Integrand size = 14, antiderivative size = 65 \[ \int x^2 \sqrt {1+x+x^2} \, dx=\frac {1}{64} (1+2 x) \sqrt {1+x+x^2}-\frac {5}{24} \left (1+x+x^2\right )^{3/2}+\frac {1}{4} x \left (1+x+x^2\right )^{3/2}+\frac {3}{128} \text {arcsinh}\left (\frac {1+2 x}{\sqrt {3}}\right ) \] Output:
-5/24*(x^2+x+1)^(3/2)+1/4*x*(x^2+x+1)^(3/2)+3/128*arcsinh(1/3*(1+2*x)*3^(1 /2))+1/64*(1+2*x)*(x^2+x+1)^(1/2)
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80 \[ \int x^2 \sqrt {1+x+x^2} \, dx=\frac {1}{192} \sqrt {1+x+x^2} \left (-37+14 x+8 x^2+48 x^3\right )-\frac {3}{128} \log \left (-1-2 x+2 \sqrt {1+x+x^2}\right ) \] Input:
Integrate[x^2*Sqrt[1 + x + x^2],x]
Output:
(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
Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1166, 27, 1160, 1087, 1090, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \sqrt {x^2+x+1} \, dx\) |
\(\Big \downarrow \) 1166 |
\(\displaystyle \frac {1}{4} \int -\frac {1}{2} (5 x+2) \sqrt {x^2+x+1}dx+\frac {1}{4} x \left (x^2+x+1\right )^{3/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} x \left (x^2+x+1\right )^{3/2}-\frac {1}{8} \int (5 x+2) \sqrt {x^2+x+1}dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \int \sqrt {x^2+x+1}dx-\frac {5}{3} \left (x^2+x+1\right )^{3/2}\right )+\frac {1}{4} x \left (x^2+x+1\right )^{3/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (\frac {3}{8} \int \frac {1}{\sqrt {x^2+x+1}}dx+\frac {1}{4} \sqrt {x^2+x+1} (2 x+1)\right )-\frac {5}{3} \left (x^2+x+1\right )^{3/2}\right )+\frac {1}{4} x \left (x^2+x+1\right )^{3/2}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (\frac {1}{8} \sqrt {3} \int \frac {1}{\sqrt {\frac {1}{3} (2 x+1)^2+1}}d(2 x+1)+\frac {1}{4} \sqrt {x^2+x+1} (2 x+1)\right )-\frac {5}{3} \left (x^2+x+1\right )^{3/2}\right )+\frac {1}{4} x \left (x^2+x+1\right )^{3/2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{2} \left (\frac {3}{8} \text {arcsinh}\left (\frac {2 x+1}{\sqrt {3}}\right )+\frac {1}{4} \sqrt {x^2+x+1} (2 x+1)\right )-\frac {5}{3} \left (x^2+x+1\right )^{3/2}\right )+\frac {1}{4} x \left (x^2+x+1\right )^{3/2}\) |
Input:
Int[x^2*Sqrt[1 + x + x^2],x]
Output:
(x*(1 + x + x^2)^(3/2))/4 + ((-5*(1 + x + x^2)^(3/2))/3 + (((1 + 2*x)*Sqrt [1 + x + x^2])/4 + (3*ArcSinh[(1 + 2*x)/Sqrt[3]])/8)/2)/8
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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] - Simp[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] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[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]
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] + Simp[(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[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[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] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.58
method | result | size |
risch | \(\frac {\left (48 x^{3}+8 x^{2}+14 x -37\right ) \sqrt {x^{2}+x +1}}{192}+\frac {3 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{128}\) | \(38\) |
trager | \(\left (\frac {1}{4} x^{3}+\frac {1}{24} x^{2}+\frac {7}{96} x -\frac {37}{192}\right ) \sqrt {x^{2}+x +1}+\frac {3 \ln \left (1+2 x +2 \sqrt {x^{2}+x +1}\right )}{128}\) | \(44\) |
default | \(\frac {x \left (x^{2}+x +1\right )^{\frac {3}{2}}}{4}-\frac {5 \left (x^{2}+x +1\right )^{\frac {3}{2}}}{24}+\frac {\left (1+2 x \right ) \sqrt {x^{2}+x +1}}{64}+\frac {3 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{128}\) | \(49\) |
Input:
int(x^2*(x^2+x+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/192*(48*x^3+8*x^2+14*x-37)*(x^2+x+1)^(1/2)+3/128*arcsinh(2/3*3^(1/2)*(x+ 1/2))
Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68 \[ \int x^2 \sqrt {1+x+x^2} \, dx=\frac {1}{192} \, {\left (48 \, x^{3} + 8 \, x^{2} + 14 \, x - 37\right )} \sqrt {x^{2} + x + 1} - \frac {3}{128} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \] Input:
integrate(x^2*(x^2+x+1)^(1/2),x, algorithm="fricas")
Output:
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)
Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int x^2 \sqrt {1+x+x^2} \, dx=\sqrt {x^{2} + x + 1} \left (\frac {x^{3}}{4} + \frac {x^{2}}{24} + \frac {7 x}{96} - \frac {37}{192}\right ) + \frac {3 \operatorname {asinh}{\left (\frac {2 \sqrt {3} \left (x + \frac {1}{2}\right )}{3} \right )}}{128} \] Input:
integrate(x**2*(x**2+x+1)**(1/2),x)
Output:
sqrt(x**2 + x + 1)*(x**3/4 + x**2/24 + 7*x/96 - 37/192) + 3*asinh(2*sqrt(3 )*(x + 1/2)/3)/128
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int x^2 \sqrt {1+x+x^2} \, dx=\frac {1}{4} \, {\left (x^{2} + x + 1\right )}^{\frac {3}{2}} x - \frac {5}{24} \, {\left (x^{2} + x + 1\right )}^{\frac {3}{2}} + \frac {1}{32} \, \sqrt {x^{2} + x + 1} x + \frac {1}{64} \, \sqrt {x^{2} + x + 1} + \frac {3}{128} \, \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) \] Input:
integrate(x^2*(x^2+x+1)^(1/2),x, algorithm="maxima")
Output:
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) + 3/128*arcsinh(1/3*sqrt(3)*(2*x + 1))
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68 \[ \int x^2 \sqrt {1+x+x^2} \, dx=\frac {1}{192} \, {\left (2 \, {\left (4 \, {\left (6 \, x + 1\right )} x + 7\right )} x - 37\right )} \sqrt {x^{2} + x + 1} - \frac {3}{128} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \] Input:
integrate(x^2*(x^2+x+1)^(1/2),x, algorithm="giac")
Output:
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)
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int x^2 \sqrt {1+x+x^2} \, dx=\frac {3\,\ln \left (x+\sqrt {x^2+x+1}+\frac {1}{2}\right )}{128}-\frac {\left (\frac {x}{2}+\frac {1}{4}\right )\,\sqrt {x^2+x+1}}{4}-\frac {5\,\left (8\,x^2+2\,x+5\right )\,\sqrt {x^2+x+1}}{192}+\frac {x\,{\left (x^2+x+1\right )}^{3/2}}{4} \] Input:
int(x^2*(x + x^2 + 1)^(1/2),x)
Output:
(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
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int x^2 \sqrt {1+x+x^2} \, dx=\frac {\sqrt {x^{2}+x +1}\, x^{3}}{4}+\frac {\sqrt {x^{2}+x +1}\, x^{2}}{24}+\frac {7 \sqrt {x^{2}+x +1}\, x}{96}-\frac {37 \sqrt {x^{2}+x +1}}{192}+\frac {3 \,\mathrm {log}\left (\frac {2 \sqrt {x^{2}+x +1}+2 x +1}{\sqrt {3}}\right )}{128} \] Input:
int(x^2*(x^2+x+1)^(1/2),x)
Output:
(96*sqrt(x**2 + x + 1)*x**3 + 16*sqrt(x**2 + x + 1)*x**2 + 28*sqrt(x**2 + x + 1)*x - 74*sqrt(x**2 + x + 1) + 9*log((2*sqrt(x**2 + x + 1) + 2*x + 1)/ sqrt(3)))/384