3.200 \(\int \frac {-3+2 x}{(3+6 x+2 x^2)^3} \, dx\)

Optimal. Leaf size=61 \[ -\frac {2 x+3}{2 \left (2 x^2+6 x+3\right )}+\frac {4 x+5}{4 \left (2 x^2+6 x+3\right )^2}+\frac {\tanh ^{-1}\left (\frac {2 x+3}{\sqrt {3}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.02, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {638, 614, 618, 206} \[ -\frac {2 x+3}{2 \left (2 x^2+6 x+3\right )}+\frac {4 x+5}{4 \left (2 x^2+6 x+3\right )^2}+\frac {\tanh ^{-1}\left (\frac {2 x+3}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(-3 + 2*x)/(3 + 6*x + 2*x^2)^3,x]

[Out]

(5 + 4*x)/(4*(3 + 6*x + 2*x^2)^2) - (3 + 2*x)/(2*(3 + 6*x + 2*x^2)) + ArcTanh[(3 + 2*x)/Sqrt[3]]/Sqrt[3]

Rule 206

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

Rule 614

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 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 638

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

Rubi steps

\begin {align*} \int \frac {-3+2 x}{\left (3+6 x+2 x^2\right )^3} \, dx &=\frac {5+4 x}{4 \left (3+6 x+2 x^2\right )^2}+3 \int \frac {1}{\left (3+6 x+2 x^2\right )^2} \, dx\\ &=\frac {5+4 x}{4 \left (3+6 x+2 x^2\right )^2}-\frac {3+2 x}{2 \left (3+6 x+2 x^2\right )}-\int \frac {1}{3+6 x+2 x^2} \, dx\\ &=\frac {5+4 x}{4 \left (3+6 x+2 x^2\right )^2}-\frac {3+2 x}{2 \left (3+6 x+2 x^2\right )}+2 \operatorname {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,6+4 x\right )\\ &=\frac {5+4 x}{4 \left (3+6 x+2 x^2\right )^2}-\frac {3+2 x}{2 \left (3+6 x+2 x^2\right )}+\frac {\tanh ^{-1}\left (\frac {3+2 x}{\sqrt {3}}\right )}{\sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 70, normalized size = 1.15 \[ \frac {1}{12} \left (-\frac {3 \left (8 x^3+36 x^2+44 x+13\right )}{\left (2 x^2+6 x+3\right )^2}-2 \sqrt {3} \log \left (-2 x+\sqrt {3}-3\right )+2 \sqrt {3} \log \left (2 x+\sqrt {3}+3\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*(13 + 44*x + 36*x^2 + 8*x^3))/(3 + 6*x + 2*x^2)^2 - 2*Sqrt[3]*Log[-3 + Sqrt[3] - 2*x] + 2*Sqrt[3]*Log[3 +
Sqrt[3] + 2*x])/12

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IntegrateAlgebraic [A]  time = 0.06, size = 53, normalized size = 0.87 \[ \frac {-8 x^3-36 x^2-44 x-13}{4 \left (2 x^2+6 x+3\right )^2}+\frac {\tanh ^{-1}\left (\frac {2 x}{\sqrt {3}}+\sqrt {3}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-3 + 2*x)/(3 + 6*x + 2*x^2)^3,x]

[Out]

(-13 - 44*x - 36*x^2 - 8*x^3)/(4*(3 + 6*x + 2*x^2)^2) + ArcTanh[Sqrt[3] + (2*x)/Sqrt[3]]/Sqrt[3]

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fricas [A]  time = 1.15, size = 97, normalized size = 1.59 \[ -\frac {24 \, x^{3} - 2 \, \sqrt {3} {\left (4 \, x^{4} + 24 \, x^{3} + 48 \, x^{2} + 36 \, x + 9\right )} \log \left (\frac {2 \, x^{2} + \sqrt {3} {\left (2 \, x + 3\right )} + 6 \, x + 6}{2 \, x^{2} + 6 \, x + 3}\right ) + 108 \, x^{2} + 132 \, x + 39}{12 \, {\left (4 \, x^{4} + 24 \, x^{3} + 48 \, x^{2} + 36 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(24*x^3 - 2*sqrt(3)*(4*x^4 + 24*x^3 + 48*x^2 + 36*x + 9)*log((2*x^2 + sqrt(3)*(2*x + 3) + 6*x + 6)/(2*x^
2 + 6*x + 3)) + 108*x^2 + 132*x + 39)/(4*x^4 + 24*x^3 + 48*x^2 + 36*x + 9)

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giac [A]  time = 1.16, size = 61, normalized size = 1.00 \[ -\frac {1}{6} \, \sqrt {3} \log \left (\frac {{\left | 4 \, x - 2 \, \sqrt {3} + 6 \right |}}{{\left | 4 \, x + 2 \, \sqrt {3} + 6 \right |}}\right ) - \frac {8 \, x^{3} + 36 \, x^{2} + 44 \, x + 13}{4 \, {\left (2 \, x^{2} + 6 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*log(abs(4*x - 2*sqrt(3) + 6)/abs(4*x + 2*sqrt(3) + 6)) - 1/4*(8*x^3 + 36*x^2 + 44*x + 13)/(2*x^2
+ 6*x + 3)^2

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maple [A]  time = 0.30, size = 56, normalized size = 0.92

method	result	size
default	\(-\frac {-24 x -30}{24 \left (2 x^{2}+6 x +3\right )^{2}}-\frac {4 x +6}{4 \left (2 x^{2}+6 x +3\right )}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (4 x +6\right ) \sqrt {3}}{6}\right )}{3}\)	\(56\)
risch	\(\frac {-2 x^{3}-9 x^{2}-11 x -\frac {13}{4}}{\left (2 x^{2}+6 x +3\right )^{2}}+\frac {\ln \left (3+2 x +\sqrt {3}\right ) \sqrt {3}}{6}-\frac {\ln \left (3+2 x -\sqrt {3}\right ) \sqrt {3}}{6}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3+2*x)/(2*x^2+6*x+3)^3,x,method=_RETURNVERBOSE)

[Out]

-1/24*(-24*x-30)/(2*x^2+6*x+3)^2-1/4*(4*x+6)/(2*x^2+6*x+3)+1/3*3^(1/2)*arctanh(1/6*(4*x+6)*3^(1/2))

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maxima [A]  time = 1.26, size = 67, normalized size = 1.10 \[ -\frac {1}{6} \, \sqrt {3} \log \left (\frac {2 \, x - \sqrt {3} + 3}{2 \, x + \sqrt {3} + 3}\right ) - \frac {8 \, x^{3} + 36 \, x^{2} + 44 \, x + 13}{4 \, {\left (4 \, x^{4} + 24 \, x^{3} + 48 \, x^{2} + 36 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*log((2*x - sqrt(3) + 3)/(2*x + sqrt(3) + 3)) - 1/4*(8*x^3 + 36*x^2 + 44*x + 13)/(4*x^4 + 24*x^3 +
48*x^2 + 36*x + 9)

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mupad [B]  time = 0.21, size = 53, normalized size = 0.87 \[ \frac {\sqrt {3}\,\mathrm {atanh}\left (\sqrt {3}\,\left (\frac {2\,x}{3}+1\right )\right )}{3}-\frac {\frac {x^3}{2}+\frac {9\,x^2}{4}+\frac {11\,x}{4}+\frac {13}{16}}{x^4+6\,x^3+12\,x^2+9\,x+\frac {9}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - 3)/(6*x + 2*x^2 + 3)^3,x)

[Out]

(3^(1/2)*atanh(3^(1/2)*((2*x)/3 + 1)))/3 - ((11*x)/4 + (9*x^2)/4 + x^3/2 + 13/16)/(9*x + 12*x^2 + 6*x^3 + x^4
+ 9/4)

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sympy [A]  time = 0.17, size = 76, normalized size = 1.25 \[ \frac {- 8 x^{3} - 36 x^{2} - 44 x - 13}{16 x^{4} + 96 x^{3} + 192 x^{2} + 144 x + 36} - \frac {\sqrt {3} \log {\left (x - \frac {\sqrt {3}}{2} + \frac {3}{2} \right )}}{6} + \frac {\sqrt {3} \log {\left (x + \frac {\sqrt {3}}{2} + \frac {3}{2} \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-8*x**3 - 36*x**2 - 44*x - 13)/(16*x**4 + 96*x**3 + 192*x**2 + 144*x + 36) - sqrt(3)*log(x - sqrt(3)/2 + 3/2)
/6 + sqrt(3)*log(x + sqrt(3)/2 + 3/2)/6

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