3.6.63 \(\int \frac {x^5}{(a+b x^3)^{2/3}} \, dx\) [563]

Optimal. Leaf size=36 \[ -\frac {a \sqrt [3]{a+b x^3}}{b^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 36, 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 {\left (a+b x^3\right )^{4/3}}{4 b^2}-\frac {a \sqrt [3]{a+b x^3}}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^3)^(2/3),x]

[Out]

-((a*(a + b*x^3)^(1/3))/b^2) + (a + b*x^3)^(4/3)/(4*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 \frac {x^5}{\left (a+b x^3\right )^{2/3}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x}{(a+b x)^{2/3}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {a}{b (a+b x)^{2/3}}+\frac {\sqrt [3]{a+b x}}{b}\right ) \, dx,x,x^3\right )\\ &=-\frac {a \sqrt [3]{a+b x^3}}{b^2}+\frac {\left (a+b x^3\right )^{4/3}}{4 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 27, normalized size = 0.75 \begin {gather*} \frac {\left (-3 a+b x^3\right ) \sqrt [3]{a+b x^3}}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^3)^(2/3),x]

[Out]

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

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Maple [A]
time = 0.14, size = 25, normalized size = 0.69

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 30, normalized size = 0.83 \begin {gather*} \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{4 \, b^{2}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} a}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 23, normalized size = 0.64 \begin {gather*} \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b x^{3} - 3 \, a\right )}}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.26, size = 44, normalized size = 1.22 \begin {gather*} \begin {cases} - \frac {3 a \sqrt [3]{a + b x^{3}}}{4 b^{2}} + \frac {x^{3} \sqrt [3]{a + b x^{3}}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 a^{\frac {2}{3}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b*x**3+a)**(2/3),x)

[Out]

Piecewise((-3*a*(a + b*x**3)**(1/3)/(4*b**2) + x**3*(a + b*x**3)**(1/3)/(4*b), Ne(b, 0)), (x**6/(6*a**(2/3)),
True))

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Giac [A]
time = 2.51, size = 30, normalized size = 0.83 \begin {gather*} \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{4 \, b^{2}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} a}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.07, size = 24, normalized size = 0.67 \begin {gather*} -\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (3\,a-b\,x^3\right )}{4\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^3)^(2/3),x)

[Out]

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

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