3.497 \(\int x^2 (1+x)^{3/2} (1-x+x^2)^{3/2} \, dx\)

Optimal. Leaf size=23 \[ \frac{2}{15} (x+1)^{5/2} \left (x^2-x+1\right )^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.0208217, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {913} \[ \frac{2}{15} (x+1)^{5/2} \left (x^2-x+1\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(1 + x)^(3/2)*(1 - x + x^2)^(3/2),x]

[Out]

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

Rule 913

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

Rubi steps

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

Mathematica [A]  time = 0.0430488, size = 23, normalized size = 1. \[ \frac{2}{15} (x+1)^{5/2} \left (x^2-x+1\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 18, normalized size = 0.8 \begin{align*}{\frac{2}{15} \left ( 1+x \right ) ^{{\frac{5}{2}}} \left ({x}^{2}-x+1 \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.44561, size = 36, normalized size = 1.57 \begin{align*} \frac{2}{15} \,{\left (x^{6} + 2 \, x^{3} + 1\right )} \sqrt{x^{2} - x + 1} \sqrt{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.7053, size = 73, normalized size = 3.17 \begin{align*} \frac{2}{15} \,{\left (x^{6} + 2 \, x^{3} + 1\right )} \sqrt{x^{2} - x + 1} \sqrt{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.19915, size = 100, normalized size = 4.35 \begin{align*} \frac{2}{45} \,{\left ({\left ({\left (3 \,{\left ({\left (x + 1\right )}{\left (x - 5\right )} + 15\right )}{\left (x + 1\right )} - 59\right )}{\left (x + 1\right )} + 42\right )}{\left (x + 1\right )} - 15\right )} \sqrt{{\left (x + 1\right )}^{2} - 3 \, x}{\left (x + 1\right )}^{\frac{3}{2}} + \frac{2}{9} \, \sqrt{{\left (x + 1\right )}^{2} - 3 \, x}{\left ({\left (x + 1\right )}{\left (x - 2\right )} + 3\right )}{\left (x + 1\right )}^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(((3*((x + 1)*(x - 5) + 15)*(x + 1) - 59)*(x + 1) + 42)*(x + 1) - 15)*sqrt((x + 1)^2 - 3*x)*(x + 1)^(3/2)
+ 2/9*sqrt((x + 1)^2 - 3*x)*((x + 1)*(x - 2) + 3)*(x + 1)^(3/2)