\(\int x^{2/3} (a+b x) \, dx\) [652]

Optimal result

Integrand size = 11, antiderivative size = 21 \[ \int x^{2/3} (a+b x) \, dx=\frac {3}{5} a x^{5/3}+\frac {3}{8} b x^{8/3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, 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} \[ \int x^{2/3} (a+b x) \, dx=\frac {3}{5} a x^{5/3}+\frac {3}{8} b x^{8/3} \]

[In]

Int[x^(2/3)*(a + b*x),x]

[Out]

(3*a*x^(5/3))/5 + (3*b*x^(8/3))/8

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 (a x^{2/3}+b x^{5/3}\right ) \, dx \\ & = \frac {3}{5} a x^{5/3}+\frac {3}{8} b x^{8/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int x^{2/3} (a+b x) \, dx=\frac {3}{40} x^{5/3} (8 a+5 b x) \]

[In]

Integrate[x^(2/3)*(a + b*x),x]

[Out]

(3*x^(5/3)*(8*a + 5*b*x))/40

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {3 x^{\frac {5}{3}} \left (5 b x +8 a \right )}{40}\)	\(14\)
derivativedivides	\(\frac {3 a \,x^{\frac {5}{3}}}{5}+\frac {3 b \,x^{\frac {8}{3}}}{8}\)	\(14\)
default	\(\frac {3 a \,x^{\frac {5}{3}}}{5}+\frac {3 b \,x^{\frac {8}{3}}}{8}\)	\(14\)
trager	\(\frac {3 x^{\frac {5}{3}} \left (5 b x +8 a \right )}{40}\)	\(14\)
risch	\(\frac {3 x^{\frac {5}{3}} \left (5 b x +8 a \right )}{40}\)	\(14\)

[In]

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

[Out]

3/40*x^(5/3)*(5*b*x+8*a)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int x^{2/3} (a+b x) \, dx=\frac {3}{40} \, {\left (5 \, b x^{2} + 8 \, a x\right )} x^{\frac {2}{3}} \]

[In]

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

[Out]

3/40*(5*b*x^2 + 8*a*x)*x^(2/3)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int x^{2/3} (a+b x) \, dx=\frac {3 a x^{\frac {5}{3}}}{5} + \frac {3 b x^{\frac {8}{3}}}{8} \]

[In]

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

[Out]

3*a*x**(5/3)/5 + 3*b*x**(8/3)/8

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x^{2/3} (a+b x) \, dx=\frac {3}{8} \, b x^{\frac {8}{3}} + \frac {3}{5} \, a x^{\frac {5}{3}} \]

[In]

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

[Out]

3/8*b*x^(8/3) + 3/5*a*x^(5/3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x^{2/3} (a+b x) \, dx=\frac {3}{8} \, b x^{\frac {8}{3}} + \frac {3}{5} \, a x^{\frac {5}{3}} \]

[In]

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

[Out]

3/8*b*x^(8/3) + 3/5*a*x^(5/3)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int x^{2/3} (a+b x) \, dx=\frac {3\,x^{5/3}\,\left (8\,a+5\,b\,x\right )}{40} \]

[In]

int(x^(2/3)*(a + b*x),x)

[Out]

(3*x^(5/3)*(8*a + 5*b*x))/40