\(\int \frac {x^{-1+4 n}}{(a+b x^n)^2} \, dx\) [2624]

Optimal result

Integrand size = 17, antiderivative size = 66 \[ \int \frac {x^{-1+4 n}}{\left (a+b x^n\right )^2} \, dx=-\frac {2 a x^n}{b^3 n}+\frac {x^{2 n}}{2 b^2 n}+\frac {a^3}{b^4 n \left (a+b x^n\right )}+\frac {3 a^2 \log \left (a+b x^n\right )}{b^4 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, 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} \[ \int \frac {x^{-1+4 n}}{\left (a+b x^n\right )^2} \, dx=\frac {a^3}{b^4 n \left (a+b x^n\right )}+\frac {3 a^2 \log \left (a+b x^n\right )}{b^4 n}-\frac {2 a x^n}{b^3 n}+\frac {x^{2 n}}{2 b^2 n} \]

[In]

Int[x^(-1 + 4*n)/(a + b*x^n)^2,x]

[Out]

(-2*a*x^n)/(b^3*n) + x^(2*n)/(2*b^2*n) + a^3/(b^4*n*(a + b*x^n)) + (3*a^2*Log[a + b*x^n])/(b^4*n)

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.12 \[ \int \frac {x^{-1+4 n}}{\left (a+b x^n\right )^2} \, dx=\frac {2 a^3-4 a^2 b x^n-3 a b^2 x^{2 n}+b^3 x^{3 n}}{2 b^4 n \left (a+b x^n\right )}+\frac {3 a^2 \log \left (a+b x^n\right )}{b^4 n} \]

[In]

Integrate[x^(-1 + 4*n)/(a + b*x^n)^2,x]

[Out]

(2*a^3 - 4*a^2*b*x^n - 3*a*b^2*x^(2*n) + b^3*x^(3*n))/(2*b^4*n*(a + b*x^n)) + (3*a^2*Log[a + b*x^n])/(b^4*n)

Maple [A] (verified)

Time = 4.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {x^{2 n}}{2 b^{2} n}-\frac {2 a \,x^{n}}{b^{3} n}+\frac {a^{3}}{b^{4} n \left (a +b \,x^{n}\right )}+\frac {3 a^{2} \ln \left (x^{n}+\frac {a}{b}\right )}{b^{4} n}\)	\(67\)
norman	\(\frac {\frac {3 a^{3}}{b^{4} n}+\frac {{\mathrm e}^{3 n \ln \left (x \right )}}{2 b n}-\frac {3 a \,{\mathrm e}^{2 n \ln \left (x \right )}}{2 b^{2} n}}{a +b \,{\mathrm e}^{n \ln \left (x \right )}}+\frac {3 a^{2} \ln \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{b^{4} n}\)	\(78\)

[In]

int(x^(-1+4*n)/(a+b*x^n)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15 \[ \int \frac {x^{-1+4 n}}{\left (a+b x^n\right )^2} \, dx=\frac {b^{3} x^{3 \, n} - 3 \, a b^{2} x^{2 \, n} - 4 \, a^{2} b x^{n} + 2 \, a^{3} + 6 \, {\left (a^{2} b x^{n} + a^{3}\right )} \log \left (b x^{n} + a\right )}{2 \, {\left (b^{5} n x^{n} + a b^{4} n\right )}} \]

[In]

integrate(x^(-1+4*n)/(a+b*x^n)^2,x, algorithm="fricas")

[Out]

1/2*(b^3*x^(3*n) - 3*a*b^2*x^(2*n) - 4*a^2*b*x^n + 2*a^3 + 6*(a^2*b*x^n + a^3)*log(b*x^n + a))/(b^5*n*x^n + a*
b^4*n)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (56) = 112\).

Time = 5.56 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.70 \[ \int \frac {x^{-1+4 n}}{\left (a+b x^n\right )^2} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{a^{2}} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x x^{4 n - 1}}{4 a^{2} n} & \text {for}\: b = 0 \\\frac {\log {\left (x \right )}}{\left (a + b\right )^{2}} & \text {for}\: n = 0 \\\frac {6 a^{3} \log {\left (\frac {a}{b} + x^{n} \right )}}{2 a b^{4} n + 2 b^{5} n x^{n}} + \frac {6 a^{3}}{2 a b^{4} n + 2 b^{5} n x^{n}} + \frac {6 a^{2} b x^{n} \log {\left (\frac {a}{b} + x^{n} \right )}}{2 a b^{4} n + 2 b^{5} n x^{n}} - \frac {3 a b^{2} x^{2 n}}{2 a b^{4} n + 2 b^{5} n x^{n}} + \frac {b^{3} x^{3 n}}{2 a b^{4} n + 2 b^{5} n x^{n}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(-1+4*n)/(a+b*x**n)**2,x)

[Out]

Piecewise((log(x)/a**2, Eq(b, 0) & Eq(n, 0)), (x*x**(4*n - 1)/(4*a**2*n), Eq(b, 0)), (log(x)/(a + b)**2, Eq(n,
0)), (6*a**3*log(a/b + x**n)/(2*a*b**4*n + 2*b**5*n*x**n) + 6*a**3/(2*a*b**4*n + 2*b**5*n*x**n) + 6*a**2*b*x*
*n*log(a/b + x**n)/(2*a*b**4*n + 2*b**5*n*x**n) - 3*a*b**2*x**(2*n)/(2*a*b**4*n + 2*b**5*n*x**n) + b**3*x**(3*
n)/(2*a*b**4*n + 2*b**5*n*x**n), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.18 \[ \int \frac {x^{-1+4 n}}{\left (a+b x^n\right )^2} \, dx=\frac {b^{3} x^{3 \, n} - 3 \, a b^{2} x^{2 \, n} - 4 \, a^{2} b x^{n} + 2 \, a^{3}}{2 \, {\left (b^{5} n x^{n} + a b^{4} n\right )}} + \frac {3 \, a^{2} \log \left (\frac {b x^{n} + a}{b}\right )}{b^{4} n} \]

[In]

integrate(x^(-1+4*n)/(a+b*x^n)^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x^{-1+4 n}}{\left (a+b x^n\right )^2} \, dx=\int { \frac {x^{4 \, n - 1}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]

[In]

integrate(x^(-1+4*n)/(a+b*x^n)^2,x, algorithm="giac")

[Out]

integrate(x^(4*n - 1)/(b*x^n + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+4 n}}{\left (a+b x^n\right )^2} \, dx=\int \frac {x^{4\,n-1}}{{\left (a+b\,x^n\right )}^2} \,d x \]

[In]

int(x^(4*n - 1)/(a + b*x^n)^2,x)

[Out]

int(x^(4*n - 1)/(a + b*x^n)^2, x)