\(\int \frac {x^{4+5 (-1+n)}}{a+b x^n} \, dx\) [2606]

Optimal result

Integrand size = 19, antiderivative size = 82 \[ \int \frac {x^{4+5 (-1+n)}}{a+b x^n} \, dx=-\frac {a^3 x^n}{b^4 n}+\frac {a^2 x^{2 n}}{2 b^3 n}-\frac {a x^{3 n}}{3 b^2 n}+\frac {x^{4 n}}{4 b n}+\frac {a^4 \log \left (a+b x^n\right )}{b^5 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {272, 45} \[ \int \frac {x^{4+5 (-1+n)}}{a+b x^n} \, dx=\frac {a^4 \log \left (a+b x^n\right )}{b^5 n}-\frac {a^3 x^n}{b^4 n}+\frac {a^2 x^{2 n}}{2 b^3 n}-\frac {a x^{3 n}}{3 b^2 n}+\frac {x^{4 n}}{4 b n} \]

[In]

Int[x^(4 + 5*(-1 + n))/(a + b*x^n),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int \frac {x^{4+5 (-1+n)}}{a+b x^n} \, dx=\frac {b x^n \left (-12 a^3+6 a^2 b x^n-4 a b^2 x^{2 n}+3 b^3 x^{3 n}\right )+12 a^4 \log \left (a+b x^n\right )}{12 b^5 n} \]

[In]

Integrate[x^(4 + 5*(-1 + n))/(a + b*x^n),x]

[Out]

(b*x^n*(-12*a^3 + 6*a^2*b*x^n - 4*a*b^2*x^(2*n) + 3*b^3*x^(3*n)) + 12*a^4*Log[a + b*x^n])/(12*b^5*n)

Maple [A] (verified)

Time = 3.87 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {x^{4 n}}{4 b n}-\frac {a \,x^{3 n}}{3 b^{2} n}+\frac {a^{2} x^{2 n}}{2 b^{3} n}-\frac {a^{3} x^{n}}{b^{4} n}+\frac {a^{4} \ln \left (x^{n}+\frac {a}{b}\right )}{b^{5} n}\)	\(79\)
norman	\(\frac {{\mathrm e}^{4 n \ln \left (x \right )}}{4 b n}-\frac {a \,{\mathrm e}^{3 n \ln \left (x \right )}}{3 b^{2} n}+\frac {a^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{2 b^{3} n}-\frac {a^{3} {\mathrm e}^{n \ln \left (x \right )}}{b^{4} n}+\frac {a^{4} \ln \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{b^{5} n}\)	\(87\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int \frac {x^{4+5 (-1+n)}}{a+b x^n} \, dx=\frac {3 \, b^{4} x^{4 \, n} - 4 \, a b^{3} x^{3 \, n} + 6 \, a^{2} b^{2} x^{2 \, n} - 12 \, a^{3} b x^{n} + 12 \, a^{4} \log \left (b x^{n} + a\right )}{12 \, b^{5} n} \]

[In]

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

[Out]

1/12*(3*b^4*x^(4*n) - 4*a*b^3*x^(3*n) + 6*a^2*b^2*x^(2*n) - 12*a^3*b*x^n + 12*a^4*log(b*x^n + a))/(b^5*n)

Sympy [A] (verification not implemented)

Time = 2.67 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10 \[ \int \frac {x^{4+5 (-1+n)}}{a+b x^n} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{a} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x x^{5 n - 1}}{5 a n} & \text {for}\: b = 0 \\\frac {\log {\left (x \right )}}{a + b} & \text {for}\: n = 0 \\\frac {a^{4} \log {\left (\frac {a}{b} + x^{n} \right )}}{b^{5} n} - \frac {a^{3} x^{n}}{b^{4} n} + \frac {a^{2} x^{2 n}}{2 b^{3} n} - \frac {a x^{3 n}}{3 b^{2} n} + \frac {x^{4 n}}{4 b n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88 \[ \int \frac {x^{4+5 (-1+n)}}{a+b x^n} \, dx=\frac {a^{4} \log \left (\frac {b x^{n} + a}{b}\right )}{b^{5} n} + \frac {3 \, b^{3} x^{4 \, n} - 4 \, a b^{2} x^{3 \, n} + 6 \, a^{2} b x^{2 \, n} - 12 \, a^{3} x^{n}}{12 \, b^{4} n} \]

[In]

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

[Out]

a^4*log((b*x^n + a)/b)/(b^5*n) + 1/12*(3*b^3*x^(4*n) - 4*a*b^2*x^(3*n) + 6*a^2*b*x^(2*n) - 12*a^3*x^n)/(b^4*n)

Giac [F]

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

[In]

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

[Out]

integrate(x^(5*n - 1)/(b*x^n + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{4+5 (-1+n)}}{a+b x^n} \, dx=\int \frac {x^{5\,n-1}}{a+b\,x^n} \,d x \]

[In]

int(x^(5*n - 1)/(a + b*x^n),x)

[Out]

int(x^(5*n - 1)/(a + b*x^n), x)