\(\int \frac {a+b x}{x^{5/3}} \, dx\) [657]

Optimal result

Integrand size = 11, antiderivative size = 19 \[ \int \frac {a+b x}{x^{5/3}} \, dx=-\frac {3 a}{2 x^{2/3}}+3 b \sqrt [3]{x} \]

[Out]

-3/2*a/x^(2/3)+3*b*x^(1/3)

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} \[ \int \frac {a+b x}{x^{5/3}} \, dx=3 b \sqrt [3]{x}-\frac {3 a}{2 x^{2/3}} \]

[In]

Int[(a + b*x)/x^(5/3),x]

[Out]

(-3*a)/(2*x^(2/3)) + 3*b*x^(1/3)

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^{5/3}}+\frac {b}{x^{2/3}}\right ) \, dx \\ & = -\frac {3 a}{2 x^{2/3}}+3 b \sqrt [3]{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {a+b x}{x^{5/3}} \, dx=-\frac {3 (a-2 b x)}{2 x^{2/3}} \]

[In]

Integrate[(a + b*x)/x^(5/3),x]

[Out]

(-3*(a - 2*b*x))/(2*x^(2/3))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63

method	result	size
gosper	\(-\frac {3 \left (-2 b x +a \right )}{2 x^{\frac {2}{3}}}\)	\(12\)
trager	\(-\frac {3 \left (-2 b x +a \right )}{2 x^{\frac {2}{3}}}\)	\(12\)
risch	\(-\frac {3 \left (-2 b x +a \right )}{2 x^{\frac {2}{3}}}\)	\(12\)
derivativedivides	\(-\frac {3 a}{2 x^{\frac {2}{3}}}+3 b \,x^{\frac {1}{3}}\)	\(14\)
default	\(-\frac {3 a}{2 x^{\frac {2}{3}}}+3 b \,x^{\frac {1}{3}}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {a+b x}{x^{5/3}} \, dx=\frac {3 \, {\left (2 \, b x - a\right )}}{2 \, x^{\frac {2}{3}}} \]

[In]

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

[Out]

3/2*(2*b*x - a)/x^(2/3)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {a+b x}{x^{5/3}} \, dx=- \frac {3 a}{2 x^{\frac {2}{3}}} + 3 b \sqrt [3]{x} \]

[In]

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

[Out]

-3*a/(2*x**(2/3)) + 3*b*x**(1/3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {a+b x}{x^{5/3}} \, dx=3 \, b x^{\frac {1}{3}} - \frac {3 \, a}{2 \, x^{\frac {2}{3}}} \]

[In]

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

[Out]

3*b*x^(1/3) - 3/2*a/x^(2/3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {a+b x}{x^{5/3}} \, dx=3 \, b x^{\frac {1}{3}} - \frac {3 \, a}{2 \, x^{\frac {2}{3}}} \]

[In]

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

[Out]

3*b*x^(1/3) - 3/2*a/x^(2/3)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {a+b x}{x^{5/3}} \, dx=-\frac {3\,a-6\,b\,x}{2\,x^{2/3}} \]

[In]

int((a + b*x)/x^(5/3),x)

[Out]

-(3*a - 6*b*x)/(2*x^(2/3))