\(\int \frac {1+5 x}{-4+5 x} \, dx\) [5431]

Optimal result

Integrand size = 13, antiderivative size = 19 \[ \int \frac {1+5 x}{-4+5 x} \, dx=x-\log \left (\frac {5-\log (4)}{4-5 x}\right ) \]

[Out]

x-ln(1/(-5*x+4)*(-2*ln(2)+5))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int \frac {1+5 x}{-4+5 x} \, dx=x+\log (4-5 x) \]

[In]

Int[(1 + 5*x)/(-4 + 5*x),x]

[Out]

x + Log[4 - 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}{-4+5 x}\right ) \, dx \\ & = x+\log (4-5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {1+5 x}{-4+5 x} \, dx=x+\log (-4+5 x) \]

[In]

Integrate[(1 + 5*x)/(-4 + 5*x),x]

[Out]

x + Log[-4 + 5*x]

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.37

method	result	size
parallelrisch	\(x +\ln \left (x -\frac {4}{5}\right )\)	\(7\)
default	\(x +\ln \left (5 x -4\right )\)	\(9\)
norman	\(x +\ln \left (5 x -4\right )\)	\(9\)
meijerg	\(\ln \left (1-\frac {5 x}{4}\right )+x\)	\(9\)
risch	\(x +\ln \left (5 x -4\right )\)	\(9\)

[In]

int((1+5*x)/(5*x-4),x,method=_RETURNVERBOSE)

[Out]

x+ln(x-4/5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {1+5 x}{-4+5 x} \, dx=x + \log \left (5 \, x - 4\right ) \]

[In]

integrate((1+5*x)/(5*x-4),x, algorithm="fricas")

[Out]

x + log(5*x - 4)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.37 \[ \int \frac {1+5 x}{-4+5 x} \, dx=x + \log {\left (5 x - 4 \right )} \]

[In]

integrate((1+5*x)/(5*x-4),x)

[Out]

x + log(5*x - 4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {1+5 x}{-4+5 x} \, dx=x + \log \left (5 \, x - 4\right ) \]

[In]

integrate((1+5*x)/(5*x-4),x, algorithm="maxima")

[Out]

x + log(5*x - 4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.47 \[ \int \frac {1+5 x}{-4+5 x} \, dx=x + \log \left ({\left | 5 \, x - 4 \right |}\right ) \]

[In]

integrate((1+5*x)/(5*x-4),x, algorithm="giac")

[Out]

x + log(abs(5*x - 4))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.32 \[ \int \frac {1+5 x}{-4+5 x} \, dx=x+\ln \left (x-\frac {4}{5}\right ) \]

[In]

int((5*x + 1)/(5*x - 4),x)

[Out]

x + log(x - 4/5)