3.264 \(\int \frac {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {x^3}{3}+\frac {1}{2} \log \left (x^2+6 x+10\right )-3 \tan ^{-1}(x+3) \]

[Out]

1/3*x^3-3*arctan(3+x)+1/2*ln(x^2+6*x+10)

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Rubi [A]  time = 0.03, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1657, 634, 618, 204, 628} \[ \frac {x^3}{3}+\frac {1}{2} \log \left (x^2+6 x+10\right )-3 \tan ^{-1}(x+3) \]

Antiderivative was successfully verified.

[In]

Int[(x + 10*x^2 + 6*x^3 + x^4)/(10 + 6*x + x^2),x]

[Out]

x^3/3 - 3*ArcTan[3 + x] + Log[10 + 6*x + x^2]/2

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 {x+10 x^2+6 x^3+x^4}{10+6 x+x^2} \, dx &=\int \left (x^2+\frac {x}{10+6 x+x^2}\right ) \, dx\\ &=\frac {x^3}{3}+\int \frac {x}{10+6 x+x^2} \, dx\\ &=\frac {x^3}{3}+\frac {1}{2} \int \frac {6+2 x}{10+6 x+x^2} \, dx-3 \int \frac {1}{10+6 x+x^2} \, dx\\ &=\frac {x^3}{3}+\frac {1}{2} \log \left (10+6 x+x^2\right )+6 \operatorname {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,6+2 x\right )\\ &=\frac {x^3}{3}-3 \tan ^{-1}(3+x)+\frac {1}{2} \log \left (10+6 x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {x^3}{3}+\frac {1}{2} \log \left (x^2+6 x+10\right )-3 \tan ^{-1}(x+3) \]

Antiderivative was successfully verified.

[In]

Integrate[(x + 10*x^2 + 6*x^3 + x^4)/(10 + 6*x + x^2),x]

[Out]

x^3/3 - 3*ArcTan[3 + x] + Log[10 + 6*x + x^2]/2

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fricas [A]  time = 1.07, size = 23, normalized size = 0.85 \[ \frac {1}{3} \, x^{3} - 3 \, \arctan \left (x + 3\right ) + \frac {1}{2} \, \log \left (x^{2} + 6 \, x + 10\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+6*x^3+10*x^2+x)/(x^2+6*x+10),x, algorithm="fricas")

[Out]

1/3*x^3 - 3*arctan(x + 3) + 1/2*log(x^2 + 6*x + 10)

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giac [A]  time = 0.28, size = 23, normalized size = 0.85 \[ \frac {1}{3} \, x^{3} - 3 \, \arctan \left (x + 3\right ) + \frac {1}{2} \, \log \left (x^{2} + 6 \, x + 10\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+6*x^3+10*x^2+x)/(x^2+6*x+10),x, algorithm="giac")

[Out]

1/3*x^3 - 3*arctan(x + 3) + 1/2*log(x^2 + 6*x + 10)

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maple [A]  time = 0.00, size = 24, normalized size = 0.89 \[ \frac {x^{3}}{3}-3 \arctan \left (x +3\right )+\frac {\ln \left (x^{2}+6 x +10\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+6*x^3+10*x^2+x)/(x^2+6*x+10),x)

[Out]

1/3*x^3-3*arctan(x+3)+1/2*ln(x^2+6*x+10)

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maxima [A]  time = 2.08, size = 23, normalized size = 0.85 \[ \frac {1}{3} \, x^{3} - 3 \, \arctan \left (x + 3\right ) + \frac {1}{2} \, \log \left (x^{2} + 6 \, x + 10\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+6*x^3+10*x^2+x)/(x^2+6*x+10),x, algorithm="maxima")

[Out]

1/3*x^3 - 3*arctan(x + 3) + 1/2*log(x^2 + 6*x + 10)

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mupad [B]  time = 2.14, size = 23, normalized size = 0.85 \[ \frac {\ln \left (x^2+6\,x+10\right )}{2}-3\,\mathrm {atan}\left (x+3\right )+\frac {x^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 10*x^2 + 6*x^3 + x^4)/(6*x + x^2 + 10),x)

[Out]

log(6*x + x^2 + 10)/2 - 3*atan(x + 3) + x^3/3

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sympy [A]  time = 0.11, size = 22, normalized size = 0.81 \[ \frac {x^{3}}{3} + \frac {\log {\left (x^{2} + 6 x + 10 \right )}}{2} - 3 \operatorname {atan}{\left (x + 3 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+6*x**3+10*x**2+x)/(x**2+6*x+10),x)

[Out]

x**3/3 + log(x**2 + 6*x + 10)/2 - 3*atan(x + 3)

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