\(\int \frac {-21+12 x}{20 x} \, dx\) [9028]

Optimal result

Integrand size = 12, antiderivative size = 20 \[ \int \frac {-21+12 x}{20 x} \, dx=4+\frac {3}{5} \left (5-e^4+x-\frac {7 \log (x)}{4}\right ) \]

[Out]

7+3/5*x-21/20*ln(x)-3/5*exp(4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 45} \[ \int \frac {-21+12 x}{20 x} \, dx=\frac {3 x}{5}-\frac {21 \log (x)}{20} \]

[In]

Int[(-21 + 12*x)/(20*x),x]

[Out]

(3*x)/5 - (21*Log[x])/20

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{align*} \text {integral}& = \frac {1}{20} \int \frac {-21+12 x}{x} \, dx \\ & = \frac {1}{20} \int \left (12-\frac {21}{x}\right ) \, dx \\ & = \frac {3 x}{5}-\frac {21 \log (x)}{20} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {-21+12 x}{20 x} \, dx=\frac {3}{20} (4 x-7 \log (x)) \]

[In]

Integrate[(-21 + 12*x)/(20*x),x]

[Out]

(3*(4*x - 7*Log[x]))/20

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.45

method	result	size
default	\(\frac {3 x}{5}-\frac {21 \ln \left (x \right )}{20}\)	\(9\)
norman	\(\frac {3 x}{5}-\frac {21 \ln \left (x \right )}{20}\)	\(9\)
risch	\(\frac {3 x}{5}-\frac {21 \ln \left (x \right )}{20}\)	\(9\)
parallelrisch	\(\frac {3 x}{5}-\frac {21 \ln \left (x \right )}{20}\)	\(9\)

[In]

int(1/20*(12*x-21)/x,x,method=_RETURNVERBOSE)

[Out]

3/5*x-21/20*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.40 \[ \int \frac {-21+12 x}{20 x} \, dx=\frac {3}{5} \, x - \frac {21}{20} \, \log \left (x\right ) \]

[In]

integrate(1/20*(12*x-21)/x,x, algorithm="fricas")

[Out]

3/5*x - 21/20*log(x)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.50 \[ \int \frac {-21+12 x}{20 x} \, dx=\frac {3 x}{5} - \frac {21 \log {\left (x \right )}}{20} \]

[In]

integrate(1/20*(12*x-21)/x,x)

[Out]

3*x/5 - 21*log(x)/20

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.40 \[ \int \frac {-21+12 x}{20 x} \, dx=\frac {3}{5} \, x - \frac {21}{20} \, \log \left (x\right ) \]

[In]

integrate(1/20*(12*x-21)/x,x, algorithm="maxima")

[Out]

3/5*x - 21/20*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.45 \[ \int \frac {-21+12 x}{20 x} \, dx=\frac {3}{5} \, x - \frac {21}{20} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/20*(12*x-21)/x,x, algorithm="giac")

[Out]

3/5*x - 21/20*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.40 \[ \int \frac {-21+12 x}{20 x} \, dx=\frac {3\,x}{5}-\frac {21\,\ln \left (x\right )}{20} \]

[In]

int(((3*x)/5 - 21/20)/x,x)

[Out]

(3*x)/5 - (21*log(x))/20