3.408 \(\int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1979, 2008, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {\frac {a}{x^2}+b x^2}}\right )}{2 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(a + b*x^4)/x^2],x]

[Out]

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

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx &=\int \frac {1}{\sqrt {\frac {a}{x^2}+b x^2}} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {\frac {a}{x^2}+b x^2}}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {\frac {a}{x^2}+b x^2}}\right )}{2 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 59, normalized size = 1.84 \[ \frac {\sqrt {a+b x^4} \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 \sqrt {b} x \sqrt {\frac {a+b x^4}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(a + b*x^4)/x^2],x]

[Out]

(Sqrt[a + b*x^4]*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(2*Sqrt[b]*x*Sqrt[(a + b*x^4)/x^2])

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fricas [A]  time = 0.42, size = 80, normalized size = 2.50 \[ \left [\frac {\log \left (-2 \, b x^{4} - 2 \, \sqrt {b} x^{3} \sqrt {\frac {b x^{4} + a}{x^{2}}} - a\right )}{4 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{3} \sqrt {\frac {b x^{4} + a}{x^{2}}}}{b x^{4} + a}\right )}{2 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x^4+a)/x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/4*log(-2*b*x^4 - 2*sqrt(b)*x^3*sqrt((b*x^4 + a)/x^2) - a)/sqrt(b), -1/2*sqrt(-b)*arctan(sqrt(-b)*x^3*sqrt((
b*x^4 + a)/x^2)/(b*x^4 + a))/b]

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giac [A]  time = 0.21, size = 40, normalized size = 1.25 \[ \frac {\log \left ({\left | a \right |}\right ) \mathrm {sgn}\relax (x)}{4 \, \sqrt {b}} - \frac {\log \left ({\left | -\sqrt {b} x^{2} + \sqrt {b x^{4} + a} \right |}\right )}{2 \, \sqrt {b} \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x^4+a)/x^2)^(1/2),x, algorithm="giac")

[Out]

1/4*log(abs(a))*sgn(x)/sqrt(b) - 1/2*log(abs(-sqrt(b)*x^2 + sqrt(b*x^4 + a)))/(sqrt(b)*sgn(x))

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maple [A]  time = 0.19, size = 49, normalized size = 1.53 \[ \frac {\sqrt {b \,x^{4}+a}\, \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {\frac {b \,x^{4}+a}{x^{2}}}\, \sqrt {b}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x^4+a)/x^2)^(1/2),x)

[Out]

1/2/((b*x^4+a)/x^2)^(1/2)/x*(b*x^4+a)^(1/2)*ln(b^(1/2)*x^2+(b*x^4+a)^(1/2))/b^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {x^{5}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}}\,{d x} + \frac {x^{2}}{2 \, \sqrt {b x^{4} + a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x^4+a)/x^2)^(1/2),x, algorithm="maxima")

[Out]

b*integrate(x^5/(b*x^4 + a)^(3/2), x) + 1/2*x^2/sqrt(b*x^4 + a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{\sqrt {\frac {b\,x^4+a}{x^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^4)/x^2)^(1/2),x)

[Out]

int(1/((a + b*x^4)/x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\frac {a + b x^{4}}{x^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x**4+a)/x**2)**(1/2),x)

[Out]

Integral(1/sqrt((a + b*x**4)/x**2), x)

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