### 3.2366 $$\int (1+x)^3 \sqrt{2+2 x+x^2} \, dx$$

Optimal. Leaf size=38 $\frac{1}{5} (x+1)^2 \left (x^2+2 x+2\right )^{3/2}-\frac{2}{15} \left (x^2+2 x+2\right )^{3/2}$

[Out]

(-2*(2 + 2*x + x^2)^(3/2))/15 + ((1 + x)^2*(2 + 2*x + x^2)^(3/2))/5

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Rubi [A]  time = 0.0118175, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.111, Rules used = {692, 629} $\frac{1}{5} (x+1)^2 \left (x^2+2 x+2\right )^{3/2}-\frac{2}{15} \left (x^2+2 x+2\right )^{3/2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(1 + x)^3*Sqrt[2 + 2*x + x^2],x]

[Out]

(-2*(2 + 2*x + x^2)^(3/2))/15 + ((1 + x)^2*(2 + 2*x + x^2)^(3/2))/5

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

Mathematica [A]  time = 0.0093214, size = 26, normalized size = 0.68 $\frac{1}{15} \left (x^2+2 x+2\right )^{3/2} \left (3 x^2+6 x+1\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2 + 2*x + x^2)^(3/2)*(1 + 6*x + 3*x^2))/15

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Maple [A]  time = 0.04, size = 23, normalized size = 0.6 \begin{align*}{\frac{3\,{x}^{2}+6\,x+1}{15} \left ({x}^{2}+2\,x+2 \right ) ^{{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(x^2+2*x+2)^(3/2)*(3*x^2+6*x+1)

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Maxima [A]  time = 1.51231, size = 55, normalized size = 1.45 \begin{align*} \frac{1}{5} \,{\left (x^{2} + 2 \, x + 2\right )}^{\frac{3}{2}} x^{2} + \frac{2}{5} \,{\left (x^{2} + 2 \, x + 2\right )}^{\frac{3}{2}} x + \frac{1}{15} \,{\left (x^{2} + 2 \, x + 2\right )}^{\frac{3}{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^2 + 2*x + 2)^(3/2)*x^2 + 2/5*(x^2 + 2*x + 2)^(3/2)*x + 1/15*(x^2 + 2*x + 2)^(3/2)

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Fricas [A]  time = 2.33861, size = 85, normalized size = 2.24 \begin{align*} \frac{1}{15} \,{\left (3 \, x^{4} + 12 \, x^{3} + 19 \, x^{2} + 14 \, x + 2\right )} \sqrt{x^{2} + 2 \, x + 2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*x^4 + 12*x^3 + 19*x^2 + 14*x + 2)*sqrt(x^2 + 2*x + 2)

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Sympy [B]  time = 0.259986, size = 85, normalized size = 2.24 \begin{align*} \frac{x^{4} \sqrt{x^{2} + 2 x + 2}}{5} + \frac{4 x^{3} \sqrt{x^{2} + 2 x + 2}}{5} + \frac{19 x^{2} \sqrt{x^{2} + 2 x + 2}}{15} + \frac{14 x \sqrt{x^{2} + 2 x + 2}}{15} + \frac{2 \sqrt{x^{2} + 2 x + 2}}{15} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*sqrt(x**2 + 2*x + 2)/5 + 4*x**3*sqrt(x**2 + 2*x + 2)/5 + 19*x**2*sqrt(x**2 + 2*x + 2)/15 + 14*x*sqrt(x**2
+ 2*x + 2)/15 + 2*sqrt(x**2 + 2*x + 2)/15

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Giac [A]  time = 1.08866, size = 38, normalized size = 1. \begin{align*} \frac{1}{15} \,{\left ({\left ({\left (3 \,{\left (x + 4\right )} x + 19\right )} x + 14\right )} x + 2\right )} \sqrt{x^{2} + 2 \, x + 2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(((3*(x + 4)*x + 19)*x + 14)*x + 2)*sqrt(x^2 + 2*x + 2)