\(\int x^{-1-4 n} (a+b x^n)^5 \, dx\) [2559]

Optimal result

Integrand size = 17, antiderivative size = 82 \[ \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx=-\frac {a^5 x^{-4 n}}{4 n}-\frac {5 a^4 b x^{-3 n}}{3 n}-\frac {5 a^3 b^2 x^{-2 n}}{n}-\frac {10 a^2 b^3 x^{-n}}{n}+\frac {b^5 x^n}{n}+5 a b^4 \log (x) \]

[Out]

-1/4*a^5/n/(x^(4*n))-5/3*a^4*b/n/(x^(3*n))-5*a^3*b^2/n/(x^(2*n))-10*a^2*b^3/n/(x^n)+b^5*x^n/n+5*a*b^4*ln(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 82, 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} \[ \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx=-\frac {a^5 x^{-4 n}}{4 n}-\frac {5 a^4 b x^{-3 n}}{3 n}-\frac {5 a^3 b^2 x^{-2 n}}{n}-\frac {10 a^2 b^3 x^{-n}}{n}+5 a b^4 \log (x)+\frac {b^5 x^n}{n} \]

[In]

Int[x^(-1 - 4*n)*(a + b*x^n)^5,x]

[Out]

-1/4*a^5/(n*x^(4*n)) - (5*a^4*b)/(3*n*x^(3*n)) - (5*a^3*b^2)/(n*x^(2*n)) - (10*a^2*b^3)/(n*x^n) + (b^5*x^n)/n
+ 5*a*b^4*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)^5}{x^5} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (b^5+\frac {a^5}{x^5}+\frac {5 a^4 b}{x^4}+\frac {10 a^3 b^2}{x^3}+\frac {10 a^2 b^3}{x^2}+\frac {5 a b^4}{x}\right ) \, dx,x,x^n\right )}{n} \\ & = -\frac {a^5 x^{-4 n}}{4 n}-\frac {5 a^4 b x^{-3 n}}{3 n}-\frac {5 a^3 b^2 x^{-2 n}}{n}-\frac {10 a^2 b^3 x^{-n}}{n}+\frac {b^5 x^n}{n}+5 a b^4 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94 \[ \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx=\frac {x^{-4 n} \left (-3 a^5-20 a^4 b x^n-60 a^3 b^2 x^{2 n}-120 a^2 b^3 x^{3 n}+12 b^5 x^{5 n}\right )}{12 n}+\frac {5 a b^4 \log \left (x^n\right )}{n} \]

[In]

Integrate[x^(-1 - 4*n)*(a + b*x^n)^5,x]

[Out]

(-3*a^5 - 20*a^4*b*x^n - 60*a^3*b^2*x^(2*n) - 120*a^2*b^3*x^(3*n) + 12*b^5*x^(5*n))/(12*n*x^(4*n)) + (5*a*b^4*
Log[x^n])/n

Maple [A] (verified)

Time = 3.96 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96

method	result	size
risch	\(5 a \,b^{4} \ln \left (x \right )+\frac {b^{5} x^{n}}{n}-\frac {10 a^{2} b^{3} x^{-n}}{n}-\frac {5 a^{3} b^{2} x^{-2 n}}{n}-\frac {5 a^{4} b \,x^{-3 n}}{3 n}-\frac {a^{5} x^{-4 n}}{4 n}\)	\(79\)
norman	\(\left (\frac {b^{5} {\mathrm e}^{5 n \ln \left (x \right )}}{n}+5 a \,b^{4} \ln \left (x \right ) {\mathrm e}^{4 n \ln \left (x \right )}-\frac {a^{5}}{4 n}-\frac {10 a^{2} b^{3} {\mathrm e}^{3 n \ln \left (x \right )}}{n}-\frac {5 a^{3} b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{n}-\frac {5 a^{4} b \,{\mathrm e}^{n \ln \left (x \right )}}{3 n}\right ) {\mathrm e}^{-4 n \ln \left (x \right )}\)	\(97\)

[In]

int(x^(-1-4*n)*(a+b*x^n)^5,x,method=_RETURNVERBOSE)

[Out]

5*a*b^4*ln(x)+b^5*x^n/n-10*a^2*b^3/n/(x^n)-5*a^3*b^2/n/(x^n)^2-5/3*a^4*b/n/(x^n)^3-1/4*a^5/n/(x^n)^4

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94 \[ \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx=\frac {60 \, a b^{4} n x^{4 \, n} \log \left (x\right ) + 12 \, b^{5} x^{5 \, n} - 120 \, a^{2} b^{3} x^{3 \, n} - 60 \, a^{3} b^{2} x^{2 \, n} - 20 \, a^{4} b x^{n} - 3 \, a^{5}}{12 \, n x^{4 \, n}} \]

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^5,x, algorithm="fricas")

[Out]

1/12*(60*a*b^4*n*x^(4*n)*log(x) + 12*b^5*x^(5*n) - 120*a^2*b^3*x^(3*n) - 60*a^3*b^2*x^(2*n) - 20*a^4*b*x^n - 3
*a^5)/(n*x^(4*n))

Sympy [A] (verification not implemented)

Time = 3.64 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx=\begin {cases} - \frac {a^{5} x^{- 4 n}}{4 n} - \frac {5 a^{4} b x^{- 3 n}}{3 n} - \frac {5 a^{3} b^{2} x^{- 2 n}}{n} - \frac {10 a^{2} b^{3} x^{- n}}{n} + 5 a b^{4} \log {\left (x \right )} + \frac {b^{5} x^{n}}{n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{5} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(-1-4*n)*(a+b*x**n)**5,x)

[Out]

Piecewise((-a**5/(4*n*x**(4*n)) - 5*a**4*b/(3*n*x**(3*n)) - 5*a**3*b**2/(n*x**(2*n)) - 10*a**2*b**3/(n*x**n) +
5*a*b**4*log(x) + b**5*x**n/n, Ne(n, 0)), ((a + b)**5*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx=5 \, a b^{4} \log \left (x\right ) + \frac {b^{5} x^{n}}{n} - \frac {a^{5}}{4 \, n x^{4 \, n}} - \frac {5 \, a^{4} b}{3 \, n x^{3 \, n}} - \frac {5 \, a^{3} b^{2}}{n x^{2 \, n}} - \frac {10 \, a^{2} b^{3}}{n x^{n}} \]

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^5,x, algorithm="maxima")

[Out]

5*a*b^4*log(x) + b^5*x^n/n - 1/4*a^5/(n*x^(4*n)) - 5/3*a^4*b/(n*x^(3*n)) - 5*a^3*b^2/(n*x^(2*n)) - 10*a^2*b^3/
(n*x^n)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94 \[ \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx=\frac {60 \, a b^{4} n x^{4 \, n} \log \left (x\right ) + 12 \, b^{5} x^{5 \, n} - 120 \, a^{2} b^{3} x^{3 \, n} - 60 \, a^{3} b^{2} x^{2 \, n} - 20 \, a^{4} b x^{n} - 3 \, a^{5}}{12 \, n x^{4 \, n}} \]

[In]

integrate(x^(-1-4*n)*(a+b*x^n)^5,x, algorithm="giac")

[Out]

1/12*(60*a*b^4*n*x^(4*n)*log(x) + 12*b^5*x^(5*n) - 120*a^2*b^3*x^(3*n) - 60*a^3*b^2*x^(2*n) - 20*a^4*b*x^n - 3
*a^5)/(n*x^(4*n))

Mupad [B] (verification not implemented)

Time = 5.80 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx=\frac {b^5\,x^n}{n}+5\,a\,b^4\,\ln \left (x\right )-\frac {a^5}{4\,n\,x^{4\,n}}-\frac {10\,a^2\,b^3}{n\,x^n}-\frac {5\,a^3\,b^2}{n\,x^{2\,n}}-\frac {5\,a^4\,b}{3\,n\,x^{3\,n}} \]

[In]

int((a + b*x^n)^5/x^(4*n + 1),x)

[Out]

(b^5*x^n)/n + 5*a*b^4*log(x) - a^5/(4*n*x^(4*n)) - (10*a^2*b^3)/(n*x^n) - (5*a^3*b^2)/(n*x^(2*n)) - (5*a^4*b)/
(3*n*x^(3*n))