\(\int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx\) [408]

Optimal result

Integrand size = 15, antiderivative size = 32 \[ \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {\frac {a}{x^2}+b x^2}}\right )}{2 \sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2004, 2033, 212} \[ \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {\frac {a}{x^2}+b x^2}}\right )}{2 \sqrt {b}} \]

[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 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 2033

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*} \text {integral}& = \int \frac {1}{\sqrt {\frac {a}{x^2}+b x^2}} \, dx \\ & = \frac {1}{2} \text {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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx=\frac {\sqrt {a+b x^4} \log \left (\sqrt {b} x^2+\sqrt {a+b x^4}\right )}{2 \sqrt {b} x \sqrt {\frac {a+b x^4}{x^2}}} \]

[In]

Integrate[1/Sqrt[(a + b*x^4)/x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53

method	result	size
default	\(\frac {\sqrt {b \,x^{4}+a}\, \ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 \sqrt {\frac {b \,x^{4}+a}{x^{2}}}\, x \sqrt {b}}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.50 \[ \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx=\left [\frac {\log \left (-2 \, b x^{4} - 2 \, \sqrt {b} x^{3} \sqrt {\frac {b x^{4} + a}{x^{2}}} - a\right )}{4 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{3} \sqrt {\frac {b x^{4} + a}{x^{2}}}}{b x^{4} + a}\right )}{2 \, b}\right ] \]

[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]

Sympy [F]

\[ \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx=\int \frac {1}{\sqrt {\frac {a + b x^{4}}{x^{2}}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx=\int { \frac {1}{\sqrt {\frac {b x^{4} + a}{x^{2}}}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx=\frac {\log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{4 \, \sqrt {b}} - \frac {\log \left ({\left | -\sqrt {b} x^{2} + \sqrt {b x^{4} + a} \right |}\right )}{2 \, \sqrt {b} \mathrm {sgn}\left (x\right )} \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {\frac {a+b x^4}{x^2}}} \, dx=\int \frac {1}{\sqrt {\frac {b\,x^4+a}{x^2}}} \,d x \]

[In]

int(1/((a + b*x^4)/x^2)^(1/2),x)

[Out]

int(1/((a + b*x^4)/x^2)^(1/2), x)