3.2560 \(\int x^{-1-5 n} (a+b x^n)^5 \, dx\)

Optimal. Leaf size=86 \[ -\frac {a^5 x^{-5 n}}{5 n}-\frac {5 a^4 b x^{-4 n}}{4 n}-\frac {10 a^3 b^2 x^{-3 n}}{3 n}-\frac {5 a^2 b^3 x^{-2 n}}{n}-\frac {5 a b^4 x^{-n}}{n}+b^5 \log (x) \]

[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)

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Rubi [A]  time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {266, 43} \[ -\frac {10 a^3 b^2 x^{-3 n}}{3 n}-\frac {5 a^2 b^3 x^{-2 n}}{n}-\frac {5 a^4 b x^{-4 n}}{4 n}-\frac {a^5 x^{-5 n}}{5 n}-\frac {5 a b^4 x^{-n}}{n}+b^5 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - 5*n)*(a + b*x^n)^5,x]

[Out]

-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]

Rule 43

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 266

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*} \int x^{-1-5 n} \left (a+b x^n\right )^5 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^5}{x^6} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.08, size = 69, normalized size = 0.80 \[ b^5 \log (x)-\frac {a x^{-5 n} \left (12 a^4+75 a^3 b x^n+200 a^2 b^2 x^{2 n}+300 a b^3 x^{3 n}+300 b^4 x^{4 n}\right )}{60 n} \]

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.57, size = 77, normalized size = 0.90 \[ \frac {60 \, b^{5} n x^{5 \, n} \log \relax (x) - 300 \, a b^{4} x^{4 \, n} - 300 \, a^{2} b^{3} x^{3 \, n} - 200 \, a^{3} b^{2} x^{2 \, n} - 75 \, a^{4} b x^{n} - 12 \, a^{5}}{60 \, n x^{5 \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.25, size = 77, normalized size = 0.90 \[ \frac {60 \, b^{5} n x^{5 \, n} \log \relax (x) - 300 \, a b^{4} x^{4 \, n} - 300 \, a^{2} b^{3} x^{3 \, n} - 200 \, a^{3} b^{2} x^{2 \, n} - 75 \, a^{4} b x^{n} - 12 \, a^{5}}{60 \, n x^{5 \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.02, size = 97, normalized size = 1.13 \[ \left (b^{5} {\mathrm e}^{5 n \ln \relax (x )} \ln \relax (x )-\frac {5 a^{4} b \,{\mathrm e}^{n \ln \relax (x )}}{4 n}-\frac {10 a^{3} b^{2} {\mathrm e}^{2 n \ln \relax (x )}}{3 n}-\frac {5 a^{2} b^{3} {\mathrm e}^{3 n \ln \relax (x )}}{n}-\frac {5 a \,b^{4} {\mathrm e}^{4 n \ln \relax (x )}}{n}-\frac {a^{5}}{5 n}\right ) {\mathrm e}^{-5 n \ln \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 88, normalized size = 1.02 \[ b^{5} \log \relax (x) - \frac {a^{5}}{5 \, n x^{5 \, n}} - \frac {5 \, a^{4} b}{4 \, n x^{4 \, n}} - \frac {10 \, a^{3} b^{2}}{3 \, n x^{3 \, n}} - \frac {5 \, a^{2} b^{3}}{n x^{2 \, n}} - \frac {5 \, a b^{4}}{n x^{n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.43, size = 88, normalized size = 1.02 \[ b^5\,\ln \relax (x)-\frac {a^5}{5\,n\,x^{5\,n}}-\frac {5\,a^2\,b^3}{n\,x^{2\,n}}-\frac {10\,a^3\,b^2}{3\,n\,x^{3\,n}}-\frac {5\,a\,b^4}{n\,x^n}-\frac {5\,a^4\,b}{4\,n\,x^{4\,n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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