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

Optimal. Leaf size=63 \[ \frac{121 (19-7 x)}{184 \left (2 x^2-x+3\right )}+\frac{55}{8} \log \left (2 x^2-x+3\right )+\frac{25 x}{4}+\frac{1859 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{92 \sqrt{23}} \]

[Out]

(25*x)/4 + (121*(19 - 7*x))/(184*(3 - x + 2*x^2)) + (1859*ArcTan[(1 - 4*x)/Sqrt[23]])/(92*Sqrt[23]) + (55*Log[
3 - x + 2*x^2])/8

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Rubi [A]  time = 0.0629492, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1660, 1657, 634, 618, 204, 628} \[ \frac{121 (19-7 x)}{184 \left (2 x^2-x+3\right )}+\frac{55}{8} \log \left (2 x^2-x+3\right )+\frac{25 x}{4}+\frac{1859 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{92 \sqrt{23}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(25*x)/4 + (121*(19 - 7*x))/(184*(3 - x + 2*x^2)) + (1859*ArcTan[(1 - 4*x)/Sqrt[23]])/(92*Sqrt[23]) + (55*Log[
3 - x + 2*x^2])/8

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
]

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^2} \, dx &=\frac{121 (19-7 x)}{184 \left (3-x+2 x^2\right )}+\frac{1}{23} \int \frac{\frac{163}{4}+\frac{1955 x}{4}+\frac{575 x^2}{2}}{3-x+2 x^2} \, dx\\ &=\frac{121 (19-7 x)}{184 \left (3-x+2 x^2\right )}+\frac{1}{23} \int \left (\frac{575}{4}-\frac{11 (71-115 x)}{2 \left (3-x+2 x^2\right )}\right ) \, dx\\ &=\frac{25 x}{4}+\frac{121 (19-7 x)}{184 \left (3-x+2 x^2\right )}-\frac{11}{46} \int \frac{71-115 x}{3-x+2 x^2} \, dx\\ &=\frac{25 x}{4}+\frac{121 (19-7 x)}{184 \left (3-x+2 x^2\right )}+\frac{55}{8} \int \frac{-1+4 x}{3-x+2 x^2} \, dx-\frac{1859}{184} \int \frac{1}{3-x+2 x^2} \, dx\\ &=\frac{25 x}{4}+\frac{121 (19-7 x)}{184 \left (3-x+2 x^2\right )}+\frac{55}{8} \log \left (3-x+2 x^2\right )+\frac{1859}{92} \operatorname{Subst}\left (\int \frac{1}{-23-x^2} \, dx,x,-1+4 x\right )\\ &=\frac{25 x}{4}+\frac{121 (19-7 x)}{184 \left (3-x+2 x^2\right )}+\frac{1859 \tan ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{92 \sqrt{23}}+\frac{55}{8} \log \left (3-x+2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0301396, size = 63, normalized size = 1. \[ -\frac{121 (7 x-19)}{184 \left (2 x^2-x+3\right )}+\frac{55}{8} \log \left (2 x^2-x+3\right )+\frac{25 x}{4}-\frac{1859 \tan ^{-1}\left (\frac{4 x-1}{\sqrt{23}}\right )}{92 \sqrt{23}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(25*x)/4 - (121*(-19 + 7*x))/(184*(3 - x + 2*x^2)) - (1859*ArcTan[(-1 + 4*x)/Sqrt[23]])/(92*Sqrt[23]) + (55*Lo
g[3 - x + 2*x^2])/8

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Maple [A]  time = 0.045, size = 51, normalized size = 0.8 \begin{align*}{\frac{25\,x}{4}}+{\frac{11}{4} \left ( -{\frac{77\,x}{92}}+{\frac{209}{92}} \right ) \left ({x}^{2}-{\frac{x}{2}}+{\frac{3}{2}} \right ) ^{-1}}+{\frac{55\,\ln \left ( 2\,{x}^{2}-x+3 \right ) }{8}}-{\frac{1859\,\sqrt{23}}{2116}\arctan \left ({\frac{ \left ( -1+4\,x \right ) \sqrt{23}}{23}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/4*x+11/4*(-77/92*x+209/92)/(x^2-1/2*x+3/2)+55/8*ln(2*x^2-x+3)-1859/2116*23^(1/2)*arctan(1/23*(-1+4*x)*23^(1
/2))

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Maxima [A]  time = 1.42417, size = 70, normalized size = 1.11 \begin{align*} -\frac{1859}{2116} \, \sqrt{23} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) + \frac{25}{4} \, x - \frac{121 \,{\left (7 \, x - 19\right )}}{184 \,{\left (2 \, x^{2} - x + 3\right )}} + \frac{55}{8} \, \log \left (2 \, x^{2} - x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1859/2116*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 25/4*x - 121/184*(7*x - 19)/(2*x^2 - x + 3) + 55/8*log(2
*x^2 - x + 3)

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Fricas [A]  time = 1.02616, size = 234, normalized size = 3.71 \begin{align*} \frac{52900 \, x^{3} - 3718 \, \sqrt{23}{\left (2 \, x^{2} - x + 3\right )} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) - 26450 \, x^{2} + 29095 \,{\left (2 \, x^{2} - x + 3\right )} \log \left (2 \, x^{2} - x + 3\right ) + 59869 \, x + 52877}{4232 \,{\left (2 \, x^{2} - x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4232*(52900*x^3 - 3718*sqrt(23)*(2*x^2 - x + 3)*arctan(1/23*sqrt(23)*(4*x - 1)) - 26450*x^2 + 29095*(2*x^2 -
x + 3)*log(2*x^2 - x + 3) + 59869*x + 52877)/(2*x^2 - x + 3)

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Sympy [A]  time = 0.170856, size = 61, normalized size = 0.97 \begin{align*} \frac{25 x}{4} - \frac{847 x - 2299}{368 x^{2} - 184 x + 552} + \frac{55 \log{\left (x^{2} - \frac{x}{2} + \frac{3}{2} \right )}}{8} - \frac{1859 \sqrt{23} \operatorname{atan}{\left (\frac{4 \sqrt{23} x}{23} - \frac{\sqrt{23}}{23} \right )}}{2116} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*x/4 - (847*x - 2299)/(368*x**2 - 184*x + 552) + 55*log(x**2 - x/2 + 3/2)/8 - 1859*sqrt(23)*atan(4*sqrt(23)*
x/23 - sqrt(23)/23)/2116

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Giac [A]  time = 1.18982, size = 70, normalized size = 1.11 \begin{align*} -\frac{1859}{2116} \, \sqrt{23} \arctan \left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) + \frac{25}{4} \, x - \frac{121 \,{\left (7 \, x - 19\right )}}{184 \,{\left (2 \, x^{2} - x + 3\right )}} + \frac{55}{8} \, \log \left (2 \, x^{2} - x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1859/2116*sqrt(23)*arctan(1/23*sqrt(23)*(4*x - 1)) + 25/4*x - 121/184*(7*x - 19)/(2*x^2 - x + 3) + 55/8*log(2
*x^2 - x + 3)