3.99.21 \(\int (1+10 \log (\log (2))) \, dx\) [9821]

Optimal. Leaf size=14 \[ x+5 \left (\frac {619}{125}+2 x\right ) \log (\log (2)) \]

[Out]

x+5*ln(ln(2))*(2*x+619/125)

________________________________________________________________________________________

Rubi [A]
time = 0.00, antiderivative size = 9, normalized size of antiderivative = 0.64, number of steps used = 1, number of rules used = 1, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {8} \begin {gather*} x (1+10 \log (\log (2))) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + 10*Log[Log[2]],x]

[Out]

x*(1 + 10*Log[Log[2]])

Rule 8

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x (1+10 \log (\log (2)))\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]
time = 0.00, size = 8, normalized size = 0.57 \begin {gather*} x+10 x \log (\log (2)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + 10*Log[Log[2]],x]

[Out]

x + 10*x*Log[Log[2]]

________________________________________________________________________________________

Maple [A]
time = 0.01, size = 10, normalized size = 0.71

method	result	size
risch	\(10 x \ln \left (\ln \left (2\right )\right )+x\)	\(9\)
default	\(x \left (10 \ln \left (\ln \left (2\right )\right )+1\right )\)	\(10\)
norman	\(x \left (10 \ln \left (\ln \left (2\right )\right )+1\right )\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*ln(ln(2))+1,x,method=_RETURNVERBOSE)

[Out]

x*(10*ln(ln(2))+1)

________________________________________________________________________________________

Maxima [A]
time = 0.25, size = 9, normalized size = 0.64 \begin {gather*} x {\left (10 \, \log \left (\log \left (2\right )\right ) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*log(log(2))+1,x, algorithm="maxima")

[Out]

x*(10*log(log(2)) + 1)

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 8, normalized size = 0.57 \begin {gather*} 10 \, x \log \left (\log \left (2\right )\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*log(log(2))+1,x, algorithm="fricas")

[Out]

10*x*log(log(2)) + x

________________________________________________________________________________________

Sympy [A]
time = 0.01, size = 8, normalized size = 0.57 \begin {gather*} x \left (10 \log {\left (\log {\left (2 \right )} \right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*ln(ln(2))+1,x)

[Out]

x*(10*log(log(2)) + 1)

________________________________________________________________________________________

Giac [A]
time = 0.40, size = 9, normalized size = 0.64 \begin {gather*} x {\left (10 \, \log \left (\log \left (2\right )\right ) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*log(log(2))+1,x, algorithm="giac")

[Out]

x*(10*log(log(2)) + 1)

________________________________________________________________________________________

Mupad [B]
time = 0.00, size = 9, normalized size = 0.64 \begin {gather*} x\,\left (10\,\ln \left (\ln \left (2\right )\right )+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*log(log(2)) + 1,x)

[Out]

x*(10*log(log(2)) + 1)

________________________________________________________________________________________