3.409 \(\int \frac{1}{\sqrt{\frac{a+b x^5}{x^3}}} \, dx\)

Optimal. Leaf size=32 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^3}+b x^2}}\right )}{5 \sqrt{b}} \]

[Out]

(2*ArcTanh[(Sqrt[b]*x)/Sqrt[a/x^3 + b*x^2]])/(5*Sqrt[b])

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Rubi [A]  time = 0.0158393, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1979, 2008, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^3}+b x^2}}\right )}{5 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(a + b*x^5)/x^3],x]

[Out]

(2*ArcTanh[(Sqrt[b]*x)/Sqrt[a/x^3 + b*x^2]])/(5*Sqrt[b])

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{\frac{a+b x^5}{x^3}}} \, dx &=\int \frac{1}{\sqrt{\frac{a}{x^3}+b x^2}} \, dx\\ &=\frac{2}{5} \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{\frac{a}{x^3}+b x^2}}\right )\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{\frac{a}{x^3}+b x^2}}\right )}{5 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0236792, size = 63, normalized size = 1.97 \[ \frac{2 \sqrt{a+b x^5} \tanh ^{-1}\left (\frac{\sqrt{b} x^{5/2}}{\sqrt{a+b x^5}}\right )}{5 \sqrt{b} x^{3/2} \sqrt{\frac{a+b x^5}{x^3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(a + b*x^5)/x^3],x]

[Out]

(2*Sqrt[a + b*x^5]*ArcTanh[(Sqrt[b]*x^(5/2))/Sqrt[a + b*x^5]])/(5*Sqrt[b]*x^(3/2)*Sqrt[(a + b*x^5)/x^3])

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt{{\frac{b{x}^{5}+a}{{x}^{3}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x^5+a)/x^3)^(1/2),x)

[Out]

int(1/((b*x^5+a)/x^3)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\frac{b x^{5} + a}{x^{3}}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x^5+a)/x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt((b*x^5 + a)/x^3), x)

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Fricas [A]  time = 2.59082, size = 244, normalized size = 7.62 \begin{align*} \left [\frac{\log \left (-8 \, b^{2} x^{10} - 8 \, a b x^{5} - a^{2} - 4 \,{\left (2 \, b x^{9} + a x^{4}\right )} \sqrt{b} \sqrt{\frac{b x^{5} + a}{x^{3}}}\right )}{10 \, \sqrt{b}}, -\frac{\sqrt{-b} \arctan \left (\frac{2 \, \sqrt{-b} x^{4} \sqrt{\frac{b x^{5} + a}{x^{3}}}}{2 \, b x^{5} + a}\right )}{5 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x^5+a)/x^3)^(1/2),x, algorithm="fricas")

[Out]

[1/10*log(-8*b^2*x^10 - 8*a*b*x^5 - a^2 - 4*(2*b*x^9 + a*x^4)*sqrt(b)*sqrt((b*x^5 + a)/x^3))/sqrt(b), -1/5*sqr
t(-b)*arctan(2*sqrt(-b)*x^4*sqrt((b*x^5 + a)/x^3)/(2*b*x^5 + a))/b]

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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/((b*x**5+a)/x**3)**(1/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x^5+a)/x^3)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError