3.49.21 \(\int \frac {e^{-18+2 \log ^{2 x}(2)} (-1+4 x \log ^{2 x}(2) \log (\log (2)))}{x^2} \, dx\)

Optimal. Leaf size=16 \[ \frac {e^{-18+2 \log ^{2 x}(2)}}{x} \]

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Rubi [A]  time = 0.08, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2288} \begin {gather*} \frac {e^{2 \log ^{2 x}(2)-18}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-18 + 2*Log[2]^(2*x))*(-1 + 4*x*Log[2]^(2*x)*Log[Log[2]]))/x^2,x]

[Out]

E^(-18 + 2*Log[2]^(2*x))/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 {e^{-18+2 \log ^{2 x}(2)}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 1.00 \begin {gather*} \frac {e^{2 \left (-9+\log ^{2 x}(2)\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-18 + 2*Log[2]^(2*x))*(-1 + 4*x*Log[2]^(2*x)*Log[Log[2]]))/x^2,x]

[Out]

E^(2*(-9 + Log[2]^(2*x)))/x

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fricas [A]  time = 1.05, size = 15, normalized size = 0.94 \begin {gather*} \frac {e^{\left (2 \, \log \relax (2)^{2 \, x} - 18\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(log(2))*exp(2*x*log(log(2)))-1)*exp(exp(2*x*log(log(2))))^2/x^2/exp(18),x, algorithm="frica
s")

[Out]

e^(2*log(2)^(2*x) - 18)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, x \log \relax (2)^{2 \, x} \log \left (\log \relax (2)\right ) - 1\right )} e^{\left (2 \, \log \relax (2)^{2 \, x} - 18\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(log(2))*exp(2*x*log(log(2)))-1)*exp(exp(2*x*log(log(2))))^2/x^2/exp(18),x, algorithm="giac"
)

[Out]

integrate((4*x*log(2)^(2*x)*log(log(2)) - 1)*e^(2*log(2)^(2*x) - 18)/x^2, x)

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maple [A]  time = 0.12, size = 16, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{-18+2 \ln \relax (2)^{2 x}}}{x}\)	\(16\)
norman	\(\frac {{\mathrm e}^{-18} {\mathrm e}^{2 \,{\mathrm e}^{2 x \ln \left (\ln \relax (2)\right )}}}{x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(ln(2))*exp(2*x*ln(ln(2)))-1)*exp(exp(2*x*ln(ln(2))))^2/x^2/exp(18),x,method=_RETURNVERBOSE)

[Out]

1/x*exp(-18+2*ln(2)^(2*x))

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maxima [A]  time = 0.53, size = 15, normalized size = 0.94 \begin {gather*} \frac {e^{\left (2 \, \log \relax (2)^{2 \, x} - 18\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(log(2))*exp(2*x*log(log(2)))-1)*exp(exp(2*x*log(log(2))))^2/x^2/exp(18),x, algorithm="maxim
a")

[Out]

e^(2*log(2)^(2*x) - 18)/x

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mupad [B]  time = 3.46, size = 15, normalized size = 0.94 \begin {gather*} \frac {{\mathrm {e}}^{-18}\,{\mathrm {e}}^{2\,{\ln \relax (2)}^{2\,x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-18)*exp(2*exp(2*x*log(log(2))))*(4*x*exp(2*x*log(log(2)))*log(log(2)) - 1))/x^2,x)

[Out]

(exp(-18)*exp(2*log(2)^(2*x)))/x

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sympy [A]  time = 0.14, size = 17, normalized size = 1.06 \begin {gather*} \frac {e^{2 e^{2 x \log {\left (\log {\relax (2 )} \right )}}}}{x e^{18}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(ln(2))*exp(2*x*ln(ln(2)))-1)*exp(exp(2*x*ln(ln(2))))**2/x**2/exp(18),x)

[Out]

exp(-18)*exp(2*exp(2*x*log(log(2))))/x

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