\(\int x^{-1-n} (a+b x^n)^5 \, dx\) [2556]

Optimal result

Integrand size = 17, antiderivative size = 83 \[ \int x^{-1-n} \left (a+b x^n\right )^5 \, dx=-\frac {a^5 x^{-n}}{n}+\frac {10 a^3 b^2 x^n}{n}+\frac {5 a^2 b^3 x^{2 n}}{n}+\frac {5 a b^4 x^{3 n}}{3 n}+\frac {b^5 x^{4 n}}{4 n}+5 a^4 b \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 83, 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} \[ \int x^{-1-n} \left (a+b x^n\right )^5 \, dx=-\frac {a^5 x^{-n}}{n}+5 a^4 b \log (x)+\frac {10 a^3 b^2 x^n}{n}+\frac {5 a^2 b^3 x^{2 n}}{n}+\frac {5 a b^4 x^{3 n}}{3 n}+\frac {b^5 x^{4 n}}{4 n} \]

[In]

Int[x^(-1 - n)*(a + b*x^n)^5,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int x^{-1-n} \left (a+b x^n\right )^5 \, dx=-\frac {12 a^5 x^{-n}-120 a^3 b^2 x^n-60 a^2 b^3 x^{2 n}-20 a b^4 x^{3 n}-3 b^5 x^{4 n}-60 a^4 b \log \left (x^n\right )}{12 n} \]

[In]

Integrate[x^(-1 - n)*(a + b*x^n)^5,x]

[Out]

-1/12*((12*a^5)/x^n - 120*a^3*b^2*x^n - 60*a^2*b^3*x^(2*n) - 20*a*b^4*x^(3*n) - 3*b^5*x^(4*n) - 60*a^4*b*Log[x
^n])/n

Maple [A] (verified)

Time = 3.82 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96

method	result	size
risch	\(5 a^{4} b \ln \left (x \right )+\frac {b^{5} x^{4 n}}{4 n}+\frac {5 a \,b^{4} x^{3 n}}{3 n}+\frac {5 a^{2} b^{3} x^{2 n}}{n}+\frac {10 a^{3} b^{2} x^{n}}{n}-\frac {a^{5} x^{-n}}{n}\)	\(80\)
norman	\(\left (5 a^{4} b \ln \left (x \right ) {\mathrm e}^{n \ln \left (x \right )}-\frac {a^{5}}{n}+\frac {b^{5} {\mathrm e}^{5 n \ln \left (x \right )}}{4 n}+\frac {5 a \,b^{4} {\mathrm e}^{4 n \ln \left (x \right )}}{3 n}+\frac {5 a^{2} b^{3} {\mathrm e}^{3 n \ln \left (x \right )}}{n}+\frac {10 a^{3} b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{n}\right ) {\mathrm e}^{-n \ln \left (x \right )}\)	\(98\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int x^{-1-n} \left (a+b x^n\right )^5 \, dx=\frac {60 \, a^{4} b n x^{n} \log \left (x\right ) + 3 \, b^{5} x^{5 \, n} + 20 \, a b^{4} x^{4 \, n} + 60 \, a^{2} b^{3} x^{3 \, n} + 120 \, a^{3} b^{2} x^{2 \, n} - 12 \, a^{5}}{12 \, n x^{n}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.70 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int x^{-1-n} \left (a+b x^n\right )^5 \, dx=\begin {cases} - \frac {a^{5} x^{- n}}{n} + 5 a^{4} b \log {\left (x \right )} + \frac {10 a^{3} b^{2} x^{n}}{n} + \frac {5 a^{2} b^{3} x^{2 n}}{n} + \frac {5 a b^{4} x^{3 n}}{3 n} + \frac {b^{5} x^{4 n}}{4 n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{5} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int x^{-1-n} \left (a+b x^n\right )^5 \, dx=5 \, a^{4} b \log \left (x\right ) + \frac {b^{5} x^{4 \, n}}{4 \, n} + \frac {5 \, a b^{4} x^{3 \, n}}{3 \, n} + \frac {5 \, a^{2} b^{3} x^{2 \, n}}{n} + \frac {10 \, a^{3} b^{2} x^{n}}{n} - \frac {a^{5}}{n x^{n}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int x^{-1-n} \left (a+b x^n\right )^5 \, dx=\frac {60 \, a^{4} b n x^{n} \log \left (x\right ) + 3 \, b^{5} x^{5 \, n} + 20 \, a b^{4} x^{4 \, n} + 60 \, a^{2} b^{3} x^{3 \, n} + 120 \, a^{3} b^{2} x^{2 \, n} - 12 \, a^{5}}{12 \, n x^{n}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int x^{-1-n} \left (a+b x^n\right )^5 \, dx=5\,a^4\,b\,\ln \left (x\right )-\frac {a^5}{n\,x^n}+\frac {b^5\,x^{4\,n}}{4\,n}+\frac {5\,a^2\,b^3\,x^{2\,n}}{n}+\frac {5\,a\,b^4\,x^{3\,n}}{3\,n}+\frac {10\,a^3\,b^2\,x^n}{n} \]

[In]

int((a + b*x^n)^5/x^(n + 1),x)

[Out]

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