3.494 \(\int \frac{x^{-1+3 n}}{b x^n+c x^{2 n}} \, dx\)

Optimal. Leaf size=28 \[ \frac{x^n}{c n}-\frac{b \log \left (b+c x^n\right )}{c^2 n} \]

[Out]

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

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Rubi [A]  time = 0.0258639, antiderivative size = 28, 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} \[ \frac{x^n}{c n}-\frac{b \log \left (b+c x^n\right )}{c^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0141799, size = 26, normalized size = 0.93 \[ \frac{\frac{x^n}{c}-\frac{b \log \left (b+c x^n\right )}{c^2}}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^n/c - (b*Log[b + c*x^n])/c^2)/n

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Maple [A]  time = 0.02, size = 33, normalized size = 1.2 \begin{align*}{\frac{{{\rm e}^{n\ln \left ( x \right ) }}}{cn}}-{\frac{b\ln \left ( c{{\rm e}^{n\ln \left ( x \right ) }}+b \right ) }{{c}^{2}n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.980079, size = 43, normalized size = 1.54 \begin{align*} \frac{x^{n}}{c n} - \frac{b \log \left (\frac{c x^{n} + b}{c}\right )}{c^{2} n} \end{align*}

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]

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

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Fricas [A]  time = 1.9112, size = 49, normalized size = 1.75 \begin{align*} \frac{c x^{n} - b \log \left (c x^{n} + b\right )}{c^{2} n} \end{align*}

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]

(c*x^n - b*log(c*x^n + b))/(c^2*n)

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Sympy [A]  time = 81.3623, size = 26, normalized size = 0.93 \begin{align*} - \frac{b \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 n} + \frac{x^{n}}{c n} \end{align*}

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]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3 \, 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+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)