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

Optimal. Leaf size=46 \[ \frac{b^2 \log \left (b+c x^n\right )}{c^3 n}-\frac{b x^n}{c^2 n}+\frac{x^{2 n}}{2 c n} \]

[Out]

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

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Rubi [A]  time = 0.036353, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {1584, 266, 43} \[ \frac{b^2 \log \left (b+c x^n\right )}{c^3 n}-\frac{b x^n}{c^2 n}+\frac{x^{2 n}}{2 c n} \]

Antiderivative was successfully verified.

[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 1584

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]

Rule 266

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 43

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])

Rubi steps

\begin{align*} \int \frac{x^{-1+4 n}}{b x^n+c x^{2 n}} \, dx &=\int \frac{x^{-1+3 n}}{b+c x^n} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{b+c x} \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{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]  time = 0.0288737, size = 38, normalized size = 0.83 \[ \frac{2 b^2 \log \left (b+c x^n\right )+c x^n \left (c x^n-2 b\right )}{2 c^3 n} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.023, size = 62, normalized size = 1.4 \begin{align*}{\frac{1}{{{\rm e}^{n\ln \left ( x \right ) }}} \left ({\frac{ \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{3}}{2\,cn}}-{\frac{b \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}}{{c}^{2}n}} \right ) }+{\frac{{b}^{2}\ln \left ( c{{\rm e}^{n\ln \left ( x \right ) }}+b \right ) }{{c}^{3}n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.975811, size = 61, normalized size = 1.33 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.93044, size = 84, normalized size = 1.83 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 31.2943, size = 42, normalized size = 0.91 \begin{align*} \frac{b^{2} \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 )}{c^{2} n} - \frac{b x^{n}}{c^{2} n} + \frac{x^{2 n}}{2 c n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4 \, n - 1}}{c x^{2 \, n} + b x^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)