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

Optimal. Leaf size=60 \[ \frac {\sqrt {a} \sqrt {\frac {b x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.01, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1089, 215} \begin {gather*} \frac {\sqrt {a} \sqrt {\frac {b x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[b]*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4))

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 1089

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 + c*x^4)^FracPart[p]
)/(1 + (2*c*x^2)/b)^(2*FracPart[p]), Int[(1 + (2*c*x^2)/b)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2
- 4*a*c, 0] &&  !IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 49, normalized size = 0.82 \begin {gather*} \frac {\sqrt {a+b x^2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b} \sqrt [4]{\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(Sqrt[b]*((a + b*x^2)^2)^(1/4))

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IntegrateAlgebraic [A]  time = 6.07, size = 52, normalized size = 0.87 \begin {gather*} -\frac {\left (\left (a+b x^2\right )^2\right )^{3/4} \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{\sqrt {b} \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((((a + b*x^2)^2)^(3/4)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(Sqrt[b]*(a + b*x^2)^(3/2)))

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fricas [A]  time = 0.99, size = 90, normalized size = 1.50 \begin {gather*} \left [\frac {\log \left (-2 \, b x^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {b} x - a\right )}{2 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {-b} x}{b x^{2} + a}\right )}{b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 24, normalized size = 0.40 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {-\frac {b x^{2} + a}{x^{2}}}}{\sqrt {b}}\right )}{\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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