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

Optimal. Leaf size=56 \[ \frac {4 x^3}{15}-\frac {16 x^2}{25}-\frac {1573 \log \left (5 x^2+3 x+2\right )}{1250}+\frac {381 x}{125}+\frac {8349 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{625 \sqrt {31}} \]

[Out]

381/125*x-16/25*x^2+4/15*x^3-1573/1250*ln(5*x^2+3*x+2)+8349/19375*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 56, 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} \[ \frac {4 x^3}{15}-\frac {16 x^2}{25}-\frac {1573 \log \left (5 x^2+3 x+2\right )}{1250}+\frac {381 x}{125}+\frac {8349 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{625 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(381*x)/125 - (16*x^2)/25 + (4*x^3)/15 + (8349*ArcTan[(3 + 10*x)/Sqrt[31]])/(625*Sqrt[31]) - (1573*Log[2 + 3*x
+ 5*x^2])/1250

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 )^2}{2+3 x+5 x^2} \, dx &=\int \left (\frac {381}{125}-\frac {32 x}{25}+\frac {4 x^2}{5}+\frac {121 (3-13 x)}{125 \left (2+3 x+5 x^2\right )}\right ) \, dx\\ &=\frac {381 x}{125}-\frac {16 x^2}{25}+\frac {4 x^3}{15}+\frac {121}{125} \int \frac {3-13 x}{2+3 x+5 x^2} \, dx\\ &=\frac {381 x}{125}-\frac {16 x^2}{25}+\frac {4 x^3}{15}-\frac {1573 \int \frac {3+10 x}{2+3 x+5 x^2} \, dx}{1250}+\frac {8349 \int \frac {1}{2+3 x+5 x^2} \, dx}{1250}\\ &=\frac {381 x}{125}-\frac {16 x^2}{25}+\frac {4 x^3}{15}-\frac {1573 \log \left (2+3 x+5 x^2\right )}{1250}-\frac {8349}{625} \operatorname {Subst}\left (\int \frac {1}{-31-x^2} \, dx,x,3+10 x\right )\\ &=\frac {381 x}{125}-\frac {16 x^2}{25}+\frac {4 x^3}{15}+\frac {8349 \tan ^{-1}\left (\frac {3+10 x}{\sqrt {31}}\right )}{625 \sqrt {31}}-\frac {1573 \log \left (2+3 x+5 x^2\right )}{1250}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 0.95 \[ \frac {10 x \left (100 x^2-240 x+1143\right )-4719 \log \left (5 x^2+3 x+2\right )}{3750}+\frac {8349 \tan ^{-1}\left (\frac {10 x+3}{\sqrt {31}}\right )}{625 \sqrt {31}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8349*ArcTan[(3 + 10*x)/Sqrt[31]])/(625*Sqrt[31]) + (10*x*(1143 - 240*x + 100*x^2) - 4719*Log[2 + 3*x + 5*x^2]
)/3750

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fricas [A]  time = 0.71, size = 43, normalized size = 0.77 \[ \frac {4}{15} \, x^{3} - \frac {16}{25} \, x^{2} + \frac {8349}{19375} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {381}{125} \, x - \frac {1573}{1250} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*x^3 - 16/25*x^2 + 8349/19375*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 381/125*x - 1573/1250*log(5*x^2
+ 3*x + 2)

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giac [A]  time = 0.21, size = 43, normalized size = 0.77 \[ \frac {4}{15} \, x^{3} - \frac {16}{25} \, x^{2} + \frac {8349}{19375} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {381}{125} \, x - \frac {1573}{1250} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*x^3 - 16/25*x^2 + 8349/19375*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 381/125*x - 1573/1250*log(5*x^2
+ 3*x + 2)

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maple [A]  time = 0.00, size = 44, normalized size = 0.79 \[ \frac {4 x^{3}}{15}-\frac {16 x^{2}}{25}+\frac {381 x}{125}+\frac {8349 \sqrt {31}\, \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right )}{19375}-\frac {1573 \ln \left (5 x^{2}+3 x +2\right )}{1250} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

381/125*x-16/25*x^2+4/15*x^3-1573/1250*ln(5*x^2+3*x+2)+8349/19375*31^(1/2)*arctan(1/31*(10*x+3)*31^(1/2))

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maxima [A]  time = 0.97, size = 43, normalized size = 0.77 \[ \frac {4}{15} \, x^{3} - \frac {16}{25} \, x^{2} + \frac {8349}{19375} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + \frac {381}{125} \, x - \frac {1573}{1250} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*x^3 - 16/25*x^2 + 8349/19375*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) + 381/125*x - 1573/1250*log(5*x^2
+ 3*x + 2)

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mupad [B]  time = 3.45, size = 45, normalized size = 0.80 \[ \frac {381\,x}{125}-\frac {1573\,\ln \left (5\,x^2+3\,x+2\right )}{1250}+\frac {8349\,\sqrt {31}\,\mathrm {atan}\left (\frac {10\,\sqrt {31}\,x}{31}+\frac {3\,\sqrt {31}}{31}\right )}{19375}-\frac {16\,x^2}{25}+\frac {4\,x^3}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(381*x)/125 - (1573*log(3*x + 5*x^2 + 2))/1250 + (8349*31^(1/2)*atan((10*31^(1/2)*x)/31 + (3*31^(1/2))/31))/19
375 - (16*x^2)/25 + (4*x^3)/15

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sympy [A]  time = 0.15, size = 63, normalized size = 1.12 \[ \frac {4 x^{3}}{15} - \frac {16 x^{2}}{25} + \frac {381 x}{125} - \frac {1573 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{1250} + \frac {8349 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{19375} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x**3/15 - 16*x**2/25 + 381*x/125 - 1573*log(x**2 + 3*x/5 + 2/5)/1250 + 8349*sqrt(31)*atan(10*sqrt(31)*x/31 +
3*sqrt(31)/31)/19375

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