3.5.39 \(\int \frac {-10-x+x \log (x)+(-1-2 x-x^2+(2 x+2 x^2) \log (x)) \log (\frac {\log (5)}{25})}{2 x \log ^2(x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {5+\frac {1}{2} \left (x+\log (x)+(1+x)^2 \log \left (\frac {\log (5)}{25}\right )\right )}{\log (x)} \]

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Rubi [F]  time = 0.38, 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 {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10 - x + x*Log[x] + (-1 - 2*x - x^2 + (2*x + 2*x^2)*Log[x])*Log[Log[5]/25])/(2*x*Log[x]^2),x]

[Out]

ExpIntegralEi[2*Log[x]]*Log[Log[5]/25] + ((1 + 2*Log[Log[5]/25])*LogIntegral[x])/2 + Defer[Int][(-10 + Log[25/
Log[5]] - x^2*Log[Log[5]/25] - x*(1 - Log[625] + 2*Log[Log[5]]))/(x*Log[x]^2), x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{x \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (\frac {1+2 \log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)}+\frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {1+2 \log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)} \, dx+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \left (\frac {2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)}+\frac {1+2 \log \left (\frac {\log (5)}{25}\right )}{\log (x)}\right ) \, dx+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx\\ &=\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx+\log \left (\frac {\log (5)}{25}\right ) \int \frac {x}{\log (x)} \, dx+\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \int \frac {1}{\log (x)} \, dx\\ &=\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \text {li}(x)+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx+\log \left (\frac {\log (5)}{25}\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\text {Ei}(2 \log (x)) \log \left (\frac {\log (5)}{25}\right )+\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \text {li}(x)+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 39, normalized size = 1.39 \begin {gather*} \frac {10+x+\log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )+x^2 \log \left (\frac {\log (5)}{25}\right )}{2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - x + x*Log[x] + (-1 - 2*x - x^2 + (2*x + 2*x^2)*Log[x])*Log[Log[5]/25])/(2*x*Log[x]^2),x]

[Out]

(10 + x + Log[Log[5]/25] + 2*x*Log[Log[5]/25] + x^2*Log[Log[5]/25])/(2*Log[x])

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fricas [A]  time = 0.92, size = 23, normalized size = 0.82 \begin {gather*} \frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {1}{25} \, \log \relax (5)\right ) + x + 10}{2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x-10)/x/log(x)^2,x, algorithm="fricas"
)

[Out]

1/2*((x^2 + 2*x + 1)*log(1/25*log(5)) + x + 10)/log(x)

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giac [B]  time = 0.74, size = 46, normalized size = 1.64 \begin {gather*} -\frac {2 \, x^{2} \log \relax (5) - x^{2} \log \left (\log \relax (5)\right ) + 4 \, x \log \relax (5) - 2 \, x \log \left (\log \relax (5)\right ) - x + 2 \, \log \relax (5) - \log \left (\log \relax (5)\right ) - 10}{2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x-10)/x/log(x)^2,x, algorithm="giac")

[Out]

-1/2*(2*x^2*log(5) - x^2*log(log(5)) + 4*x*log(5) - 2*x*log(log(5)) - x + 2*log(5) - log(log(5)) - 10)/log(x)

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maple [A]  time = 0.04, size = 42, normalized size = 1.50

method	result	size
norman	\(\frac {\left (-\ln \relax (5)+\frac {\ln \left (\ln \relax (5)\right )}{2}\right ) x^{2}+\left (-2 \ln \relax (5)+\ln \left (\ln \relax (5)\right )+\frac {1}{2}\right ) x +5-\ln \relax (5)+\frac {\ln \left (\ln \relax (5)\right )}{2}}{\ln \relax (x )}\)	\(42\)
risch	\(-\frac {2 x^{2} \ln \relax (5)-\ln \left (\ln \relax (5)\right ) x^{2}+4 x \ln \relax (5)-2 x \ln \left (\ln \relax (5)\right )+2 \ln \relax (5)-\ln \left (\ln \relax (5)\right )-x -10}{2 \ln \relax (x )}\)	\(47\)
default	\(2 \ln \relax (5) \expIntegralEi \left (1, -2 \ln \relax (x )\right )-\ln \left (\ln \relax (5)\right ) \expIntegralEi \left (1, -2 \ln \relax (x )\right )+2 \ln \relax (5) \expIntegralEi \left (1, -\ln \relax (x )\right )-\ln \left (\ln \relax (5)\right ) \expIntegralEi \left (1, -\ln \relax (x )\right )+\ln \relax (5) \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )-\frac {\ln \left (\ln \relax (5)\right ) \left (-\frac {x^{2}}{\ln \relax (x )}-2 \expIntegralEi \left (1, -2 \ln \relax (x )\right )\right )}{2}+2 \ln \relax (5) \left (-\frac {x}{\ln \relax (x )}-\expIntegralEi \left (1, -\ln \relax (x )\right )\right )-\ln \left (\ln \relax (5)\right ) \left (-\frac {x}{\ln \relax (x )}-\expIntegralEi \left (1, -\ln \relax (x )\right )\right )-\frac {\ln \relax (5)}{\ln \relax (x )}+\frac {\ln \left (\ln \relax (5)\right )}{2 \ln \relax (x )}+\frac {x}{2 \ln \relax (x )}+\frac {5}{\ln \relax (x )}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(((2*x^2+2*x)*ln(x)-x^2-2*x-1)*ln(1/25*ln(5))+x*ln(x)-x-10)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 1.00, size = 77, normalized size = 2.75 \begin {gather*} {\rm Ei}\left (2 \, \log \relax (x)\right ) \log \left (\frac {1}{25} \, \log \relax (5)\right ) + {\rm Ei}\left (\log \relax (x)\right ) \log \left (\frac {1}{25} \, \log \relax (5)\right ) - \Gamma \left (-1, -\log \relax (x)\right ) \log \left (\frac {1}{25} \, \log \relax (5)\right ) - \Gamma \left (-1, -2 \, \log \relax (x)\right ) \log \left (\frac {1}{25} \, \log \relax (5)\right ) + \frac {\log \left (\frac {1}{25} \, \log \relax (5)\right )}{2 \, \log \relax (x)} + \frac {5}{\log \relax (x)} + \frac {1}{2} \, {\rm Ei}\left (\log \relax (x)\right ) - \frac {1}{2} \, \Gamma \left (-1, -\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*x^2+2*x)*log(x)-x^2-2*x-1)*log(1/25*log(5))+x*log(x)-x-10)/x/log(x)^2,x, algorithm="maxima"
)

[Out]

Ei(2*log(x))*log(1/25*log(5)) + Ei(log(x))*log(1/25*log(5)) - gamma(-1, -log(x))*log(1/25*log(5)) - gamma(-1,
-2*log(x))*log(1/25*log(5)) + 1/2*log(1/25*log(5))/log(x) + 5/log(x) + 1/2*Ei(log(x)) - 1/2*gamma(-1, -log(x))

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mupad [B]  time = 0.46, size = 43, normalized size = 1.54 \begin {gather*} \frac {\ln \left (\frac {\ln \relax (5)}{25}\right )\,x^4+\left (2\,\ln \left (\frac {\ln \relax (5)}{25}\right )+1\right )\,x^3+\left (\ln \left (\frac {\ln \relax (5)}{25}\right )+10\right )\,x^2}{2\,x^2\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x/2 + (log(log(5)/25)*(2*x - log(x)*(2*x + 2*x^2) + x^2 + 1))/2 - (x*log(x))/2 + 5)/(x*log(x)^2),x)

[Out]

(x^3*(2*log(log(5)/25) + 1) + x^4*log(log(5)/25) + x^2*(log(log(5)/25) + 10))/(2*x^2*log(x))

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sympy [A]  time = 0.14, size = 49, normalized size = 1.75 \begin {gather*} \frac {- 2 x^{2} \log {\relax (5 )} + x^{2} \log {\left (\log {\relax (5 )} \right )} - 4 x \log {\relax (5 )} + 2 x \log {\left (\log {\relax (5 )} \right )} + x - 2 \log {\relax (5 )} + \log {\left (\log {\relax (5 )} \right )} + 10}{2 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(((2*x**2+2*x)*ln(x)-x**2-2*x-1)*ln(1/25*ln(5))+x*ln(x)-x-10)/x/ln(x)**2,x)

[Out]

(-2*x**2*log(5) + x**2*log(log(5)) - 4*x*log(5) + 2*x*log(log(5)) + x - 2*log(5) + log(log(5)) + 10)/(2*log(x)
)

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