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

Optimal. Leaf size=43 \[ -\frac{\log (x) (A b-a B)}{a^2}+\frac{(A b-a B) \log (a+b x)}{a^2}-\frac{A}{a x} \]

[Out]

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

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Rubi [A]  time = 0.0278808, antiderivative size = 43, 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} \[ -\frac{\log (x) (A b-a B)}{a^2}+\frac{(A b-a B) \log (a+b x)}{a^2}-\frac{A}{a x} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^2*(a + b*x)),x]

[Out]

-(A/(a*x)) - ((A*b - a*B)*Log[x])/a^2 + ((A*b - a*B)*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^2 (a+b x)} \, dx &=\int \left (\frac{A}{a x^2}+\frac{-A b+a B}{a^2 x}-\frac{b (-A b+a B)}{a^2 (a+b x)}\right ) \, dx\\ &=-\frac{A}{a x}-\frac{(A b-a B) \log (x)}{a^2}+\frac{(A b-a B) \log (a+b x)}{a^2}\\ \end{align*}

Mathematica [A]  time = 0.0167456, size = 42, normalized size = 0.98 \[ \frac{\log (x) (a B-A b)}{a^2}+\frac{(A b-a B) \log (a+b x)}{a^2}-\frac{A}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 51, normalized size = 1.2 \begin{align*} -{\frac{A}{ax}}-{\frac{A\ln \left ( x \right ) b}{{a}^{2}}}+{\frac{\ln \left ( x \right ) B}{a}}+{\frac{\ln \left ( bx+a \right ) Ab}{{a}^{2}}}-{\frac{\ln \left ( bx+a \right ) B}{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.99499, size = 58, normalized size = 1.35 \begin{align*} -\frac{{\left (B a - A b\right )} \log \left (b x + a\right )}{a^{2}} + \frac{{\left (B a - A b\right )} \log \left (x\right )}{a^{2}} - \frac{A}{a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.81509, size = 92, normalized size = 2.14 \begin{align*} -\frac{{\left (B a - A b\right )} x \log \left (b x + a\right ) -{\left (B a - A b\right )} x \log \left (x\right ) + A a}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.625741, size = 95, normalized size = 2.21 \begin{align*} - \frac{A}{a x} + \frac{\left (- A b + B a\right ) \log{\left (x + \frac{- A a b + B a^{2} - a \left (- A b + B a\right )}{- 2 A b^{2} + 2 B a b} \right )}}{a^{2}} - \frac{\left (- A b + B a\right ) \log{\left (x + \frac{- A a b + B a^{2} + a \left (- A b + B a\right )}{- 2 A b^{2} + 2 B a b} \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21402, size = 69, normalized size = 1.6 \begin{align*} \frac{{\left (B a - A b\right )} \log \left ({\left | x \right |}\right )}{a^{2}} - \frac{A}{a x} - \frac{{\left (B a b - A b^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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