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

Optimal. Leaf size=49 \[ \frac{2 x^{13/2}}{13}-\frac{2 x^6}{3}+\frac{8 x^{11/2}}{11}+\frac{2 x^{9/2}}{9}-x^4+\frac{8 x^{7/2}}{7} \]

[Out]

(8*x^(7/2))/7 - x^4 + (2*x^(9/2))/9 + (8*x^(11/2))/11 - (2*x^6)/3 + (2*x^(13/2))/13

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Rubi [A]  time = 0.0566495, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1584, 1820, 266, 43} \[ \frac{2 x^{13/2}}{13}-\frac{2 x^6}{3}+\frac{8 x^{11/2}}{11}+\frac{2 x^{9/2}}{9}-x^4+\frac{8 x^{7/2}}{7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*x^(7/2))/7 - x^4 + (2*x^(9/2))/9 + (8*x^(11/2))/11 - (2*x^6)/3 + (2*x^(13/2))/13

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \left (2 \sqrt{x}-x\right )^2 x^{3/2} \left (1+x^2\right ) \, dx &=\int \left (2-\sqrt{x}\right )^2 x^{5/2} \left (1+x^2\right ) \, dx\\ &=\int \left (\left (-2+\sqrt{x}\right )^2 x^{5/2}+\left (-2+\sqrt{x}\right )^2 x^{9/2}\right ) \, dx\\ &=\int \left (-2+\sqrt{x}\right )^2 x^{5/2} \, dx+\int \left (-2+\sqrt{x}\right )^2 x^{9/2} \, dx\\ &=2 \operatorname{Subst}\left (\int (-2+x)^2 x^6 \, dx,x,\sqrt{x}\right )+2 \operatorname{Subst}\left (\int (-2+x)^2 x^{10} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (4 x^6-4 x^7+x^8\right ) \, dx,x,\sqrt{x}\right )+2 \operatorname{Subst}\left (\int \left (4 x^{10}-4 x^{11}+x^{12}\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{8 x^{7/2}}{7}-x^4+\frac{2 x^{9/2}}{9}+\frac{8 x^{11/2}}{11}-\frac{2 x^6}{3}+\frac{2 x^{13/2}}{13}\\ \end{align*}

Mathematica [A]  time = 0.0218665, size = 49, normalized size = 1. \[ \frac{2 x^{13/2}}{13}-\frac{2 x^6}{3}+\frac{8 x^{11/2}}{11}+\frac{2 x^{9/2}}{9}-x^4+\frac{8 x^{7/2}}{7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*x^(7/2))/7 - x^4 + (2*x^(9/2))/9 + (8*x^(11/2))/11 - (2*x^6)/3 + (2*x^(13/2))/13

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Maple [A]  time = 0.002, size = 32, normalized size = 0.7 \begin{align*}{\frac{8}{7}{x}^{{\frac{7}{2}}}}-{x}^{4}+{\frac{2}{9}{x}^{{\frac{9}{2}}}}+{\frac{8}{11}{x}^{{\frac{11}{2}}}}-{\frac{2\,{x}^{6}}{3}}+{\frac{2}{13}{x}^{{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/7*x^(7/2)-x^4+2/9*x^(9/2)+8/11*x^(11/2)-2/3*x^6+2/13*x^(13/2)

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Maxima [A]  time = 0.957597, size = 42, normalized size = 0.86 \begin{align*} \frac{2}{13} \, x^{\frac{13}{2}} - \frac{2}{3} \, x^{6} + \frac{8}{11} \, x^{\frac{11}{2}} + \frac{2}{9} \, x^{\frac{9}{2}} - x^{4} + \frac{8}{7} \, x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/13*x^(13/2) - 2/3*x^6 + 8/11*x^(11/2) + 2/9*x^(9/2) - x^4 + 8/7*x^(7/2)

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Fricas [A]  time = 1.68706, size = 103, normalized size = 2.1 \begin{align*} -\frac{2}{3} \, x^{6} - x^{4} + \frac{2}{9009} \,{\left (693 \, x^{6} + 3276 \, x^{5} + 1001 \, x^{4} + 5148 \, x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*x^6 - x^4 + 2/9009*(693*x^6 + 3276*x^5 + 1001*x^4 + 5148*x^3)*sqrt(x)

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Sympy [A]  time = 2.23406, size = 42, normalized size = 0.86 \begin{align*} \frac{2 x^{\frac{13}{2}}}{13} + \frac{8 x^{\frac{11}{2}}}{11} + \frac{2 x^{\frac{9}{2}}}{9} + \frac{8 x^{\frac{7}{2}}}{7} - \frac{2 x^{6}}{3} - x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**(13/2)/13 + 8*x**(11/2)/11 + 2*x**(9/2)/9 + 8*x**(7/2)/7 - 2*x**6/3 - x**4

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Giac [A]  time = 1.05136, size = 42, normalized size = 0.86 \begin{align*} \frac{2}{13} \, x^{\frac{13}{2}} - \frac{2}{3} \, x^{6} + \frac{8}{11} \, x^{\frac{11}{2}} + \frac{2}{9} \, x^{\frac{9}{2}} - x^{4} + \frac{8}{7} \, x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/13*x^(13/2) - 2/3*x^6 + 8/11*x^(11/2) + 2/9*x^(9/2) - x^4 + 8/7*x^(7/2)