3.1114 \(\int \frac{a+b x+c x^2}{(b d+2 c d x)^2} \, dx\)

Optimal. Leaf size=38 \[ \frac{b^2-4 a c}{8 c^2 d^2 (b+2 c x)}+\frac{x}{4 c d^2} \]

[Out]

x/(4*c*d^2) + (b^2 - 4*a*c)/(8*c^2*d^2*(b + 2*c*x))

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Rubi [A]  time = 0.0276715, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {683} \[ \frac{b^2-4 a c}{8 c^2 d^2 (b+2 c x)}+\frac{x}{4 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(4*c*d^2) + (b^2 - 4*a*c)/(8*c^2*d^2*(b + 2*c*x))

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps

\begin{align*} \int \frac{a+b x+c x^2}{(b d+2 c d x)^2} \, dx &=\int \left (\frac{1}{4 c d^2}+\frac{-b^2+4 a c}{4 c d^2 (b+2 c x)^2}\right ) \, dx\\ &=\frac{x}{4 c d^2}+\frac{b^2-4 a c}{8 c^2 d^2 (b+2 c x)}\\ \end{align*}

Mathematica [A]  time = 0.0106645, size = 41, normalized size = 1.08 \[ \frac{\frac{b^2-4 a c}{8 c^2 (b+2 c x)}+\frac{b+2 c x}{8 c^2}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)/(8*c^2*(b + 2*c*x)) + (b + 2*c*x)/(8*c^2))/d^2

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Maple [A]  time = 0.043, size = 35, normalized size = 0.9 \begin{align*}{\frac{1}{{d}^{2}} \left ({\frac{x}{4\,c}}-{\frac{4\,ac-{b}^{2}}{8\,{c}^{2} \left ( 2\,cx+b \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^2*(1/4*x/c-1/8*(4*a*c-b^2)/c^2/(2*c*x+b))

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Maxima [A]  time = 1.34769, size = 54, normalized size = 1.42 \begin{align*} \frac{b^{2} - 4 \, a c}{8 \,{\left (2 \, c^{3} d^{2} x + b c^{2} d^{2}\right )}} + \frac{x}{4 \, c d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^2 - 4*a*c)/(2*c^3*d^2*x + b*c^2*d^2) + 1/4*x/(c*d^2)

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Fricas [A]  time = 2.20489, size = 90, normalized size = 2.37 \begin{align*} \frac{4 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 4 \, a c}{8 \,{\left (2 \, c^{3} d^{2} x + b c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*c^2*x^2 + 2*b*c*x + b^2 - 4*a*c)/(2*c^3*d^2*x + b*c^2*d^2)

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Sympy [A]  time = 0.470093, size = 36, normalized size = 0.95 \begin{align*} - \frac{4 a c - b^{2}}{8 b c^{2} d^{2} + 16 c^{3} d^{2} x} + \frac{x}{4 c d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*a*c - b**2)/(8*b*c**2*d**2 + 16*c**3*d**2*x) + x/(4*c*d**2)

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Giac [B]  time = 1.14319, size = 230, normalized size = 6.05 \begin{align*} -\frac{1}{8} \, c{\left (\frac{b^{2}}{{\left (2 \, c d x + b d\right )} c^{3} d} - \frac{2 \, b \log \left (\frac{{\left | 2 \, c d x + b d \right |}}{2 \,{\left (2 \, c d x + b d\right )}^{2}{\left | c \right |}{\left | d \right |}}\right )}{c^{3} d^{2}} - \frac{2 \, c d x + b d}{c^{3} d^{3}}\right )} + \frac{b{\left (\frac{b}{{\left (2 \, c d x + b d\right )} c} - \frac{\log \left (\frac{{\left | 2 \, c d x + b d \right |}}{2 \,{\left (2 \, c d x + b d\right )}^{2}{\left | c \right |}{\left | d \right |}}\right )}{c d}\right )}}{4 \, c d} - \frac{a}{2 \,{\left (2 \, c d x + b d\right )} c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*c*(b^2/((2*c*d*x + b*d)*c^3*d) - 2*b*log(1/2*abs(2*c*d*x + b*d)/((2*c*d*x + b*d)^2*abs(c)*abs(d)))/(c^3*d
^2) - (2*c*d*x + b*d)/(c^3*d^3)) + 1/4*b*(b/((2*c*d*x + b*d)*c) - log(1/2*abs(2*c*d*x + b*d)/((2*c*d*x + b*d)^
2*abs(c)*abs(d)))/(c*d))/(c*d) - 1/2*a/((2*c*d*x + b*d)*c*d)