\(\int \frac {A+B x}{x (a+b x)^2} \, dx\) [189]

Optimal result

Integrand size = 16, antiderivative size = 42 \[ \int \frac {A+B x}{x (a+b x)^2} \, dx=\frac {A b-a B}{a b (a+b x)}+\frac {A \log (x)}{a^2}-\frac {A \log (a+b x)}{a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {A+B x}{x (a+b x)^2} \, dx=-\frac {A \log (a+b x)}{a^2}+\frac {A \log (x)}{a^2}+\frac {A b-a B}{a b (a+b x)} \]

[In]

Int[(A + B*x)/(x*(a + b*x)^2),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {A}{a^2 x}+\frac {-A b+a B}{a (a+b x)^2}-\frac {A b}{a^2 (a+b x)}\right ) \, dx \\ & = \frac {A b-a B}{a b (a+b x)}+\frac {A \log (x)}{a^2}-\frac {A \log (a+b x)}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x}{x (a+b x)^2} \, dx=\frac {\frac {a (A b-a B)}{b (a+b x)}+A \log (x)-A \log (a+b x)}{a^2} \]

[In]

Integrate[(A + B*x)/(x*(a + b*x)^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00

method	result	size
norman	\(-\frac {\left (A b -B a \right ) x}{a^{2} \left (b x +a \right )}+\frac {A \ln \left (x \right )}{a^{2}}-\frac {A \ln \left (b x +a \right )}{a^{2}}\)	\(42\)
default	\(\frac {A \ln \left (x \right )}{a^{2}}-\frac {-A b +B a}{a b \left (b x +a \right )}-\frac {A \ln \left (b x +a \right )}{a^{2}}\)	\(44\)
risch	\(\frac {A}{a \left (b x +a \right )}-\frac {B}{\left (b x +a \right ) b}+\frac {A \ln \left (-x \right )}{a^{2}}-\frac {A \ln \left (b x +a \right )}{a^{2}}\)	\(48\)
parallelrisch	\(\frac {A \ln \left (x \right ) x b -A \ln \left (b x +a \right ) x b +a A \ln \left (x \right )-A \ln \left (b x +a \right ) a -A b x +B a x}{a^{2} \left (b x +a \right )}\)	\(54\)

[In]

int((B*x+A)/x/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.48 \[ \int \frac {A+B x}{x (a+b x)^2} \, dx=-\frac {B a^{2} - A a b + {\left (A b^{2} x + A a b\right )} \log \left (b x + a\right ) - {\left (A b^{2} x + A a b\right )} \log \left (x\right )}{a^{2} b^{2} x + a^{3} b} \]

[In]

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

[Out]

-(B*a^2 - A*a*b + (A*b^2*x + A*a*b)*log(b*x + a) - (A*b^2*x + A*a*b)*log(x))/(a^2*b^2*x + a^3*b)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x}{x (a+b x)^2} \, dx=\frac {A \left (\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{2}} + \frac {A b - B a}{a^{2} b + a b^{2} x} \]

[In]

integrate((B*x+A)/x/(b*x+a)**2,x)

[Out]

A*(log(x) - log(a/b + x))/a**2 + (A*b - B*a)/(a**2*b + a*b**2*x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x}{x (a+b x)^2} \, dx=-\frac {B a - A b}{a b^{2} x + a^{2} b} - \frac {A \log \left (b x + a\right )}{a^{2}} + \frac {A \log \left (x\right )}{a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.31 \[ \int \frac {A+B x}{x (a+b x)^2} \, dx=b {\left (\frac {A \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{2} b} - \frac {\frac {B a}{b x + a} - \frac {A b}{b x + a}}{a b^{2}}\right )} \]

[In]

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

[Out]

b*(A*log(abs(-a/(b*x + a) + 1))/(a^2*b) - (B*a/(b*x + a) - A*b/(b*x + a))/(a*b^2))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{x (a+b x)^2} \, dx=\frac {A\,b-B\,a}{a\,b\,\left (a+b\,x\right )}-\frac {2\,A\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^2} \]

[In]

int((A + B*x)/(x*(a + b*x)^2),x)

[Out]

(A*b - B*a)/(a*b*(a + b*x)) - (2*A*atanh((2*b*x)/a + 1))/a^2