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

Optimal. Leaf size=19 \[ \frac{\log \left (a+b x^n+c x^{2 n}\right )}{n} \]

[Out]

Log[a + b*x^n + c*x^(2*n)]/n

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Rubi [A]  time = 0.0269621, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {1468, 628} \[ \frac{\log \left (a+b x^n+c x^{2 n}\right )}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a + b*x^n + c*x^(2*n)]/n

Rule 1468

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x]
&& EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.109311, size = 19, normalized size = 1. \[ \frac{\log \left (a+b x^n+c x^{2 n}\right )}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[a + b*x^n + c*x^(2*n)]/n

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Maple [A]  time = 0.019, size = 24, normalized size = 1.3 \begin{align*}{\frac{\ln \left ( a+b{{\rm e}^{n\ln \left ( x \right ) }}+c \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2} \right ) }{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16091, size = 31, normalized size = 1.63 \begin{align*} \frac{\log \left (\frac{c x^{2 \, n} + b x^{n} + a}{c}\right )}{n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((c*x^(2*n) + b*x^n + a)/c)/n

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Fricas [A]  time = 1.10865, size = 41, normalized size = 2.16 \begin{align*} \frac{\log \left (c x^{2 \, n} + b x^{n} + a\right )}{n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*x^(2*n) + b*x^n + a)/n

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.08196, size = 26, normalized size = 1.37 \begin{align*} \frac{\log \left (c x^{2 \, n} + b x^{n} + a\right )}{n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*x^(2*n) + b*x^n + a)/n