\(\int \frac {1-5 x}{x} \, dx\) [3293]

Optimal result

Integrand size = 9, antiderivative size = 9 \[ \int \frac {1-5 x}{x} \, dx=3-5 (3+x)+\log (x) \]

[Out]

ln(x)-12-5*x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.67, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {45} \[ \int \frac {1-5 x}{x} \, dx=\log (x)-5 x \]

[In]

Int[(1 - 5*x)/x,x]

[Out]

-5*x + Log[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}& = \int \left (-5+\frac {1}{x}\right ) \, dx \\ & = -5 x+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.67 \[ \int \frac {1-5 x}{x} \, dx=-5 x+\log (x) \]

[In]

Integrate[(1 - 5*x)/x,x]

[Out]

-5*x + Log[x]

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78

method	result	size
default	\(\ln \left (x \right )-5 x\)	\(7\)
norman	\(\ln \left (x \right )-5 x\)	\(7\)
risch	\(\ln \left (x \right )-5 x\)	\(7\)
parallelrisch	\(\ln \left (x \right )-5 x\)	\(7\)

[In]

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

[Out]

ln(x)-5*x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.67 \[ \int \frac {1-5 x}{x} \, dx=-5 \, x + \log \left (x\right ) \]

[In]

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

[Out]

-5*x + log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.56 \[ \int \frac {1-5 x}{x} \, dx=- 5 x + \log {\left (x \right )} \]

[In]

integrate((-5*x+1)/x,x)

[Out]

-5*x + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.67 \[ \int \frac {1-5 x}{x} \, dx=-5 \, x + \log \left (x\right ) \]

[In]

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

[Out]

-5*x + log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {1-5 x}{x} \, dx=-5 \, x + \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

-5*x + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.67 \[ \int \frac {1-5 x}{x} \, dx=\ln \left (x\right )-5\,x \]

[In]

int(-(5*x - 1)/x,x)

[Out]

log(x) - 5*x