3.1.47 \(\int \frac {2+3 x+5 x^2}{(3-x+2 x^2)^2} \, dx\) [47]

Optimal. Leaf size=43 \[ -\frac {11 (5+3 x)}{46 \left (3-x+2 x^2\right )}-\frac {82 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{23 \sqrt {23}} \]

[Out]

-11/46*(5+3*x)/(2*x^2-x+3)-82/529*arctan(1/23*(1-4*x)*23^(1/2))*23^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1674, 12, 632, 210} \begin {gather*} -\frac {82 \text {ArcTan}\left (\frac {1-4 x}{\sqrt {23}}\right )}{23 \sqrt {23}}-\frac {11 (3 x+5)}{46 \left (2 x^2-x+3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x + 5*x^2)/(3 - x + 2*x^2)^2,x]

[Out]

(-11*(5 + 3*x))/(46*(3 - x + 2*x^2)) - (82*ArcTan[(1 - 4*x)/Sqrt[23]])/(23*Sqrt[23])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1674

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 {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^2} \, dx &=-\frac {11 (5+3 x)}{46 \left (3-x+2 x^2\right )}+\frac {1}{23} \int \frac {41}{3-x+2 x^2} \, dx\\ &=-\frac {11 (5+3 x)}{46 \left (3-x+2 x^2\right )}+\frac {41}{23} \int \frac {1}{3-x+2 x^2} \, dx\\ &=-\frac {11 (5+3 x)}{46 \left (3-x+2 x^2\right )}-\frac {82}{23} \text {Subst}\left (\int \frac {1}{-23-x^2} \, dx,x,-1+4 x\right )\\ &=-\frac {11 (5+3 x)}{46 \left (3-x+2 x^2\right )}-\frac {82 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{23 \sqrt {23}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 43, normalized size = 1.00 \begin {gather*} -\frac {11 (5+3 x)}{46 \left (3-x+2 x^2\right )}+\frac {82 \tan ^{-1}\left (\frac {-1+4 x}{\sqrt {23}}\right )}{23 \sqrt {23}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x + 5*x^2)/(3 - x + 2*x^2)^2,x]

[Out]

(-11*(5 + 3*x))/(46*(3 - x + 2*x^2)) + (82*ArcTan[(-1 + 4*x)/Sqrt[23]])/(23*Sqrt[23])

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Maple [A]
time = 0.11, size = 34, normalized size = 0.79

method	result	size
default	\(\frac {-\frac {33 x}{92}-\frac {55}{92}}{x^{2}-\frac {1}{2} x +\frac {3}{2}}+\frac {82 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{529}\)	\(34\)
risch	\(\frac {-\frac {33 x}{92}-\frac {55}{92}}{x^{2}-\frac {1}{2} x +\frac {3}{2}}+\frac {82 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{529}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+3*x+2)/(2*x^2-x+3)^2,x,method=_RETURNVERBOSE)

[Out]

(-33/92*x-55/92)/(x^2-1/2*x+3/2)+82/529*23^(1/2)*arctan(1/23*(4*x-1)*23^(1/2))

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Maxima [A]
time = 0.51, size = 36, normalized size = 0.84 \begin {gather*} \frac {82}{529} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {11 \, {\left (3 \, x + 5\right )}}{46 \, {\left (2 \, x^{2} - x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^2,x, algorithm="maxima")

[Out]

82/529*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 11/46*(3*x + 5)/(2*x^2 - x + 3)

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Fricas [A]
time = 1.92, size = 45, normalized size = 1.05 \begin {gather*} \frac {164 \, \sqrt {23} {\left (2 \, x^{2} - x + 3\right )} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - 759 \, x - 1265}{1058 \, {\left (2 \, x^{2} - x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^2,x, algorithm="fricas")

[Out]

1/1058*(164*sqrt(23)*(2*x^2 - x + 3)*arctan(1/23*sqrt(23)*(4*x - 1)) - 759*x - 1265)/(2*x^2 - x + 3)

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Sympy [A]
time = 0.05, size = 42, normalized size = 0.98 \begin {gather*} \frac {- 33 x - 55}{92 x^{2} - 46 x + 138} + \frac {82 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{529} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+3*x+2)/(2*x**2-x+3)**2,x)

[Out]

(-33*x - 55)/(92*x**2 - 46*x + 138) + 82*sqrt(23)*atan(4*sqrt(23)*x/23 - sqrt(23)/23)/529

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Giac [A]
time = 1.92, size = 36, normalized size = 0.84 \begin {gather*} \frac {82}{529} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {11 \, {\left (3 \, x + 5\right )}}{46 \, {\left (2 \, x^{2} - x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^2,x, algorithm="giac")

[Out]

82/529*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 11/46*(3*x + 5)/(2*x^2 - x + 3)

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Mupad [B]
time = 0.04, size = 36, normalized size = 0.84 \begin {gather*} \frac {82\,\sqrt {23}\,\mathrm {atan}\left (\frac {4\,\sqrt {23}\,x}{23}-\frac {\sqrt {23}}{23}\right )}{529}-\frac {\frac {33\,x}{92}+\frac {55}{92}}{x^2-\frac {x}{2}+\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 5*x^2 + 2)/(2*x^2 - x + 3)^2,x)

[Out]

(82*23^(1/2)*atan((4*23^(1/2)*x)/23 - 23^(1/2)/23))/529 - ((33*x)/92 + 55/92)/(x^2 - x/2 + 3/2)

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