\(\int x^{-1-5 n} (a+b x^n)^5 \, dx\) [2560]

Optimal result

Integrand size = 17, antiderivative size = 86 \[ \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx=-\frac {a^5 x^{-5 n}}{5 n}-\frac {5 a^4 b x^{-4 n}}{4 n}-\frac {10 a^3 b^2 x^{-3 n}}{3 n}-\frac {5 a^2 b^3 x^{-2 n}}{n}-\frac {5 a b^4 x^{-n}}{n}+b^5 \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 86, 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} \[ \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx=-\frac {a^5 x^{-5 n}}{5 n}-\frac {5 a^4 b x^{-4 n}}{4 n}-\frac {10 a^3 b^2 x^{-3 n}}{3 n}-\frac {5 a^2 b^3 x^{-2 n}}{n}-\frac {5 a b^4 x^{-n}}{n}+b^5 \log (x) \]

[In]

Int[x^(-1 - 5*n)*(a + b*x^n)^5,x]

[Out]

-1/5*a^5/(n*x^(5*n)) - (5*a^4*b)/(4*n*x^(4*n)) - (10*a^3*b^2)/(3*n*x^(3*n)) - (5*a^2*b^3)/(n*x^(2*n)) - (5*a*b
^4)/(n*x^n) + b^5*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^6} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a^5}{x^6}+\frac {5 a^4 b}{x^5}+\frac {10 a^3 b^2}{x^4}+\frac {10 a^2 b^3}{x^3}+\frac {5 a b^4}{x^2}+\frac {b^5}{x}\right ) \, dx,x,x^n\right )}{n} \\ & = -\frac {a^5 x^{-5 n}}{5 n}-\frac {5 a^4 b x^{-4 n}}{4 n}-\frac {10 a^3 b^2 x^{-3 n}}{3 n}-\frac {5 a^2 b^3 x^{-2 n}}{n}-\frac {5 a b^4 x^{-n}}{n}+b^5 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx=-\frac {a x^{-5 n} \left (12 a^4+75 a^3 b x^n+200 a^2 b^2 x^{2 n}+300 a b^3 x^{3 n}+300 b^4 x^{4 n}\right )}{60 n}+\frac {b^5 \log \left (x^n\right )}{n} \]

[In]

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

[Out]

-1/60*(a*(12*a^4 + 75*a^3*b*x^n + 200*a^2*b^2*x^(2*n) + 300*a*b^3*x^(3*n) + 300*b^4*x^(4*n)))/(n*x^(5*n)) + (b
^5*Log[x^n])/n

Maple [A] (verified)

Time = 3.98 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94

method	result	size
risch	\(b^{5} \ln \left (x \right )-\frac {5 a \,b^{4} x^{-n}}{n}-\frac {5 a^{2} b^{3} x^{-2 n}}{n}-\frac {10 a^{3} b^{2} x^{-3 n}}{3 n}-\frac {5 a^{4} b \,x^{-4 n}}{4 n}-\frac {a^{5} x^{-5 n}}{5 n}\)	\(81\)
norman	\(\left (b^{5} \ln \left (x \right ) {\mathrm e}^{5 n \ln \left (x \right )}-\frac {a^{5}}{5 n}-\frac {5 a \,b^{4} {\mathrm e}^{4 n \ln \left (x \right )}}{n}-\frac {5 a^{2} b^{3} {\mathrm e}^{3 n \ln \left (x \right )}}{n}-\frac {10 a^{3} b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{3 n}-\frac {5 a^{4} b \,{\mathrm e}^{n \ln \left (x \right )}}{4 n}\right ) {\mathrm e}^{-5 n \ln \left (x \right )}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx=\frac {60 \, b^{5} n x^{5 \, n} \log \left (x\right ) - 300 \, a b^{4} x^{4 \, n} - 300 \, a^{2} b^{3} x^{3 \, n} - 200 \, a^{3} b^{2} x^{2 \, n} - 75 \, a^{4} b x^{n} - 12 \, a^{5}}{60 \, n x^{5 \, n}} \]

[In]

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

[Out]

1/60*(60*b^5*n*x^(5*n)*log(x) - 300*a*b^4*x^(4*n) - 300*a^2*b^3*x^(3*n) - 200*a^3*b^2*x^(2*n) - 75*a^4*b*x^n -
12*a^5)/(n*x^(5*n))

Sympy [A] (verification not implemented)

Time = 6.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx=\begin {cases} - \frac {a^{5} x^{- 5 n}}{5 n} - \frac {5 a^{4} b x^{- 4 n}}{4 n} - \frac {10 a^{3} b^{2} x^{- 3 n}}{3 n} - \frac {5 a^{2} b^{3} x^{- 2 n}}{n} - \frac {5 a b^{4} x^{- n}}{n} + b^{5} \log {\left (x \right )} & \text {for}\: n \neq 0 \\\left (a + b\right )^{5} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02 \[ \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx=b^{5} \log \left (x\right ) - \frac {a^{5}}{5 \, n x^{5 \, n}} - \frac {5 \, a^{4} b}{4 \, n x^{4 \, n}} - \frac {10 \, a^{3} b^{2}}{3 \, n x^{3 \, n}} - \frac {5 \, a^{2} b^{3}}{n x^{2 \, n}} - \frac {5 \, a b^{4}}{n x^{n}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx=\frac {60 \, b^{5} n x^{5 \, n} \log \left (x\right ) - 300 \, a b^{4} x^{4 \, n} - 300 \, a^{2} b^{3} x^{3 \, n} - 200 \, a^{3} b^{2} x^{2 \, n} - 75 \, a^{4} b x^{n} - 12 \, a^{5}}{60 \, n x^{5 \, n}} \]

[In]

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

[Out]

1/60*(60*b^5*n*x^(5*n)*log(x) - 300*a*b^4*x^(4*n) - 300*a^2*b^3*x^(3*n) - 200*a^3*b^2*x^(2*n) - 75*a^4*b*x^n -
12*a^5)/(n*x^(5*n))

Mupad [B] (verification not implemented)

Time = 5.86 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02 \[ \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx=b^5\,\ln \left (x\right )-\frac {a^5}{5\,n\,x^{5\,n}}-\frac {5\,a^2\,b^3}{n\,x^{2\,n}}-\frac {10\,a^3\,b^2}{3\,n\,x^{3\,n}}-\frac {5\,a\,b^4}{n\,x^n}-\frac {5\,a^4\,b}{4\,n\,x^{4\,n}} \]

[In]

int((a + b*x^n)^5/x^(5*n + 1),x)

[Out]

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