\(\int \frac {(a+b x^3)^3}{x^8} \, dx\) [257]

Optimal result

Integrand size = 13, antiderivative size = 41 \[ \int \frac {\left (a+b x^3\right )^3}{x^8} \, dx=-\frac {a^3}{7 x^7}-\frac {3 a^2 b}{4 x^4}-\frac {3 a b^2}{x}+\frac {b^3 x^2}{2} \]

[Out]

-1/7*a^3/x^7-3/4*a^2*b/x^4-3*a*b^2/x+1/2*b^3*x^2

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} \[ \int \frac {\left (a+b x^3\right )^3}{x^8} \, dx=-\frac {a^3}{7 x^7}-\frac {3 a^2 b}{4 x^4}-\frac {3 a b^2}{x}+\frac {b^3 x^2}{2} \]

[In]

Int[(a + b*x^3)^3/x^8,x]

[Out]

-1/7*a^3/x^7 - (3*a^2*b)/(4*x^4) - (3*a*b^2)/x + (b^3*x^2)/2

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^8}+\frac {3 a^2 b}{x^5}+\frac {3 a b^2}{x^2}+b^3 x\right ) \, dx \\ & = -\frac {a^3}{7 x^7}-\frac {3 a^2 b}{4 x^4}-\frac {3 a b^2}{x}+\frac {b^3 x^2}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^3}{x^8} \, dx=-\frac {a^3}{7 x^7}-\frac {3 a^2 b}{4 x^4}-\frac {3 a b^2}{x}+\frac {b^3 x^2}{2} \]

[In]

Integrate[(a + b*x^3)^3/x^8,x]

[Out]

-1/7*a^3/x^7 - (3*a^2*b)/(4*x^4) - (3*a*b^2)/x + (b^3*x^2)/2

Maple [A] (verified)

Time = 3.58 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {a^{3}}{7 x^{7}}-\frac {3 a^{2} b}{4 x^{4}}-\frac {3 a \,b^{2}}{x}+\frac {b^{3} x^{2}}{2}\)	\(36\)
norman	\(\frac {\frac {1}{2} b^{3} x^{9}-3 a \,b^{2} x^{6}-\frac {3}{4} a^{2} b \,x^{3}-\frac {1}{7} a^{3}}{x^{7}}\)	\(37\)
gosper	\(-\frac {-14 b^{3} x^{9}+84 a \,b^{2} x^{6}+21 a^{2} b \,x^{3}+4 a^{3}}{28 x^{7}}\)	\(38\)
risch	\(\frac {b^{3} x^{2}}{2}+\frac {-3 a \,b^{2} x^{6}-\frac {3}{4} a^{2} b \,x^{3}-\frac {1}{7} a^{3}}{x^{7}}\)	\(38\)
parallelrisch	\(\frac {14 b^{3} x^{9}-84 a \,b^{2} x^{6}-21 a^{2} b \,x^{3}-4 a^{3}}{28 x^{7}}\)	\(38\)

[In]

int((b*x^3+a)^3/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/7*a^3/x^7-3/4*a^2*b/x^4-3*a*b^2/x+1/2*b^3*x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^3\right )^3}{x^8} \, dx=\frac {14 \, b^{3} x^{9} - 84 \, a b^{2} x^{6} - 21 \, a^{2} b x^{3} - 4 \, a^{3}}{28 \, x^{7}} \]

[In]

integrate((b*x^3+a)^3/x^8,x, algorithm="fricas")

[Out]

1/28*(14*b^3*x^9 - 84*a*b^2*x^6 - 21*a^2*b*x^3 - 4*a^3)/x^7

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^3\right )^3}{x^8} \, dx=\frac {b^{3} x^{2}}{2} + \frac {- 4 a^{3} - 21 a^{2} b x^{3} - 84 a b^{2} x^{6}}{28 x^{7}} \]

[In]

integrate((b*x**3+a)**3/x**8,x)

[Out]

b**3*x**2/2 + (-4*a**3 - 21*a**2*b*x**3 - 84*a*b**2*x**6)/(28*x**7)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^3\right )^3}{x^8} \, dx=\frac {1}{2} \, b^{3} x^{2} - \frac {84 \, a b^{2} x^{6} + 21 \, a^{2} b x^{3} + 4 \, a^{3}}{28 \, x^{7}} \]

[In]

integrate((b*x^3+a)^3/x^8,x, algorithm="maxima")

[Out]

1/2*b^3*x^2 - 1/28*(84*a*b^2*x^6 + 21*a^2*b*x^3 + 4*a^3)/x^7

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^3\right )^3}{x^8} \, dx=\frac {1}{2} \, b^{3} x^{2} - \frac {84 \, a b^{2} x^{6} + 21 \, a^{2} b x^{3} + 4 \, a^{3}}{28 \, x^{7}} \]

[In]

integrate((b*x^3+a)^3/x^8,x, algorithm="giac")

[Out]

1/2*b^3*x^2 - 1/28*(84*a*b^2*x^6 + 21*a^2*b*x^3 + 4*a^3)/x^7

Mupad [B] (verification not implemented)

Time = 5.54 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^3\right )^3}{x^8} \, dx=\frac {b^3\,x^2}{2}-\frac {\frac {a^3}{7}+\frac {3\,a^2\,b\,x^3}{4}+3\,a\,b^2\,x^6}{x^7} \]

[In]

int((a + b*x^3)^3/x^8,x)

[Out]

(b^3*x^2)/2 - (a^3/7 + (3*a^2*b*x^3)/4 + 3*a*b^2*x^6)/x^7