3.271 \(\int x^2 \sqrt{1+x+x^2} \, dx\)

Optimal. Leaf size=65 \[ \frac{1}{4} x \left (x^2+x+1\right )^{3/2}-\frac{5}{24} \left (x^2+x+1\right )^{3/2}+\frac{1}{64} (2 x+1) \sqrt{x^2+x+1}+\frac{3}{128} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

[Out]

((1 + 2*x)*Sqrt[1 + x + x^2])/64 - (5*(1 + x + x^2)^(3/2))/24 + (x*(1 + x + x^2)^(3/2))/4 + (3*ArcSinh[(1 + 2*
x)/Sqrt[3]])/128

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Rubi [A]  time = 0.0242333, antiderivative size = 65, 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 = {742, 640, 612, 619, 215} \[ \frac{1}{4} x \left (x^2+x+1\right )^{3/2}-\frac{5}{24} \left (x^2+x+1\right )^{3/2}+\frac{1}{64} (2 x+1) \sqrt{x^2+x+1}+\frac{3}{128} \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[1 + x + x^2],x]

[Out]

((1 + 2*x)*Sqrt[1 + x + x^2])/64 - (5*(1 + x + x^2)^(3/2))/24 + (x*(1 + x + x^2)^(3/2))/4 + (3*ArcSinh[(1 + 2*
x)/Sqrt[3]])/128

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 640

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

Rule 612

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

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 x^2 \sqrt{1+x+x^2} \, dx &=\frac{1}{4} x \left (1+x+x^2\right )^{3/2}+\frac{1}{4} \int \left (-1-\frac{5 x}{2}\right ) \sqrt{1+x+x^2} \, dx\\ &=-\frac{5}{24} \left (1+x+x^2\right )^{3/2}+\frac{1}{4} x \left (1+x+x^2\right )^{3/2}+\frac{1}{16} \int \sqrt{1+x+x^2} \, dx\\ &=\frac{1}{64} (1+2 x) \sqrt{1+x+x^2}-\frac{5}{24} \left (1+x+x^2\right )^{3/2}+\frac{1}{4} x \left (1+x+x^2\right )^{3/2}+\frac{3}{128} \int \frac{1}{\sqrt{1+x+x^2}} \, dx\\ &=\frac{1}{64} (1+2 x) \sqrt{1+x+x^2}-\frac{5}{24} \left (1+x+x^2\right )^{3/2}+\frac{1}{4} x \left (1+x+x^2\right )^{3/2}+\frac{1}{128} \sqrt{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{3}}} \, dx,x,1+2 x\right )\\ &=\frac{1}{64} (1+2 x) \sqrt{1+x+x^2}-\frac{5}{24} \left (1+x+x^2\right )^{3/2}+\frac{1}{4} x \left (1+x+x^2\right )^{3/2}+\frac{3}{128} \sinh ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0207481, size = 46, normalized size = 0.71 \[ \frac{1}{384} \left (2 \sqrt{x^2+x+1} \left (48 x^3+8 x^2+14 x-37\right )+9 \sinh ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[1 + x + x^2],x]

[Out]

(2*Sqrt[1 + x + x^2]*(-37 + 14*x + 8*x^2 + 48*x^3) + 9*ArcSinh[(1 + 2*x)/Sqrt[3]])/384

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Maple [A]  time = 0.005, size = 49, normalized size = 0.8 \begin{align*}{\frac{x}{4} \left ({x}^{2}+x+1 \right ) ^{{\frac{3}{2}}}}-{\frac{5}{24} \left ({x}^{2}+x+1 \right ) ^{{\frac{3}{2}}}}+{\frac{1+2\,x}{64}\sqrt{{x}^{2}+x+1}}+{\frac{3}{128}{\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^2*(x^2+x+1)^(1/2),x)

[Out]

1/4*x*(x^2+x+1)^(3/2)-5/24*(x^2+x+1)^(3/2)+1/64*(1+2*x)*(x^2+x+1)^(1/2)+3/128*arcsinh(2/3*3^(1/2)*(x+1/2))

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Maxima [A]  time = 1.43718, size = 76, normalized size = 1.17 \begin{align*} \frac{1}{4} \,{\left (x^{2} + x + 1\right )}^{\frac{3}{2}} x - \frac{5}{24} \,{\left (x^{2} + x + 1\right )}^{\frac{3}{2}} + \frac{1}{32} \, \sqrt{x^{2} + x + 1} x + \frac{1}{64} \, \sqrt{x^{2} + x + 1} + \frac{3}{128} \, \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^2*(x^2+x+1)^(1/2),x, algorithm="maxima")

[Out]

1/4*(x^2 + x + 1)^(3/2)*x - 5/24*(x^2 + x + 1)^(3/2) + 1/32*sqrt(x^2 + x + 1)*x + 1/64*sqrt(x^2 + x + 1) + 3/1
28*arcsinh(1/3*sqrt(3)*(2*x + 1))

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Fricas [A]  time = 2.02652, size = 132, normalized size = 2.03 \begin{align*} \frac{1}{192} \,{\left (48 \, x^{3} + 8 \, x^{2} + 14 \, x - 37\right )} \sqrt{x^{2} + x + 1} - \frac{3}{128} \, \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^2*(x^2+x+1)^(1/2),x, algorithm="fricas")

[Out]

1/192*(48*x^3 + 8*x^2 + 14*x - 37)*sqrt(x^2 + x + 1) - 3/128*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 x^{2} \sqrt{x^{2} + x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(x**2 + x + 1), x)

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Giac [A]  time = 1.06932, size = 59, normalized size = 0.91 \begin{align*} \frac{1}{192} \,{\left (2 \,{\left (4 \,{\left (6 \, x + 1\right )} x + 7\right )} x - 37\right )} \sqrt{x^{2} + x + 1} - \frac{3}{128} \, \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^2*(x^2+x+1)^(1/2),x, algorithm="giac")

[Out]

1/192*(2*(4*(6*x + 1)*x + 7)*x - 37)*sqrt(x^2 + x + 1) - 3/128*log(-2*x + 2*sqrt(x^2 + x + 1) - 1)