3.321 \(\int \frac {2+x+3 x^2-5 x^3+4 x^4}{(3+2 x+5 x^2)^3} \, dx\)

Optimal. Leaf size=64 \[ -\frac {423 x+1367}{7000 \left (5 x^2+2 x+3\right )^2}+\frac {11015 x+34347}{196000 \left (5 x^2+2 x+3\right )}+\frac {339 \tan ^{-1}\left (\frac {5 x+1}{\sqrt {14}}\right )}{1568 \sqrt {14}} \]

[Out]

1/7000*(-1367-423*x)/(5*x^2+2*x+3)^2+1/196000*(34347+11015*x)/(5*x^2+2*x+3)+339/21952*arctan(1/14*(1+5*x)*14^(
1/2))*14^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1660, 12, 618, 204} \[ -\frac {423 x+1367}{7000 \left (5 x^2+2 x+3\right )^2}+\frac {11015 x+34347}{196000 \left (5 x^2+2 x+3\right )}+\frac {339 \tan ^{-1}\left (\frac {5 x+1}{\sqrt {14}}\right )}{1568 \sqrt {14}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + x + 3*x^2 - 5*x^3 + 4*x^4)/(3 + 2*x + 5*x^2)^3,x]

[Out]

-(1367 + 423*x)/(7000*(3 + 2*x + 5*x^2)^2) + (34347 + 11015*x)/(196000*(3 + 2*x + 5*x^2)) + (339*ArcTan[(1 + 5
*x)/Sqrt[14]])/(1568*Sqrt[14])

Rule 12

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

Rule 204

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

Rule 618

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 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 {2+x+3 x^2-5 x^3+4 x^4}{\left (3+2 x+5 x^2\right )^3} \, dx &=-\frac {1367+423 x}{7000 \left (3+2 x+5 x^2\right )^2}+\frac {1}{112} \int \frac {\frac {6534}{125}-\frac {3696 x}{25}+\frac {448 x^2}{5}}{\left (3+2 x+5 x^2\right )^2} \, dx\\ &=-\frac {1367+423 x}{7000 \left (3+2 x+5 x^2\right )^2}+\frac {34347+11015 x}{196000 \left (3+2 x+5 x^2\right )}+\frac {\int \frac {1356}{3+2 x+5 x^2} \, dx}{6272}\\ &=-\frac {1367+423 x}{7000 \left (3+2 x+5 x^2\right )^2}+\frac {34347+11015 x}{196000 \left (3+2 x+5 x^2\right )}+\frac {339 \int \frac {1}{3+2 x+5 x^2} \, dx}{1568}\\ &=-\frac {1367+423 x}{7000 \left (3+2 x+5 x^2\right )^2}+\frac {34347+11015 x}{196000 \left (3+2 x+5 x^2\right )}-\frac {339}{784} \operatorname {Subst}\left (\int \frac {1}{-56-x^2} \, dx,x,2+10 x\right )\\ &=-\frac {1367+423 x}{7000 \left (3+2 x+5 x^2\right )^2}+\frac {34347+11015 x}{196000 \left (3+2 x+5 x^2\right )}+\frac {339 \tan ^{-1}\left (\frac {1+5 x}{\sqrt {14}}\right )}{1568 \sqrt {14}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 53, normalized size = 0.83 \[ \frac {\frac {14 \left (11015 x^3+38753 x^2+17979 x+12953\right )}{\left (5 x^2+2 x+3\right )^2}+8475 \sqrt {14} \tan ^{-1}\left (\frac {5 x+1}{\sqrt {14}}\right )}{548800} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + x + 3*x^2 - 5*x^3 + 4*x^4)/(3 + 2*x + 5*x^2)^3,x]

[Out]

((14*(12953 + 17979*x + 38753*x^2 + 11015*x^3))/(3 + 2*x + 5*x^2)^2 + 8475*Sqrt[14]*ArcTan[(1 + 5*x)/Sqrt[14]]
)/548800

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fricas [A]  time = 0.73, size = 75, normalized size = 1.17 \[ \frac {154210 \, x^{3} + 8475 \, \sqrt {14} {\left (25 \, x^{4} + 20 \, x^{3} + 34 \, x^{2} + 12 \, x + 9\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + 542542 \, x^{2} + 251706 \, x + 181342}{548800 \, {\left (25 \, x^{4} + 20 \, x^{3} + 34 \, x^{2} + 12 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3)^3,x, algorithm="fricas")

[Out]

1/548800*(154210*x^3 + 8475*sqrt(14)*(25*x^4 + 20*x^3 + 34*x^2 + 12*x + 9)*arctan(1/14*sqrt(14)*(5*x + 1)) + 5
42542*x^2 + 251706*x + 181342)/(25*x^4 + 20*x^3 + 34*x^2 + 12*x + 9)

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giac [A]  time = 0.16, size = 46, normalized size = 0.72 \[ \frac {339}{21952} \, \sqrt {14} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + \frac {11015 \, x^{3} + 38753 \, x^{2} + 17979 \, x + 12953}{39200 \, {\left (5 \, x^{2} + 2 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3)^3,x, algorithm="giac")

[Out]

339/21952*sqrt(14)*arctan(1/14*sqrt(14)*(5*x + 1)) + 1/39200*(11015*x^3 + 38753*x^2 + 17979*x + 12953)/(5*x^2
+ 2*x + 3)^2

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maple [A]  time = 0.01, size = 47, normalized size = 0.73 \[ \frac {339 \sqrt {14}\, \arctan \left (\frac {\left (10 x +2\right ) \sqrt {14}}{28}\right )}{21952}+\frac {\frac {2203}{7840} x^{3}+\frac {38753}{39200} x^{2}+\frac {17979}{39200} x +\frac {12953}{39200}}{\left (5 x^{2}+2 x +3\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3)^3,x)

[Out]

25*(2203/196000*x^3+38753/980000*x^2+17979/980000*x+12953/980000)/(5*x^2+2*x+3)^2+339/21952*14^(1/2)*arctan(1/
28*(10*x+2)*14^(1/2))

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maxima [A]  time = 0.96, size = 56, normalized size = 0.88 \[ \frac {339}{21952} \, \sqrt {14} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) + \frac {11015 \, x^{3} + 38753 \, x^{2} + 17979 \, x + 12953}{39200 \, {\left (25 \, x^{4} + 20 \, x^{3} + 34 \, x^{2} + 12 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3)^3,x, algorithm="maxima")

[Out]

339/21952*sqrt(14)*arctan(1/14*sqrt(14)*(5*x + 1)) + 1/39200*(11015*x^3 + 38753*x^2 + 17979*x + 12953)/(25*x^4
+ 20*x^3 + 34*x^2 + 12*x + 9)

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mupad [B]  time = 0.05, size = 55, normalized size = 0.86 \[ \frac {339\,\sqrt {14}\,\mathrm {atan}\left (\frac {5\,\sqrt {14}\,x}{14}+\frac {\sqrt {14}}{14}\right )}{21952}+\frac {\frac {2203\,x^3}{196000}+\frac {38753\,x^2}{980000}+\frac {17979\,x}{980000}+\frac {12953}{980000}}{x^4+\frac {4\,x^3}{5}+\frac {34\,x^2}{25}+\frac {12\,x}{25}+\frac {9}{25}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 3*x^2 - 5*x^3 + 4*x^4 + 2)/(2*x + 5*x^2 + 3)^3,x)

[Out]

(339*14^(1/2)*atan((5*14^(1/2)*x)/14 + 14^(1/2)/14))/21952 + ((17979*x)/980000 + (38753*x^2)/980000 + (2203*x^
3)/196000 + 12953/980000)/((12*x)/25 + (34*x^2)/25 + (4*x^3)/5 + x^4 + 9/25)

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sympy [A]  time = 0.20, size = 61, normalized size = 0.95 \[ \frac {11015 x^{3} + 38753 x^{2} + 17979 x + 12953}{980000 x^{4} + 784000 x^{3} + 1332800 x^{2} + 470400 x + 352800} + \frac {339 \sqrt {14} \operatorname {atan}{\left (\frac {5 \sqrt {14} x}{14} + \frac {\sqrt {14}}{14} \right )}}{21952} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**4-5*x**3+3*x**2+x+2)/(5*x**2+2*x+3)**3,x)

[Out]

(11015*x**3 + 38753*x**2 + 17979*x + 12953)/(980000*x**4 + 784000*x**3 + 1332800*x**2 + 470400*x + 352800) + 3
39*sqrt(14)*atan(5*sqrt(14)*x/14 + sqrt(14)/14)/21952

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