3.1.21 \(\int \frac {3-x+2 x^2}{(2+3 x+5 x^2)^3} \, dx\) [21]

Optimal. Leaf size=64 \[ \frac {11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac {553 (3+10 x)}{9610 \left (2+3 x+5 x^2\right )}+\frac {1106 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{961 \sqrt {31}} \]

[Out]

11/310*(7+13*x)/(5*x^2+3*x+2)^2+553/9610*(3+10*x)/(5*x^2+3*x+2)+1106/29791*arctan(1/31*(3+10*x)*31^(1/2))*31^(
1/2)

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Rubi [A]
time = 0.02, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1674, 12, 628, 632, 210} \begin {gather*} \frac {1106 \text {ArcTan}\left (\frac {10 x+3}{\sqrt {31}}\right )}{961 \sqrt {31}}+\frac {553 (10 x+3)}{9610 \left (5 x^2+3 x+2\right )}+\frac {11 (13 x+7)}{310 \left (5 x^2+3 x+2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(7 + 13*x))/(310*(2 + 3*x + 5*x^2)^2) + (553*(3 + 10*x))/(9610*(2 + 3*x + 5*x^2)) + (1106*ArcTan[(3 + 10*x
)/Sqrt[31]])/(961*Sqrt[31])

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 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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 {3-x+2 x^2}{\left (2+3 x+5 x^2\right )^3} \, dx &=\frac {11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac {1}{62} \int \frac {553}{5 \left (2+3 x+5 x^2\right )^2} \, dx\\ &=\frac {11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac {553}{310} \int \frac {1}{\left (2+3 x+5 x^2\right )^2} \, dx\\ &=\frac {11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac {553 (3+10 x)}{9610 \left (2+3 x+5 x^2\right )}+\frac {553}{961} \int \frac {1}{2+3 x+5 x^2} \, dx\\ &=\frac {11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac {553 (3+10 x)}{9610 \left (2+3 x+5 x^2\right )}-\frac {1106}{961} \text {Subst}\left (\int \frac {1}{-31-x^2} \, dx,x,3+10 x\right )\\ &=\frac {11 (7+13 x)}{310 \left (2+3 x+5 x^2\right )^2}+\frac {553 (3+10 x)}{9610 \left (2+3 x+5 x^2\right )}+\frac {1106 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{961 \sqrt {31}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 53, normalized size = 0.83 \begin {gather*} \frac {\frac {31 \left (1141+4094 x+4977 x^2+5530 x^3\right )}{\left (2+3 x+5 x^2\right )^2}+2212 \sqrt {31} \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{59582} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((31*(1141 + 4094*x + 4977*x^2 + 5530*x^3))/(2 + 3*x + 5*x^2)^2 + 2212*Sqrt[31]*ArcTan[(3 + 10*x)/Sqrt[31]])/5
9582

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Maple [A]
time = 0.10, size = 47, normalized size = 0.73

method	result	size
default	\(\frac {\frac {2765}{961} x^{3}+\frac {4977}{1922} x^{2}+\frac {2047}{961} x +\frac {1141}{1922}}{\left (5 x^{2}+3 x +2\right )^{2}}+\frac {1106 \arctan \left (\frac {\left (3+10 x \right ) \sqrt {31}}{31}\right ) \sqrt {31}}{29791}\)	\(47\)
risch	\(\frac {\frac {2765}{961} x^{3}+\frac {4977}{1922} x^{2}+\frac {2047}{961} x +\frac {1141}{1922}}{\left (5 x^{2}+3 x +2\right )^{2}}+\frac {1106 \arctan \left (\frac {\left (3+10 x \right ) \sqrt {31}}{31}\right ) \sqrt {31}}{29791}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*(553/4805*x^3+4977/48050*x^2+2047/24025*x+1141/48050)/(5*x^2+3*x+2)^2+1106/29791*arctan(1/31*(3+10*x)*31^(1
/2))*31^(1/2)

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Maxima [A]
time = 0.51, size = 56, normalized size = 0.88 \begin {gather*} \frac {1106}{29791} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {5530 \, x^{3} + 4977 \, x^{2} + 4094 \, x + 1141}{1922 \, {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1106/29791*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1/1922*(5530*x^3 + 4977*x^2 + 4094*x + 1141)/(25*x^4 +
30*x^3 + 29*x^2 + 12*x + 4)

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Fricas [A]
time = 1.81, size = 75, normalized size = 1.17 \begin {gather*} \frac {171430 \, x^{3} + 2212 \, \sqrt {31} {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + 154287 \, x^{2} + 126914 \, x + 35371}{59582 \, {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/59582*(171430*x^3 + 2212*sqrt(31)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1
54287*x^2 + 126914*x + 35371)/(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)

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Sympy [A]
time = 0.07, size = 63, normalized size = 0.98 \begin {gather*} \frac {5530 x^{3} + 4977 x^{2} + 4094 x + 1141}{48050 x^{4} + 57660 x^{3} + 55738 x^{2} + 23064 x + 7688} + \frac {1106 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{29791} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5530*x**3 + 4977*x**2 + 4094*x + 1141)/(48050*x**4 + 57660*x**3 + 55738*x**2 + 23064*x + 7688) + 1106*sqrt(31
)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/29791

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Giac [A]
time = 3.61, size = 46, normalized size = 0.72 \begin {gather*} \frac {1106}{29791} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {5530 \, x^{3} + 4977 \, x^{2} + 4094 \, x + 1141}{1922 \, {\left (5 \, x^{2} + 3 \, x + 2\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1106/29791*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 1/1922*(5530*x^3 + 4977*x^2 + 4094*x + 1141)/(5*x^2 + 3
*x + 2)^2

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Mupad [B]
time = 0.05, size = 55, normalized size = 0.86 \begin {gather*} \frac {1106\,\sqrt {31}\,\mathrm {atan}\left (\frac {10\,\sqrt {31}\,x}{31}+\frac {3\,\sqrt {31}}{31}\right )}{29791}+\frac {\frac {553\,x^3}{4805}+\frac {4977\,x^2}{48050}+\frac {2047\,x}{24025}+\frac {1141}{48050}}{x^4+\frac {6\,x^3}{5}+\frac {29\,x^2}{25}+\frac {12\,x}{25}+\frac {4}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1106*31^(1/2)*atan((10*31^(1/2)*x)/31 + (3*31^(1/2))/31))/29791 + ((2047*x)/24025 + (4977*x^2)/48050 + (553*x
^3)/4805 + 1141/48050)/((12*x)/25 + (29*x^2)/25 + (6*x^3)/5 + x^4 + 4/25)

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