3.174 $$\int \frac{1+x^2+x^3}{2 x^2+x^3+x^4} \, dx$$

Optimal. Leaf size=46 $\frac{5}{8} \log \left (x^2+x+2\right )-\frac{1}{2 x}-\frac{\log (x)}{4}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{7}}\right )}{4 \sqrt{7}}$

[Out]

-1/(2*x) + ArcTan[(1 + 2*x)/Sqrt[7]]/(4*Sqrt[7]) - Log[x]/4 + (5*Log[2 + x + x^2])/8

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Rubi [A]  time = 0.0592398, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.261, Rules used = {1594, 1628, 634, 618, 204, 628} $\frac{5}{8} \log \left (x^2+x+2\right )-\frac{1}{2 x}-\frac{\log (x)}{4}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{7}}\right )}{4 \sqrt{7}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(2*x) + ArcTan[(1 + 2*x)/Sqrt[7]]/(4*Sqrt[7]) - Log[x]/4 + (5*Log[2 + x + x^2])/8

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, 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{1+x^2+x^3}{2 x^2+x^3+x^4} \, dx &=\int \frac{1+x^2+x^3}{x^2 \left (2+x+x^2\right )} \, dx\\ &=\int \left (\frac{1}{2 x^2}-\frac{1}{4 x}+\frac{3+5 x}{4 \left (2+x+x^2\right )}\right ) \, dx\\ &=-\frac{1}{2 x}-\frac{\log (x)}{4}+\frac{1}{4} \int \frac{3+5 x}{2+x+x^2} \, dx\\ &=-\frac{1}{2 x}-\frac{\log (x)}{4}+\frac{1}{8} \int \frac{1}{2+x+x^2} \, dx+\frac{5}{8} \int \frac{1+2 x}{2+x+x^2} \, dx\\ &=-\frac{1}{2 x}-\frac{\log (x)}{4}+\frac{5}{8} \log \left (2+x+x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,1+2 x\right )\\ &=-\frac{1}{2 x}+\frac{\tan ^{-1}\left (\frac{1+2 x}{\sqrt{7}}\right )}{4 \sqrt{7}}-\frac{\log (x)}{4}+\frac{5}{8} \log \left (2+x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0253046, size = 46, normalized size = 1. $\frac{5}{8} \log \left (x^2+x+2\right )-\frac{1}{2 x}-\frac{\log (x)}{4}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{7}}\right )}{4 \sqrt{7}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(2*x) + ArcTan[(1 + 2*x)/Sqrt[7]]/(4*Sqrt[7]) - Log[x]/4 + (5*Log[2 + x + x^2])/8

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Maple [A]  time = 0.007, size = 36, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,x}}-{\frac{\ln \left ( x \right ) }{4}}+{\frac{5\,\ln \left ({x}^{2}+x+2 \right ) }{8}}+{\frac{\sqrt{7}}{28}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{7}}{7}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x-1/4*ln(x)+5/8*ln(x^2+x+2)+1/28*arctan(1/7*(1+2*x)*7^(1/2))*7^(1/2)

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Maxima [A]  time = 1.41136, size = 47, normalized size = 1.02 \begin{align*} \frac{1}{28} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (2 \, x + 1\right )}\right ) - \frac{1}{2 \, x} + \frac{5}{8} \, \log \left (x^{2} + x + 2\right ) - \frac{1}{4} \, \log \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/28*sqrt(7)*arctan(1/7*sqrt(7)*(2*x + 1)) - 1/2/x + 5/8*log(x^2 + x + 2) - 1/4*log(x)

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Fricas [A]  time = 1.9602, size = 128, normalized size = 2.78 \begin{align*} \frac{2 \, \sqrt{7} x \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (2 \, x + 1\right )}\right ) + 35 \, x \log \left (x^{2} + x + 2\right ) - 14 \, x \log \left (x\right ) - 28}{56 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/56*(2*sqrt(7)*x*arctan(1/7*sqrt(7)*(2*x + 1)) + 35*x*log(x^2 + x + 2) - 14*x*log(x) - 28)/x

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Sympy [A]  time = 0.149009, size = 46, normalized size = 1. \begin{align*} - \frac{\log{\left (x \right )}}{4} + \frac{5 \log{\left (x^{2} + x + 2 \right )}}{8} + \frac{\sqrt{7} \operatorname{atan}{\left (\frac{2 \sqrt{7} x}{7} + \frac{\sqrt{7}}{7} \right )}}{28} - \frac{1}{2 x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x)/4 + 5*log(x**2 + x + 2)/8 + sqrt(7)*atan(2*sqrt(7)*x/7 + sqrt(7)/7)/28 - 1/(2*x)

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Giac [A]  time = 1.06356, size = 49, normalized size = 1.07 \begin{align*} \frac{1}{28} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (2 \, x + 1\right )}\right ) - \frac{1}{2 \, x} + \frac{5}{8} \, \log \left (x^{2} + x + 2\right ) - \frac{1}{4} \, \log \left ({\left | x \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/28*sqrt(7)*arctan(1/7*sqrt(7)*(2*x + 1)) - 1/2/x + 5/8*log(x^2 + x + 2) - 1/4*log(abs(x))