\(\int x (a+b x)^{2/3} \, dx\) [380]

Optimal result

Integrand size = 11, antiderivative size = 34 \[ \int x (a+b x)^{2/3} \, dx=-\frac {3 a (a+b x)^{5/3}}{5 b^2}+\frac {3 (a+b x)^{8/3}}{8 b^2} \]

[Out]

-3/5*a*(b*x+a)^(5/3)/b^2+3/8*(b*x+a)^(8/3)/b^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int x (a+b x)^{2/3} \, dx=\frac {3 (a+b x)^{8/3}}{8 b^2}-\frac {3 a (a+b x)^{5/3}}{5 b^2} \]

[In]

Int[x*(a + b*x)^(2/3),x]

[Out]

(-3*a*(a + b*x)^(5/3))/(5*b^2) + (3*(a + b*x)^(8/3))/(8*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])

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {a (a+b x)^{2/3}}{b}+\frac {(a+b x)^{5/3}}{b}\right ) \, dx \\ & = -\frac {3 a (a+b x)^{5/3}}{5 b^2}+\frac {3 (a+b x)^{8/3}}{8 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int x (a+b x)^{2/3} \, dx=\frac {3 (a+b x)^{2/3} \left (-3 a^2+2 a b x+5 b^2 x^2\right )}{40 b^2} \]

[In]

Integrate[x*(a + b*x)^(2/3),x]

[Out]

(3*(a + b*x)^(2/3)*(-3*a^2 + 2*a*b*x + 5*b^2*x^2))/(40*b^2)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62

method	result	size
gosper	\(-\frac {3 \left (b x +a \right )^{\frac {5}{3}} \left (-5 b x +3 a \right )}{40 b^{2}}\)	\(21\)
pseudoelliptic	\(-\frac {3 \left (b x +a \right )^{\frac {5}{3}} \left (-5 b x +3 a \right )}{40 b^{2}}\)	\(21\)
derivativedivides	\(\frac {\frac {3 \left (b x +a \right )^{\frac {8}{3}}}{8}-\frac {3 a \left (b x +a \right )^{\frac {5}{3}}}{5}}{b^{2}}\)	\(26\)
default	\(\frac {\frac {3 \left (b x +a \right )^{\frac {8}{3}}}{8}-\frac {3 a \left (b x +a \right )^{\frac {5}{3}}}{5}}{b^{2}}\)	\(26\)
trager	\(-\frac {3 \left (-5 b^{2} x^{2}-2 a b x +3 a^{2}\right ) \left (b x +a \right )^{\frac {2}{3}}}{40 b^{2}}\)	\(32\)
risch	\(-\frac {3 \left (-5 b^{2} x^{2}-2 a b x +3 a^{2}\right ) \left (b x +a \right )^{\frac {2}{3}}}{40 b^{2}}\)	\(32\)

[In]

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

[Out]

-3/40*(b*x+a)^(5/3)*(-5*b*x+3*a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int x (a+b x)^{2/3} \, dx=\frac {3 \, {\left (5 \, b^{2} x^{2} + 2 \, a b x - 3 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {2}{3}}}{40 \, b^{2}} \]

[In]

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

[Out]

3/40*(5*b^2*x^2 + 2*a*b*x - 3*a^2)*(b*x + a)^(2/3)/b^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (31) = 62\).

Time = 0.82 (sec) , antiderivative size = 202, normalized size of antiderivative = 5.94 \[ \int x (a+b x)^{2/3} \, dx=- \frac {9 a^{\frac {14}{3}} \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{2} b^{2} + 40 a b^{3} x} + \frac {9 a^{\frac {14}{3}}}{40 a^{2} b^{2} + 40 a b^{3} x} - \frac {3 a^{\frac {11}{3}} b x \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{2} b^{2} + 40 a b^{3} x} + \frac {9 a^{\frac {11}{3}} b x}{40 a^{2} b^{2} + 40 a b^{3} x} + \frac {21 a^{\frac {8}{3}} b^{2} x^{2} \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{2} b^{2} + 40 a b^{3} x} + \frac {15 a^{\frac {5}{3}} b^{3} x^{3} \left (1 + \frac {b x}{a}\right )^{\frac {2}{3}}}{40 a^{2} b^{2} + 40 a b^{3} x} \]

[In]

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

[Out]

-9*a**(14/3)*(1 + b*x/a)**(2/3)/(40*a**2*b**2 + 40*a*b**3*x) + 9*a**(14/3)/(40*a**2*b**2 + 40*a*b**3*x) - 3*a*
*(11/3)*b*x*(1 + b*x/a)**(2/3)/(40*a**2*b**2 + 40*a*b**3*x) + 9*a**(11/3)*b*x/(40*a**2*b**2 + 40*a*b**3*x) + 2
1*a**(8/3)*b**2*x**2*(1 + b*x/a)**(2/3)/(40*a**2*b**2 + 40*a*b**3*x) + 15*a**(5/3)*b**3*x**3*(1 + b*x/a)**(2/3
)/(40*a**2*b**2 + 40*a*b**3*x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int x (a+b x)^{2/3} \, dx=\frac {3 \, {\left (b x + a\right )}^{\frac {8}{3}}}{8 \, b^{2}} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{3}} a}{5 \, b^{2}} \]

[In]

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

[Out]

3/8*(b*x + a)^(8/3)/b^2 - 3/5*(b*x + a)^(5/3)*a/b^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (26) = 52\).

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int x (a+b x)^{2/3} \, dx=\frac {3 \, {\left (\frac {4 \, {\left (2 \, {\left (b x + a\right )}^{\frac {5}{3}} - 5 \, {\left (b x + a\right )}^{\frac {2}{3}} a\right )} a}{b} + \frac {5 \, {\left (b x + a\right )}^{\frac {8}{3}} - 16 \, {\left (b x + a\right )}^{\frac {5}{3}} a + 20 \, {\left (b x + a\right )}^{\frac {2}{3}} a^{2}}{b}\right )}}{40 \, b} \]

[In]

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

[Out]

3/40*(4*(2*(b*x + a)^(5/3) - 5*(b*x + a)^(2/3)*a)*a/b + (5*(b*x + a)^(8/3) - 16*(b*x + a)^(5/3)*a + 20*(b*x +
a)^(2/3)*a^2)/b)/b

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int x (a+b x)^{2/3} \, dx=-\frac {24\,a\,{\left (a+b\,x\right )}^{5/3}-15\,{\left (a+b\,x\right )}^{8/3}}{40\,b^2} \]

[In]

int(x*(a + b*x)^(2/3),x)

[Out]

-(24*a*(a + b*x)^(5/3) - 15*(a + b*x)^(8/3))/(40*b^2)