3.47.30 \(\int \frac {1}{4} (1+\log (9 x \log (4))) \, dx\)

Optimal. Leaf size=13 \[ 81+\frac {1}{4} x \log (9 x \log (4)) \]

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Rubi [A]  time = 0.00, antiderivative size = 11, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 2295} \begin {gather*} \frac {1}{4} x \log (9 x \log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + Log[9*x*Log[4]])/4,x]

[Out]

(x*Log[9*x*Log[4]])/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int (1+\log (9 x \log (4))) \, dx\\ &=\frac {x}{4}+\frac {1}{4} \int \log (9 x \log (4)) \, dx\\ &=\frac {1}{4} x \log (9 x \log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.85 \begin {gather*} \frac {1}{4} x \log (9 x \log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + Log[9*x*Log[4]])/4,x]

[Out]

(x*Log[9*x*Log[4]])/4

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fricas [A]  time = 0.80, size = 9, normalized size = 0.69 \begin {gather*} \frac {1}{4} \, x \log \left (18 \, x \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*log(18*x*log(2))+1/4,x, algorithm="fricas")

[Out]

1/4*x*log(18*x*log(2))

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giac [B]  time = 0.12, size = 26, normalized size = 2.00 \begin {gather*} \frac {1}{4} \, x + \frac {x \log \relax (2) \log \left (18 \, x \log \relax (2)\right ) - x \log \relax (2)}{4 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*log(18*x*log(2))+1/4,x, algorithm="giac")

[Out]

1/4*x + 1/4*(x*log(2)*log(18*x*log(2)) - x*log(2))/log(2)

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maple [A]  time = 0.02, size = 10, normalized size = 0.77




method result size



derivativedivides \(\frac {\ln \left (18 x \ln \relax (2)\right ) x}{4}\) \(10\)
default \(\frac {\ln \left (18 x \ln \relax (2)\right ) x}{4}\) \(10\)
norman \(\frac {\ln \left (18 x \ln \relax (2)\right ) x}{4}\) \(10\)
risch \(\frac {\ln \left (18 x \ln \relax (2)\right ) x}{4}\) \(10\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*ln(18*x*ln(2))+1/4,x,method=_RETURNVERBOSE)

[Out]

1/4*ln(18*x*ln(2))*x

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maxima [B]  time = 0.37, size = 26, normalized size = 2.00 \begin {gather*} \frac {1}{4} \, x + \frac {x \log \relax (2) \log \left (18 \, x \log \relax (2)\right ) - x \log \relax (2)}{4 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*log(18*x*log(2))+1/4,x, algorithm="maxima")

[Out]

1/4*x + 1/4*(x*log(2)*log(18*x*log(2)) - x*log(2))/log(2)

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mupad [B]  time = 3.41, size = 11, normalized size = 0.85 \begin {gather*} \frac {x\,\left (\ln \left (18\right )+\ln \left (\ln \relax (2)\right )+\ln \relax (x)\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(18*x*log(2))/4 + 1/4,x)

[Out]

(x*(log(18) + log(log(2)) + log(x)))/4

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sympy [A]  time = 0.11, size = 10, normalized size = 0.77 \begin {gather*} \frac {x \log {\left (18 x \log {\relax (2 )} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*ln(18*x*ln(2))+1/4,x)

[Out]

x*log(18*x*log(2))/4

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