\(\int \frac {(a+b x^3)^3}{x^5} \, dx\) [255]

Optimal result

Integrand size = 13, antiderivative size = 41 \[ \int \frac {\left (a+b x^3\right )^3}{x^5} \, dx=-\frac {a^3}{4 x^4}-\frac {3 a^2 b}{x}+\frac {3}{2} a b^2 x^2+\frac {b^3 x^5}{5} \]

[Out]

-1/4*a^3/x^4-3*a^2*b/x+3/2*a*b^2*x^2+1/5*b^3*x^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {276} \[ \int \frac {\left (a+b x^3\right )^3}{x^5} \, dx=-\frac {a^3}{4 x^4}-\frac {3 a^2 b}{x}+\frac {3}{2} a b^2 x^2+\frac {b^3 x^5}{5} \]

[In]

Int[(a + b*x^3)^3/x^5,x]

[Out]

-1/4*a^3/x^4 - (3*a^2*b)/x + (3*a*b^2*x^2)/2 + (b^3*x^5)/5

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^3}{x^5}+\frac {3 a^2 b}{x^2}+3 a b^2 x+b^3 x^4\right ) \, dx \\ & = -\frac {a^3}{4 x^4}-\frac {3 a^2 b}{x}+\frac {3}{2} a b^2 x^2+\frac {b^3 x^5}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^3}{x^5} \, dx=-\frac {a^3}{4 x^4}-\frac {3 a^2 b}{x}+\frac {3}{2} a b^2 x^2+\frac {b^3 x^5}{5} \]

[In]

Integrate[(a + b*x^3)^3/x^5,x]

[Out]

-1/4*a^3/x^4 - (3*a^2*b)/x + (3*a*b^2*x^2)/2 + (b^3*x^5)/5

Maple [A] (verified)

Time = 3.62 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {a^{3}}{4 x^{4}}-\frac {3 a^{2} b}{x}+\frac {3 a \,b^{2} x^{2}}{2}+\frac {b^{3} x^{5}}{5}\)	\(36\)
norman	\(\frac {\frac {1}{5} b^{3} x^{9}+\frac {3}{2} a \,b^{2} x^{6}-3 a^{2} b \,x^{3}-\frac {1}{4} a^{3}}{x^{4}}\)	\(37\)
gosper	\(-\frac {-4 b^{3} x^{9}-30 a \,b^{2} x^{6}+60 a^{2} b \,x^{3}+5 a^{3}}{20 x^{4}}\)	\(38\)
risch	\(\frac {b^{3} x^{5}}{5}+\frac {3 a \,b^{2} x^{2}}{2}+\frac {-3 a^{2} b \,x^{3}-\frac {1}{4} a^{3}}{x^{4}}\)	\(38\)
parallelrisch	\(\frac {4 b^{3} x^{9}+30 a \,b^{2} x^{6}-60 a^{2} b \,x^{3}-5 a^{3}}{20 x^{4}}\)	\(38\)

[In]

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

[Out]

-1/4*a^3/x^4-3*a^2*b/x+3/2*a*b^2*x^2+1/5*b^3*x^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^3\right )^3}{x^5} \, dx=\frac {4 \, b^{3} x^{9} + 30 \, a b^{2} x^{6} - 60 \, a^{2} b x^{3} - 5 \, a^{3}}{20 \, x^{4}} \]

[In]

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

[Out]

1/20*(4*b^3*x^9 + 30*a*b^2*x^6 - 60*a^2*b*x^3 - 5*a^3)/x^4

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^3\right )^3}{x^5} \, dx=\frac {3 a b^{2} x^{2}}{2} + \frac {b^{3} x^{5}}{5} + \frac {- a^{3} - 12 a^{2} b x^{3}}{4 x^{4}} \]

[In]

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

[Out]

3*a*b**2*x**2/2 + b**3*x**5/5 + (-a**3 - 12*a**2*b*x**3)/(4*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^3\right )^3}{x^5} \, dx=\frac {1}{5} \, b^{3} x^{5} + \frac {3}{2} \, a b^{2} x^{2} - \frac {12 \, a^{2} b x^{3} + a^{3}}{4 \, x^{4}} \]

[In]

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

[Out]

1/5*b^3*x^5 + 3/2*a*b^2*x^2 - 1/4*(12*a^2*b*x^3 + a^3)/x^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^3\right )^3}{x^5} \, dx=\frac {1}{5} \, b^{3} x^{5} + \frac {3}{2} \, a b^{2} x^{2} - \frac {12 \, a^{2} b x^{3} + a^{3}}{4 \, x^{4}} \]

[In]

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

[Out]

1/5*b^3*x^5 + 3/2*a*b^2*x^2 - 1/4*(12*a^2*b*x^3 + a^3)/x^4

Mupad [B] (verification not implemented)

Time = 5.40 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^3\right )^3}{x^5} \, dx=\frac {b^3\,x^5}{5}-\frac {\frac {a^3}{4}+3\,b\,a^2\,x^3}{x^4}+\frac {3\,a\,b^2\,x^2}{2} \]

[In]

int((a + b*x^3)^3/x^5,x)

[Out]

(b^3*x^5)/5 - (a^3/4 + 3*a^2*b*x^3)/x^4 + (3*a*b^2*x^2)/2