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

Optimal. Leaf size=90 \[ \frac{x^n \left (a+b x^n\right )}{b n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}-\frac{a \left (a+b x^n\right ) \log \left (a+b x^n\right )}{b^2 n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

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

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Rubi [A]  time = 0.0438885, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {1355, 266, 43} \[ \frac{x^n \left (a+b x^n\right )}{b n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}-\frac{a \left (a+b x^n\right ) \log \left (a+b x^n\right )}{b^2 n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^n*(a + b*x^n))/(b*n*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) - (a*(a + b*x^n)*Log[a + b*x^n])/(b^2*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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0321583, size = 46, normalized size = 0.51 \[ \frac{\left (a+b x^n\right ) \left (\frac{x^n}{b}-\frac{a \log \left (a+b x^n\right )}{b^2}\right )}{n \sqrt{\left (a+b x^n\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 71, normalized size = 0.8 \begin{align*}{\frac{{x}^{n}}{ \left ( a+b{x}^{n} \right ) bn}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}}-{\frac{a}{ \left ( a+b{x}^{n} \right ){b}^{2}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(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.01878, size = 43, normalized size = 0.48 \begin{align*} \frac{x^{n}}{b n} - \frac{a \log \left (\frac{b x^{n} + a}{b}\right )}{b^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58961, size = 49, normalized size = 0.54 \begin{align*} \frac{b x^{n} - a \log \left (b x^{n} + a\right )}{b^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x^n - a*log(b*x^n + a))/(b^2*n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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