3.685 \(\int x (1+x^2)^3 \sqrt {2+2 x^2+x^4} \, dx\)

Optimal. Leaf size=44 \[ \frac {1}{10} \left (x^2+1\right )^2 \left (x^4+2 x^2+2\right )^{3/2}-\frac {1}{15} \left (x^4+2 x^2+2\right )^{3/2} \]

[Out]

-1/15*(x^4+2*x^2+2)^(3/2)+1/10*(x^2+1)^2*(x^4+2*x^2+2)^(3/2)

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Rubi [A]  time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1247, 692, 629} \[ \frac {1}{10} \left (x^2+1\right )^2 \left (x^4+2 x^2+2\right )^{3/2}-\frac {1}{15} \left (x^4+2 x^2+2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x*(1 + x^2)^3*Sqrt[2 + 2*x^2 + x^4],x]

[Out]

-(2 + 2*x^2 + x^4)^(3/2)/15 + ((1 + x^2)^2*(2 + 2*x^2 + x^4)^(3/2))/10

Rule 629

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

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rubi steps

\begin {align*} \int x \left (1+x^2\right )^3 \sqrt {2+2 x^2+x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int (1+x)^3 \sqrt {2+2 x+x^2} \, dx,x,x^2\right )\\ &=\frac {1}{10} \left (1+x^2\right )^2 \left (2+2 x^2+x^4\right )^{3/2}-\frac {1}{5} \operatorname {Subst}\left (\int (1+x) \sqrt {2+2 x+x^2} \, dx,x,x^2\right )\\ &=-\frac {1}{15} \left (2+2 x^2+x^4\right )^{3/2}+\frac {1}{10} \left (1+x^2\right )^2 \left (2+2 x^2+x^4\right )^{3/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.68 \[ \frac {1}{30} \left (x^4+2 x^2+2\right )^{3/2} \left (3 x^4+6 x^2+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(1 + x^2)^3*Sqrt[2 + 2*x^2 + x^4],x]

[Out]

((2 + 2*x^2 + x^4)^(3/2)*(1 + 6*x^2 + 3*x^4))/30

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fricas [A]  time = 0.43, size = 36, normalized size = 0.82 \[ \frac {1}{30} \, {\left (3 \, x^{8} + 12 \, x^{6} + 19 \, x^{4} + 14 \, x^{2} + 2\right )} \sqrt {x^{4} + 2 \, x^{2} + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)^3*(x^4+2*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

1/30*(3*x^8 + 12*x^6 + 19*x^4 + 14*x^2 + 2)*sqrt(x^4 + 2*x^2 + 2)

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giac [A]  time = 0.35, size = 29, normalized size = 0.66 \[ \frac {1}{10} \, {\left (x^{4} + 2 \, x^{2} + 2\right )}^{\frac {5}{2}} - \frac {1}{6} \, {\left (x^{4} + 2 \, x^{2} + 2\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)^3*(x^4+2*x^2+2)^(1/2),x, algorithm="giac")

[Out]

1/10*(x^4 + 2*x^2 + 2)^(5/2) - 1/6*(x^4 + 2*x^2 + 2)^(3/2)

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maple [A]  time = 0.01, size = 27, normalized size = 0.61 \[ \frac {\left (x^{4}+2 x^{2}+2\right )^{\frac {3}{2}} \left (3 x^{4}+6 x^{2}+1\right )}{30} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2+1)^3*(x^4+2*x^2+2)^(1/2),x)

[Out]

1/30*(x^4+2*x^2+2)^(3/2)*(3*x^4+6*x^2+1)

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maxima [A]  time = 1.47, size = 49, normalized size = 1.11 \[ \frac {1}{10} \, {\left (x^{4} + 2 \, x^{2} + 2\right )}^{\frac {3}{2}} x^{4} + \frac {1}{5} \, {\left (x^{4} + 2 \, x^{2} + 2\right )}^{\frac {3}{2}} x^{2} + \frac {1}{30} \, {\left (x^{4} + 2 \, x^{2} + 2\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)^3*(x^4+2*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

1/10*(x^4 + 2*x^2 + 2)^(3/2)*x^4 + 1/5*(x^4 + 2*x^2 + 2)^(3/2)*x^2 + 1/30*(x^4 + 2*x^2 + 2)^(3/2)

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mupad [B]  time = 0.09, size = 26, normalized size = 0.59 \[ \frac {{\left (x^4+2\,x^2+2\right )}^{3/2}\,\left (3\,x^4+6\,x^2+1\right )}{30} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2 + 1)^3*(2*x^2 + x^4 + 2)^(1/2),x)

[Out]

((2*x^2 + x^4 + 2)^(3/2)*(6*x^2 + 3*x^4 + 1))/30

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sympy [B]  time = 0.65, size = 94, normalized size = 2.14 \[ \frac {x^{8} \sqrt {x^{4} + 2 x^{2} + 2}}{10} + \frac {2 x^{6} \sqrt {x^{4} + 2 x^{2} + 2}}{5} + \frac {19 x^{4} \sqrt {x^{4} + 2 x^{2} + 2}}{30} + \frac {7 x^{2} \sqrt {x^{4} + 2 x^{2} + 2}}{15} + \frac {\sqrt {x^{4} + 2 x^{2} + 2}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x**2+1)**3*(x**4+2*x**2+2)**(1/2),x)

[Out]

x**8*sqrt(x**4 + 2*x**2 + 2)/10 + 2*x**6*sqrt(x**4 + 2*x**2 + 2)/5 + 19*x**4*sqrt(x**4 + 2*x**2 + 2)/30 + 7*x*
*2*sqrt(x**4 + 2*x**2 + 2)/15 + sqrt(x**4 + 2*x**2 + 2)/15

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