3.7.80 \(\int x (a+b x^2)^{2/3} \, dx\) [680]

Optimal. Leaf size=18 \[ \frac {3 \left (a+b x^2\right )^{5/3}}{10 b} \]

[Out]

3/10*(b*x^2+a)^(5/3)/b

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Rubi [A]
time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \begin {gather*} \frac {3 \left (a+b x^2\right )^{5/3}}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)^(2/3),x]

[Out]

(3*(a + b*x^2)^(5/3))/(10*b)

Rule 267

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

Rubi steps

\begin {align*} \int x \left (a+b x^2\right )^{2/3} \, dx &=\frac {3 \left (a+b x^2\right )^{5/3}}{10 b}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} \frac {3 \left (a+b x^2\right )^{5/3}}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x^2)^(2/3),x]

[Out]

(3*(a + b*x^2)^(5/3))/(10*b)

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Maple [A]
time = 0.03, size = 15, normalized size = 0.83

method	result	size
gosper	\(\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}}}{10 b}\)	\(15\)
derivativedivides	\(\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}}}{10 b}\)	\(15\)
default	\(\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}}}{10 b}\)	\(15\)
trager	\(\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}}}{10 b}\)	\(15\)
risch	\(\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}}}{10 b}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2+a)^(2/3),x,method=_RETURNVERBOSE)

[Out]

3/10*(b*x^2+a)^(5/3)/b

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Maxima [A]
time = 0.34, size = 14, normalized size = 0.78 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}}}{10 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(b*x^2 + a)^(5/3)/b

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Fricas [A]
time = 1.07, size = 14, normalized size = 0.78 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}}}{10 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(b*x^2 + a)^(5/3)/b

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (14) = 28\).
time = 0.09, size = 42, normalized size = 2.33 \begin {gather*} \begin {cases} \frac {3 a \left (a + b x^{2}\right )^{\frac {2}{3}}}{10 b} + \frac {3 x^{2} \left (a + b x^{2}\right )^{\frac {2}{3}}}{10} & \text {for}\: b \neq 0 \\\frac {a^{\frac {2}{3}} x^{2}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**2+a)**(2/3),x)

[Out]

Piecewise((3*a*(a + b*x**2)**(2/3)/(10*b) + 3*x**2*(a + b*x**2)**(2/3)/10, Ne(b, 0)), (a**(2/3)*x**2/2, True))

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Giac [A]
time = 1.92, size = 14, normalized size = 0.78 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}}}{10 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(b*x^2 + a)^(5/3)/b

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Mupad [B]
time = 4.58, size = 14, normalized size = 0.78 \begin {gather*} \frac {3\,{\left (b\,x^2+a\right )}^{5/3}}{10\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^2)^(2/3),x)

[Out]

(3*(a + b*x^2)^(5/3))/(10*b)

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