\(\int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx\) [493]

Optimal result

Integrand size = 23, antiderivative size = 46 \[ \int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx=-\frac {b x^n}{c^2 n}+\frac {x^{2 n}}{2 c n}+\frac {b^2 \log \left (b+c x^n\right )}{c^3 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1598, 272, 45} \[ \int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx=\frac {b^2 \log \left (b+c x^n\right )}{c^3 n}-\frac {b x^n}{c^2 n}+\frac {x^{2 n}}{2 c n} \]

[In]

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

[Out]

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

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^{-1+3 n}}{b+c x^n} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{b+c x} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {b}{c^2}+\frac {x}{c}+\frac {b^2}{c^2 (b+c x)}\right ) \, dx,x,x^n\right )}{n} \\ & = -\frac {b x^n}{c^2 n}+\frac {x^{2 n}}{2 c n}+\frac {b^2 \log \left (b+c x^n\right )}{c^3 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx=\frac {c x^n \left (-2 b+c x^n\right )+2 b^2 \log \left (b+c x^n\right )}{2 c^3 n} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {x^{2 n}}{2 c n}-\frac {b \,x^{n}}{c^{2} n}+\frac {b^{2} \ln \left (x^{n}+\frac {b}{c}\right )}{c^{3} n}\)	\(47\)
norman	\(\left (\frac {{\mathrm e}^{3 n \ln \left (x \right )}}{2 c n}-\frac {b \,{\mathrm e}^{2 n \ln \left (x \right )}}{c^{2} n}\right ) {\mathrm e}^{-n \ln \left (x \right )}+\frac {b^{2} \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}+b \right )}{c^{3} n}\)	\(62\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx=\frac {c^{2} x^{2 \, n} - 2 \, b c x^{n} + 2 \, b^{2} \log \left (c x^{n} + b\right )}{2 \, c^{3} n} \]

[In]

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

[Out]

1/2*(c^2*x^(2*n) - 2*b*c*x^n + 2*b^2*log(c*x^n + b))/(c^3*n)

Sympy [A] (verification not implemented)

Time = 22.44 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.37 \[ \int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{b} & \text {for}\: c = 0 \wedge n = 0 \\\frac {x x^{- n} x^{4 n - 1}}{3 b n} & \text {for}\: c = 0 \\\frac {\log {\left (x \right )}}{b + c} & \text {for}\: n = 0 \\\frac {b^{2} \log {\left (\frac {b}{c} + x^{n} \right )}}{c^{3} n} - \frac {b x^{n}}{c^{2} n} + \frac {x^{2 n}}{2 c n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98 \[ \int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx=\frac {b^{2} \log \left (\frac {c x^{n} + b}{c}\right )}{c^{3} n} + \frac {c x^{2 \, n} - 2 \, b x^{n}}{2 \, c^{2} n} \]

[In]

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

[Out]

b^2*log((c*x^n + b)/c)/(c^3*n) + 1/2*(c*x^(2*n) - 2*b*x^n)/(c^2*n)

Giac [F]

\[ \int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx=\int { \frac {x^{4 \, n - 1}}{c x^{2 \, n} + b x^{n}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx=\int \frac {x^{4\,n-1}}{b\,x^n+c\,x^{2\,n}} \,d x \]

[In]

int(x^(4*n - 1)/(b*x^n + c*x^(2*n)),x)

[Out]

int(x^(4*n - 1)/(b*x^n + c*x^(2*n)), x)