3.29.27 \(\int (1+9 \log (5)-\log ^2(\log (5))) \, dx\) [2827]

Optimal. Leaf size=20 \[ -1+x+3 (-1+3 x \log (5))-x \log ^2(\log (5)) \]

[Out]

x-x*ln(ln(5))^2+9*x*ln(5)-4

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Rubi [A]
time = 0.00, antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {8} \begin {gather*} x \left (1-\log ^2(\log (5))+9 \log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + 9*Log[5] - Log[Log[5]]^2,x]

[Out]

x*(1 + 9*Log[5] - Log[Log[5]]^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x \left (1+9 \log (5)-\log ^2(\log (5))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 15, normalized size = 0.75 \begin {gather*} x+9 x \log (5)-x \log ^2(\log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + 9*Log[5] - Log[Log[5]]^2,x]

[Out]

x + 9*x*Log[5] - x*Log[Log[5]]^2

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Maple [A]
time = 0.11, size = 16, normalized size = 0.80

method result size
default \(x \left (-\ln \left (\ln \left (5\right )\right )^{2}+9 \ln \left (5\right )+1\right )\) \(16\)
norman \(x \left (-\ln \left (\ln \left (5\right )\right )^{2}+9 \ln \left (5\right )+1\right )\) \(16\)
risch \(-x \ln \left (\ln \left (5\right )\right )^{2}+9 x \ln \left (5\right )+x\) \(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-ln(ln(5))^2+9*ln(5)+1,x,method=_RETURNVERBOSE)

[Out]

x*(-ln(ln(5))^2+9*ln(5)+1)

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Maxima [A]
time = 0.27, size = 14, normalized size = 0.70 \begin {gather*} -{\left (\log \left (\log \left (5\right )\right )^{2} - 9 \, \log \left (5\right ) - 1\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(log(5))^2+9*log(5)+1,x, algorithm="maxima")

[Out]

-(log(log(5))^2 - 9*log(5) - 1)*x

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Fricas [A]
time = 0.44, size = 15, normalized size = 0.75 \begin {gather*} -x \log \left (\log \left (5\right )\right )^{2} + 9 \, x \log \left (5\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(log(5))^2+9*log(5)+1,x, algorithm="fricas")

[Out]

-x*log(log(5))^2 + 9*x*log(5) + x

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Sympy [A]
time = 0.01, size = 14, normalized size = 0.70 \begin {gather*} x \left (- \log {\left (\log {\left (5 \right )} \right )}^{2} + 1 + 9 \log {\left (5 \right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-ln(ln(5))**2+9*ln(5)+1,x)

[Out]

x*(-log(log(5))**2 + 1 + 9*log(5))

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Giac [A]
time = 0.38, size = 14, normalized size = 0.70 \begin {gather*} -{\left (\log \left (\log \left (5\right )\right )^{2} - 9 \, \log \left (5\right ) - 1\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(log(5))^2+9*log(5)+1,x, algorithm="giac")

[Out]

-(log(log(5))^2 - 9*log(5) - 1)*x

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Mupad [B]
time = 0.00, size = 15, normalized size = 0.75 \begin {gather*} x\,\left (9\,\ln \left (5\right )-{\ln \left (\ln \left (5\right )\right )}^2+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(9*log(5) - log(log(5))^2 + 1,x)

[Out]

x*(9*log(5) - log(log(5))^2 + 1)

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