3.94.67 \(\int \frac {-64-324 x}{15 x} \, dx\)

Optimal. Leaf size=20 \[ -\frac {4}{15} x^2 \left (\frac {81}{x}+\frac {16 \log (x)}{x^2}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 0.60, 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 {108 x}{5}-\frac {64 \log (x)}{15} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64 - 324*x)/(15*x),x]

[Out]

(-108*x)/5 - (64*Log[x])/15

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}{15} \int \frac {-64-324 x}{x} \, dx\\ &=\frac {1}{15} \int \left (-324-\frac {64}{x}\right ) \, dx\\ &=-\frac {108 x}{5}-\frac {64 \log (x)}{15}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.60 \begin {gather*} -\frac {4}{15} (81 x+16 \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64 - 324*x)/(15*x),x]

[Out]

(-4*(81*x + 16*Log[x]))/15

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fricas [A]  time = 0.79, size = 8, normalized size = 0.40 \begin {gather*} -\frac {108}{5} \, x - \frac {64}{15} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-324*x-64)/x,x, algorithm="fricas")

[Out]

-108/5*x - 64/15*log(x)

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giac [A]  time = 0.18, size = 9, normalized size = 0.45 \begin {gather*} -\frac {108}{5} \, x - \frac {64}{15} \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-324*x-64)/x,x, algorithm="giac")

[Out]

-108/5*x - 64/15*log(abs(x))

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maple [A]  time = 0.03, size = 9, normalized size = 0.45

method	result	size
default	\(-\frac {108 x}{5}-\frac {64 \ln \relax (x )}{15}\)	\(9\)
norman	\(-\frac {108 x}{5}-\frac {64 \ln \relax (x )}{15}\)	\(9\)
risch	\(-\frac {108 x}{5}-\frac {64 \ln \relax (x )}{15}\)	\(9\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*(-324*x-64)/x,x,method=_RETURNVERBOSE)

[Out]

-108/5*x-64/15*ln(x)

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maxima [A]  time = 0.36, size = 8, normalized size = 0.40 \begin {gather*} -\frac {108}{5} \, x - \frac {64}{15} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-324*x-64)/x,x, algorithm="maxima")

[Out]

-108/5*x - 64/15*log(x)

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mupad [B]  time = 5.43, size = 8, normalized size = 0.40 \begin {gather*} -\frac {108\,x}{5}-\frac {64\,\ln \relax (x)}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((108*x)/5 + 64/15)/x,x)

[Out]

- (108*x)/5 - (64*log(x))/15

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sympy [A]  time = 0.07, size = 12, normalized size = 0.60 \begin {gather*} - \frac {108 x}{5} - \frac {64 \log {\relax (x )}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-324*x-64)/x,x)

[Out]

-108*x/5 - 64*log(x)/15

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