3.3.70 \(\int \frac {x^3}{(1+x+x^2)^{3/2}} \, dx\) [270]

Optimal. Leaf size=56 \[ -\frac {2 x^2 (2+x)}{3 \sqrt {1+x+x^2}}+\frac {1}{3} (5+2 x) \sqrt {1+x+x^2}-\frac {3}{2} \sinh ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right ) \]

[Out]

-3/2*arcsinh(1/3*(1+2*x)*3^(1/2))-2/3*x^2*(2+x)/(x^2+x+1)^(1/2)+1/3*(2*x+5)*(x^2+x+1)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 56, 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 = {752, 793, 633, 221} \begin {gather*} -\frac {2 (x+2) x^2}{3 \sqrt {x^2+x+1}}+\frac {1}{3} (2 x+5) \sqrt {x^2+x+1}-\frac {3}{2} \sinh ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(1 + x + x^2)^(3/2),x]

[Out]

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

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 752

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(d
*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && 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}{\left (1+x+x^2\right )^{3/2}} \, dx &=-\frac {2 x^2 (2+x)}{3 \sqrt {1+x+x^2}}+\frac {2}{3} \int \frac {x (4+2 x)}{\sqrt {1+x+x^2}} \, dx\\ &=-\frac {2 x^2 (2+x)}{3 \sqrt {1+x+x^2}}+\frac {1}{3} (5+2 x) \sqrt {1+x+x^2}-\frac {3}{2} \int \frac {1}{\sqrt {1+x+x^2}} \, dx\\ &=-\frac {2 x^2 (2+x)}{3 \sqrt {1+x+x^2}}+\frac {1}{3} (5+2 x) \sqrt {1+x+x^2}-\frac {1}{2} \sqrt {3} \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{3}}} \, dx,x,1+2 x\right )\\ &=-\frac {2 x^2 (2+x)}{3 \sqrt {1+x+x^2}}+\frac {1}{3} (5+2 x) \sqrt {1+x+x^2}-\frac {3}{2} \sinh ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 47, normalized size = 0.84 \begin {gather*} \frac {5+7 x+3 x^2}{3 \sqrt {1+x+x^2}}+\frac {3}{2} \log \left (-1-2 x+2 \sqrt {1+x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 + x + x^2)^(3/2),x]

[Out]

(5 + 7*x + 3*x^2)/(3*Sqrt[1 + x + x^2]) + (3*Log[-1 - 2*x + 2*Sqrt[1 + x + x^2]])/2

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Maple [A]
time = 0.12, size = 61, normalized size = 1.09

method	result	size
risch	\(\frac {3 x^{2}+7 x +5}{3 \sqrt {x^{2}+x +1}}-\frac {3 \arcsinh \left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{2}\)	\(33\)
trager	\(\frac {3 x^{2}+7 x +5}{3 \sqrt {x^{2}+x +1}}+\frac {3 \ln \left (2 \sqrt {x^{2}+x +1}-1-2 x \right )}{2}\)	\(40\)
default	\(\frac {x^{2}}{\sqrt {x^{2}+x +1}}+\frac {3 x}{2 \sqrt {x^{2}+x +1}}+\frac {5}{4 \sqrt {x^{2}+x +1}}+\frac {\frac {5}{12}+\frac {5 x}{6}}{\sqrt {x^{2}+x +1}}-\frac {3 \arcsinh \left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{2}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2/(x^2+x+1)^(1/2)+3/2*x/(x^2+x+1)^(1/2)+5/4/(x^2+x+1)^(1/2)+5/12*(1+2*x)/(x^2+x+1)^(1/2)-3/2*arcsinh(2/3*3^(
1/2)*(x+1/2))

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Maxima [A]
time = 2.59, size = 47, normalized size = 0.84 \begin {gather*} \frac {x^{2}}{\sqrt {x^{2} + x + 1}} + \frac {7 \, x}{3 \, \sqrt {x^{2} + x + 1}} + \frac {5}{3 \, \sqrt {x^{2} + x + 1}} - \frac {3}{2} \, \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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.43, size = 64, normalized size = 1.14 \begin {gather*} \frac {19 \, x^{2} + 18 \, {\left (x^{2} + x + 1\right )} \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) + 4 \, {\left (3 \, x^{2} + 7 \, x + 5\right )} \sqrt {x^{2} + x + 1} + 19 \, x + 19}{12 \, {\left (x^{2} + x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^2+x+1)^(3/2),x, algorithm="fricas")

[Out]

1/12*(19*x^2 + 18*(x^2 + x + 1)*log(-2*x + 2*sqrt(x^2 + x + 1) - 1) + 4*(3*x^2 + 7*x + 5)*sqrt(x^2 + x + 1) +
19*x + 19)/(x^2 + x + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (x^{2} + x + 1\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.83, size = 38, normalized size = 0.68 \begin {gather*} \frac {{\left (3 \, x + 7\right )} x + 5}{3 \, \sqrt {x^{2} + x + 1}} + \frac {3}{2} \, \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)^(3/2),x, algorithm="giac")

[Out]

1/3*((3*x + 7)*x + 5)/sqrt(x^2 + x + 1) + 3/2*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}{{\left (x^2+x+1\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x + x^2 + 1)^(3/2),x)

[Out]

int(x^3/(x + x^2 + 1)^(3/2), x)

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