3.24.3 \(\int (a+b \sqrt [3]{x})^3 x^4 \, dx\) [2303]

Optimal. Leaf size=47 \[ \frac {a^3 x^5}{5}+\frac {9}{16} a^2 b x^{16/3}+\frac {9}{17} a b^2 x^{17/3}+\frac {b^3 x^6}{6} \]

[Out]

1/5*a^3*x^5+9/16*a^2*b*x^(16/3)+9/17*a*b^2*x^(17/3)+1/6*b^3*x^6

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Rubi [A]
time = 0.02, antiderivative size = 47, 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 {a^3 x^5}{5}+\frac {9}{16} a^2 b x^{16/3}+\frac {9}{17} a b^2 x^{17/3}+\frac {b^3 x^6}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^3*x^4,x]

[Out]

(a^3*x^5)/5 + (9*a^2*b*x^(16/3))/16 + (9*a*b^2*x^(17/3))/17 + (b^3*x^6)/6

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 \left (a+b \sqrt [3]{x}\right )^3 x^4 \, dx &=3 \text {Subst}\left (\int x^{14} (a+b x)^3 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \text {Subst}\left (\int \left (a^3 x^{14}+3 a^2 b x^{15}+3 a b^2 x^{16}+b^3 x^{17}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {a^3 x^5}{5}+\frac {9}{16} a^2 b x^{16/3}+\frac {9}{17} a b^2 x^{17/3}+\frac {b^3 x^6}{6}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 43, normalized size = 0.91 \begin {gather*} \frac {816 a^3 x^5+2295 a^2 b x^{16/3}+2160 a b^2 x^{17/3}+680 b^3 x^6}{4080} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^3*x^4,x]

[Out]

(816*a^3*x^5 + 2295*a^2*b*x^(16/3) + 2160*a*b^2*x^(17/3) + 680*b^3*x^6)/4080

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Maple [A]
time = 0.19, size = 36, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {a^{3} x^{5}}{5}+\frac {9 a^{2} b \,x^{\frac {16}{3}}}{16}+\frac {9 a \,b^{2} x^{\frac {17}{3}}}{17}+\frac {b^{3} x^{6}}{6}\)	\(36\)
default	\(\frac {a^{3} x^{5}}{5}+\frac {9 a^{2} b \,x^{\frac {16}{3}}}{16}+\frac {9 a \,b^{2} x^{\frac {17}{3}}}{17}+\frac {b^{3} x^{6}}{6}\)	\(36\)
trager	\(\frac {\left (5 b^{3} x^{5}+6 a^{3} x^{4}+5 b^{3} x^{4}+6 a^{3} x^{3}+5 b^{3} x^{3}+6 a^{3} x^{2}+5 b^{3} x^{2}+6 a^{3} x +5 b^{3} x +6 a^{3}+5 b^{3}\right ) \left (x -1\right )}{30}+\frac {9 a^{2} b \,x^{\frac {16}{3}}}{16}+\frac {9 a \,b^{2} x^{\frac {17}{3}}}{17}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/3))^3*x^4,x,method=_RETURNVERBOSE)

[Out]

1/5*a^3*x^5+9/16*a^2*b*x^(16/3)+9/17*a*b^2*x^(17/3)+1/6*b^3*x^6

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (35) = 70\).
time = 0.30, size = 251, normalized size = 5.34 \begin {gather*} \frac {{\left (b x^{\frac {1}{3}} + a\right )}^{18}}{6 \, b^{15}} - \frac {42 \, {\left (b x^{\frac {1}{3}} + a\right )}^{17} a}{17 \, b^{15}} + \frac {273 \, {\left (b x^{\frac {1}{3}} + a\right )}^{16} a^{2}}{16 \, b^{15}} - \frac {364 \, {\left (b x^{\frac {1}{3}} + a\right )}^{15} a^{3}}{5 \, b^{15}} + \frac {429 \, {\left (b x^{\frac {1}{3}} + a\right )}^{14} a^{4}}{2 \, b^{15}} - \frac {462 \, {\left (b x^{\frac {1}{3}} + a\right )}^{13} a^{5}}{b^{15}} + \frac {3003 \, {\left (b x^{\frac {1}{3}} + a\right )}^{12} a^{6}}{4 \, b^{15}} - \frac {936 \, {\left (b x^{\frac {1}{3}} + a\right )}^{11} a^{7}}{b^{15}} + \frac {9009 \, {\left (b x^{\frac {1}{3}} + a\right )}^{10} a^{8}}{10 \, b^{15}} - \frac {2002 \, {\left (b x^{\frac {1}{3}} + a\right )}^{9} a^{9}}{3 \, b^{15}} + \frac {3003 \, {\left (b x^{\frac {1}{3}} + a\right )}^{8} a^{10}}{8 \, b^{15}} - \frac {156 \, {\left (b x^{\frac {1}{3}} + a\right )}^{7} a^{11}}{b^{15}} + \frac {91 \, {\left (b x^{\frac {1}{3}} + a\right )}^{6} a^{12}}{2 \, b^{15}} - \frac {42 \, {\left (b x^{\frac {1}{3}} + a\right )}^{5} a^{13}}{5 \, b^{15}} + \frac {3 \, {\left (b x^{\frac {1}{3}} + a\right )}^{4} a^{14}}{4 \, b^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^4,x, algorithm="maxima")

[Out]

1/6*(b*x^(1/3) + a)^18/b^15 - 42/17*(b*x^(1/3) + a)^17*a/b^15 + 273/16*(b*x^(1/3) + a)^16*a^2/b^15 - 364/5*(b*
x^(1/3) + a)^15*a^3/b^15 + 429/2*(b*x^(1/3) + a)^14*a^4/b^15 - 462*(b*x^(1/3) + a)^13*a^5/b^15 + 3003/4*(b*x^(
1/3) + a)^12*a^6/b^15 - 936*(b*x^(1/3) + a)^11*a^7/b^15 + 9009/10*(b*x^(1/3) + a)^10*a^8/b^15 - 2002/3*(b*x^(1
/3) + a)^9*a^9/b^15 + 3003/8*(b*x^(1/3) + a)^8*a^10/b^15 - 156*(b*x^(1/3) + a)^7*a^11/b^15 + 91/2*(b*x^(1/3) +
a)^6*a^12/b^15 - 42/5*(b*x^(1/3) + a)^5*a^13/b^15 + 3/4*(b*x^(1/3) + a)^4*a^14/b^15

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Fricas [A]
time = 0.34, size = 35, normalized size = 0.74 \begin {gather*} \frac {1}{6} \, b^{3} x^{6} + \frac {9}{17} \, a b^{2} x^{\frac {17}{3}} + \frac {9}{16} \, a^{2} b x^{\frac {16}{3}} + \frac {1}{5} \, a^{3} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^4,x, algorithm="fricas")

[Out]

1/6*b^3*x^6 + 9/17*a*b^2*x^(17/3) + 9/16*a^2*b*x^(16/3) + 1/5*a^3*x^5

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Sympy [A]
time = 1.08, size = 42, normalized size = 0.89 \begin {gather*} \frac {a^{3} x^{5}}{5} + \frac {9 a^{2} b x^{\frac {16}{3}}}{16} + \frac {9 a b^{2} x^{\frac {17}{3}}}{17} + \frac {b^{3} x^{6}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/3))**3*x**4,x)

[Out]

a**3*x**5/5 + 9*a**2*b*x**(16/3)/16 + 9*a*b**2*x**(17/3)/17 + b**3*x**6/6

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Giac [A]
time = 1.73, size = 35, normalized size = 0.74 \begin {gather*} \frac {1}{6} \, b^{3} x^{6} + \frac {9}{17} \, a b^{2} x^{\frac {17}{3}} + \frac {9}{16} \, a^{2} b x^{\frac {16}{3}} + \frac {1}{5} \, a^{3} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/3))^3*x^4,x, algorithm="giac")

[Out]

1/6*b^3*x^6 + 9/17*a*b^2*x^(17/3) + 9/16*a^2*b*x^(16/3) + 1/5*a^3*x^5

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Mupad [B]
time = 0.04, size = 35, normalized size = 0.74 \begin {gather*} \frac {a^3\,x^5}{5}+\frac {b^3\,x^6}{6}+\frac {9\,a^2\,b\,x^{16/3}}{16}+\frac {9\,a\,b^2\,x^{17/3}}{17} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*x^(1/3))^3,x)

[Out]

(a^3*x^5)/5 + (b^3*x^6)/6 + (9*a^2*b*x^(16/3))/16 + (9*a*b^2*x^(17/3))/17

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