3.6.11 \(\int \frac {x^{-1+2 n}}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx\) [511]

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

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Rubi [A]
time = 0.03, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 = {1369, 272, 45} \begin {gather*} \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 {gather*}

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 45

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])

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 {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.04, size = 47, normalized size = 0.52 \begin {gather*} \frac {\left (a+b x^n\right ) \left (b x^n-a \log \left (b n \left (a+b x^n\right )\right )\right )}{b^2 n \sqrt {\left (a+b x^n\right )^2}} \end {gather*}

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)*(b*x^n - a*Log[b*n*(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 = 0.79

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, x^{n}}{\left (a +b \,x^{n}\right ) b n}-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a \ln \left (x^{n}+\frac {a}{b}\right )}{\left (a +b \,x^{n}\right ) b^{2} n}\)	\(71\)

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,method=_RETURNVERBOSE)

[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 = 0.54, size = 32, normalized size = 0.36 \begin {gather*} \frac {x^{n}}{b n} - \frac {a \log \left (\frac {b x^{n} + a}{b}\right )}{b^{2} n} \end {gather*}

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 = 0.37, size = 24, normalized size = 0.27 \begin {gather*} \frac {b x^{n} - a \log \left (b x^{n} + a\right )}{b^{2} n} \end {gather*}

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

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 = 0.00, 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(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)

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

[Out]

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

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