\(\int \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx\) [518]

Optimal result

Integrand size = 24, antiderivative size = 88 \[ \int \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {a x \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{a+b x^n}+\frac {b^2 x^{1+n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1+n) \left (a b+b^2 x^n\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1357} \[ \int \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {b^2 x^{n+1} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(n+1) \left (a b+b^2 x^n\right )}+\frac {a x \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{a+b x^n} \]

[In]

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

[Out]

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

Rule 1357

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.44 \[ \int \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {x \sqrt {\left (a+b x^n\right )^2} \left (a+a n+b x^n\right )}{(1+n) \left (a+b x^n\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a x}{a +b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b x \,x^{n}}{\left (a +b \,x^{n}\right ) \left (1+n \right )}\)	\(56\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.23 \[ \int \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {b x x^{n} + {\left (a n + a\right )} x}{n + 1} \]

[In]

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

[Out]

(b*x*x^n + (a*n + a)*x)/(n + 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.22 \[ \int \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {a {\left (n + 1\right )} x + b x x^{n}}{n + 1} \]

[In]

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

[Out]

(a*(n + 1)*x + b*x*x^n)/(n + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.28 \[ \int \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx={\left (a x + \frac {b x^{n + 1}}{n + 1}\right )} \mathrm {sgn}\left (b x^{n} + a\right ) \]

[In]

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

[Out]

(a*x + b*x^(n + 1)/(n + 1))*sgn(b*x^n + a)

Mupad [F(-1)]

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

[In]

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

[Out]

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