3.11.77 \(\int \frac {e^{\frac {x+\log ^2(-\log (5 x^2))}{x}} (4 \log (-\log (5 x^2))-\log (5 x^2) \log ^2(-\log (5 x^2)))}{x^2 \log (5 x^2)} \, dx\)

Optimal. Leaf size=19 \[ e^{\frac {x+\log ^2\left (-\log \left (5 x^2\right )\right )}{x}} \]

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Rubi [A]  time = 0.41, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6706} \begin {gather*} e^{\frac {\log ^2\left (-\log \left (5 x^2\right )\right )+x}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^((x + Log[-Log[5*x^2]]^2)/x)

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{\frac {x+\log ^2\left (-\log \left (5 x^2\right )\right )}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 19, normalized size = 1.00 \begin {gather*} e^{1+\frac {\log ^2\left (-\log \left (5 x^2\right )\right )}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(1 + Log[-Log[5*x^2]]^2/x)

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fricas [A]  time = 0.67, size = 18, normalized size = 0.95 \begin {gather*} e^{\left (\frac {\log \left (-\log \left (5 \, x^{2}\right )\right )^{2} + x}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^((log(-log(5*x^2))^2 + x)/x)

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giac [A]  time = 0.49, size = 18, normalized size = 0.95 \begin {gather*} e^{\left (\frac {\log \left (-\log \left (5 \, x^{2}\right )\right )^{2}}{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(log(-log(5*x^2))^2/x + 1)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (-\ln \left (5 x^{2}\right ) \ln \left (-\ln \left (5 x^{2}\right )\right )^{2}+4 \ln \left (-\ln \left (5 x^{2}\right )\right )\right ) {\mathrm e}^{\frac {\ln \left (-\ln \left (5 x^{2}\right )\right )^{2}+x}{x}}}{x^{2} \ln \left (5 x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(5*x^2)*ln(-ln(5*x^2))^2+4*ln(-ln(5*x^2)))*exp((ln(-ln(5*x^2))^2+x)/x)/x^2/ln(5*x^2),x)

[Out]

int((-ln(5*x^2)*ln(-ln(5*x^2))^2+4*ln(-ln(5*x^2)))*exp((ln(-ln(5*x^2))^2+x)/x)/x^2/ln(5*x^2),x)

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maxima [A]  time = 0.90, size = 19, normalized size = 1.00 \begin {gather*} e^{\left (\frac {\log \left (-\log \relax (5) - 2 \, \log \relax (x)\right )^{2}}{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(log(-log(5) - 2*log(x))^2/x + 1)

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mupad [B]  time = 0.83, size = 22, normalized size = 1.16 \begin {gather*} {\mathrm {e}}^{\frac {{\ln \left (-\ln \left (x^2\right )-\ln \relax (5)\right )}^2}{x}}\,\mathrm {e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(log(- log(x^2) - log(5))^2/x)*exp(1)

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sympy [A]  time = 0.46, size = 15, normalized size = 0.79 \begin {gather*} e^{\frac {x + \log {\left (- \log {\left (5 x^{2} \right )} \right )}^{2}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((x + log(-log(5*x**2))**2)/x)

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