3.3.67 \(\int \frac {x^3}{\sqrt {1+x+x^2}} \, dx\) [267]

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

[Out]

7/16*arcsinh(1/3*(1+2*x)*3^(1/2))+1/3*x^2*(x^2+x+1)^(1/2)-1/24*(1+10*x)*(x^2+x+1)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 = {756, 793, 633, 221} \begin {gather*} \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 ) \end {gather*}

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 221

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 633

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 756

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 793

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]

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 \text {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*}

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Mathematica [A]
time = 0.07, size = 47, normalized size = 0.89 \begin {gather*} \frac {1}{24} \sqrt {1+x+x^2} \left (-1-10 x+8 x^2\right )-\frac {7}{16} \log \left (-1-2 x+2 \sqrt {1+x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[1 + x + x^2],x]

[Out]

(Sqrt[1 + x + x^2]*(-1 - 10*x + 8*x^2))/24 - (7*Log[-1 - 2*x + 2*Sqrt[1 + x + x^2]])/16

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Maple [A]
time = 0.12, size = 47, normalized size = 0.89

method	result	size
risch	\(\frac {\left (8 x^{2}-10 x -1\right ) \sqrt {x^{2}+x +1}}{24}+\frac {7 \arcsinh \left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{16}\)	\(33\)
trager	\(\left (\frac {1}{3} x^{2}-\frac {5}{12} x -\frac {1}{24}\right ) \sqrt {x^{2}+x +1}+\frac {7 \ln \left (1+2 x +2 \sqrt {x^{2}+x +1}\right )}{16}\)	\(39\)
default	\(\frac {x^{2} \sqrt {x^{2}+x +1}}{3}-\frac {5 x \sqrt {x^{2}+x +1}}{12}-\frac {\sqrt {x^{2}+x +1}}{24}+\frac {7 \arcsinh \left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{16}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^2+x+1)^(1/2),x,method=_RETURNVERBOSE)

[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 = 2.07, size = 48, normalized size = 0.91 \begin {gather*} \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 {gather*}

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 = 0.43, size = 39, normalized size = 0.74 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\sqrt {x^{2} + x + 1}}\, dx \end {gather*}

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 = 1.00, size = 39, normalized size = 0.74 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^3}{\sqrt {x^2+x+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x + x^2 + 1)^(1/2),x)

[Out]

int(x^3/(x + x^2 + 1)^(1/2), x)

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