\(\int \frac {a c-b c x}{(a+b x)^2} \, dx\) [1058]

Optimal result

Integrand size = 17, antiderivative size = 27 \[ \int \frac {a c-b c x}{(a+b x)^2} \, dx=-\frac {2 a c}{b (a+b x)}-\frac {c \log (a+b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, 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} \[ \int \frac {a c-b c x}{(a+b x)^2} \, dx=-\frac {2 a c}{b (a+b x)}-\frac {c \log (a+b x)}{b} \]

[In]

Int[(a*c - b*c*x)/(a + b*x)^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {a c-b c x}{(a+b x)^2} \, dx=-\frac {c \left (\frac {2 a}{a+b x}+\log (a+b x)\right )}{b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93

method	result	size
norman	\(\frac {2 c x}{b x +a}-\frac {c \ln \left (b x +a \right )}{b}\)	\(25\)
default	\(c \left (-\frac {\ln \left (b x +a \right )}{b}-\frac {2 a}{b \left (b x +a \right )}\right )\)	\(28\)
risch	\(-\frac {2 a c}{b \left (b x +a \right )}-\frac {c \ln \left (b x +a \right )}{b}\)	\(28\)
parallelrisch	\(-\frac {\ln \left (b x +a \right ) x b c +a c \ln \left (b x +a \right )+2 a c}{\left (b x +a \right ) b}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {a c-b c x}{(a+b x)^2} \, dx=-\frac {2 \, a c + {\left (b c x + a c\right )} \log \left (b x + a\right )}{b^{2} x + a b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {a c-b c x}{(a+b x)^2} \, dx=- \frac {2 a c}{a b + b^{2} x} - \frac {c \log {\left (a + b x \right )}}{b} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {a c-b c x}{(a+b x)^2} \, dx=-\frac {2 \, a c}{b^{2} x + a b} - \frac {c \log \left (b x + a\right )}{b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {a c-b c x}{(a+b x)^2} \, dx=c {\left (\frac {\log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x + a\right )} b}\right )} - \frac {a c}{{\left (b x + a\right )} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {a c-b c x}{(a+b x)^2} \, dx=-\frac {c\,\ln \left (a+b\,x\right )}{b}-\frac {2\,a\,c}{b\,\left (a+b\,x\right )} \]

[In]

int((a*c - b*c*x)/(a + b*x)^2,x)

[Out]

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