3.540 \(\int \frac{1}{x (a^2+2 a b x^n+b^2 x^{2 n})^{3/2}} \, dx\)

Optimal. Leaf size=159 \[ \frac{1}{a^2 n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}+\frac{1}{2 a n \left (a+b x^n\right ) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}+\frac{\log (x) \left (a+b x^n\right )}{a^3 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}-\frac{\left (a+b x^n\right ) \log \left (a+b x^n\right )}{a^3 n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

1/(a^2*n*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) + 1/(2*a*n*(a + b*x^n)*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) + ((
a + b*x^n)*Log[x])/(a^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) - ((a + b*x^n)*Log[a + b*x^n])/(a^3*n*Sqrt[a^2 +
2*a*b*x^n + b^2*x^(2*n)])

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Rubi [A]  time = 0.0830614, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {1355, 266, 44} \[ \frac{1}{a^2 n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}+\frac{1}{2 a n \left (a+b x^n\right ) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}+\frac{\log (x) \left (a+b x^n\right )}{a^3 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}-\frac{\left (a+b x^n\right ) \log \left (a+b x^n\right )}{a^3 n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2)),x]

[Out]

1/(a^2*n*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) + 1/(2*a*n*(a + b*x^n)*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) + ((
a + b*x^n)*Log[x])/(a^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) - ((a + b*x^n)*Log[a + b*x^n])/(a^3*n*Sqrt[a^2 +
2*a*b*x^n + b^2*x^(2*n)])

Rule 1355

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]

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]]

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])

Rubi steps

\begin{align*} \int \frac{1}{x \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x^n\right )\right ) \int \frac{1}{x \left (a b+b^2 x^n\right )^3} \, dx}{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac{\left (b^2 \left (a b+b^2 x^n\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (a b+b^2 x\right )^3} \, dx,x,x^n\right )}{n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac{\left (b^2 \left (a b+b^2 x^n\right )\right ) \operatorname{Subst}\left (\int \left (\frac{1}{a^3 b^3 x}-\frac{1}{a b^2 (a+b x)^3}-\frac{1}{a^2 b^2 (a+b x)^2}-\frac{1}{a^3 b^2 (a+b x)}\right ) \, dx,x,x^n\right )}{n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac{1}{a^2 n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}+\frac{1}{2 a n \left (a+b x^n\right ) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}+\frac{\left (a+b x^n\right ) \log (x)}{a^3 \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^3 n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ \end{align*}

Mathematica [A]  time = 0.0673037, size = 76, normalized size = 0.48 \[ \frac{\left (a+b x^n\right )^3 \left (\frac{1}{a^2 \left (a+b x^n\right )}-\frac{\log \left (a+b x^n\right )}{a^3}+\frac{n \log (x)}{a^3}+\frac{1}{2 a \left (a+b x^n\right )^2}\right )}{n \left (\left (a+b x^n\right )^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2)),x]

[Out]

((a + b*x^n)^3*(1/(2*a*(a + b*x^n)^2) + 1/(a^2*(a + b*x^n)) + (n*Log[x])/a^3 - Log[a + b*x^n]/a^3))/(n*((a + b
*x^n)^2)^(3/2))

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Maple [A]  time = 0.02, size = 104, normalized size = 0.7 \begin{align*}{\frac{\ln \left ( x \right ) }{ \left ( a+b{x}^{n} \right ){a}^{3}}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}+{\frac{2\,b{x}^{n}+3\,a}{2\, \left ( a+b{x}^{n} \right ) ^{3}{a}^{2}n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}-{\frac{1}{ \left ( a+b{x}^{n} \right ){a}^{3}n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}\ln \left ({x}^{n}+{\frac{a}{b}} \right ) } \end{align*}

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))^(3/2),x)

[Out]

((a+b*x^n)^2)^(1/2)/(a+b*x^n)*ln(x)/a^3+1/2*((a+b*x^n)^2)^(1/2)/(a+b*x^n)^3*(2*b*x^n+3*a)/a^2/n-((a+b*x^n)^2)^
(1/2)/(a+b*x^n)/a^3/n*ln(x^n+a/b)

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Maxima [A]  time = 0.962489, size = 95, normalized size = 0.6 \begin{align*} \frac{2 \, b x^{n} + 3 \, a}{2 \,{\left (a^{2} b^{2} n x^{2 \, n} + 2 \, a^{3} b n x^{n} + a^{4} n\right )}} + \frac{\log \left (x\right )}{a^{3}} - \frac{\log \left (\frac{b x^{n} + a}{b}\right )}{a^{3} n} \end{align*}

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))^(3/2),x, algorithm="maxima")

[Out]

1/2*(2*b*x^n + 3*a)/(a^2*b^2*n*x^(2*n) + 2*a^3*b*n*x^n + a^4*n) + log(x)/a^3 - log((b*x^n + a)/b)/(a^3*n)

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Fricas [A]  time = 1.86974, size = 244, normalized size = 1.53 \begin{align*} \frac{2 \, b^{2} n x^{2 \, n} \log \left (x\right ) + 2 \, a^{2} n \log \left (x\right ) + 3 \, a^{2} + 2 \,{\left (2 \, a b n \log \left (x\right ) + a b\right )} x^{n} - 2 \,{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )} \log \left (b x^{n} + a\right )}{2 \,{\left (a^{3} b^{2} n x^{2 \, n} + 2 \, a^{4} b n x^{n} + a^{5} n\right )}} \end{align*}

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))^(3/2),x, algorithm="fricas")

[Out]

1/2*(2*b^2*n*x^(2*n)*log(x) + 2*a^2*n*log(x) + 3*a^2 + 2*(2*a*b*n*log(x) + a*b)*x^n - 2*(b^2*x^(2*n) + 2*a*b*x
^n + a^2)*log(b*x^n + a))/(a^3*b^2*n*x^(2*n) + 2*a^4*b*n*x^n + a^5*n)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (\left (a + b x^{n}\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

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))**(3/2),x)

[Out]

Integral(1/(x*((a + b*x**n)**2)**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

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))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b^2*x^(2*n) + 2*a*b*x^n + a^2)^(3/2)*x), x)