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

Optimal. Leaf size=61 \[ \frac {2 x \sqrt {\frac {a}{x^2}+b x^{n-2}}}{n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a}}{x \sqrt {\frac {a}{x^2}+b x^{n-2}}}\right )}{n} \]

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Rubi [A]  time = 0.08, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1979, 2007, 2029, 206} \begin {gather*} \frac {2 x \sqrt {\frac {a}{x^2}+b x^{n-2}}}{n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a}}{x \sqrt {\frac {a}{x^2}+b x^{n-2}}}\right )}{n} \end {gather*}

Antiderivative was successfully verified.

[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 206

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

Rule 1979

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

Rule 2007

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 2029

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*} \int \sqrt {\frac {a+b x^n}{x^2}} \, dx &=\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) \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 70, normalized size = 1.15 \begin {gather*} \frac {x \sqrt {\frac {a+b x^n}{x^2}} \left (2 \sqrt {a+b x^n}-2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )\right )}{n \sqrt {a+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(a + b*x^n)/x^2],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.04, size = 73, normalized size = 1.20 \begin {gather*} \frac {x \sqrt {\frac {a+b x^n}{x^2}} \left (\frac {2 \sqrt {a+b x^n}}{n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{n}\right )}{\sqrt {a+b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 112, normalized size = 1.84 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {\frac {b x^{n} + a}{x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((b*x^n + a)/x^2), x)

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maple [A]  time = 1.87, size = 74, normalized size = 1.21 \begin {gather*} -\frac {2 \sqrt {\frac {b \,{\mathrm e}^{n \ln \relax (x )}+a}{x^{2}}}\, \sqrt {a}\, x \arctanh \left (\frac {\sqrt {b \,{\mathrm e}^{n \ln \relax (x )}+a}}{\sqrt {a}}\right )}{\sqrt {b \,{\mathrm e}^{n \ln \relax (x )}+a}\, n}+\frac {2 \sqrt {\frac {b \,{\mathrm e}^{n \ln \relax (x )}+a}{x^{2}}}\, x}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {\frac {b x^{n} + a}{x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((b*x^n + a)/x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \sqrt {\frac {a+b\,x^n}{x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^n)/x^2)^(1/2),x)

[Out]

int(((a + b*x^n)/x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {\frac {a + b x^{n}}{x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((a + b*x**n)/x**2), x)

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