\(\int \frac {x}{-5+x} \, dx\) [177]

Optimal result

Integrand size = 7, antiderivative size = 10 \[ \int \frac {x}{-5+x} \, dx=x+5 \log (5-x) \]

[Out]

x+5*ln(5-x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {45} \[ \int \frac {x}{-5+x} \, dx=x+5 \log (5-x) \]

[In]

Int[x/(-5 + x),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {x}{-5+x} \, dx=x+5 \log (-5+x) \]

[In]

Integrate[x/(-5 + x),x]

[Out]

x + 5*Log[-5 + x]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90

method	result	size
default	\(x +5 \ln \left (x -5\right )\)	\(9\)
norman	\(x +5 \ln \left (x -5\right )\)	\(9\)
risch	\(x +5 \ln \left (x -5\right )\)	\(9\)
parallelrisch	\(x +5 \ln \left (x -5\right )\)	\(9\)
meijerg	\(x +5 \ln \left (1-\frac {x}{5}\right )\)	\(11\)

[In]

int(x/(x-5),x,method=_RETURNVERBOSE)

[Out]

x+5*ln(x-5)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {x}{-5+x} \, dx=x + 5 \, \log \left (x - 5\right ) \]

[In]

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

[Out]

x + 5*log(x - 5)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.70 \[ \int \frac {x}{-5+x} \, dx=x + 5 \log {\left (x - 5 \right )} \]

[In]

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

[Out]

x + 5*log(x - 5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {x}{-5+x} \, dx=x + 5 \, \log \left (x - 5\right ) \]

[In]

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

[Out]

x + 5*log(x - 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {x}{-5+x} \, dx=x + 5 \, \log \left ({\left | x - 5 \right |}\right ) \]

[In]

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

[Out]

x + 5*log(abs(x - 5))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {x}{-5+x} \, dx=x+5\,\ln \left (x-5\right ) \]

[In]

int(x/(x - 5),x)

[Out]

x + 5*log(x - 5)