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

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

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Rubi [A]  time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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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IntegrateAlgebraic [A]  time = 0.09, size = 52, normalized size = 0.80 \[ \frac {1}{192} \sqrt {x^2+x+1} \left (48 x^3+8 x^2+14 x-37\right )-\frac {3}{128} \log \left (2 \sqrt {x^2+x+1}-2 x-1\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*Sqrt[1 + x + x^2],x]

[Out]

(Sqrt[1 + x + x^2]*(-37 + 14*x + 8*x^2 + 48*x^3))/192 - (3*Log[-1 - 2*x + 2*Sqrt[1 + x + x^2]])/128

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fricas [A]  time = 0.71, size = 44, normalized size = 0.68 \[ \frac {1}{192} \, {\left (48 \, x^{3} + 8 \, x^{2} + 14 \, x - 37\right )} \sqrt {x^{2} + x + 1} - \frac {3}{128} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]

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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giac [A]  time = 0.65, size = 44, normalized size = 0.68 \[ \frac {1}{192} \, {\left (2 \, {\left (4 \, {\left (6 \, x + 1\right )} x + 7\right )} x - 37\right )} \sqrt {x^{2} + x + 1} - \frac {3}{128} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]

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)

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maple [A]  time = 0.35, size = 38, normalized size = 0.58

method	result	size
risch	\(\frac {\left (48 x^{3}+8 x^{2}+14 x -37\right ) \sqrt {x^{2}+x +1}}{192}+\frac {3 \arcsinh \left (\frac {2 \left (\frac {1}{2}+x \right ) \sqrt {3}}{3}\right )}{128}\)	\(38\)
trager	\(\left (\frac {1}{4} x^{3}+\frac {1}{24} x^{2}+\frac {7}{96} x -\frac {37}{192}\right ) \sqrt {x^{2}+x +1}-\frac {3 \ln \left (2 \sqrt {x^{2}+x +1}-1-2 x \right )}{128}\)	\(44\)
default	\(\frac {x \left (x^{2}+x +1\right )^{\frac {3}{2}}}{4}-\frac {5 \left (x^{2}+x +1\right )^{\frac {3}{2}}}{24}+\frac {\left (1+2 x \right ) \sqrt {x^{2}+x +1}}{64}+\frac {3 \arcsinh \left (\frac {2 \left (\frac {1}{2}+x \right ) \sqrt {3}}{3}\right )}{128}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(48*x^3+8*x^2+14*x-37)*(x^2+x+1)^(1/2)+3/128*arcsinh(2/3*(1/2+x)*3^(1/2))

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maxima [A]  time = 1.16, size = 56, normalized size = 0.86 \[ \frac {1}{4} \, {\left (x^{2} + x + 1\right )}^{\frac {3}{2}} x - \frac {5}{24} \, {\left (x^{2} + x + 1\right )}^{\frac {3}{2}} + \frac {1}{32} \, \sqrt {x^{2} + x + 1} x + \frac {1}{64} \, \sqrt {x^{2} + x + 1} + \frac {3}{128} \, \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) \]

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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mupad [B]  time = 0.13, size = 61, normalized size = 0.94 \[ \frac {3\,\ln \left (x+\sqrt {x^2+x+1}+\frac {1}{2}\right )}{128}-\frac {\left (\frac {x}{2}+\frac {1}{4}\right )\,\sqrt {x^2+x+1}}{4}-\frac {5\,\left (8\,x^2+2\,x+5\right )\,\sqrt {x^2+x+1}}{192}+\frac {x\,{\left (x^2+x+1\right )}^{3/2}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x + x^2 + 1)^(1/2),x)

[Out]

(3*log(x + (x + x^2 + 1)^(1/2) + 1/2))/128 - ((x/2 + 1/4)*(x + x^2 + 1)^(1/2))/4 - (5*(2*x + 8*x^2 + 5)*(x + x
^2 + 1)^(1/2))/192 + (x*(x + x^2 + 1)^(3/2))/4

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sqrt {x^{2} + x + 1}\, dx \]

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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