3.25.6 \(\int (a+\frac {b}{\sqrt [3]{x}})^2 x \, dx\) [2406]

Optimal. Leaf size=34 \[ \frac {3}{4} b^2 x^{4/3}+\frac {6}{5} a b x^{5/3}+\frac {a^2 x^2}{2} \]

[Out]

3/4*b^2*x^(4/3)+6/5*a*b*x^(5/3)+1/2*a^2*x^2

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Rubi [A]
time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {269, 272, 45} \begin {gather*} \frac {a^2 x^2}{2}+\frac {6}{5} a b x^{5/3}+\frac {3}{4} b^2 x^{4/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^(1/3))^2*x,x]

[Out]

(3*b^2*x^(4/3))/4 + (6*a*b*x^(5/3))/5 + (a^2*x^2)/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 269

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

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 \left (a+\frac {b}{\sqrt [3]{x}}\right )^2 x \, dx &=\int \left (b+a \sqrt [3]{x}\right )^2 \sqrt [3]{x} \, dx\\ &=3 \text {Subst}\left (\int x^3 (b+a x)^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \text {Subst}\left (\int \left (b^2 x^3+2 a b x^4+a^2 x^5\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {3}{4} b^2 x^{4/3}+\frac {6}{5} a b x^{5/3}+\frac {a^2 x^2}{2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 34, normalized size = 1.00 \begin {gather*} \frac {1}{20} \left (15 b^2+24 a b \sqrt [3]{x}+10 a^2 x^{2/3}\right ) x^{4/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^(1/3))^2*x,x]

[Out]

((15*b^2 + 24*a*b*x^(1/3) + 10*a^2*x^(2/3))*x^(4/3))/20

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Maple [A]
time = 0.20, size = 25, normalized size = 0.74

method	result	size
derivativedivides	\(\frac {3 b^{2} x^{\frac {4}{3}}}{4}+\frac {6 a b \,x^{\frac {5}{3}}}{5}+\frac {a^{2} x^{2}}{2}\)	\(25\)
default	\(\frac {3 b^{2} x^{\frac {4}{3}}}{4}+\frac {6 a b \,x^{\frac {5}{3}}}{5}+\frac {a^{2} x^{2}}{2}\)	\(25\)
trager	\(\frac {\left (x -1\right ) a^{2} \left (x +1\right )}{2}+\frac {3 b^{2} x^{\frac {4}{3}}}{4}+\frac {6 a b \,x^{\frac {5}{3}}}{5}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*b^2*x^(4/3)+6/5*a*b*x^(5/3)+1/2*a^2*x^2

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Maxima [A]
time = 0.30, size = 26, normalized size = 0.76 \begin {gather*} \frac {1}{20} \, {\left (10 \, a^{2} + \frac {24 \, a b}{x^{\frac {1}{3}}} + \frac {15 \, b^{2}}{x^{\frac {2}{3}}}\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(10*a^2 + 24*a*b/x^(1/3) + 15*b^2/x^(2/3))*x^2

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Fricas [A]
time = 0.36, size = 24, normalized size = 0.71 \begin {gather*} \frac {1}{2} \, a^{2} x^{2} + \frac {6}{5} \, a b x^{\frac {5}{3}} + \frac {3}{4} \, b^{2} x^{\frac {4}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*x^2 + 6/5*a*b*x^(5/3) + 3/4*b^2*x^(4/3)

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Sympy [A]
time = 0.10, size = 31, normalized size = 0.91 \begin {gather*} \frac {a^{2} x^{2}}{2} + \frac {6 a b x^{\frac {5}{3}}}{5} + \frac {3 b^{2} x^{\frac {4}{3}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**2/2 + 6*a*b*x**(5/3)/5 + 3*b**2*x**(4/3)/4

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Giac [A]
time = 0.60, size = 24, normalized size = 0.71 \begin {gather*} \frac {1}{2} \, a^{2} x^{2} + \frac {6}{5} \, a b x^{\frac {5}{3}} + \frac {3}{4} \, b^{2} x^{\frac {4}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*x^2 + 6/5*a*b*x^(5/3) + 3/4*b^2*x^(4/3)

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Mupad [B]
time = 0.04, size = 24, normalized size = 0.71 \begin {gather*} \frac {a^2\,x^2}{2}+\frac {3\,b^2\,x^{4/3}}{4}+\frac {6\,a\,b\,x^{5/3}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/x^(1/3))^2,x)

[Out]

(a^2*x^2)/2 + (3*b^2*x^(4/3))/4 + (6*a*b*x^(5/3))/5

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