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

Optimal. Leaf size=77 \[ \frac {125 x^3}{12}+\frac {175 x^2}{4}-\frac {1331 (17-45 x)}{736 \left (2 x^2-x+3\right )}-\frac {2057}{32} \log \left (2 x^2-x+3\right )+\frac {915 x}{16}+\frac {223971 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{368 \sqrt {23}} \]

[Out]

915/16*x+175/4*x^2+125/12*x^3-1331/736*(17-45*x)/(2*x^2-x+3)-2057/32*ln(2*x^2-x+3)+223971/8464*arctan(1/23*(1-
4*x)*23^(1/2))*23^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1660, 1657, 634, 618, 204, 628} \[ \frac {125 x^3}{12}+\frac {175 x^2}{4}-\frac {1331 (17-45 x)}{736 \left (2 x^2-x+3\right )}-\frac {2057}{32} \log \left (2 x^2-x+3\right )+\frac {915 x}{16}+\frac {223971 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{368 \sqrt {23}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(915*x)/16 + (175*x^2)/4 + (125*x^3)/12 - (1331*(17 - 45*x))/(736*(3 - x + 2*x^2)) + (223971*ArcTan[(1 - 4*x)/
Sqrt[23]])/(368*Sqrt[23]) - (2057*Log[3 - x + 2*x^2])/32

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]

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 {\left (2+3 x+5 x^2\right )^3}{\left (3-x+2 x^2\right )^2} \, dx &=-\frac {1331 (17-45 x)}{736 \left (3-x+2 x^2\right )}+\frac {1}{23} \int \frac {-\frac {25195}{16}-\frac {19067 x}{16}+\frac {22195 x^2}{8}+\frac {13225 x^3}{4}+\frac {2875 x^4}{2}}{3-x+2 x^2} \, dx\\ &=-\frac {1331 (17-45 x)}{736 \left (3-x+2 x^2\right )}+\frac {1}{23} \int \left (\frac {21045}{16}+\frac {4025 x}{2}+\frac {2875 x^2}{4}-\frac {121 (365+391 x)}{8 \left (3-x+2 x^2\right )}\right ) \, dx\\ &=\frac {915 x}{16}+\frac {175 x^2}{4}+\frac {125 x^3}{12}-\frac {1331 (17-45 x)}{736 \left (3-x+2 x^2\right )}-\frac {121}{184} \int \frac {365+391 x}{3-x+2 x^2} \, dx\\ &=\frac {915 x}{16}+\frac {175 x^2}{4}+\frac {125 x^3}{12}-\frac {1331 (17-45 x)}{736 \left (3-x+2 x^2\right )}-\frac {2057}{32} \int \frac {-1+4 x}{3-x+2 x^2} \, dx-\frac {223971}{736} \int \frac {1}{3-x+2 x^2} \, dx\\ &=\frac {915 x}{16}+\frac {175 x^2}{4}+\frac {125 x^3}{12}-\frac {1331 (17-45 x)}{736 \left (3-x+2 x^2\right )}-\frac {2057}{32} \log \left (3-x+2 x^2\right )+\frac {223971}{368} \operatorname {Subst}\left (\int \frac {1}{-23-x^2} \, dx,x,-1+4 x\right )\\ &=\frac {915 x}{16}+\frac {175 x^2}{4}+\frac {125 x^3}{12}-\frac {1331 (17-45 x)}{736 \left (3-x+2 x^2\right )}+\frac {223971 \tan ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{368 \sqrt {23}}-\frac {2057}{32} \log \left (3-x+2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 77, normalized size = 1.00 \[ \frac {125 x^3}{12}+\frac {175 x^2}{4}+\frac {1331 (45 x-17)}{736 \left (2 x^2-x+3\right )}-\frac {2057}{32} \log \left (2 x^2-x+3\right )+\frac {915 x}{16}-\frac {223971 \tan ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{368 \sqrt {23}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(915*x)/16 + (175*x^2)/4 + (125*x^3)/12 + (1331*(-17 + 45*x))/(736*(3 - x + 2*x^2)) - (223971*ArcTan[(-1 + 4*x
)/Sqrt[23]])/(368*Sqrt[23]) - (2057*Log[3 - x + 2*x^2])/32

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fricas [A]  time = 0.73, size = 88, normalized size = 1.14 \[ \frac {1058000 \, x^{5} + 3914600 \, x^{4} + 5173620 \, x^{3} - 1343826 \, \sqrt {23} {\left (2 \, x^{2} - x + 3\right )} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + 3761190 \, x^{2} - 3264459 \, {\left (2 \, x^{2} - x + 3\right )} \log \left (2 \, x^{2} - x + 3\right ) + 12845385 \, x - 1561263}{50784 \, {\left (2 \, x^{2} - x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50784*(1058000*x^5 + 3914600*x^4 + 5173620*x^3 - 1343826*sqrt(23)*(2*x^2 - x + 3)*arctan(1/23*sqrt(23)*(4*x
- 1)) + 3761190*x^2 - 3264459*(2*x^2 - x + 3)*log(2*x^2 - x + 3) + 12845385*x - 1561263)/(2*x^2 - x + 3)

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giac [A]  time = 0.19, size = 62, normalized size = 0.81 \[ \frac {125}{12} \, x^{3} + \frac {175}{4} \, x^{2} - \frac {223971}{8464} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {915}{16} \, x + \frac {1331 \, {\left (45 \, x - 17\right )}}{736 \, {\left (2 \, x^{2} - x + 3\right )}} - \frac {2057}{32} \, \log \left (2 \, x^{2} - x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/12*x^3 + 175/4*x^2 - 223971/8464*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 915/16*x + 1331/736*(45*x - 17
)/(2*x^2 - x + 3) - 2057/32*log(2*x^2 - x + 3)

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maple [A]  time = 0.01, size = 61, normalized size = 0.79 \[ \frac {125 x^{3}}{12}+\frac {175 x^{2}}{4}+\frac {915 x}{16}-\frac {223971 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{8464}-\frac {2057 \ln \left (2 x^{2}-x +3\right )}{32}-\frac {121 \left (-\frac {495 x}{92}+\frac {187}{92}\right )}{16 \left (x^{2}-\frac {1}{2} x +\frac {3}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/12*x^3+175/4*x^2+915/16*x-121/16*(-495/92*x+187/92)/(x^2-1/2*x+3/2)-2057/32*ln(2*x^2-x+3)-223971/8464*23^(
1/2)*arctan(1/23*(4*x-1)*23^(1/2))

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maxima [A]  time = 0.96, size = 62, normalized size = 0.81 \[ \frac {125}{12} \, x^{3} + \frac {175}{4} \, x^{2} - \frac {223971}{8464} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {915}{16} \, x + \frac {1331 \, {\left (45 \, x - 17\right )}}{736 \, {\left (2 \, x^{2} - x + 3\right )}} - \frac {2057}{32} \, \log \left (2 \, x^{2} - x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/12*x^3 + 175/4*x^2 - 223971/8464*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 915/16*x + 1331/736*(45*x - 17
)/(2*x^2 - x + 3) - 2057/32*log(2*x^2 - x + 3)

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mupad [B]  time = 3.42, size = 61, normalized size = 0.79 \[ \frac {915\,x}{16}-\frac {2057\,\ln \left (2\,x^2-x+3\right )}{32}+\frac {\frac {59895\,x}{1472}-\frac {22627}{1472}}{x^2-\frac {x}{2}+\frac {3}{2}}-\frac {223971\,\sqrt {23}\,\mathrm {atan}\left (\frac {4\,\sqrt {23}\,x}{23}-\frac {\sqrt {23}}{23}\right )}{8464}+\frac {175\,x^2}{4}+\frac {125\,x^3}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(915*x)/16 - (2057*log(2*x^2 - x + 3))/32 + ((59895*x)/1472 - 22627/1472)/(x^2 - x/2 + 3/2) - (223971*23^(1/2)
*atan((4*23^(1/2)*x)/23 - 23^(1/2)/23))/8464 + (175*x^2)/4 + (125*x^3)/12

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sympy [A]  time = 0.19, size = 75, normalized size = 0.97 \[ \frac {125 x^{3}}{12} + \frac {175 x^{2}}{4} + \frac {915 x}{16} + \frac {59895 x - 22627}{1472 x^{2} - 736 x + 2208} - \frac {2057 \log {\left (x^{2} - \frac {x}{2} + \frac {3}{2} \right )}}{32} - \frac {223971 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{8464} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125*x**3/12 + 175*x**2/4 + 915*x/16 + (59895*x - 22627)/(1472*x**2 - 736*x + 2208) - 2057*log(x**2 - x/2 + 3/2
)/32 - 223971*sqrt(23)*atan(4*sqrt(23)*x/23 - sqrt(23)/23)/8464

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