\(\int x^{-1-4 n} (a+b x^n)^3 \, dx\) [2547]

Optimal result

Integrand size = 17, antiderivative size = 24 \[ \int x^{-1-4 n} \left (a+b x^n\right )^3 \, dx=-\frac {x^{-4 n} \left (a+b x^n\right )^4}{4 a n} \]

[Out]

-1/4*(a+b*x^n)^4/a/n/(x^(4*n))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {270} \[ \int x^{-1-4 n} \left (a+b x^n\right )^3 \, dx=-\frac {x^{-4 n} \left (a+b x^n\right )^4}{4 a n} \]

[In]

Int[x^(-1 - 4*n)*(a + b*x^n)^3,x]

[Out]

-1/4*(a + b*x^n)^4/(a*n*x^(4*n))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^{-4 n} \left (a+b x^n\right )^4}{4 a n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int x^{-1-4 n} \left (a+b x^n\right )^3 \, dx=\frac {x^{-4 n} \left (-a^3-4 a^2 b x^n-6 a b^2 x^{2 n}-4 b^3 x^{3 n}\right )}{4 n} \]

[In]

Integrate[x^(-1 - 4*n)*(a + b*x^n)^3,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(24)=48\).

Time = 3.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33

method	result	size
risch	\(-\frac {b^{3} x^{-n}}{n}-\frac {3 a \,b^{2} x^{-2 n}}{2 n}-\frac {a^{2} b \,x^{-3 n}}{n}-\frac {a^{3} x^{-4 n}}{4 n}\)	\(56\)
norman	\(\left (-\frac {a^{3}}{4 n}-\frac {b^{3} {\mathrm e}^{3 n \ln \left (x \right )}}{n}-\frac {3 a \,b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{2 n}-\frac {a^{2} b \,{\mathrm e}^{n \ln \left (x \right )}}{n}\right ) {\mathrm e}^{-4 n \ln \left (x \right )}\)	\(63\)
parallelrisch	\(\frac {-4 x \,x^{3 n} x^{-1-4 n} b^{3}-6 x \,x^{2 n} x^{-1-4 n} a \,b^{2}-4 x \,x^{n} x^{-1-4 n} a^{2} b -x \,x^{-1-4 n} a^{3}}{4 n}\)	\(74\)

[In]

int(x^(-1-4*n)*(a+b*x^n)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int x^{-1-4 n} \left (a+b x^n\right )^3 \, dx=-\frac {4 \, b^{3} x^{3 \, n} + 6 \, a b^{2} x^{2 \, n} + 4 \, a^{2} b x^{n} + a^{3}}{4 \, n x^{4 \, n}} \]

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^3,x, algorithm="fricas")

[Out]

-1/4*(4*b^3*x^(3*n) + 6*a*b^2*x^(2*n) + 4*a^2*b*x^n + a^3)/(n*x^(4*n))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (19) = 38\).

Time = 0.39 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.83 \[ \int x^{-1-4 n} \left (a+b x^n\right )^3 \, dx=\begin {cases} - \frac {a^{3} x x^{- 4 n - 1}}{4 n} - \frac {a^{2} b x x^{n} x^{- 4 n - 1}}{n} - \frac {3 a b^{2} x x^{2 n} x^{- 4 n - 1}}{2 n} - \frac {b^{3} x x^{3 n} x^{- 4 n - 1}}{n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{3} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(-1-4*n)*(a+b*x**n)**3,x)

[Out]

Piecewise((-a**3*x*x**(-4*n - 1)/(4*n) - a**2*b*x*x**n*x**(-4*n - 1)/n - 3*a*b**2*x*x**(2*n)*x**(-4*n - 1)/(2*
n) - b**3*x*x**(3*n)*x**(-4*n - 1)/n, Ne(n, 0)), ((a + b)**3*log(x), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (24) = 48\).

Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.54 \[ \int x^{-1-4 n} \left (a+b x^n\right )^3 \, dx=-\frac {a^{3}}{4 \, n x^{4 \, n}} - \frac {a^{2} b}{n x^{3 \, n}} - \frac {3 \, a b^{2}}{2 \, n x^{2 \, n}} - \frac {b^{3}}{n x^{n}} \]

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int x^{-1-4 n} \left (a+b x^n\right )^3 \, dx=-\frac {4 \, b^{3} x^{3 \, n} + 6 \, a b^{2} x^{2 \, n} + 4 \, a^{2} b x^{n} + a^{3}}{4 \, n x^{4 \, n}} \]

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^3,x, algorithm="giac")

[Out]

-1/4*(4*b^3*x^(3*n) + 6*a*b^2*x^(2*n) + 4*a^2*b*x^n + a^3)/(n*x^(4*n))

Mupad [B] (verification not implemented)

Time = 5.77 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.54 \[ \int x^{-1-4 n} \left (a+b x^n\right )^3 \, dx=-\frac {a^3}{4\,n\,x^{4\,n}}-\frac {b^3}{n\,x^n}-\frac {3\,a\,b^2}{2\,n\,x^{2\,n}}-\frac {a^2\,b}{n\,x^{3\,n}} \]

[In]

int((a + b*x^n)^3/x^(4*n + 1),x)

[Out]

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