\(\int x^{-1-3 n} (a+b x^n)^2 \, dx\) [2535]

Optimal result

Integrand size = 17, antiderivative size = 24 \[ \int x^{-1-3 n} \left (a+b x^n\right )^2 \, dx=-\frac {x^{-3 n} \left (a+b x^n\right )^3}{3 a n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (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} \[ \int x^{-1-3 n} \left (a+b x^n\right )^2 \, dx=-\frac {x^{-3 n} \left (a+b x^n\right )^3}{3 a n} \]

[In]

Int[x^(-1 - 3*n)*(a + b*x^n)^2,x]

[Out]

-1/3*(a + b*x^n)^3/(a*n*x^(3*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^{-3 n} \left (a+b x^n\right )^3}{3 a n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int x^{-1-3 n} \left (a+b x^n\right )^2 \, dx=\frac {x^{-3 n} \left (-a^2-3 a b x^n-3 b^2 x^{2 n}\right )}{3 n} \]

[In]

Integrate[x^(-1 - 3*n)*(a + b*x^n)^2,x]

[Out]

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

Maple [A] (verified)

Time = 3.75 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67

method	result	size
risch	\(-\frac {b^{2} x^{-n}}{n}-\frac {a b \,x^{-2 n}}{n}-\frac {a^{2} x^{-3 n}}{3 n}\)	\(40\)
norman	\(\left (-\frac {a^{2}}{3 n}-\frac {b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{n}-\frac {a b \,{\mathrm e}^{n \ln \left (x \right )}}{n}\right ) {\mathrm e}^{-3 n \ln \left (x \right )}\)	\(45\)
parallelrisch	\(\frac {-3 x \,x^{2 n} x^{-1-3 n} b^{2}-3 x \,x^{n} x^{-1-3 n} a b -x \,x^{-1-3 n} a^{2}}{3 n}\)	\(53\)

[In]

int(x^(-1-3*n)*(a+b*x^n)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int x^{-1-3 n} \left (a+b x^n\right )^2 \, dx=-\frac {3 \, b^{2} x^{2 \, n} + 3 \, a b x^{n} + a^{2}}{3 \, n x^{3 \, n}} \]

[In]

integrate(x^(-1-3*n)*(a+b*x^n)^2,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (19) = 38\).

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.71 \[ \int x^{-1-3 n} \left (a+b x^n\right )^2 \, dx=\begin {cases} - \frac {a^{2} x x^{- 3 n - 1}}{3 n} - \frac {a b x x^{n} x^{- 3 n - 1}}{n} - \frac {b^{2} x x^{2 n} x^{- 3 n - 1}}{n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{2} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(-1-3*n)*(a+b*x**n)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int x^{-1-3 n} \left (a+b x^n\right )^2 \, dx=-\frac {a^{2}}{3 \, n x^{3 \, n}} - \frac {a b}{n x^{2 \, n}} - \frac {b^{2}}{n x^{n}} \]

[In]

integrate(x^(-1-3*n)*(a+b*x^n)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int x^{-1-3 n} \left (a+b x^n\right )^2 \, dx=-\frac {3 \, b^{2} x^{2 \, n} + 3 \, a b x^{n} + a^{2}}{3 \, n x^{3 \, n}} \]

[In]

integrate(x^(-1-3*n)*(a+b*x^n)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.93 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int x^{-1-3 n} \left (a+b x^n\right )^2 \, dx=-\frac {a^2+3\,b^2\,x^{2\,n}+3\,a\,b\,x^n}{3\,n\,x^{3\,n}} \]

[In]

int((a + b*x^n)^2/x^(3*n + 1),x)

[Out]

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