3.210 \(\int \frac{c+d x}{x^2 (a+b x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0290796, 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} \[ -\frac{\log (x) (b c-a d)}{a^2}+\frac{(b c-a d) \log (a+b x)}{a^2}-\frac{c}{a x} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(x^2*(a + b*x)),x]

[Out]

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

Mathematica [A]  time = 0.0164427, size = 42, normalized size = 0.98 \[ \frac{\log (x) (a d-b c)}{a^2}+\frac{(b c-a d) \log (a+b x)}{a^2}-\frac{c}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/x^2/(b*x+a),x)

[Out]

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

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Maxima [A]  time = 1.04522, size = 58, normalized size = 1.35 \begin{align*} \frac{{\left (b c - a d\right )} \log \left (b x + a\right )}{a^{2}} - \frac{{\left (b c - a d\right )} \log \left (x\right )}{a^{2}} - \frac{c}{a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.97221, size = 90, normalized size = 2.09 \begin{align*} \frac{{\left (b c - a d\right )} x \log \left (b x + a\right ) -{\left (b c - a d\right )} x \log \left (x\right ) - a c}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.620552, size = 95, normalized size = 2.21 \begin{align*} - \frac{c}{a x} + \frac{\left (a d - b c\right ) \log{\left (x + \frac{a^{2} d - a b c - a \left (a d - b c\right )}{2 a b d - 2 b^{2} c} \right )}}{a^{2}} - \frac{\left (a d - b c\right ) \log{\left (x + \frac{a^{2} d - a b c + a \left (a d - b c\right )}{2 a b d - 2 b^{2} c} \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20716, size = 69, normalized size = 1.6 \begin{align*} -\frac{{\left (b c - a d\right )} \log \left ({\left | x \right |}\right )}{a^{2}} - \frac{c}{a x} + \frac{{\left (b^{2} c - a b d\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((d*x+c)/x^2/(b*x+a),x, algorithm="giac")

[Out]

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