3.6.33 \(\int \frac {1}{x \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx\) [533]

Optimal. Leaf size=85 \[ \frac {\left (a+b x^n\right ) \log (x)}{a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {\left (a+b x^n\right ) \log \left (a+b x^n\right )}{a n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1369, 272, 36, 29, 31} \begin {gather*} \frac {\log (x) \left (a+b x^n\right )}{a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {\left (a+b x^n\right ) \log \left (a+b x^n\right )}{a n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^n)*Log[x])/(a*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) - ((a + b*x^n)*Log[a + b*x^n])/(a*n*Sqrt[a^2 + 2*
a*b*x^n + b^2*x^(2*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 272

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]]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 45, normalized size = 0.53 \begin {gather*} \frac {\left (a+b x^n\right ) \left (\log \left (x^n\right )-\log \left (a n \left (a+b x^n\right )\right )\right )}{a n \sqrt {\left (a+b x^n\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 66, normalized size = 0.78

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, \ln \left (x \right )}{\left (a +b \,x^{n}\right ) a}-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, \ln \left (x^{n}+\frac {a}{b}\right )}{\left (a +b \,x^{n}\right ) a n}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 27, normalized size = 0.32 \begin {gather*} \frac {\log \left (x\right )}{a} - \frac {\log \left (\frac {b x^{n} + a}{b}\right )}{a n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/a - log((b*x^n + a)/b)/(a*n)

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Fricas [A]
time = 0.40, size = 22, normalized size = 0.26 \begin {gather*} \frac {n \log \left (x\right ) - \log \left (b x^{n} + a\right )}{a n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(n*log(x) - log(b*x^n + a))/(a*n)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \sqrt {\left (a + b x^{n}\right )^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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