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]

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

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Rubi [A]  time = 0.030516, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0140994, 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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Maple [A]  time = 0.006, size = 27, normalized size = 0.6 \begin{align*}{\frac{3\,{x}^{4}+6\,{x}^{2}+1}{30} \left ({x}^{4}+2\,{x}^{2}+2 \right ) ^{{\frac{3}{2}}}} \end{align*}

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 = 1.69889, size = 66, normalized size = 1.5 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 1.62559, size = 90, normalized size = 2.05 \begin{align*} \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} \end{align*}

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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Sympy [B]  time = 0.634387, size = 94, normalized size = 2.14 \begin{align*} \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} \end{align*}

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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Giac [A]  time = 1.16469, size = 51, normalized size = 1.16 \begin{align*} \frac{1}{30} \, \sqrt{x^{4} + 2 \, x^{2} + 2}{\left ({\left ({\left (3 \,{\left (x^{2} + 4\right )} x^{2} + 19\right )} x^{2} + 14\right )} x^{2} + 2\right )} \end{align*}

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]

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