3.2.58 \(\int \frac {2-3 x+4 x^2}{3-4 x+4 x^2} \, dx\) [158]

Optimal. Leaf size=38 \[ x+\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {1}{8} \log \left (3-4 x+4 x^2\right ) \]

[Out]

x+1/8*ln(4*x^2-4*x+3)+1/8*arctan(1/2*(1-2*x)*2^(1/2))*2^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1671, 648, 632, 210, 642} \begin {gather*} \frac {1}{8} \log \left (4 x^2-4 x+3\right )+x+\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {2}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + ArcTan[(1 - 2*x)/Sqrt[2]]/(4*Sqrt[2]) + Log[3 - 4*x + 4*x^2]/8

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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 642

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 648

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 1671

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 {2-3 x+4 x^2}{3-4 x+4 x^2} \, dx &=\int \left (1-\frac {1-x}{3-4 x+4 x^2}\right ) \, dx\\ &=x-\int \frac {1-x}{3-4 x+4 x^2} \, dx\\ &=x+\frac {1}{8} \int \frac {-4+8 x}{3-4 x+4 x^2} \, dx-\frac {1}{2} \int \frac {1}{3-4 x+4 x^2} \, dx\\ &=x+\frac {1}{8} \log \left (3-4 x+4 x^2\right )+\text {Subst}\left (\int \frac {1}{-32-x^2} \, dx,x,-4+8 x\right )\\ &=x-\frac {\tan ^{-1}\left (\frac {-1+2 x}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {1}{8} \log \left (3-4 x+4 x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 38, normalized size = 1.00 \begin {gather*} x-\frac {\tan ^{-1}\left (\frac {-1+2 x}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {1}{8} \log \left (3-4 x+4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - ArcTan[(-1 + 2*x)/Sqrt[2]]/(4*Sqrt[2]) + Log[3 - 4*x + 4*x^2]/8

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Mathics [A]
time = 1.94, size = 29, normalized size = 0.76 \begin {gather*} x-\frac {\sqrt {2} \text {ArcTan}\left [\frac {\sqrt {2} \left (-1+2 x\right )}{2}\right ]}{8}+\frac {\text {Log}\left [\frac {3}{4}-x+x^2\right ]}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(2 - 3*x + 4*x^2)/(3 - 4*x + 4*x^2),x]')

[Out]

x - Sqrt[2] ArcTan[Sqrt[2] (-1 + 2 x) / 2] / 8 + Log[3 / 4 - x + x ^ 2] / 8

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Maple [A]
time = 0.11, size = 32, normalized size = 0.84

method	result	size
default	\(x +\frac {\ln \left (4 x^{2}-4 x +3\right )}{8}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (8 x -4\right ) \sqrt {2}}{8}\right )}{8}\)	\(32\)
risch	\(x +\frac {\ln \left (4 x^{2}-4 x +3\right )}{8}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {2}}{2}\right )}{8}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2-3*x+2)/(4*x^2-4*x+3),x,method=_RETURNVERBOSE)

[Out]

x+1/8*ln(4*x^2-4*x+3)-1/8*2^(1/2)*arctan(1/8*(8*x-4)*2^(1/2))

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Maxima [A]
time = 0.35, size = 31, normalized size = 0.82 \begin {gather*} -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - 1\right )}\right ) + x + \frac {1}{8} \, \log \left (4 \, x^{2} - 4 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - 1)) + x + 1/8*log(4*x^2 - 4*x + 3)

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Fricas [A]
time = 0.33, size = 31, normalized size = 0.82 \begin {gather*} -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - 1\right )}\right ) + x + \frac {1}{8} \, \log \left (4 \, x^{2} - 4 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - 1)) + x + 1/8*log(4*x^2 - 4*x + 3)

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Sympy [A]
time = 0.06, size = 34, normalized size = 0.89 \begin {gather*} x + \frac {\log {\left (x^{2} - x + \frac {3}{4} \right )}}{8} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - \frac {\sqrt {2}}{2} \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + log(x**2 - x + 3/4)/8 - sqrt(2)*atan(sqrt(2)*x - sqrt(2)/2)/8

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Giac [A]
time = 0.00, size = 44, normalized size = 1.16 \begin {gather*} \frac {\ln \left (4 x^{2}-4 x+3\right )}{8}-\frac {\arctan \left (\frac {2 x-1}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {4}{4} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2-3*x+2)/(4*x^2-4*x+3),x)

[Out]

-1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - 1)) + x + 1/8*log(4*x^2 - 4*x + 3)

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Mupad [B]
time = 0.17, size = 30, normalized size = 0.79 \begin {gather*} x+\frac {\ln \left (x^2-x+\frac {3}{4}\right )}{8}-\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x-\frac {\sqrt {2}}{2}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + log(x^2 - x + 3/4)/8 - (2^(1/2)*atan(2^(1/2)*x - 2^(1/2)/2))/8

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