3.3.76 \(\int \frac {1}{x^2 (1+x+x^2)^{3/2}} \, dx\) [276]

Optimal. Leaf size=62 \[ \frac {2 (1-x)}{3 x \sqrt {1+x+x^2}}-\frac {5 \sqrt {1+x+x^2}}{3 x}+\frac {3}{2} \tanh ^{-1}\left (\frac {2+x}{2 \sqrt {1+x+x^2}}\right ) \]

[Out]

3/2*arctanh(1/2*(2+x)/(x^2+x+1)^(1/2))+2/3*(1-x)/x/(x^2+x+1)^(1/2)-5/3*(x^2+x+1)^(1/2)/x

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Rubi [A]
time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {754, 820, 738, 212} \begin {gather*} \frac {2 (1-x)}{3 x \sqrt {x^2+x+1}}-\frac {5 \sqrt {x^2+x+1}}{3 x}+\frac {3}{2} \tanh ^{-1}\left (\frac {x+2}{2 \sqrt {x^2+x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(1 + x + x^2)^(3/2)),x]

[Out]

(2*(1 - x))/(3*x*Sqrt[1 + x + x^2]) - (5*Sqrt[1 + x + x^2])/(3*x) + (3*ArcTanh[(2 + x)/(2*Sqrt[1 + x + x^2])])
/2

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 738

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

Rule 754

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

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Dist[
(b*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[S
implify[m + 2*p + 3], 0]

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 45, normalized size = 0.73 \begin {gather*} \frac {-3-7 x-5 x^2}{3 x \sqrt {1+x+x^2}}-3 \tanh ^{-1}\left (x-\sqrt {1+x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 - 7*x - 5*x^2)/(3*x*Sqrt[1 + x + x^2]) - 3*ArcTanh[x - Sqrt[1 + x + x^2]]

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Maple [A]
time = 0.12, size = 56, normalized size = 0.90

method	result	size
risch	\(-\frac {5 x^{2}+7 x +3}{3 x \sqrt {x^{2}+x +1}}+\frac {3 \arctanh \left (\frac {2+x}{2 \sqrt {x^{2}+x +1}}\right )}{2}\)	\(41\)
trager	\(-\frac {5 x^{2}+7 x +3}{3 x \sqrt {x^{2}+x +1}}+\frac {3 \ln \left (\frac {2 \sqrt {x^{2}+x +1}+2+x}{x}\right )}{2}\)	\(45\)
default	\(-\frac {1}{x \sqrt {x^{2}+x +1}}-\frac {3}{2 \sqrt {x^{2}+x +1}}-\frac {5 \left (1+2 x \right )}{6 \sqrt {x^{2}+x +1}}+\frac {3 \arctanh \left (\frac {2+x}{2 \sqrt {x^{2}+x +1}}\right )}{2}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x/(x^2+x+1)^(1/2)-3/2/(x^2+x+1)^(1/2)-5/6*(1+2*x)/(x^2+x+1)^(1/2)+3/2*arctanh(1/2*(2+x)/(x^2+x+1)^(1/2))

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Maxima [A]
time = 2.35, size = 58, normalized size = 0.94 \begin {gather*} -\frac {5 \, x}{3 \, \sqrt {x^{2} + x + 1}} - \frac {7}{3 \, \sqrt {x^{2} + x + 1}} - \frac {1}{\sqrt {x^{2} + x + 1} x} + \frac {3}{2} \, \operatorname {arsinh}\left (\frac {\sqrt {3} x}{3 \, {\left | x \right |}} + \frac {2 \, \sqrt {3}}{3 \, {\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/3*x/sqrt(x^2 + x + 1) - 7/3/sqrt(x^2 + x + 1) - 1/(sqrt(x^2 + x + 1)*x) + 3/2*arcsinh(1/3*sqrt(3)*x/abs(x)
+ 2/3*sqrt(3)/abs(x))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (46) = 92\).
time = 0.51, size = 94, normalized size = 1.52 \begin {gather*} -\frac {10 \, x^{3} + 10 \, x^{2} - 9 \, {\left (x^{3} + x^{2} + x\right )} \log \left (-x + \sqrt {x^{2} + x + 1} + 1\right ) + 9 \, {\left (x^{3} + x^{2} + x\right )} \log \left (-x + \sqrt {x^{2} + x + 1} - 1\right ) + 2 \, {\left (5 \, x^{2} + 7 \, x + 3\right )} \sqrt {x^{2} + x + 1} + 10 \, x}{6 \, {\left (x^{3} + x^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(10*x^3 + 10*x^2 - 9*(x^3 + x^2 + x)*log(-x + sqrt(x^2 + x + 1) + 1) + 9*(x^3 + x^2 + x)*log(-x + sqrt(x^
2 + x + 1) - 1) + 2*(5*x^2 + 7*x + 3)*sqrt(x^2 + x + 1) + 10*x)/(x^3 + x^2 + x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (x^{2} + x + 1\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*(x**2 + x + 1)**(3/2)), x)

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Giac [A]
time = 1.51, size = 80, normalized size = 1.29 \begin {gather*} -\frac {2 \, {\left (x + 2\right )}}{3 \, \sqrt {x^{2} + x + 1}} + \frac {x - \sqrt {x^{2} + x + 1} + 2}{{\left (x - \sqrt {x^{2} + x + 1}\right )}^{2} - 1} + \frac {3}{2} \, \log \left ({\left | -x + \sqrt {x^{2} + x + 1} + 1 \right |}\right ) - \frac {3}{2} \, \log \left ({\left | -x + \sqrt {x^{2} + x + 1} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(x + 2)/sqrt(x^2 + x + 1) + (x - sqrt(x^2 + x + 1) + 2)/((x - sqrt(x^2 + x + 1))^2 - 1) + 3/2*log(abs(-x
+ sqrt(x^2 + x + 1) + 1)) - 3/2*log(abs(-x + sqrt(x^2 + x + 1) - 1))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x^2\,{\left (x^2+x+1\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(x + x^2 + 1)^(3/2)),x)

[Out]

int(1/(x^2*(x + x^2 + 1)^(3/2)), x)

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