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]

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

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, 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 8

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

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
]

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*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \[ \frac {-3 x-2}{2 \left (2 x^2+7 x+4\right )} \]

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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fricas [A]  time = 0.78, size = 19, normalized size = 0.90 \[ -\frac {3 \, x + 2}{2 \, {\left (2 \, x^{2} + 7 \, x + 4\right )}} \]

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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giac [A]  time = 0.16, size = 19, normalized size = 0.90 \[ -\frac {3 \, x + 2}{2 \, {\left (2 \, x^{2} + 7 \, x + 4\right )}} \]

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)

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maple [A]  time = 0.00, size = 17, normalized size = 0.81 \[ \frac {-\frac {3 x}{4}-\frac {1}{2}}{x^{2}+\frac {7}{2} x +2} \]

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 = 0.43, size = 19, normalized size = 0.90 \[ -\frac {3 \, x + 2}{2 \, {\left (2 \, x^{2} + 7 \, x + 4\right )}} \]

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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mupad [B]  time = 3.84, size = 17, normalized size = 0.81 \[ -\frac {\frac {3\,x}{4}+\frac {1}{2}}{x^2+\frac {7\,x}{2}+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((3*x)/4 + 1/2)/((7*x)/2 + x^2 + 2)

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sympy [A]  time = 0.12, size = 15, normalized size = 0.71 \[ \frac {- 3 x - 2}{4 x^{2} + 14 x + 8} \]

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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