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

Optimal. Leaf size=27 \[ 5 e^{-\left (3-\log \left (\frac {1}{16} \left (4 x^2-\log (x)\right )\right )\right )^2} \]

________________________________________________________________________________________

Rubi [B]  time = 0.18, antiderivative size = 68, normalized size of antiderivative = 2.52, number of steps used = 2, number of rules used = 2, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {12, 2288} \begin {gather*} \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 )} \end {gather*}

Antiderivative was successfully verified.

[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 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} &=\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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 37, normalized size = 1.37 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 1.11, size = 33, normalized size = 1.22 \begin {gather*} 5 \, e^{\left (-\log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \relax (x)\right )^{2} + 6 \, \log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \relax (x)\right ) - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.62, size = 33, normalized size = 1.22 \begin {gather*} 5 \, e^{\left (-\log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \relax (x)\right )^{2} + 6 \, \log \left (\frac {1}{4} \, x^{2} - \frac {1}{16} \, \log \relax (x)\right ) - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 33, normalized size = 1.22

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [B]  time = 0.68, size = 94, normalized size = 3.48 \begin {gather*} \frac {5}{16777216} \, {\left (4096 \, x^{12} - 6144 \, x^{10} \log \relax (x) + 3840 \, x^{8} \log \relax (x)^{2} - 1280 \, x^{6} \log \relax (x)^{3} + 240 \, x^{4} \log \relax (x)^{4} - 24 \, x^{2} \log \relax (x)^{5} + \log \relax (x)^{6}\right )} e^{\left (-16 \, \log \relax (2)^{2} + 8 \, \log \relax (2) \log \left (4 \, x^{2} - \log \relax (x)\right ) - \log \left (4 \, x^{2} - \log \relax (x)\right )^{2} - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 5.50, size = 181, normalized size = 6.70 \begin {gather*} \frac {5\,x^{12}\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \relax (x)}{16}\right )}^2-9}}{4096}+\frac {5\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \relax (x)}{16}\right )}^2-9}\,{\ln \relax (x)}^6}{16777216}-\frac {15\,x^2\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \relax (x)}{16}\right )}^2-9}\,{\ln \relax (x)}^5}{2097152}+\frac {75\,x^4\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \relax (x)}{16}\right )}^2-9}\,{\ln \relax (x)}^4}{1048576}-\frac {25\,x^6\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \relax (x)}{16}\right )}^2-9}\,{\ln \relax (x)}^3}{65536}+\frac {75\,x^8\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \relax (x)}{16}\right )}^2-9}\,{\ln \relax (x)}^2}{65536}-\frac {15\,x^{10}\,{\mathrm {e}}^{-{\ln \left (\frac {x^2}{4}-\frac {\ln \relax (x)}{16}\right )}^2-9}\,\ln \relax (x)}{8192} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [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(((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]

Timed out

________________________________________________________________________________________