3.2628 \(\int \frac {1}{x (a+b x^n)^2} \, dx\)

Optimal. Leaf size=39 \[ -\frac {\log \left (a+b x^n\right )}{a^2 n}+\frac {\log (x)}{a^2}+\frac {1}{a n \left (a+b x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 44} \[ -\frac {\log \left (a+b x^n\right )}{a^2 n}+\frac {\log (x)}{a^2}+\frac {1}{a n \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^n)^2),x]

[Out]

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x (a+b x)^2} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a^2 x}-\frac {b}{a (a+b x)^2}-\frac {b}{a^2 (a+b x)}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {1}{a n \left (a+b x^n\right )}+\frac {\log (x)}{a^2}-\frac {\log \left (a+b x^n\right )}{a^2 n}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 33, normalized size = 0.85 \[ \frac {\frac {a}{a+b x^n}-\log \left (a+b x^n\right )+n \log (x)}{a^2 n} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^n)^2),x]

[Out]

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

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fricas [A]  time = 0.53, size = 50, normalized size = 1.28 \[ \frac {b n x^{n} \log \relax (x) + a n \log \relax (x) - {\left (b x^{n} + a\right )} \log \left (b x^{n} + a\right ) + a}{a^{2} b n x^{n} + a^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*n*x^n*log(x) + a*n*log(x) - (b*x^n + a)*log(b*x^n + a) + a)/(a^2*b*n*x^n + a^3*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{n} + a\right )}^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^n + a)^2*x), x)

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maple [A]  time = 0.00, size = 45, normalized size = 1.15 \[ \frac {1}{\left (b \,x^{n}+a \right ) a n}+\frac {\ln \left (x^{n}\right )}{a^{2} n}-\frac {\ln \left (b \,x^{n}+a \right )}{a^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 43, normalized size = 1.10 \[ \frac {1}{a b n x^{n} + a^{2} n} - \frac {\log \left (b x^{n} + a\right )}{a^{2} n} + \frac {\log \left (x^{n}\right )}{a^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.19, size = 39, normalized size = 1.00 \[ \frac {\ln \relax (x)}{a^2}+\frac {1}{a\,n\,\left (a+b\,x^n\right )}-\frac {\ln \left (a+b\,x^n\right )}{a^2\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^n)^2),x)

[Out]

log(x)/a^2 + 1/(a*n*(a + b*x^n)) - log(a + b*x^n)/(a^2*n)

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sympy [A]  time = 1.45, size = 160, normalized size = 4.10 \[ \begin {cases} \tilde {\infty } \log {\relax (x )} & \text {for}\: a = 0 \wedge b = 0 \wedge n = 0 \\- \frac {x^{- 2 n}}{2 b^{2} n} & \text {for}\: a = 0 \\\tilde {\infty } \log {\relax (x )} & \text {for}\: b = - a x^{- n} \\\frac {\log {\relax (x )}}{\left (a + b\right )^{2}} & \text {for}\: n = 0 \\\frac {\log {\relax (x )}}{a^{2}} & \text {for}\: b = 0 \\\frac {a n \log {\relax (x )}}{a^{3} n + a^{2} b n x^{n}} - \frac {a \log {\left (\frac {a}{b} + x^{n} \right )}}{a^{3} n + a^{2} b n x^{n}} + \frac {a}{a^{3} n + a^{2} b n x^{n}} + \frac {b n x^{n} \log {\relax (x )}}{a^{3} n + a^{2} b n x^{n}} - \frac {b x^{n} \log {\left (\frac {a}{b} + x^{n} \right )}}{a^{3} n + a^{2} b n x^{n}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*log(x), Eq(a, 0) & Eq(b, 0) & Eq(n, 0)), (-x**(-2*n)/(2*b**2*n), Eq(a, 0)), (zoo*log(x), Eq(b,
-a*x**(-n))), (log(x)/(a + b)**2, Eq(n, 0)), (log(x)/a**2, Eq(b, 0)), (a*n*log(x)/(a**3*n + a**2*b*n*x**n) - a
*log(a/b + x**n)/(a**3*n + a**2*b*n*x**n) + a/(a**3*n + a**2*b*n*x**n) + b*n*x**n*log(x)/(a**3*n + a**2*b*n*x*
*n) - b*x**n*log(a/b + x**n)/(a**3*n + a**2*b*n*x**n), True))

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