3.39.13 \(\int \frac {-e^{10} x+2 e^{10} x \log (x) \log (\log (x))-2 e^{10} \log (x) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=23 \[ e^{10} \left (-2 x+\frac {3}{\log (4)}+\frac {x^2}{\log (\log (x))}\right ) \]

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Rubi [F]  time = 0.22, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-e^{10} x+2 e^{10} x \log (x) \log (\log (x))-2 e^{10} \log (x) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^10*x) + 2*E^10*x*Log[x]*Log[Log[x]] - 2*E^10*Log[x]*Log[Log[x]]^2)/(Log[x]*Log[Log[x]]^2),x]

[Out]

-2*E^10*x - E^10*Defer[Int][x/(Log[x]*Log[Log[x]]^2), x] + 2*E^10*Defer[Int][x/Log[Log[x]], x]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 19, normalized size = 0.83 \begin {gather*} -2 e^{10} x+\frac {e^{10} x^2}{\log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^10*x) + 2*E^10*x*Log[x]*Log[Log[x]] - 2*E^10*Log[x]*Log[Log[x]]^2)/(Log[x]*Log[Log[x]]^2),x]

[Out]

-2*E^10*x + (E^10*x^2)/Log[Log[x]]

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fricas [A]  time = 0.68, size = 21, normalized size = 0.91 \begin {gather*} \frac {x^{2} e^{10} - 2 \, x e^{10} \log \left (\log \relax (x)\right )}{\log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*e^10 - 2*x*e^10*log(log(x)))/log(log(x))

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giac [A]  time = 0.25, size = 17, normalized size = 0.74 \begin {gather*} -2 \, x e^{10} + \frac {x^{2} e^{10}}{\log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*e^10 + x^2*e^10/log(log(x))

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maple [A]  time = 0.05, size = 18, normalized size = 0.78

method	result	size
risch	\(-2 x \,{\mathrm e}^{10}+\frac {x^{2} {\mathrm e}^{10}}{\ln \left (\ln \relax (x )\right )}\)	\(18\)
norman	\(\frac {x^{2} {\mathrm e}^{10}-2 x \,{\mathrm e}^{10} \ln \left (\ln \relax (x )\right )}{\ln \left (\ln \relax (x )\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(5)^2*ln(x)*ln(ln(x))^2+2*x*exp(5)^2*ln(x)*ln(ln(x))-x*exp(5)^2)/ln(x)/ln(ln(x))^2,x,method=_RETURN
VERBOSE)

[Out]

-2*x*exp(10)+x^2*exp(10)/ln(ln(x))

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maxima [A]  time = 0.39, size = 17, normalized size = 0.74 \begin {gather*} -2 \, x e^{10} + \frac {x^{2} e^{10}}{\log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*e^10 + x^2*e^10/log(log(x))

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mupad [B]  time = 2.41, size = 17, normalized size = 0.74 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{10}}{\ln \left (\ln \relax (x)\right )}-2\,x\,{\mathrm {e}}^{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*exp(10) + 2*log(log(x))^2*exp(10)*log(x) - 2*x*log(log(x))*exp(10)*log(x))/(log(log(x))^2*log(x)),x)

[Out]

(x^2*exp(10))/log(log(x)) - 2*x*exp(10)

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sympy [A]  time = 0.31, size = 17, normalized size = 0.74 \begin {gather*} \frac {x^{2} e^{10}}{\log {\left (\log {\relax (x )} \right )}} - 2 x e^{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*exp(10)/log(log(x)) - 2*x*exp(10)

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