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

Optimal. Leaf size=56 \[ -\frac{2 (x+2) x^2}{3 \sqrt{x^2+x+1}}+\frac{1}{3} (2 x+5) \sqrt{x^2+x+1}-\frac{3}{2} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0240697, antiderivative size = 56, normalized size of antiderivative = 1., 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 = {738, 779, 619, 215} \[ -\frac{2 (x+2) x^2}{3 \sqrt{x^2+x+1}}+\frac{1}{3} (2 x+5) \sqrt{x^2+x+1}-\frac{3}{2} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3/(1 + x + x^2)^(3/2),x]

[Out]

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

Rule 738

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

Rule 779

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

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

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

Rubi steps

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

Mathematica [A]  time = 0.0118343, size = 48, normalized size = 0.86 \[ \frac{6 x^2-9 \sqrt{x^2+x+1} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )+14 x+10}{6 \sqrt{x^2+x+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 + x + x^2)^(3/2),x]

[Out]

(10 + 14*x + 6*x^2 - 9*Sqrt[1 + x + x^2]*ArcSinh[(1 + 2*x)/Sqrt[3]])/(6*Sqrt[1 + x + x^2])

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Maple [A]  time = 0.004, size = 61, normalized size = 1.1 \begin{align*}{{x}^{2}{\frac{1}{\sqrt{{x}^{2}+x+1}}}}+{\frac{3\,x}{2}{\frac{1}{\sqrt{{x}^{2}+x+1}}}}+{\frac{5}{4}{\frac{1}{\sqrt{{x}^{2}+x+1}}}}+{\frac{5+10\,x}{12}{\frac{1}{\sqrt{{x}^{2}+x+1}}}}-{\frac{3}{2}{\it Arcsinh} \left ({\frac{2\,\sqrt{3}}{3} \left ( x+{\frac{1}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2/(x^2+x+1)^(1/2)+3/2*x/(x^2+x+1)^(1/2)+5/4/(x^2+x+1)^(1/2)+5/12*(1+2*x)/(x^2+x+1)^(1/2)-3/2*arcsinh(2/3*3^(
1/2)*(x+1/2))

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Maxima [A]  time = 1.43428, size = 63, normalized size = 1.12 \begin{align*} \frac{x^{2}}{\sqrt{x^{2} + x + 1}} + \frac{7 \, x}{3 \, \sqrt{x^{2} + x + 1}} + \frac{5}{3 \, \sqrt{x^{2} + x + 1}} - \frac{3}{2} \, \operatorname{arsinh}\left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.04525, size = 184, normalized size = 3.29 \begin{align*} \frac{19 \, x^{2} + 18 \,{\left (x^{2} + x + 1\right )} \log \left (-2 \, x + 2 \, \sqrt{x^{2} + x + 1} - 1\right ) + 4 \,{\left (3 \, x^{2} + 7 \, x + 5\right )} \sqrt{x^{2} + x + 1} + 19 \, x + 19}{12 \,{\left (x^{2} + x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(19*x^2 + 18*(x^2 + x + 1)*log(-2*x + 2*sqrt(x^2 + x + 1) - 1) + 4*(3*x^2 + 7*x + 5)*sqrt(x^2 + x + 1) +
19*x + 19)/(x^2 + x + 1)

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

[Out]

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

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Giac [A]  time = 1.07698, size = 51, normalized size = 0.91 \begin{align*} \frac{{\left (3 \, x + 7\right )} x + 5}{3 \, \sqrt{x^{2} + x + 1}} + \frac{3}{2} \, \log \left (-2 \, x + 2 \, \sqrt{x^{2} + x + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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