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

Optimal. Leaf size=59 \[ -\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {a^2 x^4+b^2}+a x^2+b}\right )}{\sqrt {a} \sqrt {b}} \]

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Rubi [A]  time = 0.08, antiderivative size = 50, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1699, 205} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {a^2 x^4+b^2}}\right )}{\sqrt {2} \sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 1699

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rubi steps

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

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Mathematica [C]  time = 0.22, size = 95, normalized size = 1.61 \begin {gather*} -\frac {i \sqrt {\frac {a^2 x^4}{b^2}+1} \left (F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i a}{b}} x\right )\right |-1\right )-2 \Pi \left (-i;\left .i \sinh ^{-1}\left (\sqrt {\frac {i a}{b}} x\right )\right |-1\right )\right )}{\sqrt {\frac {i a}{b}} \sqrt {a^2 x^4+b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + (a^2*x^4)/b^2]*(EllipticF[I*ArcSinh[Sqrt[(I*a)/b]*x], -1] - 2*EllipticPi[-I, I*ArcSinh[Sqrt[(I*
a)/b]*x], -1]))/(Sqrt[(I*a)/b]*Sqrt[b^2 + a^2*x^4])

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IntegrateAlgebraic [A]  time = 0.48, size = 50, normalized size = 0.85 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {b} x}{\sqrt {b^2+a^2 x^4}}\right )}{\sqrt {2} \sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 132, normalized size = 2.24 \begin {gather*} \left [\frac {1}{4} \, \sqrt {2} \sqrt {-\frac {1}{a b}} \log \left (\frac {a^{2} x^{4} + 2 \, \sqrt {2} \sqrt {a^{2} x^{4} + b^{2}} a b x \sqrt {-\frac {1}{a b}} - 2 \, a b x^{2} + b^{2}}{a^{2} x^{4} + 2 \, a b x^{2} + b^{2}}\right ), \frac {1}{2} \, \sqrt {2} \sqrt {\frac {1}{a b}} \arctan \left (\frac {\sqrt {2} \sqrt {a^{2} x^{4} + b^{2}} \sqrt {\frac {1}{a b}}}{2 \, x}\right )\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 38, normalized size = 0.64

method	result	size
elliptic	\(\frac {\arctan \left (\frac {\sqrt {a^{2} x^{4}+b^{2}}\, \sqrt {2}}{2 x \sqrt {a b}}\right ) \sqrt {2}}{2 \sqrt {a b}}\)	\(38\)
default	\(\frac {\sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \EllipticF \left (x \sqrt {\frac {i a}{b}}, i\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}-\frac {2 \sqrt {1-\frac {i a \,x^{2}}{b}}\, \sqrt {1+\frac {i a \,x^{2}}{b}}\, \EllipticPi \left (x \sqrt {\frac {i a}{b}}, i, \frac {\sqrt {-\frac {i a}{b}}}{\sqrt {\frac {i a}{b}}}\right )}{\sqrt {\frac {i a}{b}}\, \sqrt {a^{2} x^{4}+b^{2}}}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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