3.81.11 \(\int \frac {4 x+24 \log (3)}{3 x} \, dx\) [8011]

Optimal. Leaf size=14 \[ \frac {4 x}{3}+4 \log (3) \log \left (x^2\right ) \]

[Out]

4/3*x+4*ln(3)*ln(x^2)

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Rubi [A]
time = 0.00, antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 45} \begin {gather*} \frac {4 x}{3}+8 \log (3) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x + 24*Log[3])/(3*x),x]

[Out]

(4*x)/3 + 8*Log[3]*Log[x]

Rule 12

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {4 x+24 \log (3)}{x} \, dx\\ &=\frac {1}{3} \int \left (4+\frac {24 \log (3)}{x}\right ) \, dx\\ &=\frac {4 x}{3}+8 \log (3) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 12, normalized size = 0.86 \begin {gather*} \frac {4 x}{3}+8 \log (3) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x + 24*Log[3])/(3*x),x]

[Out]

(4*x)/3 + 8*Log[3]*Log[x]

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Maple [A]
time = 0.29, size = 11, normalized size = 0.79

method	result	size
default	\(\frac {4 x}{3}+8 \ln \left (3\right ) \ln \left (x \right )\)	\(11\)
norman	\(\frac {4 x}{3}+8 \ln \left (3\right ) \ln \left (x \right )\)	\(11\)
risch	\(\frac {4 x}{3}+8 \ln \left (3\right ) \ln \left (x \right )\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(24*ln(3)+4*x)/x,x,method=_RETURNVERBOSE)

[Out]

4/3*x+8*ln(3)*ln(x)

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Maxima [A]
time = 0.27, size = 10, normalized size = 0.71 \begin {gather*} 8 \, \log \left (3\right ) \log \left (x\right ) + \frac {4}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(24*log(3)+4*x)/x,x, algorithm="maxima")

[Out]

8*log(3)*log(x) + 4/3*x

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Fricas [A]
time = 0.37, size = 10, normalized size = 0.71 \begin {gather*} 8 \, \log \left (3\right ) \log \left (x\right ) + \frac {4}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(24*log(3)+4*x)/x,x, algorithm="fricas")

[Out]

8*log(3)*log(x) + 4/3*x

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Sympy [A]
time = 0.03, size = 12, normalized size = 0.86 \begin {gather*} \frac {4 x}{3} + 8 \log {\left (3 \right )} \log {\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(24*ln(3)+4*x)/x,x)

[Out]

4*x/3 + 8*log(3)*log(x)

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Giac [A]
time = 0.41, size = 11, normalized size = 0.79 \begin {gather*} 8 \, \log \left (3\right ) \log \left ({\left | x \right |}\right ) + \frac {4}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(24*log(3)+4*x)/x,x, algorithm="giac")

[Out]

8*log(3)*log(abs(x)) + 4/3*x

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Mupad [B]
time = 5.48, size = 10, normalized size = 0.71 \begin {gather*} \frac {4\,x}{3}+8\,\ln \left (3\right )\,\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x)/3 + 8*log(3))/x,x)

[Out]

(4*x)/3 + 8*log(3)*log(x)

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