\(\int \frac {x^{-1+4 n}}{a+b x^n} \, dx\) [2598]

Optimal result

Integrand size = 17, antiderivative size = 64 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\frac {a^2 x^n}{b^3 n}-\frac {a x^{2 n}}{2 b^2 n}+\frac {x^{3 n}}{3 b n}-\frac {a^3 \log \left (a+b x^n\right )}{b^4 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, 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} \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=-\frac {a^3 \log \left (a+b x^n\right )}{b^4 n}+\frac {a^2 x^n}{b^3 n}-\frac {a x^{2 n}}{2 b^2 n}+\frac {x^{3 n}}{3 b n} \]

[In]

Int[x^(-1 + 4*n)/(a + b*x^n),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\frac {b x^n \left (6 a^2-3 a b x^n+2 b^2 x^{2 n}\right )-6 a^3 \log \left (a+b x^n\right )}{6 b^4 n} \]

[In]

Integrate[x^(-1 + 4*n)/(a + b*x^n),x]

[Out]

(b*x^n*(6*a^2 - 3*a*b*x^n + 2*b^2*x^(2*n)) - 6*a^3*Log[a + b*x^n])/(6*b^4*n)

Maple [A] (verified)

Time = 4.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {x^{3 n}}{3 b n}-\frac {a \,x^{2 n}}{2 b^{2} n}+\frac {a^{2} x^{n}}{b^{3} n}-\frac {a^{3} \ln \left (x^{n}+\frac {a}{b}\right )}{b^{4} n}\)	\(63\)
norman	\(\frac {a^{2} {\mathrm e}^{n \ln \left (x \right )}}{b^{3} n}+\frac {{\mathrm e}^{3 n \ln \left (x \right )}}{3 b n}-\frac {a \,{\mathrm e}^{2 n \ln \left (x \right )}}{2 b^{2} n}-\frac {a^{3} \ln \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{b^{4} n}\)	\(69\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\frac {2 \, b^{3} x^{3 \, n} - 3 \, a b^{2} x^{2 \, n} + 6 \, a^{2} b x^{n} - 6 \, a^{3} \log \left (b x^{n} + a\right )}{6 \, b^{4} n} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{a} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x x^{4 n - 1}}{4 a n} & \text {for}\: b = 0 \\\frac {\log {\left (x \right )}}{a + b} & \text {for}\: n = 0 \\- \frac {a^{3} \log {\left (\frac {a}{b} + x^{n} \right )}}{b^{4} n} + \frac {a^{2} x^{n}}{b^{3} n} - \frac {a x^{2 n}}{2 b^{2} n} + \frac {x^{3 n}}{3 b n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=-\frac {a^{3} \log \left (\frac {b x^{n} + a}{b}\right )}{b^{4} n} + \frac {2 \, b^{2} x^{3 \, n} - 3 \, a b x^{2 \, n} + 6 \, a^{2} x^{n}}{6 \, b^{3} n} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(x^(4*n - 1)/(b*x^n + a), x)

Mupad [F(-1)]

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

[In]

int(x^(4*n - 1)/(a + b*x^n),x)

[Out]

int(x^(4*n - 1)/(a + b*x^n), x)