3.5.99 \(\int \frac {x^{-1-3 n}}{b x^n+c x^{2 n}} \, dx\) [499]

Optimal. Leaf size=93 \[ -\frac {x^{-4 n}}{4 b n}+\frac {c x^{-3 n}}{3 b^2 n}-\frac {c^2 x^{-2 n}}{2 b^3 n}+\frac {c^3 x^{-n}}{b^4 n}+\frac {c^4 \log (x)}{b^5}-\frac {c^4 \log \left (b+c x^n\right )}{b^5 n} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1598, 272, 46} \begin {gather*} -\frac {c^4 \log \left (b+c x^n\right )}{b^5 n}+\frac {c^4 \log (x)}{b^5}+\frac {c^3 x^{-n}}{b^4 n}-\frac {c^2 x^{-2 n}}{2 b^3 n}+\frac {c x^{-3 n}}{3 b^2 n}-\frac {x^{-4 n}}{4 b n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*1/(b*n*x^(4*n)) + c/(3*b^2*n*x^(3*n)) - c^2/(2*b^3*n*x^(2*n)) + c^3/(b^4*n*x^n) + (c^4*Log[x])/b^5 - (c^4
*Log[b + c*x^n])/(b^5*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*} \int \frac {x^{-1-3 n}}{b x^n+c x^{2 n}} \, dx &=\int \frac {x^{-1-4 n}}{b+c x^n} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {1}{x^5 (b+c x)} \, dx,x,x^n\right )}{n}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{b x^5}-\frac {c}{b^2 x^4}+\frac {c^2}{b^3 x^3}-\frac {c^3}{b^4 x^2}+\frac {c^4}{b^5 x}-\frac {c^5}{b^5 (b+c x)}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac {x^{-4 n}}{4 b n}+\frac {c x^{-3 n}}{3 b^2 n}-\frac {c^2 x^{-2 n}}{2 b^3 n}+\frac {c^3 x^{-n}}{b^4 n}+\frac {c^4 \log (x)}{b^5}-\frac {c^4 \log \left (b+c x^n\right )}{b^5 n}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 76, normalized size = 0.82 \begin {gather*} -\frac {b x^{-4 n} \left (3 b^3-4 b^2 c x^n+6 b c^2 x^{2 n}-12 c^3 x^{3 n}\right )-12 c^4 \log \left (x^n\right )+12 c^4 \log \left (b+c x^n\right )}{12 b^5 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((b*(3*b^3 - 4*b^2*c*x^n + 6*b*c^2*x^(2*n) - 12*c^3*x^(3*n)))/x^(4*n) - 12*c^4*Log[x^n] + 12*c^4*Log[b +
c*x^n])/(b^5*n)

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Maple [A]
time = 0.20, size = 90, normalized size = 0.97

method	result	size
risch	\(\frac {c^{3} x^{-n}}{b^{4} n}-\frac {c^{2} x^{-2 n}}{2 b^{3} n}+\frac {c \,x^{-3 n}}{3 b^{2} n}-\frac {x^{-4 n}}{4 b n}+\frac {c^{4} \ln \left (x \right )}{b^{5}}-\frac {c^{4} \ln \left (x^{n}+\frac {b}{c}\right )}{b^{5} n}\)	\(90\)
norman	\(\left (\frac {c^{3} {\mathrm e}^{3 n \ln \left (x \right )}}{b^{4} n}-\frac {1}{4 b n}+\frac {c \,{\mathrm e}^{n \ln \left (x \right )}}{3 b^{2} n}-\frac {c^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{2 b^{3} n}+\frac {c^{4} \ln \left (x \right ) {\mathrm e}^{4 n \ln \left (x \right )}}{b^{5}}\right ) {\mathrm e}^{-4 n \ln \left (x \right )}-\frac {c^{4} \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}+b \right )}{b^{5} n}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1-3*n)/(b*x^n+c*x^(2*n)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 84, normalized size = 0.90 \begin {gather*} \frac {c^{4} \log \left (x\right )}{b^{5}} - \frac {c^{4} \log \left (\frac {c x^{n} + b}{c}\right )}{b^{5} n} + \frac {12 \, c^{3} x^{3 \, n} - 6 \, b c^{2} x^{2 \, n} + 4 \, b^{2} c x^{n} - 3 \, b^{3}}{12 \, b^{4} n x^{4 \, n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-3*n)/(b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 85, normalized size = 0.91 \begin {gather*} \frac {12 \, c^{4} n x^{4 \, n} \log \left (x\right ) - 12 \, c^{4} x^{4 \, n} \log \left (c x^{n} + b\right ) + 12 \, b c^{3} x^{3 \, n} - 6 \, b^{2} c^{2} x^{2 \, n} + 4 \, b^{3} c x^{n} - 3 \, b^{4}}{12 \, b^{5} n x^{4 \, n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-3*n)/(b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 40.51, size = 88, normalized size = 0.95 \begin {gather*} - \frac {x^{- 4 n}}{4 b n} + \frac {c x^{- 3 n}}{3 b^{2} n} - \frac {c^{2} x^{- 2 n}}{2 b^{3} n} + \frac {c^{3} x^{- n}}{b^{4} n} - \frac {c^{5} \left (\begin {cases} \frac {x^{n}}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + c x^{n} \right )}}{c} & \text {otherwise} \end {cases}\right )}{b^{5} n} + \frac {c^{4} \log {\left (x^{n} \right )}}{b^{5} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-3*n)/(b*x**n+c*x**(2*n)),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-3*n)/(b*x^n+c*x^(2*n)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{3\,n+1}\,\left (b\,x^n+c\,x^{2\,n}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(3*n + 1)*(b*x^n + c*x^(2*n))),x)

[Out]

int(1/(x^(3*n + 1)*(b*x^n + c*x^(2*n))), x)

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