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

Optimal. Leaf size=42 \[ -\frac{A \log (a+b x)}{a^2}+\frac{A \log (x)}{a^2}+\frac{A b-a B}{a b (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0249285, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {77} \[ -\frac{A \log (a+b x)}{a^2}+\frac{A \log (x)}{a^2}+\frac{A b-a B}{a b (a+b x)} \]

Antiderivative was successfully verified.

[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 77

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*} \int \frac{A+B x}{x (a+b x)^2} \, dx &=\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]  time = 0.0259676, size = 38, normalized size = 0.9 \[ \frac{\frac{a (A b-a B)}{b (a+b x)}-A \log (a+b x)+A \log (x)}{a^2} \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.007, size = 46, normalized size = 1.1 \begin{align*}{\frac{A\ln \left ( x \right ) }{{a}^{2}}}+{\frac{A}{a \left ( bx+a \right ) }}-{\frac{B}{b \left ( bx+a \right ) }}-{\frac{A\ln \left ( bx+a \right ) }{{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03312, size = 59, normalized size = 1.4 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.96464, size = 132, normalized size = 3.14 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.618404, size = 32, normalized size = 0.76 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.16804, size = 74, normalized size = 1.76 \begin{align*} 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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))