3.932 \(\int x \sqrt{x+x^{3/2}} \, dx\)

Optimal. Leaf size=94 \[ \frac{4}{11} \sqrt{x} \left (x^{3/2}+x\right )^{3/2}+\frac{64 \left (x^{3/2}+x\right )^{3/2}}{231 \sqrt{x}}-\frac{256 \left (x^{3/2}+x\right )^{3/2}}{1155 x}+\frac{512 \left (x^{3/2}+x\right )^{3/2}}{3465 x^{3/2}}-\frac{32}{99} \left (x^{3/2}+x\right )^{3/2} \]

[Out]

(-32*(x + x^(3/2))^(3/2))/99 + (512*(x + x^(3/2))^(3/2))/(3465*x^(3/2)) - (256*(x + x^(3/2))^(3/2))/(1155*x) +
(64*(x + x^(3/2))^(3/2))/(231*Sqrt[x]) + (4*Sqrt[x]*(x + x^(3/2))^(3/2))/11

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Rubi [A]  time = 0.0907418, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2016, 2002, 2014} \[ \frac{4}{11} \sqrt{x} \left (x^{3/2}+x\right )^{3/2}+\frac{64 \left (x^{3/2}+x\right )^{3/2}}{231 \sqrt{x}}-\frac{256 \left (x^{3/2}+x\right )^{3/2}}{1155 x}+\frac{512 \left (x^{3/2}+x\right )^{3/2}}{3465 x^{3/2}}-\frac{32}{99} \left (x^{3/2}+x\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[x + x^(3/2)],x]

[Out]

(-32*(x + x^(3/2))^(3/2))/99 + (512*(x + x^(3/2))^(3/2))/(3465*x^(3/2)) - (256*(x + x^(3/2))^(3/2))/(1155*x) +
(64*(x + x^(3/2))^(3/2))/(231*Sqrt[x]) + (4*Sqrt[x]*(x + x^(3/2))^(3/2))/11

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int x \sqrt{x+x^{3/2}} \, dx &=\frac{4}{11} \sqrt{x} \left (x+x^{3/2}\right )^{3/2}-\frac{8}{11} \int \sqrt{x} \sqrt{x+x^{3/2}} \, dx\\ &=-\frac{32}{99} \left (x+x^{3/2}\right )^{3/2}+\frac{4}{11} \sqrt{x} \left (x+x^{3/2}\right )^{3/2}+\frac{16}{33} \int \sqrt{x+x^{3/2}} \, dx\\ &=-\frac{32}{99} \left (x+x^{3/2}\right )^{3/2}+\frac{64 \left (x+x^{3/2}\right )^{3/2}}{231 \sqrt{x}}+\frac{4}{11} \sqrt{x} \left (x+x^{3/2}\right )^{3/2}-\frac{64}{231} \int \frac{\sqrt{x+x^{3/2}}}{\sqrt{x}} \, dx\\ &=-\frac{32}{99} \left (x+x^{3/2}\right )^{3/2}-\frac{256 \left (x+x^{3/2}\right )^{3/2}}{1155 x}+\frac{64 \left (x+x^{3/2}\right )^{3/2}}{231 \sqrt{x}}+\frac{4}{11} \sqrt{x} \left (x+x^{3/2}\right )^{3/2}+\frac{128 \int \frac{\sqrt{x+x^{3/2}}}{x} \, dx}{1155}\\ &=-\frac{32}{99} \left (x+x^{3/2}\right )^{3/2}+\frac{512 \left (x+x^{3/2}\right )^{3/2}}{3465 x^{3/2}}-\frac{256 \left (x+x^{3/2}\right )^{3/2}}{1155 x}+\frac{64 \left (x+x^{3/2}\right )^{3/2}}{231 \sqrt{x}}+\frac{4}{11} \sqrt{x} \left (x+x^{3/2}\right )^{3/2}\\ \end{align*}

Mathematica [A]  time = 0.0303241, size = 51, normalized size = 0.54 \[ \frac{4 \left (\sqrt{x}+1\right ) \sqrt{x^{3/2}+x} \left (315 x^2-280 x^{3/2}+240 x-192 \sqrt{x}+128\right )}{3465 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[x + x^(3/2)],x]

[Out]

(4*(1 + Sqrt[x])*Sqrt[x + x^(3/2)]*(128 - 192*Sqrt[x] + 240*x - 280*x^(3/2) + 315*x^2))/(3465*Sqrt[x])

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Maple [A]  time = 0.003, size = 38, normalized size = 0.4 \begin{align*}{\frac{4}{3465}\sqrt{x+{x}^{{\frac{3}{2}}}} \left ( 1+\sqrt{x} \right ) \left ( 315\,{x}^{2}-280\,{x}^{3/2}+240\,x-192\,\sqrt{x}+128 \right ){\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3465*(x+x^(3/2))^(1/2)*(1+x^(1/2))*(315*x^2-280*x^(3/2)+240*x-192*x^(1/2)+128)/x^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{\frac{3}{2}} + x} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^(3/2) + x)*x, x)

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Fricas [A]  time = 1.9217, size = 116, normalized size = 1.23 \begin{align*} \frac{4 \,{\left (315 \, x^{3} - 40 \, x^{2} +{\left (35 \, x^{2} + 48 \, x + 128\right )} \sqrt{x} - 64 \, x\right )} \sqrt{x^{\frac{3}{2}} + x}}{3465 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3465*(315*x^3 - 40*x^2 + (35*x^2 + 48*x + 128)*sqrt(x) - 64*x)*sqrt(x^(3/2) + x)/x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{x^{\frac{3}{2}} + x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(x**(3/2) + x), x)

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Giac [A]  time = 1.10183, size = 69, normalized size = 0.73 \begin{align*} \frac{4}{3465} \,{\left (315 \,{\left (\sqrt{x} + 1\right )}^{\frac{11}{2}} - 1540 \,{\left (\sqrt{x} + 1\right )}^{\frac{9}{2}} + 2970 \,{\left (\sqrt{x} + 1\right )}^{\frac{7}{2}} - 2772 \,{\left (\sqrt{x} + 1\right )}^{\frac{5}{2}} + 1155 \,{\left (\sqrt{x} + 1\right )}^{\frac{3}{2}} - 128\right )} \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3465*(315*(sqrt(x) + 1)^(11/2) - 1540*(sqrt(x) + 1)^(9/2) + 2970*(sqrt(x) + 1)^(7/2) - 2772*(sqrt(x) + 1)^(5
/2) + 1155*(sqrt(x) + 1)^(3/2) - 128)*sgn(x)