x/ln(x),x)

[Out]

exp(8*(3*x + (2*x - 20)*log(2) + (-3 - 2*log(2))*log(log(x)) - 34)/(2*log(2) + 3))

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