\(\int \frac {x^{-1-n}}{b x^n+c x^{2 n}} \, dx\) [497]

Optimal result

Integrand size = 23, antiderivative size = 57 \[ \int \frac {x^{-1-n}}{b x^n+c x^{2 n}} \, dx=-\frac {x^{-2 n}}{2 b n}+\frac {c x^{-n}}{b^2 n}+\frac {c^2 \log (x)}{b^3}-\frac {c^2 \log \left (b+c x^n\right )}{b^3 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, 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, 46} \[ \int \frac {x^{-1-n}}{b x^n+c x^{2 n}} \, dx=-\frac {c^2 \log \left (b+c x^n\right )}{b^3 n}+\frac {c^2 \log (x)}{b^3}+\frac {c x^{-n}}{b^2 n}-\frac {x^{-2 n}}{2 b n} \]

[In]

Int[x^(-1 - n)/(b*x^n + c*x^(2*n)),x]

[Out]

-1/2*1/(b*n*x^(2*n)) + c/(b^2*n*x^n) + (c^2*Log[x])/b^3 - (c^2*Log[b + c*x^n])/(b^3*n)

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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^{-1-2 n}}{b+c x^n} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {1}{x^3 (b+c x)} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{b x^3}-\frac {c}{b^2 x^2}+\frac {c^2}{b^3 x}-\frac {c^3}{b^3 (b+c x)}\right ) \, dx,x,x^n\right )}{n} \\ & = -\frac {x^{-2 n}}{2 b n}+\frac {c x^{-n}}{b^2 n}+\frac {c^2 \log (x)}{b^3}-\frac {c^2 \log \left (b+c x^n\right )}{b^3 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int \frac {x^{-1-n}}{b x^n+c x^{2 n}} \, dx=-\frac {b x^{-2 n} \left (b-2 c x^n\right )-2 c^2 \log \left (x^n\right )+2 c^2 \log \left (b+c x^n\right )}{2 b^3 n} \]

[In]

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

[Out]

-1/2*((b*(b - 2*c*x^n))/x^(2*n) - 2*c^2*Log[x^n] + 2*c^2*Log[b + c*x^n])/(b^3*n)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {c \,x^{-n}}{b^{2} n}-\frac {x^{-2 n}}{2 b n}+\frac {c^{2} \ln \left (x \right )}{b^{3}}-\frac {c^{2} \ln \left (x^{n}+\frac {b}{c}\right )}{b^{3} n}\)	\(58\)
norman	\(\left (\frac {c \,{\mathrm e}^{n \ln \left (x \right )}}{b^{2} n}-\frac {1}{2 b n}+\frac {c^{2} \ln \left (x \right ) {\mathrm e}^{2 n \ln \left (x \right )}}{b^{3}}\right ) {\mathrm e}^{-2 n \ln \left (x \right )}-\frac {c^{2} \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}+b \right )}{b^{3} n}\)	\(69\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04 \[ \int \frac {x^{-1-n}}{b x^n+c x^{2 n}} \, dx=\frac {2 \, c^{2} n x^{2 \, n} \log \left (x\right ) - 2 \, c^{2} x^{2 \, n} \log \left (c x^{n} + b\right ) + 2 \, b c x^{n} - b^{2}}{2 \, b^{3} n x^{2 \, n}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (48) = 96\).

Time = 29.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.75 \[ \int \frac {x^{-1-n}}{b x^n+c x^{2 n}} \, dx=\begin {cases} \tilde {\infty } \log {\left (x \right )} & \text {for}\: b = 0 \wedge c = 0 \wedge n = 0 \\- \frac {x x^{- 2 n} x^{- n - 1}}{3 c n} & \text {for}\: b = 0 \\- \frac {x x^{- n} x^{- n - 1}}{2 b n} & \text {for}\: c = 0 \\\frac {\log {\left (x \right )}}{b + c} & \text {for}\: n = 0 \\- \frac {x^{- 2 n}}{2 b n} + \frac {c x^{- n}}{b^{2} n} + \frac {c^{2} \log {\left (x^{n} \right )}}{b^{3} n} - \frac {c^{2} \log {\left (\frac {b}{c} + x^{n} \right )}}{b^{3} n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02 \[ \int \frac {x^{-1-n}}{b x^n+c x^{2 n}} \, dx=\frac {c^{2} \log \left (x\right )}{b^{3}} - \frac {c^{2} \log \left (\frac {c x^{n} + b}{c}\right )}{b^{3} n} + \frac {2 \, c x^{n} - b}{2 \, b^{2} n x^{2 \, n}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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