\(\int (a+b \sqrt [3]{x})^5 x^2 \, dx\) [2314]

Optimal result

Integrand size = 15, antiderivative size = 77 \[ \int \left (a+b \sqrt [3]{x}\right )^5 x^2 \, dx=\frac {a^5 x^3}{3}+\frac {3}{2} a^4 b x^{10/3}+\frac {30}{11} a^3 b^2 x^{11/3}+\frac {5}{2} a^2 b^3 x^4+\frac {15}{13} a b^4 x^{13/3}+\frac {3}{14} b^5 x^{14/3} \]

[Out]

1/3*a^5*x^3+3/2*a^4*b*x^(10/3)+30/11*a^3*b^2*x^(11/3)+5/2*a^2*b^3*x^4+15/13*a*b^4*x^(13/3)+3/14*b^5*x^(14/3)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \left (a+b \sqrt [3]{x}\right )^5 x^2 \, dx=\frac {a^5 x^3}{3}+\frac {3}{2} a^4 b x^{10/3}+\frac {30}{11} a^3 b^2 x^{11/3}+\frac {5}{2} a^2 b^3 x^4+\frac {15}{13} a b^4 x^{13/3}+\frac {3}{14} b^5 x^{14/3} \]

[In]

Int[(a + b*x^(1/3))^5*x^2,x]

[Out]

(a^5*x^3)/3 + (3*a^4*b*x^(10/3))/2 + (30*a^3*b^2*x^(11/3))/11 + (5*a^2*b^3*x^4)/2 + (15*a*b^4*x^(13/3))/13 + (
3*b^5*x^(14/3))/14

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*} \text {integral}& = 3 \text {Subst}\left (\int x^8 (a+b x)^5 \, dx,x,\sqrt [3]{x}\right ) \\ & = 3 \text {Subst}\left (\int \left (a^5 x^8+5 a^4 b x^9+10 a^3 b^2 x^{10}+10 a^2 b^3 x^{11}+5 a b^4 x^{12}+b^5 x^{13}\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {a^5 x^3}{3}+\frac {3}{2} a^4 b x^{10/3}+\frac {30}{11} a^3 b^2 x^{11/3}+\frac {5}{2} a^2 b^3 x^4+\frac {15}{13} a b^4 x^{13/3}+\frac {3}{14} b^5 x^{14/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \left (a+b \sqrt [3]{x}\right )^5 x^2 \, dx=\frac {2002 a^5 x^3+9009 a^4 b x^{10/3}+16380 a^3 b^2 x^{11/3}+15015 a^2 b^3 x^4+6930 a b^4 x^{13/3}+1287 b^5 x^{14/3}}{6006} \]

[In]

Integrate[(a + b*x^(1/3))^5*x^2,x]

[Out]

(2002*a^5*x^3 + 9009*a^4*b*x^(10/3) + 16380*a^3*b^2*x^(11/3) + 15015*a^2*b^3*x^4 + 6930*a*b^4*x^(13/3) + 1287*
b^5*x^(14/3))/6006

Maple [A] (verified)

Time = 3.63 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {a^{5} x^{3}}{3}+\frac {3 a^{4} b \,x^{\frac {10}{3}}}{2}+\frac {30 a^{3} b^{2} x^{\frac {11}{3}}}{11}+\frac {5 a^{2} b^{3} x^{4}}{2}+\frac {15 a \,b^{4} x^{\frac {13}{3}}}{13}+\frac {3 b^{5} x^{\frac {14}{3}}}{14}\)	\(58\)
default	\(\frac {a^{5} x^{3}}{3}+\frac {3 a^{4} b \,x^{\frac {10}{3}}}{2}+\frac {30 a^{3} b^{2} x^{\frac {11}{3}}}{11}+\frac {5 a^{2} b^{3} x^{4}}{2}+\frac {15 a \,b^{4} x^{\frac {13}{3}}}{13}+\frac {3 b^{5} x^{\frac {14}{3}}}{14}\)	\(58\)
trager	\(\frac {a^{2} \left (15 b^{3} x^{3}+2 a^{3} x^{2}+15 b^{3} x^{2}+2 a^{3} x +15 b^{3} x +2 a^{3}+15 b^{3}\right ) \left (-1+x \right )}{6}+\frac {3 a b \,x^{\frac {10}{3}} \left (10 b^{3} x +13 a^{3}\right )}{26}+\frac {3 b^{2} x^{\frac {11}{3}} \left (11 b^{3} x +140 a^{3}\right )}{154}\)	\(96\)

[In]

int((a+b*x^(1/3))^5*x^2,x,method=_RETURNVERBOSE)

[Out]

1/3*a^5*x^3+3/2*a^4*b*x^(10/3)+30/11*a^3*b^2*x^(11/3)+5/2*a^2*b^3*x^4+15/13*a*b^4*x^(13/3)+3/14*b^5*x^(14/3)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \left (a+b \sqrt [3]{x}\right )^5 x^2 \, dx=\frac {5}{2} \, a^{2} b^{3} x^{4} + \frac {1}{3} \, a^{5} x^{3} + \frac {3}{154} \, {\left (11 \, b^{5} x^{4} + 140 \, a^{3} b^{2} x^{3}\right )} x^{\frac {2}{3}} + \frac {3}{26} \, {\left (10 \, a b^{4} x^{4} + 13 \, a^{4} b x^{3}\right )} x^{\frac {1}{3}} \]

[In]

integrate((a+b*x^(1/3))^5*x^2,x, algorithm="fricas")

[Out]

5/2*a^2*b^3*x^4 + 1/3*a^5*x^3 + 3/154*(11*b^5*x^4 + 140*a^3*b^2*x^3)*x^(2/3) + 3/26*(10*a*b^4*x^4 + 13*a^4*b*x
^3)*x^(1/3)

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \left (a+b \sqrt [3]{x}\right )^5 x^2 \, dx=\frac {a^{5} x^{3}}{3} + \frac {3 a^{4} b x^{\frac {10}{3}}}{2} + \frac {30 a^{3} b^{2} x^{\frac {11}{3}}}{11} + \frac {5 a^{2} b^{3} x^{4}}{2} + \frac {15 a b^{4} x^{\frac {13}{3}}}{13} + \frac {3 b^{5} x^{\frac {14}{3}}}{14} \]

[In]

integrate((a+b*x**(1/3))**5*x**2,x)

[Out]

a**5*x**3/3 + 3*a**4*b*x**(10/3)/2 + 30*a**3*b**2*x**(11/3)/11 + 5*a**2*b**3*x**4/2 + 15*a*b**4*x**(13/3)/13 +
3*b**5*x**(14/3)/14

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (57) = 114\).

Time = 0.21 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.94 \[ \int \left (a+b \sqrt [3]{x}\right )^5 x^2 \, dx=\frac {3 \, {\left (b x^{\frac {1}{3}} + a\right )}^{14}}{14 \, b^{9}} - \frac {24 \, {\left (b x^{\frac {1}{3}} + a\right )}^{13} a}{13 \, b^{9}} + \frac {7 \, {\left (b x^{\frac {1}{3}} + a\right )}^{12} a^{2}}{b^{9}} - \frac {168 \, {\left (b x^{\frac {1}{3}} + a\right )}^{11} a^{3}}{11 \, b^{9}} + \frac {21 \, {\left (b x^{\frac {1}{3}} + a\right )}^{10} a^{4}}{b^{9}} - \frac {56 \, {\left (b x^{\frac {1}{3}} + a\right )}^{9} a^{5}}{3 \, b^{9}} + \frac {21 \, {\left (b x^{\frac {1}{3}} + a\right )}^{8} a^{6}}{2 \, b^{9}} - \frac {24 \, {\left (b x^{\frac {1}{3}} + a\right )}^{7} a^{7}}{7 \, b^{9}} + \frac {{\left (b x^{\frac {1}{3}} + a\right )}^{6} a^{8}}{2 \, b^{9}} \]

[In]

integrate((a+b*x^(1/3))^5*x^2,x, algorithm="maxima")

[Out]

3/14*(b*x^(1/3) + a)^14/b^9 - 24/13*(b*x^(1/3) + a)^13*a/b^9 + 7*(b*x^(1/3) + a)^12*a^2/b^9 - 168/11*(b*x^(1/3
) + a)^11*a^3/b^9 + 21*(b*x^(1/3) + a)^10*a^4/b^9 - 56/3*(b*x^(1/3) + a)^9*a^5/b^9 + 21/2*(b*x^(1/3) + a)^8*a^
6/b^9 - 24/7*(b*x^(1/3) + a)^7*a^7/b^9 + 1/2*(b*x^(1/3) + a)^6*a^8/b^9

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.74 \[ \int \left (a+b \sqrt [3]{x}\right )^5 x^2 \, dx=\frac {3}{14} \, b^{5} x^{\frac {14}{3}} + \frac {15}{13} \, a b^{4} x^{\frac {13}{3}} + \frac {5}{2} \, a^{2} b^{3} x^{4} + \frac {30}{11} \, a^{3} b^{2} x^{\frac {11}{3}} + \frac {3}{2} \, a^{4} b x^{\frac {10}{3}} + \frac {1}{3} \, a^{5} x^{3} \]

[In]

integrate((a+b*x^(1/3))^5*x^2,x, algorithm="giac")

[Out]

3/14*b^5*x^(14/3) + 15/13*a*b^4*x^(13/3) + 5/2*a^2*b^3*x^4 + 30/11*a^3*b^2*x^(11/3) + 3/2*a^4*b*x^(10/3) + 1/3
*a^5*x^3

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.74 \[ \int \left (a+b \sqrt [3]{x}\right )^5 x^2 \, dx=\frac {a^5\,x^3}{3}+\frac {3\,b^5\,x^{14/3}}{14}+\frac {3\,a^4\,b\,x^{10/3}}{2}+\frac {15\,a\,b^4\,x^{13/3}}{13}+\frac {5\,a^2\,b^3\,x^4}{2}+\frac {30\,a^3\,b^2\,x^{11/3}}{11} \]

[In]

int(x^2*(a + b*x^(1/3))^5,x)

[Out]

(a^5*x^3)/3 + (3*b^5*x^(14/3))/14 + (3*a^4*b*x^(10/3))/2 + (15*a*b^4*x^(13/3))/13 + (5*a^2*b^3*x^4)/2 + (30*a^
3*b^2*x^(11/3))/11