\(\int \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x^2} \, dx\) [520]

Optimal result

Integrand size = 28, antiderivative size = 94 \[ \int \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x^2} \, dx=-\frac {a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x \left (a+b x^n\right )}-\frac {b^2 x^{-1+n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1-n) \left (a b+b^2 x^n\right )} \]

[Out]

-a*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/x/(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 = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1369, 14} \[ \int \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x^2} \, dx=-\frac {b^2 x^{n-1} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1-n) \left (a b+b^2 x^n\right )}-\frac {a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x \left (a+b x^n\right )} \]

[In]

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

[Out]

-((a*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(x*(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 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 \frac {a b+b^2 x^n}{x^2} \, dx}{a b+b^2 x^n} \\ & = \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (\frac {a b}{x^2}+b^2 x^{-2+n}\right ) \, dx}{a b+b^2 x^n} \\ & = -\frac {a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x \left (a+b x^n\right )}-\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 = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x^2} \, dx=\frac {\sqrt {\left (a+b x^n\right )^2} \left (a-a n+b x^n\right )}{(-1+n) x \left (a+b x^n\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.65

method	result	size
risch	\(-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a}{\left (a +b \,x^{n}\right ) x}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b \,x^{n}}{\left (a +b \,x^{n}\right ) \left (-1+n \right ) x}\)	\(61\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.24 \[ \int \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x^2} \, dx=-\frac {a n - b x^{n} - a}{{\left (n - 1\right )} x} \]

[In]

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

[Out]

-(a*n - b*x^n - a)/((n - 1)*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.23 \[ \int \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x^2} \, dx=-\frac {a {\left (n - 1\right )} - b x^{n}}{{\left (n - 1\right )} x} \]

[In]

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

[Out]

-(a*(n - 1) - b*x^n)/((n - 1)*x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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