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

Optimal. Leaf size=10 \[ \log \left (a x+b x^n\right ) \]

[Out]

Log[a*x + b*x^n]

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Rubi [A]  time = 0.0523407, antiderivative size = 17, normalized size of antiderivative = 1.7, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1593, 514, 446, 72} \[ \log \left (a x^{1-n}+b\right )+n \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

n*Log[x] + Log[b + a*x^(1 - n)]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 446

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.0272881, size = 17, normalized size = 1.7 \[ \log \left (a x^{1-n}+b\right )+n \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

n*Log[x] + Log[b + a*x^(1 - n)]

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Maple [A]  time = 0.016, size = 13, normalized size = 1.3 \begin{align*} \ln \left ( ax+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(a*x+b*exp(n*ln(x)))

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Maxima [A]  time = 1.10843, size = 14, normalized size = 1.4 \begin{align*} \log \left (a x + b x^{n}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*x + b*x^n)

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Fricas [A]  time = 1.83338, size = 24, normalized size = 2.4 \begin{align*} \log \left (a x + b x^{n}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*x + b*x^n)

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Sympy [A]  time = 8.93697, size = 32, normalized size = 3.2 \begin{align*} \begin{cases} \log{\left (x + \frac{b x^{n}}{a} \right )} & \text{for}\: a \neq 0 \\n \left (\frac{n^{2} \log{\left (x \right )}}{n^{2} - n} - \frac{n \log{\left (x \right )}}{n^{2} - n}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((log(x + b*x**n/a), Ne(a, 0)), (n*(n**2*log(x)/(n**2 - n) - n*log(x)/(n**2 - n)), True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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