\(\int \sqrt {\frac {a+b x^n}{x^2}} \, dx\) [385]

Optimal result

Integrand size = 15, antiderivative size = 61 \[ \int \sqrt {\frac {a+b x^n}{x^2}} \, dx=\frac {2 x \sqrt {\frac {a}{x^2}+b x^{-2+n}}}{n}-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a}}{x \sqrt {\frac {a}{x^2}+b x^{-2+n}}}\right )}{n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2004, 2032, 2054, 212} \[ \int \sqrt {\frac {a+b x^n}{x^2}} \, dx=\frac {2 x \sqrt {\frac {a}{x^2}+b x^{n-2}}}{n}-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a}}{x \sqrt {\frac {a}{x^2}+b x^{n-2}}}\right )}{n} \]

[In]

Int[Sqrt[(a + b*x^n)/x^2],x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2004

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2032

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(p*(n - j))), x] + Dist
[a, Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, j, n}, x] && IGtQ[p + 1/2, 0] && NeQ[n, j] && EqQ[
Simplify[j*p + 1], 0]

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13 \[ \int \sqrt {\frac {a+b x^n}{x^2}} \, dx=\frac {2 x \sqrt {\frac {a+b x^n}{x^2}} \left (\sqrt {a+b x^n}-\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )\right )}{n \sqrt {a+b x^n}} \]

[In]

Integrate[Sqrt[(a + b*x^n)/x^2],x]

[Out]

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

Maple [A] (verified)

Time = 2.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.21

method	result	size
risch	\(\frac {2 \sqrt {\frac {a +b \,{\mathrm e}^{n \ln \left (x \right )}}{x^{2}}}\, x}{n}-\frac {2 \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{\sqrt {a}}\right ) \sqrt {\frac {a +b \,{\mathrm e}^{n \ln \left (x \right )}}{x^{2}}}\, x}{n \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.84 \[ \int \sqrt {\frac {a+b x^n}{x^2}} \, dx=\left [\frac {2 \, x \sqrt {\frac {b x^{n} + a}{x^{2}}} + \sqrt {a} \log \left (\frac {b x^{n} - 2 \, \sqrt {a} x \sqrt {\frac {b x^{n} + a}{x^{2}}} + 2 \, a}{x^{n}}\right )}{n}, \frac {2 \, {\left (x \sqrt {\frac {b x^{n} + a}{x^{2}}} + \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x \sqrt {\frac {b x^{n} + a}{x^{2}}}}{a}\right )\right )}}{n}\right ] \]

[In]

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

[Out]

[(2*x*sqrt((b*x^n + a)/x^2) + sqrt(a)*log((b*x^n - 2*sqrt(a)*x*sqrt((b*x^n + a)/x^2) + 2*a)/x^n))/n, 2*(x*sqrt
((b*x^n + a)/x^2) + sqrt(-a)*arctan(sqrt(-a)*x*sqrt((b*x^n + a)/x^2)/a))/n]

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt((a + b*x**n)/x**2), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt((b*x^n + a)/x^2), x)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt((b*x^n + a)/x^2), x)

Mupad [F(-1)]

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

[In]

int(((a + b*x^n)/x^2)^(1/2),x)

[Out]

int(((a + b*x^n)/x^2)^(1/2), x)