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

Optimal. Leaf size=70 \[ \frac {25 x^5}{2}+\frac {575 x^4}{16}+\frac {965 x^3}{24}-\frac {829 x^2}{32}+\frac {1331}{128} \log \left (2 x^2-x+3\right )-\frac {4795 x}{32}-\frac {59895 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{64 \sqrt {23}} \]

[Out]

-4795/32*x-829/32*x^2+965/24*x^3+575/16*x^4+25/2*x^5+1331/128*ln(2*x^2-x+3)-59895/1472*arctan(1/23*(1-4*x)*23^
(1/2))*23^(1/2)

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Rubi [A]  time = 0.06, 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} \[ \frac {25 x^5}{2}+\frac {575 x^4}{16}+\frac {965 x^3}{24}-\frac {829 x^2}{32}+\frac {1331}{128} \log \left (2 x^2-x+3\right )-\frac {4795 x}{32}-\frac {59895 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{64 \sqrt {23}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4795*x)/32 - (829*x^2)/32 + (965*x^3)/24 + (575*x^4)/16 + (25*x^5)/2 - (59895*ArcTan[(1 - 4*x)/Sqrt[23]])/(6
4*Sqrt[23]) + (1331*Log[3 - x + 2*x^2])/128

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 (2+3 x+5 x^2\right )^3}{3-x+2 x^2} \, dx &=\int \left (-\frac {4795}{32}-\frac {829 x}{16}+\frac {965 x^2}{8}+\frac {575 x^3}{4}+\frac {125 x^4}{2}+\frac {1331 (11+x)}{32 \left (3-x+2 x^2\right )}\right ) \, dx\\ &=-\frac {4795 x}{32}-\frac {829 x^2}{32}+\frac {965 x^3}{24}+\frac {575 x^4}{16}+\frac {25 x^5}{2}+\frac {1331}{32} \int \frac {11+x}{3-x+2 x^2} \, dx\\ &=-\frac {4795 x}{32}-\frac {829 x^2}{32}+\frac {965 x^3}{24}+\frac {575 x^4}{16}+\frac {25 x^5}{2}+\frac {1331}{128} \int \frac {-1+4 x}{3-x+2 x^2} \, dx+\frac {59895}{128} \int \frac {1}{3-x+2 x^2} \, dx\\ &=-\frac {4795 x}{32}-\frac {829 x^2}{32}+\frac {965 x^3}{24}+\frac {575 x^4}{16}+\frac {25 x^5}{2}+\frac {1331}{128} \log \left (3-x+2 x^2\right )-\frac {59895}{64} \operatorname {Subst}\left (\int \frac {1}{-23-x^2} \, dx,x,-1+4 x\right )\\ &=-\frac {4795 x}{32}-\frac {829 x^2}{32}+\frac {965 x^3}{24}+\frac {575 x^4}{16}+\frac {25 x^5}{2}-\frac {59895 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{64 \sqrt {23}}+\frac {1331}{128} \log \left (3-x+2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.90 \[ \frac {1}{384} \left (3993 \log \left (2 x^2-x+3\right )+4 x \left (1200 x^4+3450 x^3+3860 x^2-2487 x-14385\right )\right )+\frac {59895 \tan ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{64 \sqrt {23}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(59895*ArcTan[(-1 + 4*x)/Sqrt[23]])/(64*Sqrt[23]) + (4*x*(-14385 - 2487*x + 3860*x^2 + 3450*x^3 + 1200*x^4) +
3993*Log[3 - x + 2*x^2])/384

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fricas [A]  time = 1.02, size = 53, normalized size = 0.76 \[ \frac {25}{2} \, x^{5} + \frac {575}{16} \, x^{4} + \frac {965}{24} \, x^{3} - \frac {829}{32} \, x^{2} + \frac {59895}{1472} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {4795}{32} \, x + \frac {1331}{128} \, \log \left (2 \, x^{2} - x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/2*x^5 + 575/16*x^4 + 965/24*x^3 - 829/32*x^2 + 59895/1472*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 4795/3
2*x + 1331/128*log(2*x^2 - x + 3)

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giac [A]  time = 0.21, size = 53, normalized size = 0.76 \[ \frac {25}{2} \, x^{5} + \frac {575}{16} \, x^{4} + \frac {965}{24} \, x^{3} - \frac {829}{32} \, x^{2} + \frac {59895}{1472} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {4795}{32} \, x + \frac {1331}{128} \, \log \left (2 \, x^{2} - x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/2*x^5 + 575/16*x^4 + 965/24*x^3 - 829/32*x^2 + 59895/1472*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 4795/3
2*x + 1331/128*log(2*x^2 - x + 3)

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maple [A]  time = 0.00, size = 54, normalized size = 0.77 \[ \frac {25 x^{5}}{2}+\frac {575 x^{4}}{16}+\frac {965 x^{3}}{24}-\frac {829 x^{2}}{32}-\frac {4795 x}{32}+\frac {59895 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{1472}+\frac {1331 \ln \left (2 x^{2}-x +3\right )}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/2*x^5+575/16*x^4+965/24*x^3-829/32*x^2-4795/32*x+1331/128*ln(2*x^2-x+3)+59895/1472*23^(1/2)*arctan(1/23*(4*
x-1)*23^(1/2))

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maxima [A]  time = 0.97, size = 53, normalized size = 0.76 \[ \frac {25}{2} \, x^{5} + \frac {575}{16} \, x^{4} + \frac {965}{24} \, x^{3} - \frac {829}{32} \, x^{2} + \frac {59895}{1472} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {4795}{32} \, x + \frac {1331}{128} \, \log \left (2 \, x^{2} - x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/2*x^5 + 575/16*x^4 + 965/24*x^3 - 829/32*x^2 + 59895/1472*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) - 4795/3
2*x + 1331/128*log(2*x^2 - x + 3)

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mupad [B]  time = 0.04, size = 55, normalized size = 0.79 \[ \frac {1331\,\ln \left (2\,x^2-x+3\right )}{128}-\frac {4795\,x}{32}+\frac {59895\,\sqrt {23}\,\mathrm {atan}\left (\frac {4\,\sqrt {23}\,x}{23}-\frac {\sqrt {23}}{23}\right )}{1472}-\frac {829\,x^2}{32}+\frac {965\,x^3}{24}+\frac {575\,x^4}{16}+\frac {25\,x^5}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1331*log(2*x^2 - x + 3))/128 - (4795*x)/32 + (59895*23^(1/2)*atan((4*23^(1/2)*x)/23 - 23^(1/2)/23))/1472 - (8
29*x^2)/32 + (965*x^3)/24 + (575*x^4)/16 + (25*x^5)/2

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sympy [A]  time = 0.15, size = 73, normalized size = 1.04 \[ \frac {25 x^{5}}{2} + \frac {575 x^{4}}{16} + \frac {965 x^{3}}{24} - \frac {829 x^{2}}{32} - \frac {4795 x}{32} + \frac {1331 \log {\left (x^{2} - \frac {x}{2} + \frac {3}{2} \right )}}{128} + \frac {59895 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{1472} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*x**5/2 + 575*x**4/16 + 965*x**3/24 - 829*x**2/32 - 4795*x/32 + 1331*log(x**2 - x/2 + 3/2)/128 + 59895*sqrt(
23)*atan(4*sqrt(23)*x/23 - sqrt(23)/23)/1472

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