\(\int \frac {x^{-1-2 n}}{b x^n+c x^{2 n}} \, dx\) [498]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int \frac {x^{-1-2 n}}{b x^n+c x^{2 n}} \, dx=-\frac {x^{-3 n}}{3 b n}+\frac {c x^{-2 n}}{2 b^2 n}-\frac {c^2 x^{-n}}{b^3 n}-\frac {c^3 \log (x)}{b^4}+\frac {c^3 \log \left (b+c x^n\right )}{b^4 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, 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} \[ \int \frac {x^{-1-2 n}}{b x^n+c x^{2 n}} \, dx=\frac {c^3 \log \left (b+c x^n\right )}{b^4 n}-\frac {c^3 \log (x)}{b^4}-\frac {c^2 x^{-n}}{b^3 n}+\frac {c x^{-2 n}}{2 b^2 n}-\frac {x^{-3 n}}{3 b n} \]

[In]

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

[Out]

-1/3*1/(b*n*x^(3*n)) + c/(2*b^2*n*x^(2*n)) - c^2/(b^3*n*x^n) - (c^3*Log[x])/b^4 + (c^3*Log[b + c*x^n])/(b^4*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^{-1-3 n}}{b+c x^n} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {1}{x^4 (b+c x)} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{b x^4}-\frac {c}{b^2 x^3}+\frac {c^2}{b^3 x^2}-\frac {c^3}{b^4 x}+\frac {c^4}{b^4 (b+c x)}\right ) \, dx,x,x^n\right )}{n} \\ & = -\frac {x^{-3 n}}{3 b n}+\frac {c x^{-2 n}}{2 b^2 n}-\frac {c^2 x^{-n}}{b^3 n}-\frac {c^3 \log (x)}{b^4}+\frac {c^3 \log \left (b+c x^n\right )}{b^4 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.83 \[ \int \frac {x^{-1-2 n}}{b x^n+c x^{2 n}} \, dx=-\frac {b x^{-3 n} \left (2 b^2-3 b c x^n+6 c^2 x^{2 n}\right )+6 c^3 \log \left (x^n\right )-6 c^3 \log \left (b+c x^n\right )}{6 b^4 n} \]

[In]

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

[Out]

-1/6*((b*(2*b^2 - 3*b*c*x^n + 6*c^2*x^(2*n)))/x^(3*n) + 6*c^3*Log[x^n] - 6*c^3*Log[b + c*x^n])/(b^4*n)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99

method	result	size
risch	\(-\frac {c^{2} x^{-n}}{b^{3} n}+\frac {c \,x^{-2 n}}{2 b^{2} n}-\frac {x^{-3 n}}{3 b n}-\frac {c^{3} \ln \left (x \right )}{b^{4}}+\frac {c^{3} \ln \left (x^{n}+\frac {b}{c}\right )}{b^{4} n}\)	\(75\)
norman	\(\left (-\frac {1}{3 b n}+\frac {c \,{\mathrm e}^{n \ln \left (x \right )}}{2 b^{2} n}-\frac {c^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{b^{3} n}-\frac {c^{3} \ln \left (x \right ) {\mathrm e}^{3 n \ln \left (x \right )}}{b^{4}}\right ) {\mathrm e}^{-3 n \ln \left (x \right )}+\frac {c^{3} \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}+b \right )}{b^{4} n}\)	\(88\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.95 \[ \int \frac {x^{-1-2 n}}{b x^n+c x^{2 n}} \, dx=-\frac {6 \, c^{3} n x^{3 \, n} \log \left (x\right ) - 6 \, c^{3} x^{3 \, n} \log \left (c x^{n} + b\right ) + 6 \, b c^{2} x^{2 \, n} - 3 \, b^{2} c x^{n} + 2 \, b^{3}}{6 \, b^{4} n x^{3 \, n}} \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {x^{-1-2 n}}{b x^n+c x^{2 n}} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93 \[ \int \frac {x^{-1-2 n}}{b x^n+c x^{2 n}} \, dx=-\frac {c^{3} \log \left (x\right )}{b^{4}} + \frac {c^{3} \log \left (\frac {c x^{n} + b}{c}\right )}{b^{4} n} - \frac {6 \, c^{2} x^{2 \, n} - 3 \, b c x^{n} + 2 \, b^{2}}{6 \, b^{3} n x^{3 \, n}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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