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]

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

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Rubi [A]  time = 0.05, antiderivative size = 17, normalized size of antiderivative = 1.70, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 1.70 \[ \log \left (a x^{1-n}+b\right )+n \log (x) \]

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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fricas [A]  time = 0.42, size = 10, normalized size = 1.00 \[ \log \left (a x + b x^{n}\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b n x^{n - 1} + a}{a x + b x^{n}}\,{d x} \]

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)

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maple [A]  time = 0.02, size = 13, normalized size = 1.30 \[ \ln \left (a x +b \,{\mathrm e}^{n \ln \relax (x )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.87, size = 10, normalized size = 1.00 \[ \log \left (a x + b x^{n}\right ) \]

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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mupad [B]  time = 3.27, size = 10, normalized size = 1.00 \[ \ln \left (b\,x^n+a\,x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b*x^n + a*x)

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sympy [A]  time = 6.96, size = 32, normalized size = 3.20 \[ \begin {cases} \log {\left (x + \frac {b x^{n}}{a} \right )} & \text {for}\: a \neq 0 \\n \left (\frac {n^{2} \log {\relax (x )}}{n^{2} - n} - \frac {n \log {\relax (x )}}{n^{2} - n}\right ) & \text {otherwise} \end {cases} \]

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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