\(\int \frac {a+b x}{(a c-b c x)^2} \, dx\) [1033]

Optimal result

Integrand size = 17, antiderivative size = 32 \[ \int \frac {a+b x}{(a c-b c x)^2} \, dx=\frac {2 a}{b c^2 (a-b x)}+\frac {\log (a-b x)}{b c^2} \]

[Out]

2*a/b/c^2/(-b*x+a)+ln(-b*x+a)/b/c^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {45} \[ \int \frac {a+b x}{(a c-b c x)^2} \, dx=\frac {2 a}{b c^2 (a-b x)}+\frac {\log (a-b x)}{b c^2} \]

[In]

Int[(a + b*x)/(a*c - b*c*x)^2,x]

[Out]

(2*a)/(b*c^2*(a - b*x)) + Log[a - b*x]/(b*c^2)

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 (\frac {2 a}{c^2 (a-b x)^2}-\frac {1}{c^2 (a-b x)}\right ) \, dx \\ & = \frac {2 a}{b c^2 (a-b x)}+\frac {\log (a-b x)}{b c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {a+b x}{(a c-b c x)^2} \, dx=\frac {\frac {2 a}{a-b x}+\log (c (a-b x))}{b c^2} \]

[In]

Integrate[(a + b*x)/(a*c - b*c*x)^2,x]

[Out]

((2*a)/(a - b*x) + Log[c*(a - b*x)])/(b*c^2)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\frac {\ln \left (-b x +a \right )}{b}+\frac {2 a}{b \left (-b x +a \right )}}{c^{2}}\)	\(31\)
norman	\(\frac {2 a}{b \,c^{2} \left (-b x +a \right )}+\frac {\ln \left (-b x +a \right )}{b \,c^{2}}\)	\(33\)
risch	\(\frac {2 a}{b \,c^{2} \left (-b x +a \right )}+\frac {\ln \left (-b x +a \right )}{b \,c^{2}}\)	\(33\)
parallelrisch	\(\frac {\ln \left (b x -a \right ) x b -a \ln \left (b x -a \right )-2 a}{b \,c^{2} \left (b x -a \right )}\)	\(43\)

[In]

int((b*x+a)/(-b*c*x+a*c)^2,x,method=_RETURNVERBOSE)

[Out]

1/c^2*(1/b*ln(-b*x+a)+2*a/b/(-b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {a+b x}{(a c-b c x)^2} \, dx=\frac {{\left (b x - a\right )} \log \left (b x - a\right ) - 2 \, a}{b^{2} c^{2} x - a b c^{2}} \]

[In]

integrate((b*x+a)/(-b*c*x+a*c)^2,x, algorithm="fricas")

[Out]

((b*x - a)*log(b*x - a) - 2*a)/(b^2*c^2*x - a*b*c^2)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {a+b x}{(a c-b c x)^2} \, dx=- \frac {2 a}{- a b c^{2} + b^{2} c^{2} x} + \frac {\log {\left (- a + b x \right )}}{b c^{2}} \]

[In]

integrate((b*x+a)/(-b*c*x+a*c)**2,x)

[Out]

-2*a/(-a*b*c**2 + b**2*c**2*x) + log(-a + b*x)/(b*c**2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {a+b x}{(a c-b c x)^2} \, dx=-\frac {2 \, a}{b^{2} c^{2} x - a b c^{2}} + \frac {\log \left (b x - a\right )}{b c^{2}} \]

[In]

integrate((b*x+a)/(-b*c*x+a*c)^2,x, algorithm="maxima")

[Out]

-2*a/(b^2*c^2*x - a*b*c^2) + log(b*x - a)/(b*c^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (33) = 66\).

Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.53 \[ \int \frac {a+b x}{(a c-b c x)^2} \, dx=-\frac {\frac {a}{{\left (b c x - a c\right )} b} + \frac {\log \left (\frac {{\left | b c x - a c \right |}}{{\left (b c x - a c\right )}^{2} {\left | b \right |} {\left | c \right |}}\right )}{b c}}{c} - \frac {a}{{\left (b c x - a c\right )} b c} \]

[In]

integrate((b*x+a)/(-b*c*x+a*c)^2,x, algorithm="giac")

[Out]

-(a/((b*c*x - a*c)*b) + log(abs(b*c*x - a*c)/((b*c*x - a*c)^2*abs(b)*abs(c)))/(b*c))/c - a/((b*c*x - a*c)*b*c)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {a+b x}{(a c-b c x)^2} \, dx=\frac {\ln \left (b\,x-a\right )}{b\,c^2}+\frac {2\,a}{b\,\left (a\,c^2-b\,c^2\,x\right )} \]

[In]

int((a + b*x)/(a*c - b*c*x)^2,x)

[Out]

log(b*x - a)/(b*c^2) + (2*a)/(b*(a*c^2 - b*c^2*x))