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

Optimal. Leaf size=70 \[ \frac {8 x^5}{25}-\frac {21 x^4}{25}+\frac {1222 x^3}{375}-\frac {7451 x^2}{1250}-\frac {158389 \log \left (5 x^2+3 x+2\right )}{31250}+\frac {49508 x}{3125}+\frac {328757 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{15625 \sqrt {31}} \]

[Out]

49508/3125*x-7451/1250*x^2+1222/375*x^3-21/25*x^4+8/25*x^5-158389/31250*ln(5*x^2+3*x+2)+328757/484375*arctan(1
/31*(3+10*x)*31^(1/2))*31^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1657, 634, 618, 204, 628} \[ \frac {8 x^5}{25}-\frac {21 x^4}{25}+\frac {1222 x^3}{375}-\frac {7451 x^2}{1250}-\frac {158389 \log \left (5 x^2+3 x+2\right )}{31250}+\frac {49508 x}{3125}+\frac {328757 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{15625 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(49508*x)/3125 - (7451*x^2)/1250 + (1222*x^3)/375 - (21*x^4)/25 + (8*x^5)/25 + (328757*ArcTan[(3 + 10*x)/Sqrt[
31]])/(15625*Sqrt[31]) - (158389*Log[2 + 3*x + 5*x^2])/31250

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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {\left (3-x+2 x^2\right )^3}{2+3 x+5 x^2} \, dx &=\int \left (\frac {49508}{3125}-\frac {7451 x}{625}+\frac {1222 x^2}{125}-\frac {84 x^3}{25}+\frac {8 x^4}{5}-\frac {1331 (11+119 x)}{3125 \left (2+3 x+5 x^2\right )}\right ) \, dx\\ &=\frac {49508 x}{3125}-\frac {7451 x^2}{1250}+\frac {1222 x^3}{375}-\frac {21 x^4}{25}+\frac {8 x^5}{25}-\frac {1331 \int \frac {11+119 x}{2+3 x+5 x^2} \, dx}{3125}\\ &=\frac {49508 x}{3125}-\frac {7451 x^2}{1250}+\frac {1222 x^3}{375}-\frac {21 x^4}{25}+\frac {8 x^5}{25}-\frac {158389 \int \frac {3+10 x}{2+3 x+5 x^2} \, dx}{31250}+\frac {328757 \int \frac {1}{2+3 x+5 x^2} \, dx}{31250}\\ &=\frac {49508 x}{3125}-\frac {7451 x^2}{1250}+\frac {1222 x^3}{375}-\frac {21 x^4}{25}+\frac {8 x^5}{25}-\frac {158389 \log \left (2+3 x+5 x^2\right )}{31250}-\frac {328757 \operatorname {Subst}\left (\int \frac {1}{-31-x^2} \, dx,x,3+10 x\right )}{15625}\\ &=\frac {49508 x}{3125}-\frac {7451 x^2}{1250}+\frac {1222 x^3}{375}-\frac {21 x^4}{25}+\frac {8 x^5}{25}+\frac {328757 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{15625 \sqrt {31}}-\frac {158389 \log \left (2+3 x+5 x^2\right )}{31250}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.90 \[ \frac {31 \left (5 x \left (6000 x^4-15750 x^3+61100 x^2-111765 x+297048\right )-475167 \log \left (5 x^2+3 x+2\right )\right )+1972542 \sqrt {31} \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{2906250} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1972542*Sqrt[31]*ArcTan[(3 + 10*x)/Sqrt[31]] + 31*(5*x*(297048 - 111765*x + 61100*x^2 - 15750*x^3 + 6000*x^4)
 - 475167*Log[2 + 3*x + 5*x^2]))/2906250

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fricas [A]  time = 0.87, size = 53, normalized size = 0.76 \[ \frac {8}{25} \, x^{5} - \frac {21}{25} \, x^{4} + \frac {1222}{375} \, x^{3} - \frac {7451}{1250} \, x^{2} + \frac {328757}{484375} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {49508}{3125} \, x - \frac {158389}{31250} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/25*x^5 - 21/25*x^4 + 1222/375*x^3 - 7451/1250*x^2 + 328757/484375*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3))
+ 49508/3125*x - 158389/31250*log(5*x^2 + 3*x + 2)

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giac [A]  time = 0.21, size = 53, normalized size = 0.76 \[ \frac {8}{25} \, x^{5} - \frac {21}{25} \, x^{4} + \frac {1222}{375} \, x^{3} - \frac {7451}{1250} \, x^{2} + \frac {328757}{484375} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {49508}{3125} \, x - \frac {158389}{31250} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/25*x^5 - 21/25*x^4 + 1222/375*x^3 - 7451/1250*x^2 + 328757/484375*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3))
+ 49508/3125*x - 158389/31250*log(5*x^2 + 3*x + 2)

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maple [A]  time = 0.00, size = 54, normalized size = 0.77 \[ \frac {8 x^{5}}{25}-\frac {21 x^{4}}{25}+\frac {1222 x^{3}}{375}-\frac {7451 x^{2}}{1250}+\frac {49508 x}{3125}+\frac {328757 \sqrt {31}\, \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right )}{484375}-\frac {158389 \ln \left (5 x^{2}+3 x +2\right )}{31250} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

49508/3125*x-7451/1250*x^2+1222/375*x^3-21/25*x^4+8/25*x^5-158389/31250*ln(5*x^2+3*x+2)+328757/484375*31^(1/2)
*arctan(1/31*(10*x+3)*31^(1/2))

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maxima [A]  time = 0.96, size = 53, normalized size = 0.76 \[ \frac {8}{25} \, x^{5} - \frac {21}{25} \, x^{4} + \frac {1222}{375} \, x^{3} - \frac {7451}{1250} \, x^{2} + \frac {328757}{484375} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {49508}{3125} \, x - \frac {158389}{31250} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/25*x^5 - 21/25*x^4 + 1222/375*x^3 - 7451/1250*x^2 + 328757/484375*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3))
+ 49508/3125*x - 158389/31250*log(5*x^2 + 3*x + 2)

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mupad [B]  time = 0.04, size = 55, normalized size = 0.79 \[ \frac {49508\,x}{3125}-\frac {158389\,\ln \left (5\,x^2+3\,x+2\right )}{31250}+\frac {328757\,\sqrt {31}\,\mathrm {atan}\left (\frac {10\,\sqrt {31}\,x}{31}+\frac {3\,\sqrt {31}}{31}\right )}{484375}-\frac {7451\,x^2}{1250}+\frac {1222\,x^3}{375}-\frac {21\,x^4}{25}+\frac {8\,x^5}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(49508*x)/3125 - (158389*log(3*x + 5*x^2 + 2))/31250 + (328757*31^(1/2)*atan((10*31^(1/2)*x)/31 + (3*31^(1/2))
/31))/484375 - (7451*x^2)/1250 + (1222*x^3)/375 - (21*x^4)/25 + (8*x^5)/25

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sympy [A]  time = 0.16, size = 76, normalized size = 1.09 \[ \frac {8 x^{5}}{25} - \frac {21 x^{4}}{25} + \frac {1222 x^{3}}{375} - \frac {7451 x^{2}}{1250} + \frac {49508 x}{3125} - \frac {158389 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{31250} + \frac {328757 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{484375} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*x**5/25 - 21*x**4/25 + 1222*x**3/375 - 7451*x**2/1250 + 49508*x/3125 - 158389*log(x**2 + 3*x/5 + 2/5)/31250
+ 328757*sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/484375

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