\(\int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx\) [510]

Optimal result

Integrand size = 32, antiderivative size = 99 \[ \int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {a x^{2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a+b x^n\right )}+\frac {b^2 x^{3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1369, 14} \[ \int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {a x^{2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a+b x^n\right )}+\frac {b^2 x^{3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )} \]

[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 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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.44 \[ \int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {x^{2 n} \sqrt {\left (a+b x^n\right )^2} \left (3 a+2 b x^n\right )}{6 n \left (a+b x^n\right )} \]

[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))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b \,x^{3 n}}{3 \left (a +b \,x^{n}\right ) n}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a \,x^{2 n}}{2 \left (a +b \,x^{n}\right ) n}\)	\(64\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.22 \[ \int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {2 \, b x^{3 \, n} + 3 \, a x^{2 \, n}}{6 \, n} \]

[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

Sympy [F]

\[ \int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\int x^{2 n - 1} \sqrt {\left (a + b x^{n}\right )^{2}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.22 \[ \int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {2 \, b x^{3 \, n} + 3 \, a x^{2 \, n}}{6 \, n} \]

[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

Giac [F]

\[ \int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\int { \sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}} x^{2 \, n - 1} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\int x^{2\,n-1}\,\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n} \,d x \]

[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)