\(\int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx\) [7773]

Optimal result

Integrand size = 46, antiderivative size = 20 \[ \int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx=\frac {e^{-80+8 x-\frac {32}{3+\log (4)}}}{\log ^8(x)} \]

[Out]

exp(x-ln(ln(x))-4/(2*ln(2)+3)-10)^8

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2306, 2326} \[ \int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx=\frac {4^{-\frac {8 (10-x)}{3+\log (4)}} e^{-\frac {8 (34-3 x)}{3+\log (4)}}}{\log ^8(x)} \]

[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 2306

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 2326

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx=\frac {2^{-\frac {160}{3+\log (4)}} e^{8 \left (x-\frac {34}{3+\log (4)}\right )}}{\log ^8(x)} \]

[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 - 34/(3 + Log[4])))/(2^(160/(3 + Log[4]))*Log[x]^8)

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.75

method	result	size
default	\({\mathrm e}^{\frac {8 \left (-2 \ln \left (2\right )-3\right ) \ln \left (\ln \left (x \right )\right )+8 \left (2 x -20\right ) \ln \left (2\right )+24 x -272}{2 \ln \left (2\right )+3}}\)	\(35\)
parallelrisch	\({\mathrm e}^{\frac {8 \left (-2 \ln \left (2\right )-3\right ) \ln \left (\ln \left (x \right )\right )+8 \left (2 x -20\right ) \ln \left (2\right )+24 x -272}{2 \ln \left (2\right )+3}}\)	\(35\)
risch	\({\mathrm e}^{\frac {-16 \ln \left (2\right ) \ln \left (\ln \left (x \right )\right )-24 \ln \left (\ln \left (x \right )\right )+16 x \ln \left (2\right )-160 \ln \left (2\right )+24 x -272}{2 \ln \left (2\right )+3}}\)	\(38\)

[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(((-2*ln(2)-3)*ln(ln(x))+2*(x-10)*ln(2)+3*x-34)/(2*ln(2)+3))^8

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx=e^{\left (\frac {8 \, {\left (2 \, {\left (x - 10\right )} \log \left (2\right ) - {\left (2 \, \log \left (2\right ) + 3\right )} \log \left (\log \left (x\right )\right ) + 3 \, x - 34\right )}}{2 \, \log \left (2\right ) + 3}\right )} \]

[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))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85 \[ \int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx=e^{\frac {8 \cdot \left (3 x + \left (2 x - 20\right ) \log {\left (2 \right )} + \left (-3 - 2 \log {\left (2 \right )}\right ) \log {\left (\log {\left (x \right )} \right )} - 34\right )}{2 \log {\left (2 \right )} + 3}} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (21) = 42\).

Time = 0.43 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.95 \[ \int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx=\frac {e^{\left (\frac {16 \, x \log \left (2\right )}{2 \, \log \left (2\right ) + 3} - \frac {16 \, \log \left (2\right ) \log \left (\log \left (x\right )\right )}{2 \, \log \left (2\right ) + 3} + \frac {24 \, x}{2 \, \log \left (2\right ) + 3} - \frac {24 \, \log \left (\log \left (x\right )\right )}{2 \, \log \left (2\right ) + 3} - \frac {272}{2 \, \log \left (2\right ) + 3}\right )}}{2^{\frac {160}{2 \, \log \left (2\right ) + 3}}} \]

[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))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (21) = 42\).

Time = 0.40 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.80 \[ \int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx=e^{\left (\frac {16 \, x \log \left (2\right )}{2 \, \log \left (2\right ) + 3} - \frac {16 \, \log \left (2\right ) \log \left (\log \left (x\right )\right )}{2 \, \log \left (2\right ) + 3} + \frac {24 \, x}{2 \, \log \left (2\right ) + 3} - \frac {160 \, \log \left (2\right )}{2 \, \log \left (2\right ) + 3} - \frac {24 \, \log \left (\log \left (x\right )\right )}{2 \, \log \left (2\right ) + 3} - \frac {272}{2 \, \log \left (2\right ) + 3}\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 12.69 (sec) , antiderivative size = 83, normalized size of antiderivative = 4.15 \[ \int \frac {e^{\frac {8 (-34+3 x+(-10+x) \log (4)+(-3-\log (4)) \log (\log (x)))}{3+\log (4)}} (-8+8 x \log (x))}{x \log (x)} \, dx=\frac {2^{\frac {16\,x}{2\,\ln \left (2\right )+3}}\,{\mathrm {e}}^{-\frac {272}{2\,\ln \left (2\right )+3}}\,{\mathrm {e}}^{\frac {24\,x}{2\,\ln \left (2\right )+3}}}{2^{\frac {160}{2\,\ln \left (2\right )+3}}\,{\ln \left (x\right )}^{\frac {24}{2\,\ln \left (2\right )+3}}\,{\ln \left (x\right )}^{\frac {16\,\ln \left (2\right )}{2\,\ln \left (2\right )+3}}} \]

[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)))