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

Optimal. Leaf size=30 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {b}} \]

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3, 2008, 206} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4],x]

[Out]

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

Rule 3

Int[(u_.)*((a_) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n + c*x^(2*n))^p, x] /;
FreeQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[a, 0]

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

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Mathematica [A]  time = 0.01, size = 52, normalized size = 1.73 \begin {gather*} -\frac {x \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.04, size = 30, normalized size = 1.00 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4],x]

[Out]

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

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fricas [A]  time = 0.99, size = 80, normalized size = 2.67 \begin {gather*} \left [\frac {\log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right )}{2 \, \sqrt {b}}, \frac {\sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right )}{b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 46, normalized size = 1.53 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b}} + \frac {\arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} \mathrm {sgn}\relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(sqrt(b)/sqrt(-b))*sgn(x)/sqrt(-b) + arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*sgn(x))

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maple [B]  time = 0.00, size = 50, normalized size = 1.67 \begin {gather*} -\frac {\sqrt {c \,x^{2}+b}\, x \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )}{\sqrt {c \,x^{4}+b \,x^{2}}\, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {c x^{4} + b x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(c*x^4 + b*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2 + c*x^4)^(1/2),x)

[Out]

int(1/(b*x^2 + c*x^4)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b*x**2 + c*x**4), x)

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