3.7.79 \(\int x^3 (a+b x^2)^{2/3} \, dx\) [679]

Optimal. Leaf size=38 \[ -\frac {3 a \left (a+b x^2\right )^{5/3}}{10 b^2}+\frac {3 \left (a+b x^2\right )^{8/3}}{16 b^2} \]

[Out]

-3/10*a*(b*x^2+a)^(5/3)/b^2+3/16*(b*x^2+a)^(8/3)/b^2

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Rubi [A]
time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \begin {gather*} \frac {3 \left (a+b x^2\right )^{8/3}}{16 b^2}-\frac {3 a \left (a+b x^2\right )^{5/3}}{10 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^(2/3),x]

[Out]

(-3*a*(a + b*x^2)^(5/3))/(10*b^2) + (3*(a + b*x^2)^(8/3))/(16*b^2)

Rule 45

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])

Rule 272

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]]

Rubi steps

\begin {align*} \int x^3 \left (a+b x^2\right )^{2/3} \, dx &=\frac {1}{2} \text {Subst}\left (\int x (a+b x)^{2/3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (-\frac {a (a+b x)^{2/3}}{b}+\frac {(a+b x)^{5/3}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac {3 a \left (a+b x^2\right )^{5/3}}{10 b^2}+\frac {3 \left (a+b x^2\right )^{8/3}}{16 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 39, normalized size = 1.03 \begin {gather*} \frac {3 \left (a+b x^2\right )^{2/3} \left (-3 a^2+2 a b x^2+5 b^2 x^4\right )}{80 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)^(2/3),x]

[Out]

(3*(a + b*x^2)^(2/3)*(-3*a^2 + 2*a*b*x^2 + 5*b^2*x^4))/(80*b^2)

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Maple [A]
time = 0.04, size = 25, normalized size = 0.66

method	result	size
gosper	\(-\frac {3 \left (b \,x^{2}+a \right )^{\frac {5}{3}} \left (-5 b \,x^{2}+3 a \right )}{80 b^{2}}\)	\(25\)
trager	\(-\frac {3 \left (-5 b^{2} x^{4}-2 a b \,x^{2}+3 a^{2}\right ) \left (b \,x^{2}+a \right )^{\frac {2}{3}}}{80 b^{2}}\)	\(36\)
risch	\(-\frac {3 \left (-5 b^{2} x^{4}-2 a b \,x^{2}+3 a^{2}\right ) \left (b \,x^{2}+a \right )^{\frac {2}{3}}}{80 b^{2}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/80*(b*x^2+a)^(5/3)*(-5*b*x^2+3*a)/b^2

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Maxima [A]
time = 0.28, size = 30, normalized size = 0.79 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {8}{3}}}{16 \, b^{2}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} a}{10 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*(b*x^2 + a)^(8/3)/b^2 - 3/10*(b*x^2 + a)^(5/3)*a/b^2

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Fricas [A]
time = 1.06, size = 35, normalized size = 0.92 \begin {gather*} \frac {3 \, {\left (5 \, b^{2} x^{4} + 2 \, a b x^{2} - 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{80 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/80*(5*b^2*x^4 + 2*a*b*x^2 - 3*a^2)*(b*x^2 + a)^(2/3)/b^2

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Sympy [A]
time = 0.15, size = 66, normalized size = 1.74 \begin {gather*} \begin {cases} - \frac {9 a^{2} \left (a + b x^{2}\right )^{\frac {2}{3}}}{80 b^{2}} + \frac {3 a x^{2} \left (a + b x^{2}\right )^{\frac {2}{3}}}{40 b} + \frac {3 x^{4} \left (a + b x^{2}\right )^{\frac {2}{3}}}{16} & \text {for}\: b \neq 0 \\\frac {a^{\frac {2}{3}} x^{4}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-9*a**2*(a + b*x**2)**(2/3)/(80*b**2) + 3*a*x**2*(a + b*x**2)**(2/3)/(40*b) + 3*x**4*(a + b*x**2)**
(2/3)/16, Ne(b, 0)), (a**(2/3)*x**4/4, True))

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Giac [A]
time = 2.48, size = 29, normalized size = 0.76 \begin {gather*} \frac {3 \, {\left (5 \, {\left (b x^{2} + a\right )}^{\frac {8}{3}} - 8 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} a\right )}}{80 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/80*(5*(b*x^2 + a)^(8/3) - 8*(b*x^2 + a)^(5/3)*a)/b^2

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Mupad [B]
time = 4.71, size = 33, normalized size = 0.87 \begin {gather*} {\left (b\,x^2+a\right )}^{2/3}\,\left (\frac {3\,x^4}{16}-\frac {9\,a^2}{80\,b^2}+\frac {3\,a\,x^2}{40\,b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2)^(2/3),x)

[Out]

(a + b*x^2)^(2/3)*((3*x^4)/16 - (9*a^2)/(80*b^2) + (3*a*x^2)/(40*b))

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