3.95 \(\int \frac{4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx\)

Optimal. Leaf size=47 \[ x^3-\frac{x^2}{2}+\frac{1}{4} \log \left (2 x^2-x+1\right )-\frac{\tan ^{-1}\left (\frac{1-4 x}{\sqrt{7}}\right )}{2 \sqrt{7}} \]

[Out]

-x^2/2 + x^3 - ArcTan[(1 - 4*x)/Sqrt[7]]/(2*Sqrt[7]) + Log[1 - x + 2*x^2]/4

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Rubi [A]  time = 0.0623759, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {1594, 1628, 634, 618, 204, 628} \[ x^3-\frac{x^2}{2}+\frac{1}{4} \log \left (2 x^2-x+1\right )-\frac{\tan ^{-1}\left (\frac{1-4 x}{\sqrt{7}}\right )}{2 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

Int[(4*x^2 - 5*x^3 + 6*x^4)/(1 - x + 2*x^2),x]

[Out]

-x^2/2 + x^3 - ArcTan[(1 - 4*x)/Sqrt[7]]/(2*Sqrt[7]) + Log[1 - x + 2*x^2]/4

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{4 x^2-5 x^3+6 x^4}{1-x+2 x^2} \, dx &=\int \frac{x^2 \left (4-5 x+6 x^2\right )}{1-x+2 x^2} \, dx\\ &=\int \left (-x+3 x^2+\frac{x}{1-x+2 x^2}\right ) \, dx\\ &=-\frac{x^2}{2}+x^3+\int \frac{x}{1-x+2 x^2} \, dx\\ &=-\frac{x^2}{2}+x^3+\frac{1}{4} \int \frac{1}{1-x+2 x^2} \, dx+\frac{1}{4} \int \frac{-1+4 x}{1-x+2 x^2} \, dx\\ &=-\frac{x^2}{2}+x^3+\frac{1}{4} \log \left (1-x+2 x^2\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,-1+4 x\right )\\ &=-\frac{x^2}{2}+x^3-\frac{\tan ^{-1}\left (\frac{1-4 x}{\sqrt{7}}\right )}{2 \sqrt{7}}+\frac{1}{4} \log \left (1-x+2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0159531, size = 47, normalized size = 1. \[ x^3-\frac{x^2}{2}+\frac{1}{4} \log \left (2 x^2-x+1\right )+\frac{\tan ^{-1}\left (\frac{4 x-1}{\sqrt{7}}\right )}{2 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 - 5*x^3 + 6*x^4)/(1 - x + 2*x^2),x]

[Out]

-x^2/2 + x^3 + ArcTan[(-1 + 4*x)/Sqrt[7]]/(2*Sqrt[7]) + Log[1 - x + 2*x^2]/4

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Maple [A]  time = 0.004, size = 39, normalized size = 0.8 \begin{align*}{x}^{3}-{\frac{{x}^{2}}{2}}+{\frac{\ln \left ( 2\,{x}^{2}-x+1 \right ) }{4}}+{\frac{\sqrt{7}}{14}\arctan \left ({\frac{ \left ( -1+4\,x \right ) \sqrt{7}}{7}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^4-5*x^3+4*x^2)/(2*x^2-x+1),x)

[Out]

x^3-1/2*x^2+1/4*ln(2*x^2-x+1)+1/14*7^(1/2)*arctan(1/7*(-1+4*x)*7^(1/2))

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Maxima [A]  time = 1.41874, size = 51, normalized size = 1.09 \begin{align*} x^{3} - \frac{1}{2} \, x^{2} + \frac{1}{14} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x - 1\right )}\right ) + \frac{1}{4} \, \log \left (2 \, x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4-5*x^3+4*x^2)/(2*x^2-x+1),x, algorithm="maxima")

[Out]

x^3 - 1/2*x^2 + 1/14*sqrt(7)*arctan(1/7*sqrt(7)*(4*x - 1)) + 1/4*log(2*x^2 - x + 1)

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Fricas [A]  time = 1.85802, size = 115, normalized size = 2.45 \begin{align*} x^{3} - \frac{1}{2} \, x^{2} + \frac{1}{14} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x - 1\right )}\right ) + \frac{1}{4} \, \log \left (2 \, x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4-5*x^3+4*x^2)/(2*x^2-x+1),x, algorithm="fricas")

[Out]

x^3 - 1/2*x^2 + 1/14*sqrt(7)*arctan(1/7*sqrt(7)*(4*x - 1)) + 1/4*log(2*x^2 - x + 1)

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Sympy [A]  time = 0.10346, size = 46, normalized size = 0.98 \begin{align*} x^{3} - \frac{x^{2}}{2} + \frac{\log{\left (x^{2} - \frac{x}{2} + \frac{1}{2} \right )}}{4} + \frac{\sqrt{7} \operatorname{atan}{\left (\frac{4 \sqrt{7} x}{7} - \frac{\sqrt{7}}{7} \right )}}{14} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x**4-5*x**3+4*x**2)/(2*x**2-x+1),x)

[Out]

x**3 - x**2/2 + log(x**2 - x/2 + 1/2)/4 + sqrt(7)*atan(4*sqrt(7)*x/7 - sqrt(7)/7)/14

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Giac [A]  time = 1.04706, size = 51, normalized size = 1.09 \begin{align*} x^{3} - \frac{1}{2} \, x^{2} + \frac{1}{14} \, \sqrt{7} \arctan \left (\frac{1}{7} \, \sqrt{7}{\left (4 \, x - 1\right )}\right ) + \frac{1}{4} \, \log \left (2 \, x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4-5*x^3+4*x^2)/(2*x^2-x+1),x, algorithm="giac")

[Out]

x^3 - 1/2*x^2 + 1/14*sqrt(7)*arctan(1/7*sqrt(7)*(4*x - 1)) + 1/4*log(2*x^2 - x + 1)