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

Optimal. Leaf size=82 \[ -\frac {a^5 x^{-4 n}}{4 n}-\frac {5 a^4 b x^{-3 n}}{3 n}-\frac {5 a^3 b^2 x^{-2 n}}{n}-\frac {10 a^2 b^3 x^{-n}}{n}+\frac {b^5 x^n}{n}+5 a b^4 \log (x) \]

[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)

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Rubi [A]
time = 0.03, antiderivative size = 82, 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 = {272, 45} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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*} \int x^{-1-4 n} \left (a+b x^n\right )^5 \, dx &=\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*}

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Mathematica [A]
time = 0.06, size = 77, normalized size = 0.94 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 0.23, size = 79, normalized size = 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	\(\left (\frac {b^{5} {\mathrm e}^{5 n \ln \left (x \right )}}{n}+5 a \,b^{4} \ln \left (x \right ) {\mathrm e}^{4 n \ln \left (x \right )}-\frac {a^{5}}{4 n}-\frac {10 a^{2} b^{3} {\mathrm e}^{3 n \ln \left (x \right )}}{n}-\frac {5 a^{3} b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{n}-\frac {5 a^{4} b \,{\mathrm e}^{n \ln \left (x \right )}}{3 n}\right ) {\mathrm e}^{-4 n \ln \left (x \right )}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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Maxima [A]
time = 0.29, size = 84, normalized size = 1.02 \begin {gather*} 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 0.41, size = 77, normalized size = 0.94 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 486 vs. \(2 (73) = 146\).
time = 70.20, size = 486, normalized size = 5.93 \begin {gather*} \begin {cases} a^{5} x + \frac {20 a^{4} b x^{\frac {3}{4}}}{3} + 20 a^{3} b^{2} \sqrt {x} + 40 a^{2} b^{3} \sqrt [4]{x} - 20 a b^{4} \log {\left (\frac {1}{\sqrt [4]{x}} \right )} - \frac {4 b^{5}}{\sqrt [4]{x}} & \text {for}\: n = - \frac {1}{4} \\\left (a + b\right )^{5} \log {\left (x \right )} & \text {for}\: n = 0 \\- \frac {12 a^{5} n}{48 n^{2} x^{4 n} + 12 n x^{4 n}} - \frac {3 a^{5}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} - \frac {80 a^{4} b n x^{n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} - \frac {20 a^{4} b x^{n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} - \frac {240 a^{3} b^{2} n x^{2 n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} - \frac {60 a^{3} b^{2} x^{2 n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} - \frac {480 a^{2} b^{3} n x^{3 n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} - \frac {120 a^{2} b^{3} x^{3 n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} + \frac {240 a b^{4} n x^{4 n} \log {\left (x^{n} \right )}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} + \frac {60 a b^{4} n x^{4 n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} + \frac {60 a b^{4} x^{4 n} \log {\left (x^{n} \right )}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} + \frac {48 b^{5} n x^{5 n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} + \frac {12 b^{5} x^{5 n}}{48 n^{2} x^{4 n} + 12 n x^{4 n}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**5*x + 20*a**4*b*x**(3/4)/3 + 20*a**3*b**2*sqrt(x) + 40*a**2*b**3*x**(1/4) - 20*a*b**4*log(x**(-1
/4)) - 4*b**5/x**(1/4), Eq(n, -1/4)), ((a + b)**5*log(x), Eq(n, 0)), (-12*a**5*n/(48*n**2*x**(4*n) + 12*n*x**(
4*n)) - 3*a**5/(48*n**2*x**(4*n) + 12*n*x**(4*n)) - 80*a**4*b*n*x**n/(48*n**2*x**(4*n) + 12*n*x**(4*n)) - 20*a
**4*b*x**n/(48*n**2*x**(4*n) + 12*n*x**(4*n)) - 240*a**3*b**2*n*x**(2*n)/(48*n**2*x**(4*n) + 12*n*x**(4*n)) -
60*a**3*b**2*x**(2*n)/(48*n**2*x**(4*n) + 12*n*x**(4*n)) - 480*a**2*b**3*n*x**(3*n)/(48*n**2*x**(4*n) + 12*n*x
**(4*n)) - 120*a**2*b**3*x**(3*n)/(48*n**2*x**(4*n) + 12*n*x**(4*n)) + 240*a*b**4*n*x**(4*n)*log(x**n)/(48*n**
2*x**(4*n) + 12*n*x**(4*n)) + 60*a*b**4*n*x**(4*n)/(48*n**2*x**(4*n) + 12*n*x**(4*n)) + 60*a*b**4*x**(4*n)*log
(x**n)/(48*n**2*x**(4*n) + 12*n*x**(4*n)) + 48*b**5*n*x**(5*n)/(48*n**2*x**(4*n) + 12*n*x**(4*n)) + 12*b**5*x*
*(5*n)/(48*n**2*x**(4*n) + 12*n*x**(4*n)), True))

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Giac [A]
time = 1.08, size = 77, normalized size = 0.94 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Mupad [B]
time = 1.41, size = 84, normalized size = 1.02 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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