3.12.59 \(\int \frac {-e^5 \log (\frac {5}{2})+e^5 \log (\frac {5}{2}) \log (x) \log (\log (x))+(1+e^5 (1+2 x)) \log (\frac {5}{2}) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=29 \[ \log \left (\frac {5}{2}\right ) \left (-2+\frac {1}{2} \left (-3+x+\frac {x}{e^5}+x^2+\frac {x}{\log (\log (x))}\right )\right ) \]

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Rubi [F]  time = 0.20, 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^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{2 e^5 \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^5*Log[5/2]) + E^5*Log[5/2]*Log[x]*Log[Log[x]] + (1 + E^5*(1 + 2*x))*Log[5/2]*Log[x]*Log[Log[x]]^2)/(2
*E^5*Log[x]*Log[Log[x]]^2),x]

[Out]

((1 + E^5 + 2*E^5*x)^2*Log[5/2])/(8*E^10) - (Log[5/2]*Defer[Int][1/(Log[x]*Log[Log[x]]^2), x])/2 + (Log[5/2]*D
efer[Int][Log[Log[x]]^(-1), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-e^5 \log \left (\frac {5}{2}\right )+e^5 \log \left (\frac {5}{2}\right ) \log (x) \log (\log (x))+\left (1+e^5 (1+2 x)\right ) \log \left (\frac {5}{2}\right ) \log (x) \log ^2(\log (x))}{\log (x) \log ^2(\log (x))} \, dx}{2 e^5}\\ &=\frac {\int \left (\left (1+e^5+2 e^5 x\right ) \log \left (\frac {5}{2}\right )-\frac {e^5 \log \left (\frac {5}{2}\right )}{\log (x) \log ^2(\log (x))}+\frac {e^5 \log \left (\frac {5}{2}\right )}{\log (\log (x))}\right ) \, dx}{2 e^5}\\ &=\frac {\left (1+e^5+2 e^5 x\right )^2 \log \left (\frac {5}{2}\right )}{8 e^{10}}-\frac {1}{2} \log \left (\frac {5}{2}\right ) \int \frac {1}{\log (x) \log ^2(\log (x))} \, dx+\frac {1}{2} \log \left (\frac {5}{2}\right ) \int \frac {1}{\log (\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 30, normalized size = 1.03 \begin {gather*} \frac {x \log \left (\frac {5}{2}\right ) \left (1+e^5 (1+x)+\frac {e^5}{\log (\log (x))}\right )}{2 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^5*Log[5/2]) + E^5*Log[5/2]*Log[x]*Log[Log[x]] + (1 + E^5*(1 + 2*x))*Log[5/2]*Log[x]*Log[Log[x]]
^2)/(2*E^5*Log[x]*Log[Log[x]]^2),x]

[Out]

(x*Log[5/2]*(1 + E^5*(1 + x) + E^5/Log[Log[x]]))/(2*E^5)

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fricas [A]  time = 0.67, size = 32, normalized size = 1.10 \begin {gather*} \frac {{\left (x e^{5} \log \left (\frac {5}{2}\right ) + {\left ({\left (x^{2} + x\right )} e^{5} + x\right )} \log \left (\frac {5}{2}\right ) \log \left (\log \relax (x)\right )\right )} e^{\left (-5\right )}}{2 \, \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*x+1)*exp(5)+1)*log(5/2)*log(x)*log(log(x))^2+exp(5)*log(5/2)*log(x)*log(log(x))-exp(5)*log(
5/2))/exp(5)/log(x)/log(log(x))^2,x, algorithm="fricas")

[Out]

1/2*(x*e^5*log(5/2) + ((x^2 + x)*e^5 + x)*log(5/2)*log(log(x)))*e^(-5)/log(log(x))

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giac [B]  time = 0.77, size = 67, normalized size = 2.31 \begin {gather*} \frac {1}{2} \, {\left (x^{2} e^{5} \log \relax (5) - x^{2} e^{5} \log \relax (2) + x e^{5} \log \relax (5) - x e^{5} \log \relax (2) + x \log \relax (5) - x \log \relax (2) + \frac {x e^{5} \log \relax (5)}{\log \left (\log \relax (x)\right )} - \frac {x e^{5} \log \relax (2)}{\log \left (\log \relax (x)\right )}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*x+1)*exp(5)+1)*log(5/2)*log(x)*log(log(x))^2+exp(5)*log(5/2)*log(x)*log(log(x))-exp(5)*log(
5/2))/exp(5)/log(x)/log(log(x))^2,x, algorithm="giac")

[Out]

1/2*(x^2*e^5*log(5) - x^2*e^5*log(2) + x*e^5*log(5) - x*e^5*log(2) + x*log(5) - x*log(2) + x*e^5*log(5)/log(lo
g(x)) - x*e^5*log(2)/log(log(x)))*e^(-5)

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maple [A]  time = 0.14, size = 55, normalized size = 1.90

method	result	size
risch	\(-\frac {x^{2} \ln \relax (2)}{2}+\frac {x^{2} \ln \relax (5)}{2}-\frac {x \ln \relax (2)}{2}+\frac {x \ln \relax (5)}{2}-\frac {{\mathrm e}^{-5} x \ln \relax (2)}{2}+\frac {x \ln \relax (5) {\mathrm e}^{-5}}{2}-\frac {x \left (\ln \relax (2)-\ln \relax (5)\right )}{2 \ln \left (\ln \relax (x )\right )}\)	\(55\)
norman	\(\frac {\left (-\frac {\ln \relax (2)}{2}+\frac {\ln \relax (5)}{2}\right ) x +\left (-\frac {\ln \relax (2)}{2}+\frac {\ln \relax (5)}{2}\right ) x^{2} \ln \left (\ln \relax (x )\right )-\frac {{\mathrm e}^{-5} \left ({\mathrm e}^{5} \ln \relax (2)-{\mathrm e}^{5} \ln \relax (5)+\ln \relax (2)-\ln \relax (5)\right ) x \ln \left (\ln \relax (x )\right )}{2}}{\ln \left (\ln \relax (x )\right )}\)	\(63\)
default	\(\frac {{\mathrm e}^{-5} \left (\frac {\left (-{\mathrm e}^{5} \ln \relax (2)-\ln \relax (2)\right ) x \ln \left (\ln \relax (x )\right )-x \,{\mathrm e}^{5} \ln \relax (2)-{\mathrm e}^{5} \ln \relax (2) x^{2} \ln \left (\ln \relax (x )\right )}{\ln \left (\ln \relax (x )\right )}+\frac {\left ({\mathrm e}^{5} \ln \relax (5)+\ln \relax (5)\right ) x \ln \left (\ln \relax (x )\right )+x \,{\mathrm e}^{5} \ln \relax (5)+{\mathrm e}^{5} \ln \relax (5) x^{2} \ln \left (\ln \relax (x )\right )}{\ln \left (\ln \relax (x )\right )}\right )}{2}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(((2*x+1)*exp(5)+1)*ln(5/2)*ln(x)*ln(ln(x))^2+exp(5)*ln(5/2)*ln(x)*ln(ln(x))-exp(5)*ln(5/2))/exp(5)/ln
(x)/ln(ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.60, size = 39, normalized size = 1.34 \begin {gather*} \frac {1}{2} \, {\left (x^{2} e^{5} \log \left (\frac {5}{2}\right ) + x e^{5} \log \left (\frac {5}{2}\right ) + x \log \left (\frac {5}{2}\right ) + \frac {x {\left (\log \relax (5) - \log \relax (2)\right )} e^{5}}{\log \left (\log \relax (x)\right )}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*x+1)*exp(5)+1)*log(5/2)*log(x)*log(log(x))^2+exp(5)*log(5/2)*log(x)*log(log(x))-exp(5)*log(
5/2))/exp(5)/log(x)/log(log(x))^2,x, algorithm="maxima")

[Out]

1/2*(x^2*e^5*log(5/2) + x*e^5*log(5/2) + x*log(5/2) + x*(log(5) - log(2))*e^5/log(log(x)))*e^(-5)

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mupad [B]  time = 0.96, size = 31, normalized size = 1.07 \begin {gather*} x\,\left (\frac {\ln \left (\frac {5}{2}\right )}{2}+\frac {{\mathrm {e}}^{-5}\,\ln \left (\frac {5}{2}\right )}{2}\right )+\frac {x^2\,\ln \left (\frac {5}{2}\right )}{2}+\frac {x\,\ln \left (\frac {5}{2}\right )}{2\,\ln \left (\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-5)*((log(log(x))*exp(5)*log(5/2)*log(x))/2 - (exp(5)*log(5/2))/2 + (log(log(x))^2*log(5/2)*log(x)*(e
xp(5)*(2*x + 1) + 1))/2))/(log(log(x))^2*log(x)),x)

[Out]

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

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sympy [A]  time = 0.28, size = 56, normalized size = 1.93 \begin {gather*} x^{2} \left (- \frac {\log {\relax (2 )}}{2} + \frac {\log {\relax (5 )}}{2}\right ) + \frac {x \left (- e^{5} \log {\relax (2 )} - \log {\relax (2 )} + \log {\relax (5 )} + e^{5} \log {\relax (5 )}\right )}{2 e^{5}} + \frac {- x \log {\relax (2 )} + x \log {\relax (5 )}}{2 \log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*x+1)*exp(5)+1)*ln(5/2)*ln(x)*ln(ln(x))**2+exp(5)*ln(5/2)*ln(x)*ln(ln(x))-exp(5)*ln(5/2))/ex
p(5)/ln(x)/ln(ln(x))**2,x)

[Out]

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

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