\(\int \frac {a+b x}{x^{2/3}} \, dx\) [655]

Optimal result

Integrand size = 11, antiderivative size = 19 \[ \int \frac {a+b x}{x^{2/3}} \, dx=3 a \sqrt [3]{x}+\frac {3}{4} b x^{4/3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, 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} \[ \int \frac {a+b x}{x^{2/3}} \, dx=3 a \sqrt [3]{x}+\frac {3}{4} b x^{4/3} \]

[In]

Int[(a + b*x)/x^(2/3),x]

[Out]

3*a*x^(1/3) + (3*b*x^(4/3))/4

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}{x^{2/3}}+b \sqrt [3]{x}\right ) \, dx \\ & = 3 a \sqrt [3]{x}+\frac {3}{4} b x^{4/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {a+b x}{x^{2/3}} \, dx=\frac {3}{4} \sqrt [3]{x} (4 a+b x) \]

[In]

Integrate[(a + b*x)/x^(2/3),x]

[Out]

(3*x^(1/3)*(4*a + b*x))/4

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68

method	result	size
gosper	\(\frac {3 x^{\frac {1}{3}} \left (b x +4 a \right )}{4}\)	\(13\)
trager	\(\left (\frac {3 b x}{4}+3 a \right ) x^{\frac {1}{3}}\)	\(13\)
risch	\(\frac {3 x^{\frac {1}{3}} \left (b x +4 a \right )}{4}\)	\(13\)
derivativedivides	\(3 a \,x^{\frac {1}{3}}+\frac {3 b \,x^{\frac {4}{3}}}{4}\)	\(14\)
default	\(3 a \,x^{\frac {1}{3}}+\frac {3 b \,x^{\frac {4}{3}}}{4}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {a+b x}{x^{2/3}} \, dx=\frac {3}{4} \, {\left (b x + 4 \, a\right )} x^{\frac {1}{3}} \]

[In]

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

[Out]

3/4*(b*x + 4*a)*x^(1/3)

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {a+b x}{x^{2/3}} \, dx=3 a \sqrt [3]{x} + \frac {3 b x^{\frac {4}{3}}}{4} \]

[In]

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

[Out]

3*a*x**(1/3) + 3*b*x**(4/3)/4

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {a+b x}{x^{2/3}} \, dx=\frac {3}{4} \, b x^{\frac {4}{3}} + 3 \, a x^{\frac {1}{3}} \]

[In]

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

[Out]

3/4*b*x^(4/3) + 3*a*x^(1/3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {a+b x}{x^{2/3}} \, dx=\frac {3}{4} \, b x^{\frac {4}{3}} + 3 \, a x^{\frac {1}{3}} \]

[In]

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

[Out]

3/4*b*x^(4/3) + 3*a*x^(1/3)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {a+b x}{x^{2/3}} \, dx=\frac {3\,x^{1/3}\,\left (4\,a+b\,x\right )}{4} \]

[In]

int((a + b*x)/x^(2/3),x)

[Out]

(3*x^(1/3)*(4*a + b*x))/4