3.9.24 \(\int \frac {40+5 x}{24 x} \, dx\)

Optimal. Leaf size=17 \[ -2+\frac {5 x}{24}+\frac {5}{3} \log \left (\frac {x}{3}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 0.71, 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, 43} \begin {gather*} \frac {5 x}{24}+\frac {5 \log (x)}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 + 5*x)/(24*x),x]

[Out]

(5*x)/24 + (5*Log[x])/3

Rule 12

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

Rule 43

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}{24} \int \frac {40+5 x}{x} \, dx\\ &=\frac {1}{24} \int \left (5+\frac {40}{x}\right ) \, dx\\ &=\frac {5 x}{24}+\frac {5 \log (x)}{3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 0.59 \begin {gather*} \frac {5}{24} (x+8 \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 + 5*x)/(24*x),x]

[Out]

(5*(x + 8*Log[x]))/24

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fricas [A]  time = 0.59, size = 8, normalized size = 0.47 \begin {gather*} \frac {5}{24} \, x + \frac {5}{3} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/24*(5*x+40)/x,x, algorithm="fricas")

[Out]

5/24*x + 5/3*log(x)

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giac [A]  time = 0.34, size = 9, normalized size = 0.53 \begin {gather*} \frac {5}{24} \, x + \frac {5}{3} \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/24*(5*x+40)/x,x, algorithm="giac")

[Out]

5/24*x + 5/3*log(abs(x))

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maple [A]  time = 0.02, size = 9, normalized size = 0.53

method	result	size
default	\(\frac {5 x}{24}+\frac {5 \ln \relax (x )}{3}\)	\(9\)
norman	\(\frac {5 x}{24}+\frac {5 \ln \relax (x )}{3}\)	\(9\)
risch	\(\frac {5 x}{24}+\frac {5 \ln \relax (x )}{3}\)	\(9\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/24*(5*x+40)/x,x,method=_RETURNVERBOSE)

[Out]

5/24*x+5/3*ln(x)

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maxima [A]  time = 0.37, size = 8, normalized size = 0.47 \begin {gather*} \frac {5}{24} \, x + \frac {5}{3} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/24*(5*x+40)/x,x, algorithm="maxima")

[Out]

5/24*x + 5/3*log(x)

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mupad [B]  time = 0.02, size = 8, normalized size = 0.47 \begin {gather*} \frac {5\,x}{24}+\frac {5\,\ln \relax (x)}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x)/24 + 5/3)/x,x)

[Out]

(5*x)/24 + (5*log(x))/3

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sympy [A]  time = 0.07, size = 10, normalized size = 0.59 \begin {gather*} \frac {5 x}{24} + \frac {5 \log {\relax (x )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/24*(5*x+40)/x,x)

[Out]

5*x/24 + 5*log(x)/3

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