3.2602 \(\int \frac {1}{x (a+b x^n)} \, dx\)

Optimal. Leaf size=23 \[ \frac {\log (x)}{a}-\frac {\log \left (a+b x^n\right )}{a n} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 36, 29, 31} \[ \frac {\log (x)}{a}-\frac {\log \left (a+b x^n\right )}{a n} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^n)),x]

[Out]

Log[x]/a - Log[a + b*x^n]/(a*n)

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.96 \[ \frac {n \log (x)-\log \left (a+b x^n\right )}{a n} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^n)),x]

[Out]

(n*Log[x] - Log[a + b*x^n])/(a*n)

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fricas [A]  time = 0.69, size = 22, normalized size = 0.96 \[ \frac {n \log \relax (x) - \log \left (b x^{n} + a\right )}{a n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(n*log(x) - log(b*x^n + a))/(a*n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^n + a)*x), x)

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maple [A]  time = 0.00, size = 29, normalized size = 1.26 \[ \frac {\ln \left (x^{n}\right )}{a n}-\frac {\ln \left (b \,x^{n}+a \right )}{a n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.60, size = 28, normalized size = 1.22 \[ -\frac {\log \left (b x^{n} + a\right )}{a n} + \frac {\log \left (x^{n}\right )}{a n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(b*x^n + a)/(a*n) + log(x^n)/(a*n)

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mupad [B]  time = 1.18, size = 22, normalized size = 0.96 \[ -\frac {\ln \left (a+b\,x^n\right )-n\,\ln \relax (x)}{a\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^n)),x)

[Out]

-(log(a + b*x^n) - n*log(x))/(a*n)

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sympy [A]  time = 0.74, size = 41, normalized size = 1.78 \[ \begin {cases} \tilde {\infty } \log {\relax (x )} & \text {for}\: a = 0 \wedge b = 0 \wedge n = 0 \\\frac {\log {\relax (x )}}{a} & \text {for}\: b = 0 \\- \frac {x^{- n}}{b n} & \text {for}\: a = 0 \\\frac {\log {\relax (x )}}{a + b} & \text {for}\: n = 0 \\\frac {\log {\relax (x )}}{a} - \frac {\log {\left (\frac {a}{b} + x^{n} \right )}}{a n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*log(x), Eq(a, 0) & Eq(b, 0) & Eq(n, 0)), (log(x)/a, Eq(b, 0)), (-x**(-n)/(b*n), Eq(a, 0)), (log
(x)/(a + b), Eq(n, 0)), (log(x)/a - log(a/b + x**n)/(a*n), True))

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