3.1803 \(\int \frac{(a+b x)^2}{a c+(b c+a d) x+b d x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac{b x}{d}-\frac{(b c-a d) \log (c+d x)}{d^2} \]

[Out]

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

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Rubi [A]  time = 0.0220228, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {626, 43} \[ \frac{b x}{d}-\frac{(b c-a d) \log (c+d x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

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

Mathematica [A]  time = 0.0070554, size = 25, normalized size = 0.96 \[ \frac{(a d-b c) \log (c+d x)}{d^2}+\frac{b x}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 32, normalized size = 1.2 \begin{align*}{\frac{bx}{d}}+{\frac{\ln \left ( dx+c \right ) a}{d}}-{\frac{\ln \left ( dx+c \right ) bc}{{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06554, size = 35, normalized size = 1.35 \begin{align*} \frac{b x}{d} - \frac{{\left (b c - a d\right )} \log \left (d x + c\right )}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/d - (b*c - a*d)*log(d*x + c)/d^2

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Fricas [A]  time = 1.50057, size = 54, normalized size = 2.08 \begin{align*} \frac{b d x -{\left (b c - a d\right )} \log \left (d x + c\right )}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.506721, size = 20, normalized size = 0.77 \begin{align*} \frac{b x}{d} + \frac{\left (a d - b c\right ) \log{\left (c + d x \right )}}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/d + (a*d - b*c)*log(c + d*x)/d**2

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Giac [A]  time = 1.1909, size = 36, normalized size = 1.38 \begin{align*} \frac{b x}{d} - \frac{{\left (b c - a d\right )} \log \left ({\left | d x + c \right |}\right )}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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