3.175 \(\int \frac{1+4 x+3 x^2}{(4+7 x+2 x^2)^2} \, dx\)

Optimal. Leaf size=21 \[ -\frac{3 x+2}{2 \left (2 x^2+7 x+4\right )} \]

[Out]

-(2 + 3*x)/(2*(4 + 7*x + 2*x^2))

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Rubi [A]  time = 0.0119356, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {1660, 8} \[ -\frac{3 x+2}{2 \left (2 x^2+7 x+4\right )} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 4*x + 3*x^2)/(4 + 7*x + 2*x^2)^2,x]

[Out]

-(2 + 3*x)/(2*(4 + 7*x + 2*x^2))

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1+4 x+3 x^2}{\left (4+7 x+2 x^2\right )^2} \, dx &=-\frac{2+3 x}{2 \left (4+7 x+2 x^2\right )}-\frac{\int 0 \, dx}{17}\\ &=-\frac{2+3 x}{2 \left (4+7 x+2 x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0074776, size = 21, normalized size = 1. \[ \frac{-3 x-2}{2 \left (2 x^2+7 x+4\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4*x + 3*x^2)/(4 + 7*x + 2*x^2)^2,x]

[Out]

(-2 - 3*x)/(2*(4 + 7*x + 2*x^2))

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Maple [A]  time = 0.048, size = 17, normalized size = 0.8 \begin{align*}{ \left ( -{\frac{3\,x}{4}}-{\frac{1}{2}} \right ) \left ({x}^{2}+{\frac{7\,x}{2}}+2 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+4*x+1)/(2*x^2+7*x+4)^2,x)

[Out]

(-3/4*x-1/2)/(x^2+7/2*x+2)

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Maxima [A]  time = 1.00437, size = 26, normalized size = 1.24 \begin{align*} -\frac{3 \, x + 2}{2 \,{\left (2 \, x^{2} + 7 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+4*x+1)/(2*x^2+7*x+4)^2,x, algorithm="maxima")

[Out]

-1/2*(3*x + 2)/(2*x^2 + 7*x + 4)

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Fricas [A]  time = 1.6484, size = 46, normalized size = 2.19 \begin{align*} -\frac{3 \, x + 2}{2 \,{\left (2 \, x^{2} + 7 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+4*x+1)/(2*x^2+7*x+4)^2,x, algorithm="fricas")

[Out]

-1/2*(3*x + 2)/(2*x^2 + 7*x + 4)

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Sympy [A]  time = 0.112659, size = 15, normalized size = 0.71 \begin{align*} - \frac{3 x + 2}{4 x^{2} + 14 x + 8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+4*x+1)/(2*x**2+7*x+4)**2,x)

[Out]

-(3*x + 2)/(4*x**2 + 14*x + 8)

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Giac [A]  time = 1.22712, size = 26, normalized size = 1.24 \begin{align*} -\frac{3 \, x + 2}{2 \,{\left (2 \, x^{2} + 7 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+4*x+1)/(2*x^2+7*x+4)^2,x, algorithm="giac")

[Out]

-1/2*(3*x + 2)/(2*x^2 + 7*x + 4)