3.267 \(\int \frac{x^3}{\sqrt{1+x+x^2}} \, dx\)

Optimal. Leaf size=53 \[ \frac{1}{3} \sqrt{x^2+x+1} x^2-\frac{1}{24} (10 x+1) \sqrt{x^2+x+1}+\frac{7}{16} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

[Out]

(x^2*Sqrt[1 + x + x^2])/3 - ((1 + 10*x)*Sqrt[1 + x + x^2])/24 + (7*ArcSinh[(1 + 2*x)/Sqrt[3]])/16

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Rubi [A]  time = 0.023292, antiderivative size = 53, 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 = {742, 779, 619, 215} \[ \frac{1}{3} \sqrt{x^2+x+1} x^2-\frac{1}{24} (10 x+1) \sqrt{x^2+x+1}+\frac{7}{16} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[1 + x + x^2],x]

[Out]

(x^2*Sqrt[1 + x + x^2])/3 - ((1 + 10*x)*Sqrt[1 + x + x^2])/24 + (7*ArcSinh[(1 + 2*x)/Sqrt[3]])/16

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, 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]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && 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}{\sqrt{1+x+x^2}} \, dx &=\frac{1}{3} x^2 \sqrt{1+x+x^2}+\frac{1}{3} \int \frac{\left (-2-\frac{5 x}{2}\right ) x}{\sqrt{1+x+x^2}} \, dx\\ &=\frac{1}{3} x^2 \sqrt{1+x+x^2}-\frac{1}{24} (1+10 x) \sqrt{1+x+x^2}+\frac{7}{16} \int \frac{1}{\sqrt{1+x+x^2}} \, dx\\ &=\frac{1}{3} x^2 \sqrt{1+x+x^2}-\frac{1}{24} (1+10 x) \sqrt{1+x+x^2}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 x\right )}{16 \sqrt{3}}\\ &=\frac{1}{3} x^2 \sqrt{1+x+x^2}-\frac{1}{24} (1+10 x) \sqrt{1+x+x^2}+\frac{7}{16} \sinh ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0165027, size = 41, normalized size = 0.77 \[ \frac{1}{48} \left (2 \sqrt{x^2+x+1} \left (8 x^2-10 x-1\right )+21 \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[1 + x + x^2],x]

[Out]

(2*Sqrt[1 + x + x^2]*(-1 - 10*x + 8*x^2) + 21*ArcSinh[(1 + 2*x)/Sqrt[3]])/48

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Maple [A]  time = 0.005, size = 47, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{3}\sqrt{{x}^{2}+x+1}}-{\frac{5\,x}{12}\sqrt{{x}^{2}+x+1}}-{\frac{1}{24}\sqrt{{x}^{2}+x+1}}+{\frac{7}{16}{\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)^(1/2),x)

[Out]

1/3*x^2*(x^2+x+1)^(1/2)-5/12*x*(x^2+x+1)^(1/2)-1/24*(x^2+x+1)^(1/2)+7/16*arcsinh(2/3*3^(1/2)*(x+1/2))

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Maxima [A]  time = 1.44799, size = 65, normalized size = 1.23 \begin{align*} \frac{1}{3} \, \sqrt{x^{2} + x + 1} x^{2} - \frac{5}{12} \, \sqrt{x^{2} + x + 1} x - \frac{1}{24} \, \sqrt{x^{2} + x + 1} + \frac{7}{16} \, \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)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(x^2 + x + 1)*x^2 - 5/12*sqrt(x^2 + x + 1)*x - 1/24*sqrt(x^2 + x + 1) + 7/16*arcsinh(1/3*sqrt(3)*(2*x
+ 1))

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Fricas [A]  time = 2.12527, size = 116, normalized size = 2.19 \begin{align*} \frac{1}{24} \,{\left (8 \, x^{2} - 10 \, x - 1\right )} \sqrt{x^{2} + x + 1} - \frac{7}{16} \, \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)^(1/2),x, algorithm="fricas")

[Out]

1/24*(8*x^2 - 10*x - 1)*sqrt(x^2 + x + 1) - 7/16*log(-2*x + 2*sqrt(x^2 + x + 1) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/sqrt(x**2 + x + 1), x)

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Giac [A]  time = 1.07601, size = 53, normalized size = 1. \begin{align*} \frac{1}{24} \,{\left (2 \,{\left (4 \, x - 5\right )} x - 1\right )} \sqrt{x^{2} + x + 1} - \frac{7}{16} \, \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)^(1/2),x, algorithm="giac")

[Out]

1/24*(2*(4*x - 5)*x - 1)*sqrt(x^2 + x + 1) - 7/16*log(-2*x + 2*sqrt(x^2 + x + 1) - 1)