Integrand size = 81, antiderivative size = 27 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=5 e^{-\left (3-\log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )^2} \] Output:
5/exp((3-ln(1/4*x^2-1/16*ln(x)))^2)
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {5 e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (-4 x^2+\log (x)\right )^6}{16777216} \] Input:
Integrate[(E^(-9 - Log[(4*x^2 - Log[x])/16]^2)*(4*x^2 - Log[x])^6*(30 - 24 0*x^2 + (-10 + 80*x^2)*Log[(4*x^2 - Log[x])/16]))/(16777216*(-4*x^3 + x*Lo g[x])),x]
Output:
(5*E^(-9 - Log[(4*x^2 - Log[x])/16]^2)*(-4*x^2 + Log[x])^6)/16777216
Leaf count is larger than twice the leaf count of optimal. \(68\) vs. \(2(27)=54\).
Time = 0.58 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {27, 27, 2726}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )-9} \left (4 x^2-\log (x)\right )^6 \left (-240 x^2+\left (80 x^2-10\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )+30\right )}{16777216 \left (x \log (x)-4 x^3\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {10 e^{-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )-9} \left (4 x^2-\log (x)\right )^6 \left (-24 x^2-\left (1-8 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )+3\right )}{4 x^3-x \log (x)}dx}{16777216}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5 \int \frac {e^{-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )-9} \left (4 x^2-\log (x)\right )^6 \left (-24 x^2-\left (1-8 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )+3\right )}{4 x^3-x \log (x)}dx}{8388608}\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {5 \left (1-8 x^2\right ) e^{-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )-9} \left (4 x^2-\log (x)\right )^7}{16777216 \left (\frac {1}{x}-8 x\right ) \left (4 x^3-x \log (x)\right )}\) |
Input:
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]
Output:
(5*E^(-9 - Log[(4*x^2 - Log[x])/16]^2)*(1 - 8*x^2)*(4*x^2 - Log[x])^7)/(16 777216*(x^(-1) - 8*x)*(4*x^3 - x*Log[x]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Time = 0.56 (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 | \(5 \left (\frac {x^{2}}{4}+\ln \left (\frac {1}{x^{\frac {1}{16}}}\right )\right )^{6} {\mathrm e}^{-\ln \left (\frac {x^{2}}{4}+\ln \left (\frac {1}{x^{\frac {1}{16}}}\right )\right )^{2}-9}\) | \(34\) |
Input:
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 )
Output:
5*(1/4*x^2-1/16*ln(x))^6*exp(-ln(1/4*x^2-1/16*ln(x))^2-9)
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=5 \, e^{\left (-\log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \left (x\right )\right )^{2} + 6 \, \log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \left (x\right )\right ) - 9\right )} \] Input:
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, algorit hm="fricas")
Output:
5*e^(-log(1/4*x^2 - 1/16*log(x))^2 + 6*log(1/4*x^2 - 1/16*log(x)) - 9)
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (17) = 34\).
Time = 5.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.96 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {\left (20480 x^{12} - 30720 x^{10} \log {\left (x \right )} + 19200 x^{8} \log {\left (x \right )}^{2} - 6400 x^{6} \log {\left (x \right )}^{3} + 1200 x^{4} \log {\left (x \right )}^{4} - 120 x^{2} \log {\left (x \right )}^{5} + 5 \log {\left (x \right )}^{6}\right ) e^{- \log {\left (\frac {x^{2}}{4} - \frac {\log {\left (x \right )}}{16} \right )}^{2} - 9}}{16777216} \] Input:
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)
Output:
(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
Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (20) = 40\).
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.48 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {5}{16777216} \, {\left (4096 \, x^{12} - 6144 \, x^{10} \log \left (x\right ) + 3840 \, x^{8} \log \left (x\right )^{2} - 1280 \, x^{6} \log \left (x\right )^{3} + 240 \, x^{4} \log \left (x\right )^{4} - 24 \, x^{2} \log \left (x\right )^{5} + \log \left (x\right )^{6}\right )} e^{\left (-16 \, \log \left (2\right )^{2} + 8 \, \log \left (2\right ) \log \left (4 \, x^{2} - \log \left (x\right )\right ) - \log \left (4 \, x^{2} - \log \left (x\right )\right )^{2} - 9\right )} \] Input:
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, algorit hm="maxima")
Output:
5/16777216*(4096*x^12 - 6144*x^10*log(x) + 3840*x^8*log(x)^2 - 1280*x^6*lo g(x)^3 + 240*x^4*log(x)^4 - 24*x^2*log(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)
Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=5 \, e^{\left (-\log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \left (x\right )\right )^{2} + 6 \, \log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \left (x\right )\right ) - 9\right )} \] Input:
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, algorit hm="giac")
Output:
5*e^(-log(1/4*x^2 - 1/16*log(x))^2 + 6*log(1/4*x^2 - 1/16*log(x)) - 9)
Time = 3.24 (sec) , antiderivative size = 181, normalized size of antiderivative = 6.70 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {5\,x^{12}\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}}{4096}+\frac {5\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^6}{16777216}-\frac {15\,x^2\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^5}{2097152}+\frac {75\,x^4\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^4}{1048576}-\frac {25\,x^6\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^3}{65536}+\frac {75\,x^8\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,{\ln \left (x\right )}^2}{65536}-\frac {15\,x^{10}\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \left (x\right )}{16}\right )}^2-9}\,\ln \left (x\right )}{8192} \] Input:
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)
Output:
(5*x^12*exp(- log(x^2/4 - log(x)/16)^2 - 9))/4096 + (5*exp(- log(x^2/4 - l og(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)*lo g(x)^4)/1048576 - (25*x^6*exp(- log(x^2/4 - log(x)/16)^2 - 9)*log(x)^3)/65 536 + (75*x^8*exp(- log(x^2/4 - log(x)/16)^2 - 9)*log(x)^2)/65536 - (15*x^ 10*exp(- log(x^2/4 - log(x)/16)^2 - 9)*log(x))/8192
Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.78 \[ \int \frac {e^{-9-\log ^2\left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )} \left (4 x^2-\log (x)\right )^6 \left (30-240 x^2+\left (-10+80 x^2\right ) \log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )}{16777216 \left (-4 x^3+x \log (x)\right )} \, dx=\frac {\frac {5 \mathrm {log}\left (x \right )^{6}}{16777216}-\frac {15 \mathrm {log}\left (x \right )^{5} x^{2}}{2097152}+\frac {75 \mathrm {log}\left (x \right )^{4} x^{4}}{1048576}-\frac {25 \mathrm {log}\left (x \right )^{3} x^{6}}{65536}+\frac {75 \mathrm {log}\left (x \right )^{2} x^{8}}{65536}-\frac {15 \,\mathrm {log}\left (x \right ) x^{10}}{8192}+\frac {5 x^{12}}{4096}}{e^{\mathrm {log}\left (-\frac {\mathrm {log}\left (x \right )}{16}+\frac {x^{2}}{4}\right )^{2}} e^{9}} \] Input:
int(((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)
Output:
(5*(log(x)**6 - 24*log(x)**5*x**2 + 240*log(x)**4*x**4 - 1280*log(x)**3*x* *6 + 3840*log(x)**2*x**8 - 6144*log(x)*x**10 + 4096*x**12))/(16777216*e**( log(( - log(x) + 4*x**2)/16)**2)*e**9)