\(\int \frac {e^{-9-\log ^2(\frac {1}{16} (4 x^2-\log (x)))} (4 x^2-\log (x))^6 (30-240 x^2+(-10+80 x^2) \log (\frac {1}{16} (4 x^2-\log (x))))}{16777216 (-4 x^3+x \log (x))} \, dx\) [7854]

Optimal result

Integrand size = 81, antiderivative size = 27 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=5 e^{-\left (3-\log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )^2} \]

[Out]

5/exp((3-ln(1/4*x^2-1/16*ln(x)))^2)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(68\) vs. \(2(27)=54\).

Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {12, 2326} \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {5 \left (1-8 x^2\right ) e^{-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )-9} \left (4 x^2-\log (x)\right )^7}{16777216 \left (\frac {1}{x}-8 x\right ) \left (4 x^3-x \log (x)\right )} \]

[In]

Int[(E^(-9 - Log[(4*x^2 - Log[x])/16]^2)*(4*x^2 - Log[x])^6*(30 - 240*x^2 + (-10 + 80*x^2)*Log[(4*x^2 - Log[x]
)/16]))/(16777216*(-4*x^3 + x*Log[x])),x]

[Out]

(5*E^(-9 - Log[(4*x^2 - Log[x])/16]^2)*(1 - 8*x^2)*(4*x^2 - Log[x])^7)/(16777216*(x^(-1) - 8*x)*(4*x^3 - x*Log
[x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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}& = \frac {\int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{-4 x^3+x \log (x)} \, dx}{16777216} \\ & = \frac {5 e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (1-8 x^2\right ) \left (4 x^2-\log (x)\right )^7}{16777216 \left (\frac {1}{x}-8 x\right ) \left (4 x^3-x \log (x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {5 e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (-4 x^2+\log (x)\right )^6}{16777216} \]

[In]

Integrate[(E^(-9 - Log[(4*x^2 - Log[x])/16]^2)*(4*x^2 - Log[x])^6*(30 - 240*x^2 + (-10 + 80*x^2)*Log[(4*x^2 -
Log[x])/16]))/(16777216*(-4*x^3 + x*Log[x])),x]

[Out]

(5*E^(-9 - Log[(4*x^2 - Log[x])/16]^2)*(-4*x^2 + Log[x])^6)/16777216

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22

method	result	size
risch	\(5 \left (\frac {x^{2}}{4}-\frac {\ln \left (x \right )}{16}\right )^{6} {\mathrm e}^{-\ln \left (\frac {x^{2}}{4}-\frac {\ln \left (x \right )}{16}\right )^{2}-9}\)	\(33\)
parallelrisch	\(5 \left (\frac {x^{2}}{4}+\ln \left (\frac {1}{x^{\frac {1}{16}}}\right )\right )^{6} {\mathrm e}^{-\ln \left (\frac {x^{2}}{4}+\ln \left (\frac {1}{x^{\frac {1}{16}}}\right )\right )^{2}-9}\)	\(34\)

[In]

int(((80*x^2-10)*ln(1/4*x^2-1/16*ln(x))-240*x^2+30)/(x*ln(x)-4*x^3)/exp(ln(1/4*x^2-1/16*ln(x))^2-6*ln(1/4*x^2-
1/16*ln(x))+9),x,method=_RETURNVERBOSE)

[Out]

5*(1/4*x^2-1/16*ln(x))^6*exp(-ln(1/4*x^2-1/16*ln(x))^2-9)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=5 \, e^{\left (-\log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \left (x\right )\right )^{2} + 6 \, \log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \left (x\right )\right ) - 9\right )} \]

[In]

integrate(((80*x^2-10)*log(1/4*x^2-1/16*log(x))-240*x^2+30)/(x*log(x)-4*x^3)/exp(log(1/4*x^2-1/16*log(x))^2-6*
log(1/4*x^2-1/16*log(x))+9),x, algorithm="fricas")

[Out]

5*e^(-log(1/4*x^2 - 1/16*log(x))^2 + 6*log(1/4*x^2 - 1/16*log(x)) - 9)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (17) = 34\).

Time = 4.88 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.96 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {\left (20480 x^{12} - 30720 x^{10} \log {\left (x \right )} + 19200 x^{8} \log {\left (x \right )}^{2} - 6400 x^{6} \log {\left (x \right )}^{3} + 1200 x^{4} \log {\left (x \right )}^{4} - 120 x^{2} \log {\left (x \right )}^{5} + 5 \log {\left (x \right )}^{6}\right ) e^{- \log {\left (\frac {x^{2}}{4} - \frac {\log {\left (x \right )}}{16} \right )}^{2} - 9}}{16777216} \]

[In]

integrate(((80*x**2-10)*ln(1/4*x**2-1/16*ln(x))-240*x**2+30)/(x*ln(x)-4*x**3)/exp(ln(1/4*x**2-1/16*ln(x))**2-6
*ln(1/4*x**2-1/16*ln(x))+9),x)

[Out]

(20480*x**12 - 30720*x**10*log(x) + 19200*x**8*log(x)**2 - 6400*x**6*log(x)**3 + 1200*x**4*log(x)**4 - 120*x**
2*log(x)**5 + 5*log(x)**6)*exp(-log(x**2/4 - log(x)/16)**2 - 9)/16777216

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (20) = 40\).

Time = 0.48 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.48 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {5}{16777216} \, {\left (4096 \, x^{12} - 6144 \, x^{10} \log \left (x\right ) + 3840 \, x^{8} \log \left (x\right )^{2} - 1280 \, x^{6} \log \left (x\right )^{3} + 240 \, x^{4} \log \left (x\right )^{4} - 24 \, x^{2} \log \left (x\right )^{5} + \log \left (x\right )^{6}\right )} e^{\left (-16 \, \log \left (2\right )^{2} + 8 \, \log \left (2\right ) \log \left (4 \, x^{2} - \log \left (x\right )\right ) - \log \left (4 \, x^{2} - \log \left (x\right )\right )^{2} - 9\right )} \]

[In]

integrate(((80*x^2-10)*log(1/4*x^2-1/16*log(x))-240*x^2+30)/(x*log(x)-4*x^3)/exp(log(1/4*x^2-1/16*log(x))^2-6*
log(1/4*x^2-1/16*log(x))+9),x, algorithm="maxima")

[Out]

5/16777216*(4096*x^12 - 6144*x^10*log(x) + 3840*x^8*log(x)^2 - 1280*x^6*log(x)^3 + 240*x^4*log(x)^4 - 24*x^2*l
og(x)^5 + log(x)^6)*e^(-16*log(2)^2 + 8*log(2)*log(4*x^2 - log(x)) - log(4*x^2 - log(x))^2 - 9)

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=5 \, e^{\left (-\log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \left (x\right )\right )^{2} + 6 \, \log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \left (x\right )\right ) - 9\right )} \]

[In]

integrate(((80*x^2-10)*log(1/4*x^2-1/16*log(x))-240*x^2+30)/(x*log(x)-4*x^3)/exp(log(1/4*x^2-1/16*log(x))^2-6*
log(1/4*x^2-1/16*log(x))+9),x, algorithm="giac")

[Out]

5*e^(-log(1/4*x^2 - 1/16*log(x))^2 + 6*log(1/4*x^2 - 1/16*log(x)) - 9)

Mupad [B] (verification not implemented)

Time = 12.27 (sec) , antiderivative size = 181, normalized size of antiderivative = 6.70 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {5\,x^{12}\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}}{4096}+\frac {5\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^6}{16777216}-\frac {15\,x^2\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^5}{2097152}+\frac {75\,x^4\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^4}{1048576}-\frac {25\,x^6\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^3}{65536}+\frac {75\,x^8\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^2}{65536}-\frac {15\,x^{10}\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,\ln \left (x\right )}{8192} \]

[In]

int((exp(6*log(x^2/4 - log(x)/16) - log(x^2/4 - log(x)/16)^2 - 9)*(log(x^2/4 - log(x)/16)*(80*x^2 - 10) - 240*
x^2 + 30))/(x*log(x) - 4*x^3),x)

[Out]

(5*x^12*exp(- log(x^2/4 - log(x)/16)^2 - 9))/4096 + (5*exp(- log(x^2/4 - log(x)/16)^2 - 9)*log(x)^6)/16777216
- (15*x^2*exp(- log(x^2/4 - log(x)/16)^2 - 9)*log(x)^5)/2097152 + (75*x^4*exp(- log(x^2/4 - log(x)/16)^2 - 9)*
log(x)^4)/1048576 - (25*x^6*exp(- log(x^2/4 - log(x)/16)^2 - 9)*log(x)^3)/65536 + (75*x^8*exp(- log(x^2/4 - lo
g(x)/16)^2 - 9)*log(x)^2)/65536 - (15*x^10*exp(- log(x^2/4 - log(x)/16)^2 - 9)*log(x))/8192