3.7.67 \(\int x^3 \sqrt [3]{a+b x^2} \, dx\) [667]

Optimal. Leaf size=38 \[ -\frac {3 a \left (a+b x^2\right )^{4/3}}{8 b^2}+\frac {3 \left (a+b x^2\right )^{7/3}}{14 b^2} \]

[Out]

-3/8*a*(b*x^2+a)^(4/3)/b^2+3/14*(b*x^2+a)^(7/3)/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 )^{7/3}}{14 b^2}-\frac {3 a \left (a+b x^2\right )^{4/3}}{8 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^(1/3),x]

[Out]

(-3*a*(a + b*x^2)^(4/3))/(8*b^2) + (3*(a + b*x^2)^(7/3))/(14*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 \sqrt [3]{a+b x^2} \, dx &=\frac {1}{2} \text {Subst}\left (\int x \sqrt [3]{a+b x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (-\frac {a \sqrt [3]{a+b x}}{b}+\frac {(a+b x)^{4/3}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac {3 a \left (a+b x^2\right )^{4/3}}{8 b^2}+\frac {3 \left (a+b x^2\right )^{7/3}}{14 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 38, normalized size = 1.00 \begin {gather*} \frac {3 \sqrt [3]{a+b x^2} \left (-3 a^2+a b x^2+4 b^2 x^4\right )}{56 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)^(1/3),x]

[Out]

(3*(a + b*x^2)^(1/3)*(-3*a^2 + a*b*x^2 + 4*b^2*x^4))/(56*b^2)

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

method	result	size
gosper	\(-\frac {3 \left (b \,x^{2}+a \right )^{\frac {4}{3}} \left (-4 b \,x^{2}+3 a \right )}{56 b^{2}}\)	\(25\)
trager	\(-\frac {3 \left (-4 b^{2} x^{4}-a b \,x^{2}+3 a^{2}\right ) \left (b \,x^{2}+a \right )^{\frac {1}{3}}}{56 b^{2}}\)	\(36\)
risch	\(-\frac {3 \left (-4 b^{2} x^{4}-a b \,x^{2}+3 a^{2}\right ) \left (b \,x^{2}+a \right )^{\frac {1}{3}}}{56 b^{2}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^(1/3),x,method=_RETURNVERBOSE)

[Out]

-3/56*(b*x^2+a)^(4/3)*(-4*b*x^2+3*a)/b^2

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Maxima [A]
time = 0.33, size = 30, normalized size = 0.79 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{3}}}{14 \, b^{2}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} a}{8 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/14*(b*x^2 + a)^(7/3)/b^2 - 3/8*(b*x^2 + a)^(4/3)*a/b^2

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Fricas [A]
time = 1.95, size = 34, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (4 \, b^{2} x^{4} + a b x^{2} - 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{56 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/56*(4*b^2*x^4 + a*b*x^2 - 3*a^2)*(b*x^2 + a)^(1/3)/b^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (34) = 68\).
time = 0.59, size = 223, normalized size = 5.87 \begin {gather*} - \frac {9 a^{\frac {13}{3}} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + \frac {9 a^{\frac {13}{3}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} - \frac {6 a^{\frac {10}{3}} b x^{2} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + \frac {9 a^{\frac {10}{3}} b x^{2}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + \frac {15 a^{\frac {7}{3}} b^{2} x^{4} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} + \frac {12 a^{\frac {4}{3}} b^{3} x^{6} \sqrt [3]{1 + \frac {b x^{2}}{a}}}{56 a^{2} b^{2} + 56 a b^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*a**(13/3)*(1 + b*x**2/a)**(1/3)/(56*a**2*b**2 + 56*a*b**3*x**2) + 9*a**(13/3)/(56*a**2*b**2 + 56*a*b**3*x**
2) - 6*a**(10/3)*b*x**2*(1 + b*x**2/a)**(1/3)/(56*a**2*b**2 + 56*a*b**3*x**2) + 9*a**(10/3)*b*x**2/(56*a**2*b*
*2 + 56*a*b**3*x**2) + 15*a**(7/3)*b**2*x**4*(1 + b*x**2/a)**(1/3)/(56*a**2*b**2 + 56*a*b**3*x**2) + 12*a**(4/
3)*b**3*x**6*(1 + b*x**2/a)**(1/3)/(56*a**2*b**2 + 56*a*b**3*x**2)

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Giac [A]
time = 1.13, size = 29, normalized size = 0.76 \begin {gather*} \frac {3 \, {\left (4 \, {\left (b x^{2} + a\right )}^{\frac {7}{3}} - 7 \, {\left (b x^{2} + a\right )}^{\frac {4}{3}} a\right )}}{56 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/56*(4*(b*x^2 + a)^(7/3) - 7*(b*x^2 + a)^(4/3)*a)/b^2

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Mupad [B]
time = 4.72, size = 33, normalized size = 0.87 \begin {gather*} {\left (b\,x^2+a\right )}^{1/3}\,\left (\frac {3\,x^4}{14}-\frac {9\,a^2}{56\,b^2}+\frac {3\,a\,x^2}{56\,b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2)^(1/3),x)

[Out]

(a + b*x^2)^(1/3)*((3*x^4)/14 - (9*a^2)/(56*b^2) + (3*a*x^2)/(56*b))

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