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

Optimal. Leaf size=99 \[ \frac {a x^{2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a+b x^n\right )}+\frac {b^2 x^{3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1355, 14} \[ \frac {a x^{2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a+b x^n\right )}+\frac {b^2 x^{3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )} \]

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]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 44, normalized size = 0.44 \[ \frac {x^{2 n} \sqrt {\left (a+b x^n\right )^2} \left (3 a+2 b x^n\right )}{6 n \left (a+b x^n\right )} \]

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]

(x^(2*n)*Sqrt[(a + b*x^n)^2]*(3*a + 2*b*x^n))/(6*n*(a + b*x^n))

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fricas [A]  time = 0.72, size = 22, normalized size = 0.22 \[ \frac {2 \, b x^{3 \, n} + 3 \, a x^{2 \, n}}{6 \, n} \]

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]

1/6*(2*b*x^(3*n) + 3*a*x^(2*n))/n

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}} x^{2 \, n - 1}\,{d x} \]

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

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maple [A]  time = 0.02, size = 64, normalized size = 0.65 \[ \frac {\sqrt {\left (b \,x^{n}+a \right )^{2}}\, a \,x^{2 n}}{2 \left (b \,x^{n}+a \right ) n}+\frac {\sqrt {\left (b \,x^{n}+a \right )^{2}}\, b \,x^{3 n}}{3 \left (b \,x^{n}+a \right ) n} \]

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]

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

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maxima [A]  time = 0.89, size = 22, normalized size = 0.22 \[ \frac {2 \, b x^{3 \, n} + 3 \, a x^{2 \, n}}{6 \, n} \]

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]

1/6*(2*b*x^(3*n) + 3*a*x^(2*n))/n

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{2\,n-1}\,\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n} \,d x \]

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2 n - 1} \sqrt {\left (a + b x^{n}\right )^{2}}\, dx \]

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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