3.25.49 \(\int \frac {-20 x+6 e^{3+3 e^{e^2}} \log (\log (5))}{x^4 \log (\log (5))} \, dx\)

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

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {12, 43} \begin {gather*} \frac {10}{x^2 \log (\log (5))}-\frac {2 e^{3+3 e^{e^2}}}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*x + 6*E^(3 + 3*E^E^2)*Log[Log[5]])/(x^4*Log[Log[5]]),x]

[Out]

(-2*E^(3 + 3*E^E^2))/x^3 + 10/(x^2*Log[Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-20 x+6 e^{3+3 e^{e^2}} \log (\log (5))}{x^4} \, dx}{\log (\log (5))}\\ &=\frac {\int \left (-\frac {20}{x^3}+\frac {6 e^{3+3 e^{e^2}} \log (\log (5))}{x^4}\right ) \, dx}{\log (\log (5))}\\ &=-\frac {2 e^{3+3 e^{e^2}}}{x^3}+\frac {10}{x^2 \log (\log (5))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.07 \begin {gather*} 2 \left (-\frac {e^{3+3 e^{e^2}}}{x^3}+\frac {5}{x^2 \log (\log (5))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x + 6*E^(3 + 3*E^E^2)*Log[Log[5]])/(x^4*Log[Log[5]]),x]

[Out]

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

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fricas [A]  time = 0.58, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (e^{\left (3 \, e^{\left (e^{2}\right )} + 3\right )} \log \left (\log \relax (5)\right ) - 5 \, x\right )}}{x^{3} \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(log(5))*exp(3*exp(exp(2))+3)-20*x)/x^4/log(log(5)),x, algorithm="fricas")

[Out]

-2*(e^(3*e^(e^2) + 3)*log(log(5)) - 5*x)/(x^3*log(log(5)))

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giac [A]  time = 0.41, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (e^{\left (3 \, e^{\left (e^{2}\right )} + 3\right )} \log \left (\log \relax (5)\right ) - 5 \, x\right )}}{x^{3} \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(log(5))*exp(3*exp(exp(2))+3)-20*x)/x^4/log(log(5)),x, algorithm="giac")

[Out]

-2*(e^(3*e^(e^2) + 3)*log(log(5)) - 5*x)/(x^3*log(log(5)))

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maple [A]  time = 0.06, size = 24, normalized size = 0.89

method	result	size
norman	\(\frac {\frac {10 x}{\ln \left (\ln \relax (5)\right )}-2 \,{\mathrm e}^{3 \,{\mathrm e}^{{\mathrm e}^{2}}} {\mathrm e}^{3}}{x^{3}}\)	\(24\)
gosper	\(-\frac {2 \left (\ln \left (\ln \relax (5)\right ) {\mathrm e}^{3 \,{\mathrm e}^{{\mathrm e}^{2}}+3}-5 x \right )}{\ln \left (\ln \relax (5)\right ) x^{3}}\)	\(27\)
risch	\(\frac {-2 \ln \left (\ln \relax (5)\right ) {\mathrm e}^{3 \,{\mathrm e}^{{\mathrm e}^{2}}+3}+10 x}{\ln \left (\ln \relax (5)\right ) x^{3}}\)	\(27\)
default	\(\frac {-\frac {2 \ln \left (\ln \relax (5)\right ) {\mathrm e}^{3 \,{\mathrm e}^{{\mathrm e}^{2}}+3}}{x^{3}}+\frac {10}{x^{2}}}{\ln \left (\ln \relax (5)\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*ln(ln(5))*exp(3*exp(exp(2))+3)-20*x)/x^4/ln(ln(5)),x,method=_RETURNVERBOSE)

[Out]

(10*x/ln(ln(5))-2*exp(exp(exp(2)))^3*exp(3))/x^3

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maxima [A]  time = 0.66, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (e^{\left (3 \, e^{\left (e^{2}\right )} + 3\right )} \log \left (\log \relax (5)\right ) - 5 \, x\right )}}{x^{3} \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(log(5))*exp(3*exp(exp(2))+3)-20*x)/x^4/log(log(5)),x, algorithm="maxima")

[Out]

-2*(e^(3*e^(e^2) + 3)*log(log(5)) - 5*x)/(x^3*log(log(5)))

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mupad [B]  time = 1.34, size = 24, normalized size = 0.89 \begin {gather*} \frac {10}{x^2\,\ln \left (\ln \relax (5)\right )}-\frac {2\,{\mathrm {e}}^{3\,{\mathrm {e}}^{{\mathrm {e}}^2}+3}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x - 6*exp(3*exp(exp(2)) + 3)*log(log(5)))/(x^4*log(log(5))),x)

[Out]

10/(x^2*log(log(5))) - (2*exp(3*exp(exp(2)) + 3))/x^3

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sympy [A]  time = 0.14, size = 31, normalized size = 1.15 \begin {gather*} - \frac {- 10 x + 2 e^{3} e^{3 e^{e^{2}}} \log {\left (\log {\relax (5 )} \right )}}{x^{3} \log {\left (\log {\relax (5 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*ln(ln(5))*exp(3*exp(exp(2))+3)-20*x)/x**4/ln(ln(5)),x)

[Out]

-(-10*x + 2*exp(3)*exp(3*exp(exp(2)))*log(log(5)))/(x**3*log(log(5)))

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