3.1.92 \(\int \frac {1}{(2+4 x-3 x^2)^2} \, dx\) [92]

Optimal. Leaf size=43 \[ -\frac {2-3 x}{20 \left (2+4 x-3 x^2\right )}-\frac {3 \tanh ^{-1}\left (\frac {2-3 x}{\sqrt {10}}\right )}{20 \sqrt {10}} \]

[Out]

1/20*(-2+3*x)/(-3*x^2+4*x+2)-3/200*arctanh(1/10*(2-3*x)*10^(1/2))*10^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {628, 632, 212} \begin {gather*} -\frac {2-3 x}{20 \left (-3 x^2+4 x+2\right )}-\frac {3 \tanh ^{-1}\left (\frac {2-3 x}{\sqrt {10}}\right )}{20 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 4*x - 3*x^2)^(-2),x]

[Out]

-1/20*(2 - 3*x)/(2 + 4*x - 3*x^2) - (3*ArcTanh[(2 - 3*x)/Sqrt[10]])/(20*Sqrt[10])

Rule 212

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

Rule 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (2+4 x-3 x^2\right )^2} \, dx &=-\frac {2-3 x}{20 \left (2+4 x-3 x^2\right )}+\frac {3}{20} \int \frac {1}{2+4 x-3 x^2} \, dx\\ &=-\frac {2-3 x}{20 \left (2+4 x-3 x^2\right )}-\frac {3}{10} \text {Subst}\left (\int \frac {1}{40-x^2} \, dx,x,4-6 x\right )\\ &=-\frac {2-3 x}{20 \left (2+4 x-3 x^2\right )}-\frac {3 \tanh ^{-1}\left (\frac {2-3 x}{\sqrt {10}}\right )}{20 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 62, normalized size = 1.44 \begin {gather*} \frac {2-3 x}{20 \left (-2-4 x+3 x^2\right )}-\frac {3 \log \left (2+\sqrt {10}-3 x\right )}{40 \sqrt {10}}+\frac {3 \log \left (-2+\sqrt {10}+3 x\right )}{40 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 4*x - 3*x^2)^(-2),x]

[Out]

(2 - 3*x)/(20*(-2 - 4*x + 3*x^2)) - (3*Log[2 + Sqrt[10] - 3*x])/(40*Sqrt[10]) + (3*Log[-2 + Sqrt[10] + 3*x])/(
40*Sqrt[10])

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Maple [A]
time = 0.56, size = 37, normalized size = 0.86

method	result	size
default	\(-\frac {6 x -4}{40 \left (3 x^{2}-4 x -2\right )}+\frac {3 \sqrt {10}\, \arctanh \left (\frac {\left (6 x -4\right ) \sqrt {10}}{20}\right )}{200}\)	\(37\)
risch	\(\frac {-\frac {x}{20}+\frac {1}{30}}{x^{2}-\frac {4}{3} x -\frac {2}{3}}+\frac {3 \sqrt {10}\, \ln \left (3 x -2+\sqrt {10}\right )}{400}-\frac {3 \sqrt {10}\, \ln \left (3 x -2-\sqrt {10}\right )}{400}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-3*x^2+4*x+2)^2,x,method=_RETURNVERBOSE)

[Out]

-1/40*(6*x-4)/(3*x^2-4*x-2)+3/200*10^(1/2)*arctanh(1/20*(6*x-4)*10^(1/2))

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Maxima [A]
time = 0.49, size = 47, normalized size = 1.09 \begin {gather*} -\frac {3}{400} \, \sqrt {10} \log \left (\frac {3 \, x - \sqrt {10} - 2}{3 \, x + \sqrt {10} - 2}\right ) - \frac {3 \, x - 2}{20 \, {\left (3 \, x^{2} - 4 \, x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/400*sqrt(10)*log((3*x - sqrt(10) - 2)/(3*x + sqrt(10) - 2)) - 1/20*(3*x - 2)/(3*x^2 - 4*x - 2)

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Fricas [A]
time = 1.71, size = 68, normalized size = 1.58 \begin {gather*} \frac {3 \, \sqrt {10} {\left (3 \, x^{2} - 4 \, x - 2\right )} \log \left (\frac {9 \, x^{2} + 2 \, \sqrt {10} {\left (3 \, x - 2\right )} - 12 \, x + 14}{3 \, x^{2} - 4 \, x - 2}\right ) - 60 \, x + 40}{400 \, {\left (3 \, x^{2} - 4 \, x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/400*(3*sqrt(10)*(3*x^2 - 4*x - 2)*log((9*x^2 + 2*sqrt(10)*(3*x - 2) - 12*x + 14)/(3*x^2 - 4*x - 2)) - 60*x +
40)/(3*x^2 - 4*x - 2)

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Sympy [A]
time = 0.07, size = 58, normalized size = 1.35 \begin {gather*} \frac {2 - 3 x}{60 x^{2} - 80 x - 40} + \frac {3 \sqrt {10} \log {\left (x - \frac {2}{3} + \frac {\sqrt {10}}{3} \right )}}{400} - \frac {3 \sqrt {10} \log {\left (x - \frac {\sqrt {10}}{3} - \frac {2}{3} \right )}}{400} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2 - 3*x)/(60*x**2 - 80*x - 40) + 3*sqrt(10)*log(x - 2/3 + sqrt(10)/3)/400 - 3*sqrt(10)*log(x - sqrt(10)/3 - 2
/3)/400

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Giac [A]
time = 0.64, size = 51, normalized size = 1.19 \begin {gather*} -\frac {3}{400} \, \sqrt {10} \log \left (\frac {{\left | 6 \, x - 2 \, \sqrt {10} - 4 \right |}}{{\left | 6 \, x + 2 \, \sqrt {10} - 4 \right |}}\right ) - \frac {3 \, x - 2}{20 \, {\left (3 \, x^{2} - 4 \, x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/400*sqrt(10)*log(abs(6*x - 2*sqrt(10) - 4)/abs(6*x + 2*sqrt(10) - 4)) - 1/20*(3*x - 2)/(3*x^2 - 4*x - 2)

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Mupad [B]
time = 0.16, size = 34, normalized size = 0.79 \begin {gather*} \frac {3\,\sqrt {10}\,\mathrm {atanh}\left (\sqrt {10}\,\left (\frac {3\,x}{10}-\frac {1}{5}\right )\right )}{200}+\frac {\frac {x}{20}-\frac {1}{30}}{-x^2+\frac {4\,x}{3}+\frac {2}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*x - 3*x^2 + 2)^2,x)

[Out]

(3*10^(1/2)*atanh(10^(1/2)*((3*x)/10 - 1/5)))/200 + (x/20 - 1/30)/((4*x)/3 - x^2 + 2/3)

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