3.277 \(\int \frac{1}{x^3 (1+x+x^2)^{3/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac{2 (1-x)}{3 x^2 \sqrt{x^2+x+1}}+\frac{37 \sqrt{x^2+x+1}}{12 x}-\frac{7 \sqrt{x^2+x+1}}{6 x^2}-\frac{3}{8} \tanh ^{-1}\left (\frac{x+2}{2 \sqrt{x^2+x+1}}\right ) \]

[Out]

(2*(1 - x))/(3*x^2*Sqrt[1 + x + x^2]) - (7*Sqrt[1 + x + x^2])/(6*x^2) + (37*Sqrt[1 + x + x^2])/(12*x) - (3*Arc
Tanh[(2 + x)/(2*Sqrt[1 + x + x^2])])/8

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Rubi [A]  time = 0.0371738, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {740, 834, 806, 724, 206} \[ \frac{2 (1-x)}{3 x^2 \sqrt{x^2+x+1}}+\frac{37 \sqrt{x^2+x+1}}{12 x}-\frac{7 \sqrt{x^2+x+1}}{6 x^2}-\frac{3}{8} \tanh ^{-1}\left (\frac{x+2}{2 \sqrt{x^2+x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + x + x^2)^(3/2)),x]

[Out]

(2*(1 - x))/(3*x^2*Sqrt[1 + x + x^2]) - (7*Sqrt[1 + x + x^2])/(6*x^2) + (37*Sqrt[1 + x + x^2])/(12*x) - (3*Arc
Tanh[(2 + x)/(2*Sqrt[1 + x + x^2])])/8

Rule 740

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 834

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))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((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[Sim
plify[m + 2*p + 3], 0]

Rule 724

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 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])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (1+x+x^2\right )^{3/2}} \, dx &=\frac{2 (1-x)}{3 x^2 \sqrt{1+x+x^2}}+\frac{2}{3} \int \frac{\frac{7}{2}-2 x}{x^3 \sqrt{1+x+x^2}} \, dx\\ &=\frac{2 (1-x)}{3 x^2 \sqrt{1+x+x^2}}-\frac{7 \sqrt{1+x+x^2}}{6 x^2}-\frac{1}{3} \int \frac{\frac{37}{4}+\frac{7 x}{2}}{x^2 \sqrt{1+x+x^2}} \, dx\\ &=\frac{2 (1-x)}{3 x^2 \sqrt{1+x+x^2}}-\frac{7 \sqrt{1+x+x^2}}{6 x^2}+\frac{37 \sqrt{1+x+x^2}}{12 x}+\frac{3}{8} \int \frac{1}{x \sqrt{1+x+x^2}} \, dx\\ &=\frac{2 (1-x)}{3 x^2 \sqrt{1+x+x^2}}-\frac{7 \sqrt{1+x+x^2}}{6 x^2}+\frac{37 \sqrt{1+x+x^2}}{12 x}-\frac{3}{4} \operatorname{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^2 \sqrt{1+x+x^2}}-\frac{7 \sqrt{1+x+x^2}}{6 x^2}+\frac{37 \sqrt{1+x+x^2}}{12 x}-\frac{3}{8} \tanh ^{-1}\left (\frac{2+x}{2 \sqrt{1+x+x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0130602, size = 65, normalized size = 0.82 \[ \frac{74 x^3+46 x^2-9 \sqrt{x^2+x+1} x^2 \tanh ^{-1}\left (\frac{x+2}{2 \sqrt{x^2+x+1}}\right )+30 x-12}{24 x^2 \sqrt{x^2+x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12 + 30*x + 46*x^2 + 74*x^3 - 9*x^2*Sqrt[1 + x + x^2]*ArcTanh[(2 + x)/(2*Sqrt[1 + x + x^2])])/(24*x^2*Sqrt[1
+ x + x^2])

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Maple [A]  time = 0.005, size = 69, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{x}^{2}}{\frac{1}{\sqrt{{x}^{2}+x+1}}}}+{\frac{5}{4\,x}{\frac{1}{\sqrt{{x}^{2}+x+1}}}}+{\frac{3}{8}{\frac{1}{\sqrt{{x}^{2}+x+1}}}}+{\frac{37+74\,x}{24}{\frac{1}{\sqrt{{x}^{2}+x+1}}}}-{\frac{3}{8}{\it Artanh} \left ({\frac{2+x}{2}{\frac{1}{\sqrt{{x}^{2}+x+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(x^2+x+1)^(3/2),x)

[Out]

-1/2/x^2/(x^2+x+1)^(1/2)+5/4/x/(x^2+x+1)^(1/2)+3/8/(x^2+x+1)^(1/2)+37/24*(1+2*x)/(x^2+x+1)^(1/2)-3/8*arctanh(1
/2*(2+x)/(x^2+x+1)^(1/2))

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Maxima [A]  time = 1.66985, size = 96, normalized size = 1.22 \begin{align*} \frac{37 \, x}{12 \, \sqrt{x^{2} + x + 1}} + \frac{23}{12 \, \sqrt{x^{2} + x + 1}} + \frac{5}{4 \, \sqrt{x^{2} + x + 1} x} - \frac{1}{2 \, \sqrt{x^{2} + x + 1} x^{2}} - \frac{3}{8} \, \operatorname{arsinh}\left (\frac{\sqrt{3} x}{3 \,{\left | x \right |}} + \frac{2 \, \sqrt{3}}{3 \,{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

37/12*x/sqrt(x^2 + x + 1) + 23/12/sqrt(x^2 + x + 1) + 5/4/(sqrt(x^2 + x + 1)*x) - 1/2/(sqrt(x^2 + x + 1)*x^2)
- 3/8*arcsinh(1/3*sqrt(3)*x/abs(x) + 2/3*sqrt(3)/abs(x))

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Fricas [A]  time = 2.11412, size = 284, normalized size = 3.59 \begin{align*} \frac{74 \, x^{4} + 74 \, x^{3} + 74 \, x^{2} - 9 \,{\left (x^{4} + x^{3} + x^{2}\right )} \log \left (-x + \sqrt{x^{2} + x + 1} + 1\right ) + 9 \,{\left (x^{4} + x^{3} + x^{2}\right )} \log \left (-x + \sqrt{x^{2} + x + 1} - 1\right ) + 2 \,{\left (37 \, x^{3} + 23 \, x^{2} + 15 \, x - 6\right )} \sqrt{x^{2} + x + 1}}{24 \,{\left (x^{4} + x^{3} + x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(74*x^4 + 74*x^3 + 74*x^2 - 9*(x^4 + x^3 + x^2)*log(-x + sqrt(x^2 + x + 1) + 1) + 9*(x^4 + x^3 + x^2)*log
(-x + sqrt(x^2 + x + 1) - 1) + 2*(37*x^3 + 23*x^2 + 15*x - 6)*sqrt(x^2 + x + 1))/(x^4 + x^3 + x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09846, size = 158, normalized size = 2. \begin{align*} \frac{2 \,{\left (2 \, x + 1\right )}}{3 \, \sqrt{x^{2} + x + 1}} - \frac{3 \,{\left (x - \sqrt{x^{2} + x + 1}\right )}^{3} + 8 \,{\left (x - \sqrt{x^{2} + x + 1}\right )}^{2} - 13 \, x + 13 \, \sqrt{x^{2} + x + 1} - 16}{4 \,{\left ({\left (x - \sqrt{x^{2} + x + 1}\right )}^{2} - 1\right )}^{2}} - \frac{3}{8} \, \log \left ({\left | -x + \sqrt{x^{2} + x + 1} + 1 \right |}\right ) + \frac{3}{8} \, \log \left ({\left | -x + \sqrt{x^{2} + x + 1} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(2*x + 1)/sqrt(x^2 + x + 1) - 1/4*(3*(x - sqrt(x^2 + x + 1))^3 + 8*(x - sqrt(x^2 + x + 1))^2 - 13*x + 13*s
qrt(x^2 + x + 1) - 16)/((x - sqrt(x^2 + x + 1))^2 - 1)^2 - 3/8*log(abs(-x + sqrt(x^2 + x + 1) + 1)) + 3/8*log(
abs(-x + sqrt(x^2 + x + 1) - 1))