\(\int x^{-1-3 n} (a+b x^n)^3 \, dx\) [2546]

Optimal result

Integrand size = 17, antiderivative size = 52 \[ \int x^{-1-3 n} \left (a+b x^n\right )^3 \, dx=-\frac {a^3 x^{-3 n}}{3 n}-\frac {3 a^2 b x^{-2 n}}{2 n}-\frac {3 a b^2 x^{-n}}{n}+b^3 \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {272, 45} \[ \int x^{-1-3 n} \left (a+b x^n\right )^3 \, dx=-\frac {a^3 x^{-3 n}}{3 n}-\frac {3 a^2 b x^{-2 n}}{2 n}-\frac {3 a b^2 x^{-n}}{n}+b^3 \log (x) \]

[In]

Int[x^(-1 - 3*n)*(a + b*x^n)^3,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.92 \[ \int x^{-1-3 n} \left (a+b x^n\right )^3 \, dx=-\frac {a x^{-3 n} \left (2 a^2+9 a b x^n+18 b^2 x^{2 n}\right )}{6 n}+\frac {b^3 \log \left (x^n\right )}{n} \]

[In]

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

[Out]

-1/6*(a*(2*a^2 + 9*a*b*x^n + 18*b^2*x^(2*n)))/(n*x^(3*n)) + (b^3*Log[x^n])/n

Maple [A] (verified)

Time = 3.83 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94

method	result	size
risch	\(b^{3} \ln \left (x \right )-\frac {3 a \,b^{2} x^{-n}}{n}-\frac {3 a^{2} b \,x^{-2 n}}{2 n}-\frac {a^{3} x^{-3 n}}{3 n}\)	\(49\)
norman	\(\left (b^{3} \ln \left (x \right ) {\mathrm e}^{3 n \ln \left (x \right )}-\frac {a^{3}}{3 n}-\frac {3 a \,b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{n}-\frac {3 a^{2} b \,{\mathrm e}^{n \ln \left (x \right )}}{2 n}\right ) {\mathrm e}^{-3 n \ln \left (x \right )}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int x^{-1-3 n} \left (a+b x^n\right )^3 \, dx=\frac {6 \, b^{3} n x^{3 \, n} \log \left (x\right ) - 18 \, a b^{2} x^{2 \, n} - 9 \, a^{2} b x^{n} - 2 \, a^{3}}{6 \, n x^{3 \, n}} \]

[In]

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

[Out]

1/6*(6*b^3*n*x^(3*n)*log(x) - 18*a*b^2*x^(2*n) - 9*a^2*b*x^n - 2*a^3)/(n*x^(3*n))

Sympy [A] (verification not implemented)

Time = 1.87 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int x^{-1-3 n} \left (a+b x^n\right )^3 \, dx=\begin {cases} - \frac {a^{3} x^{- 3 n}}{3 n} - \frac {3 a^{2} b x^{- 2 n}}{2 n} - \frac {3 a b^{2} x^{- n}}{n} + b^{3} \log {\left (x \right )} & \text {for}\: n \neq 0 \\\left (a + b\right )^{3} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int x^{-1-3 n} \left (a+b x^n\right )^3 \, dx=b^{3} \log \left (x\right ) - \frac {a^{3}}{3 \, n x^{3 \, n}} - \frac {3 \, a^{2} b}{2 \, n x^{2 \, n}} - \frac {3 \, a b^{2}}{n x^{n}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int x^{-1-3 n} \left (a+b x^n\right )^3 \, dx=\frac {6 \, b^{3} n x^{3 \, n} \log \left (x\right ) - 18 \, a b^{2} x^{2 \, n} - 9 \, a^{2} b x^{n} - 2 \, a^{3}}{6 \, n x^{3 \, n}} \]

[In]

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

[Out]

1/6*(6*b^3*n*x^(3*n)*log(x) - 18*a*b^2*x^(2*n) - 9*a^2*b*x^n - 2*a^3)/(n*x^(3*n))

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int x^{-1-3 n} \left (a+b x^n\right )^3 \, dx=b^3\,\ln \left (x\right )-\frac {a^3}{3\,n\,x^{3\,n}}-\frac {3\,a\,b^2}{n\,x^n}-\frac {3\,a^2\,b}{2\,n\,x^{2\,n}} \]

[In]

int((a + b*x^n)^3/x^(3*n + 1),x)

[Out]

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