3.1242 \(\int \frac{c+d x}{(a+b x)^2} \, dx\)

Optimal. Leaf size=32 \[ \frac{d \log (a+b x)}{b^2}-\frac{b c-a d}{b^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0193538, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {43} \[ \frac{d \log (a+b x)}{b^2}-\frac{b c-a d}{b^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*x)^2,x]

[Out]

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

Rule 43

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

Mathematica [A]  time = 0.0107254, size = 31, normalized size = 0.97 \[ \frac{a d-b c}{b^2 (a+b x)}+\frac{d \log (a+b x)}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + b*x)^2,x]

[Out]

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

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Maple [A]  time = 0.006, size = 39, normalized size = 1.2 \begin{align*}{\frac{ad}{{b}^{2} \left ( bx+a \right ) }}-{\frac{c}{b \left ( bx+a \right ) }}+{\frac{d\ln \left ( bx+a \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05097, size = 47, normalized size = 1.47 \begin{align*} -\frac{b c - a d}{b^{3} x + a b^{2}} + \frac{d \log \left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*c - a*d)/(b^3*x + a*b^2) + d*log(b*x + a)/b^2

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Fricas [A]  time = 1.91972, size = 80, normalized size = 2.5 \begin{align*} -\frac{b c - a d -{\left (b d x + a d\right )} \log \left (b x + a\right )}{b^{3} x + a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*c - a*d - (b*d*x + a*d)*log(b*x + a))/(b^3*x + a*b^2)

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Sympy [A]  time = 0.341905, size = 27, normalized size = 0.84 \begin{align*} \frac{a d - b c}{a b^{2} + b^{3} x} + \frac{d \log{\left (a + b x \right )}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*d - b*c)/(a*b**2 + b**3*x) + d*log(a + b*x)/b**2

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Giac [A]  time = 1.06566, size = 77, normalized size = 2.41 \begin{align*} -\frac{d{\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 )}}{b} - \frac{c}{{\left (b x + a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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