3.365 \(\int \frac{1+x^3}{(13+4 x+x^2)^2} \, dx\)

Optimal. Leaf size=45 \[ \frac{47 x+67}{18 \left (x^2+4 x+13\right )}+\frac{1}{2} \log \left (x^2+4 x+13\right )-\frac{61}{54} \tan ^{-1}\left (\frac{x+2}{3}\right ) \]

[Out]

(67 + 47*x)/(18*(13 + 4*x + x^2)) - (61*ArcTan[(2 + x)/3])/54 + Log[13 + 4*x + x^2]/2

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Rubi [A]  time = 0.0248464, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {1660, 634, 618, 204, 628} \[ \frac{47 x+67}{18 \left (x^2+4 x+13\right )}+\frac{1}{2} \log \left (x^2+4 x+13\right )-\frac{61}{54} \tan ^{-1}\left (\frac{x+2}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)/(13 + 4*x + x^2)^2,x]

[Out]

(67 + 47*x)/(18*(13 + 4*x + x^2)) - (61*ArcTan[(2 + x)/3])/54 + Log[13 + 4*x + x^2]/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
]

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{1+x^3}{\left (13+4 x+x^2\right )^2} \, dx &=\frac{67+47 x}{18 \left (13+4 x+x^2\right )}+\frac{1}{36} \int \frac{-50+36 x}{13+4 x+x^2} \, dx\\ &=\frac{67+47 x}{18 \left (13+4 x+x^2\right )}+\frac{1}{2} \int \frac{4+2 x}{13+4 x+x^2} \, dx-\frac{61}{18} \int \frac{1}{13+4 x+x^2} \, dx\\ &=\frac{67+47 x}{18 \left (13+4 x+x^2\right )}+\frac{1}{2} \log \left (13+4 x+x^2\right )+\frac{61}{9} \operatorname{Subst}\left (\int \frac{1}{-36-x^2} \, dx,x,4+2 x\right )\\ &=\frac{67+47 x}{18 \left (13+4 x+x^2\right )}-\frac{61}{54} \tan ^{-1}\left (\frac{2+x}{3}\right )+\frac{1}{2} \log \left (13+4 x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0121463, size = 45, normalized size = 1. \[ \frac{47 x+67}{18 \left (x^2+4 x+13\right )}+\frac{1}{2} \log \left (x^2+4 x+13\right )-\frac{61}{54} \tan ^{-1}\left (\frac{x+2}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)/(13 + 4*x + x^2)^2,x]

[Out]

(67 + 47*x)/(18*(13 + 4*x + x^2)) - (61*ArcTan[(2 + x)/3])/54 + Log[13 + 4*x + x^2]/2

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Maple [A]  time = 0.006, size = 37, normalized size = 0.8 \begin{align*}{\frac{1}{{x}^{2}+4\,x+13} \left ({\frac{47\,x}{18}}+{\frac{67}{18}} \right ) }+{\frac{\ln \left ({x}^{2}+4\,x+13 \right ) }{2}}-{\frac{61}{54}\arctan \left ({\frac{2}{3}}+{\frac{x}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)/(x^2+4*x+13)^2,x)

[Out]

(47/18*x+67/18)/(x^2+4*x+13)+1/2*ln(x^2+4*x+13)-61/54*arctan(2/3+1/3*x)

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Maxima [A]  time = 1.6549, size = 50, normalized size = 1.11 \begin{align*} \frac{47 \, x + 67}{18 \,{\left (x^{2} + 4 \, x + 13\right )}} - \frac{61}{54} \, \arctan \left (\frac{1}{3} \, x + \frac{2}{3}\right ) + \frac{1}{2} \, \log \left (x^{2} + 4 \, x + 13\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)/(x^2+4*x+13)^2,x, algorithm="maxima")

[Out]

1/18*(47*x + 67)/(x^2 + 4*x + 13) - 61/54*arctan(1/3*x + 2/3) + 1/2*log(x^2 + 4*x + 13)

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Fricas [A]  time = 1.40957, size = 165, normalized size = 3.67 \begin{align*} -\frac{61 \,{\left (x^{2} + 4 \, x + 13\right )} \arctan \left (\frac{1}{3} \, x + \frac{2}{3}\right ) - 27 \,{\left (x^{2} + 4 \, x + 13\right )} \log \left (x^{2} + 4 \, x + 13\right ) - 141 \, x - 201}{54 \,{\left (x^{2} + 4 \, x + 13\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)/(x^2+4*x+13)^2,x, algorithm="fricas")

[Out]

-1/54*(61*(x^2 + 4*x + 13)*arctan(1/3*x + 2/3) - 27*(x^2 + 4*x + 13)*log(x^2 + 4*x + 13) - 141*x - 201)/(x^2 +
4*x + 13)

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Sympy [A]  time = 0.128042, size = 37, normalized size = 0.82 \begin{align*} \frac{47 x + 67}{18 x^{2} + 72 x + 234} + \frac{\log{\left (x^{2} + 4 x + 13 \right )}}{2} - \frac{61 \operatorname{atan}{\left (\frac{x}{3} + \frac{2}{3} \right )}}{54} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+1)/(x**2+4*x+13)**2,x)

[Out]

(47*x + 67)/(18*x**2 + 72*x + 234) + log(x**2 + 4*x + 13)/2 - 61*atan(x/3 + 2/3)/54

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Giac [A]  time = 1.11191, size = 50, normalized size = 1.11 \begin{align*} \frac{47 \, x + 67}{18 \,{\left (x^{2} + 4 \, x + 13\right )}} - \frac{61}{54} \, \arctan \left (\frac{1}{3} \, x + \frac{2}{3}\right ) + \frac{1}{2} \, \log \left (x^{2} + 4 \, x + 13\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)/(x^2+4*x+13)^2,x, algorithm="giac")

[Out]

1/18*(47*x + 67)/(x^2 + 4*x + 13) - 61/54*arctan(1/3*x + 2/3) + 1/2*log(x^2 + 4*x + 13)