3.100.100 \(\int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+(-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)) \log (\log (5)))}{2 x \log ^5(x)} \, dx\)

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

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Rubi [B]  time = 0.55, antiderivative size = 141, normalized size of antiderivative = 5.22, number of steps used = 2, number of rules used = 2, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {12, 2288} \begin {gather*} -\frac {2 \left (20 x \log ^4(2)+15 x \log ^2(2) \log ^2(x)+\log (\log (5)) \left (3 \log ^2(2) \log ^2(x)+4 \log ^4(2)\right )\right ) \exp \left (\frac {9 \log ^4(x)+24 \log ^2(2) \log ^2(x)+16 \log ^4(2)}{\log ^4(x)}\right )}{x \log ^5(x) \left (\frac {3 \left (\frac {3 \log ^3(x)}{x}+\frac {4 \log ^2(2) \log (x)}{x}\right )}{\log ^4(x)}-\frac {9 \log ^4(x)+24 \log ^2(2) \log ^2(x)+16 \log ^4(2)}{x \log ^5(x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((16*Log[2]^4 + 24*Log[2]^2*Log[x]^2 + 9*Log[x]^4)/Log[x]^4)*(-320*x*Log[2]^4 - 240*x*Log[2]^2*Log[x]^2
+ 5*x*Log[x]^5 + (-64*Log[2]^4 - 48*Log[2]^2*Log[x]^2)*Log[Log[5]]))/(2*x*Log[x]^5),x]

[Out]

(-2*E^((16*Log[2]^4 + 24*Log[2]^2*Log[x]^2 + 9*Log[x]^4)/Log[x]^4)*(20*x*Log[2]^4 + 15*x*Log[2]^2*Log[x]^2 + (
4*Log[2]^4 + 3*Log[2]^2*Log[x]^2)*Log[Log[5]]))/(x*Log[x]^5*((3*((4*Log[2]^2*Log[x])/x + (3*Log[x]^3)/x))/Log[
x]^4 - (16*Log[2]^4 + 24*Log[2]^2*Log[x]^2 + 9*Log[x]^4)/(x*Log[x]^5)))

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 {1}{2} \int \frac {\exp \left (\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}\right ) \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{x \log ^5(x)} \, dx\\ &=-\frac {2 \exp \left (\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}\right ) \left (20 x \log ^4(2)+15 x \log ^2(2) \log ^2(x)+\left (4 \log ^4(2)+3 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{x \log ^5(x) \left (\frac {3 \left (\frac {4 \log ^2(2) \log (x)}{x}+\frac {3 \log ^3(x)}{x}\right )}{\log ^4(x)}-\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{x \log ^5(x)}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 33, normalized size = 1.22 \begin {gather*} \frac {1}{2} e^{\frac {\left (4 \log ^2(2)+3 \log ^2(x)\right )^2}{\log ^4(x)}} (5 x+\log (\log (5))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((16*Log[2]^4 + 24*Log[2]^2*Log[x]^2 + 9*Log[x]^4)/Log[x]^4)*(-320*x*Log[2]^4 - 240*x*Log[2]^2*Lo
g[x]^2 + 5*x*Log[x]^5 + (-64*Log[2]^4 - 48*Log[2]^2*Log[x]^2)*Log[Log[5]]))/(2*x*Log[x]^5),x]

[Out]

(E^((4*Log[2]^2 + 3*Log[x]^2)^2/Log[x]^4)*(5*x + Log[Log[5]]))/2

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fricas [A]  time = 1.05, size = 38, normalized size = 1.41 \begin {gather*} \frac {1}{2} \, {\left (5 \, x + \log \left (\log \relax (5)\right )\right )} e^{\left (\frac {16 \, \log \relax (2)^{4} + 24 \, \log \relax (2)^{2} \log \relax (x)^{2} + 9 \, \log \relax (x)^{4}}{\log \relax (x)^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-48*log(2)^2*log(x)^2-64*log(2)^4)*log(log(5))+5*x*log(x)^5-240*x*log(2)^2*log(x)^2-320*x*log(
2)^4)*exp((9*log(x)^4+24*log(2)^2*log(x)^2+16*log(2)^4)/log(x)^4)/x/log(x)^5,x, algorithm="fricas")

[Out]

1/2*(5*x + log(log(5)))*e^((16*log(2)^4 + 24*log(2)^2*log(x)^2 + 9*log(x)^4)/log(x)^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-48*log(2)^2*log(x)^2-64*log(2)^4)*log(log(5))+5*x*log(x)^5-240*x*log(2)^2*log(x)^2-320*x*log(
2)^4)*exp((9*log(x)^4+24*log(2)^2*log(x)^2+16*log(2)^4)/log(x)^4)/x/log(x)^5,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.04, size = 31, normalized size = 1.15

method	result	size
risch	\(\frac {\left (5 x +\ln \left (\ln \relax (5)\right )\right ) {\mathrm e}^{\frac {\left (3 \ln \relax (x )^{2}+4 \ln \relax (2)^{2}\right )^{2}}{\ln \relax (x )^{4}}}}{2}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-48*ln(2)^2*ln(x)^2-64*ln(2)^4)*ln(ln(5))+5*x*ln(x)^5-240*x*ln(2)^2*ln(x)^2-320*x*ln(2)^4)*exp((9*ln
(x)^4+24*ln(2)^2*ln(x)^2+16*ln(2)^4)/ln(x)^4)/x/ln(x)^5,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{2} \, \int \frac {{\left (5 \, x \log \relax (x)^{5} - 320 \, x \log \relax (2)^{4} - 240 \, x \log \relax (2)^{2} \log \relax (x)^{2} - 16 \, {\left (4 \, \log \relax (2)^{4} + 3 \, \log \relax (2)^{2} \log \relax (x)^{2}\right )} \log \left (\log \relax (5)\right )\right )} e^{\left (\frac {16 \, \log \relax (2)^{4} + 24 \, \log \relax (2)^{2} \log \relax (x)^{2} + 9 \, \log \relax (x)^{4}}{\log \relax (x)^{4}}\right )}}{x \log \relax (x)^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-48*log(2)^2*log(x)^2-64*log(2)^4)*log(log(5))+5*x*log(x)^5-240*x*log(2)^2*log(x)^2-320*x*log(
2)^4)*exp((9*log(x)^4+24*log(2)^2*log(x)^2+16*log(2)^4)/log(x)^4)/x/log(x)^5,x, algorithm="maxima")

[Out]

1/2*integrate((5*x*log(x)^5 - 320*x*log(2)^4 - 240*x*log(2)^2*log(x)^2 - 16*(4*log(2)^4 + 3*log(2)^2*log(x)^2)
*log(log(5)))*e^((16*log(2)^4 + 24*log(2)^2*log(x)^2 + 9*log(x)^4)/log(x)^4)/(x*log(x)^5), x)

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mupad [B]  time = 6.70, size = 57, normalized size = 2.11 \begin {gather*} \frac {{\mathrm {e}}^9\,{\mathrm {e}}^{\frac {16\,{\ln \relax (2)}^4}{{\ln \relax (x)}^4}}\,{\mathrm {e}}^{\frac {24\,{\ln \relax (2)}^2}{{\ln \relax (x)}^2}}\,\ln \left (\ln \relax (5)\right )}{2}+\frac {5\,x\,{\mathrm {e}}^9\,{\mathrm {e}}^{\frac {16\,{\ln \relax (2)}^4}{{\ln \relax (x)}^4}}\,{\mathrm {e}}^{\frac {24\,{\ln \relax (2)}^2}{{\ln \relax (x)}^2}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((9*log(x)^4 + 24*log(2)^2*log(x)^2 + 16*log(2)^4)/log(x)^4)*(log(log(5))*(48*log(2)^2*log(x)^2 + 64*
log(2)^4) - 5*x*log(x)^5 + 320*x*log(2)^4 + 240*x*log(2)^2*log(x)^2))/(2*x*log(x)^5),x)

[Out]

(exp(9)*exp((16*log(2)^4)/log(x)^4)*exp((24*log(2)^2)/log(x)^2)*log(log(5)))/2 + (5*x*exp(9)*exp((16*log(2)^4)
/log(x)^4)*exp((24*log(2)^2)/log(x)^2))/2

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sympy [A]  time = 2.42, size = 41, normalized size = 1.52 \begin {gather*} \frac {\left (5 x + \log {\left (\log {\relax (5 )} \right )}\right ) e^{\frac {9 \log {\relax (x )}^{4} + 24 \log {\relax (2 )}^{2} \log {\relax (x )}^{2} + 16 \log {\relax (2 )}^{4}}{\log {\relax (x )}^{4}}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-48*ln(2)**2*ln(x)**2-64*ln(2)**4)*ln(ln(5))+5*x*ln(x)**5-240*x*ln(2)**2*ln(x)**2-320*x*ln(2)*
*4)*exp((9*ln(x)**4+24*ln(2)**2*ln(x)**2+16*ln(2)**4)/ln(x)**4)/x/ln(x)**5,x)

[Out]

(5*x + log(log(5)))*exp((9*log(x)**4 + 24*log(2)**2*log(x)**2 + 16*log(2)**4)/log(x)**4)/2

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