3.410 \(\int \frac{1}{\sqrt{x^{2-n} (a+b x^n)}} \, dx\)

Optimal. Leaf size=37 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x^{2-n}+b x^2}}\right )}{\sqrt{b} n} \]

[Out]

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

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Rubi [A]  time = 0.0239837, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1979, 2008, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x^{2-n}+b x^2}}\right )}{\sqrt{b} n} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[x^(2 - n)*(a + b*x^n)],x]

[Out]

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

Rule 1979

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

Rule 2008

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

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])

Rubi steps

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

Mathematica [B]  time = 0.0387811, size = 76, normalized size = 2.05 \[ \frac{2 \sqrt{a} x^{1-\frac{n}{2}} \sqrt{\frac{b x^n}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x^{n/2}}{\sqrt{a}}\right )}{\sqrt{b} n \sqrt{x^{2-n} \left (a+b x^n\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.358, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt{{x}^{2-n} \left ( a+b{x}^{n} \right ) }}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{{\left (b x^{n} + a\right )} x^{-n + 2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.911825, size = 216, normalized size = 5.84 \begin{align*} \left [\frac{\log \left (\frac{2 \, b x x^{n} + a x + 2 \, \sqrt{b} x^{n} \sqrt{\frac{b x^{2} x^{n} + a x^{2}}{x^{n}}}}{x}\right )}{\sqrt{b} n}, -\frac{2 \, \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{b x^{2} x^{n} + a x^{2}}{x^{n}}}}{b x}\right )}{b n}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{{\left (b x^{n} + a\right )} x^{-n + 2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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