\(\int \frac {(a+b x^4)^3}{x^4} \, dx\) [642]

Optimal result

Integrand size = 13, antiderivative size = 39 \[ \int \frac {\left (a+b x^4\right )^3}{x^4} \, dx=-\frac {a^3}{3 x^3}+3 a^2 b x+\frac {3}{5} a b^2 x^5+\frac {b^3 x^9}{9} \]

[Out]

-1/3*a^3/x^3+3*a^2*b*x+3/5*a*b^2*x^5+1/9*b^3*x^9

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, 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} \[ \int \frac {\left (a+b x^4\right )^3}{x^4} \, dx=-\frac {a^3}{3 x^3}+3 a^2 b x+\frac {3}{5} a b^2 x^5+\frac {b^3 x^9}{9} \]

[In]

Int[(a + b*x^4)^3/x^4,x]

[Out]

-1/3*a^3/x^3 + 3*a^2*b*x + (3*a*b^2*x^5)/5 + (b^3*x^9)/9

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 (3 a^2 b+\frac {a^3}{x^4}+3 a b^2 x^4+b^3 x^8\right ) \, dx \\ & = -\frac {a^3}{3 x^3}+3 a^2 b x+\frac {3}{5} a b^2 x^5+\frac {b^3 x^9}{9} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^4\right )^3}{x^4} \, dx=-\frac {a^3}{3 x^3}+3 a^2 b x+\frac {3}{5} a b^2 x^5+\frac {b^3 x^9}{9} \]

[In]

Integrate[(a + b*x^4)^3/x^4,x]

[Out]

-1/3*a^3/x^3 + 3*a^2*b*x + (3*a*b^2*x^5)/5 + (b^3*x^9)/9

Maple [A] (verified)

Time = 3.80 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {a^{3}}{3 x^{3}}+3 a^{2} b x +\frac {3 a \,b^{2} x^{5}}{5}+\frac {b^{3} x^{9}}{9}\)	\(34\)
risch	\(-\frac {a^{3}}{3 x^{3}}+3 a^{2} b x +\frac {3 a \,b^{2} x^{5}}{5}+\frac {b^{3} x^{9}}{9}\)	\(34\)
norman	\(\frac {\frac {1}{9} b^{3} x^{12}+\frac {3}{5} a \,b^{2} x^{8}+3 a^{2} b \,x^{4}-\frac {1}{3} a^{3}}{x^{3}}\)	\(37\)
gosper	\(-\frac {-5 b^{3} x^{12}-27 a \,b^{2} x^{8}-135 a^{2} b \,x^{4}+15 a^{3}}{45 x^{3}}\)	\(38\)
parallelrisch	\(\frac {5 b^{3} x^{12}+27 a \,b^{2} x^{8}+135 a^{2} b \,x^{4}-15 a^{3}}{45 x^{3}}\)	\(38\)

[In]

int((b*x^4+a)^3/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*a^3/x^3+3*a^2*b*x+3/5*a*b^2*x^5+1/9*b^3*x^9

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^4\right )^3}{x^4} \, dx=\frac {5 \, b^{3} x^{12} + 27 \, a b^{2} x^{8} + 135 \, a^{2} b x^{4} - 15 \, a^{3}}{45 \, x^{3}} \]

[In]

integrate((b*x^4+a)^3/x^4,x, algorithm="fricas")

[Out]

1/45*(5*b^3*x^12 + 27*a*b^2*x^8 + 135*a^2*b*x^4 - 15*a^3)/x^3

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^4\right )^3}{x^4} \, dx=- \frac {a^{3}}{3 x^{3}} + 3 a^{2} b x + \frac {3 a b^{2} x^{5}}{5} + \frac {b^{3} x^{9}}{9} \]

[In]

integrate((b*x**4+a)**3/x**4,x)

[Out]

-a**3/(3*x**3) + 3*a**2*b*x + 3*a*b**2*x**5/5 + b**3*x**9/9

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^4\right )^3}{x^4} \, dx=\frac {1}{9} \, b^{3} x^{9} + \frac {3}{5} \, a b^{2} x^{5} + 3 \, a^{2} b x - \frac {a^{3}}{3 \, x^{3}} \]

[In]

integrate((b*x^4+a)^3/x^4,x, algorithm="maxima")

[Out]

1/9*b^3*x^9 + 3/5*a*b^2*x^5 + 3*a^2*b*x - 1/3*a^3/x^3

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^4\right )^3}{x^4} \, dx=\frac {1}{9} \, b^{3} x^{9} + \frac {3}{5} \, a b^{2} x^{5} + 3 \, a^{2} b x - \frac {a^{3}}{3 \, x^{3}} \]

[In]

integrate((b*x^4+a)^3/x^4,x, algorithm="giac")

[Out]

1/9*b^3*x^9 + 3/5*a*b^2*x^5 + 3*a^2*b*x - 1/3*a^3/x^3

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^4\right )^3}{x^4} \, dx=\frac {b^3\,x^9}{9}-\frac {a^3}{3\,x^3}+\frac {3\,a\,b^2\,x^5}{5}+3\,a^2\,b\,x \]

[In]

int((a + b*x^4)^3/x^4,x)

[Out]

(b^3*x^9)/9 - a^3/(3*x^3) + (3*a*b^2*x^5)/5 + 3*a^2*b*x