3.20.33 \(\int \frac {b+a x^2}{(-b+a x^2) \sqrt {b^2+a^2 x^4}} \, dx\) [1933]

Optimal. Leaf size=134 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {6-4 \sqrt {2}} \sqrt {a} \sqrt {b} x}{b+a x^2+\sqrt {b^2+a^2 x^4}}\right )}{\sqrt {2} \sqrt {a} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {6+4 \sqrt {2}} \sqrt {a} \sqrt {b} x}{b+a x^2+\sqrt {b^2+a^2 x^4}}\right )}{\sqrt {2} \sqrt {a} \sqrt {b}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 50, normalized size of antiderivative = 0.37, 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 = {1713, 214} \begin {gather*} -\frac {\tanh ^{-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]

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

Rule 214

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

Rule 1713

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 &=b \text {Subst}\left (\int \frac {1}{-b+2 a b^2 x^2} \, dx,x,\frac {x}{\sqrt {b^2+a^2 x^4}}\right )\\ &=-\frac {\tanh ^{-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 [A]
time = 0.37, size = 50, normalized size = 0.37 \begin {gather*} -\frac {\tanh ^{-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]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.08, size = 152, normalized size = 1.13

method	result	size
elliptic	\(-\frac {\arctanh \left (\frac {\sqrt {a^{2} x^{4}+b^{2}}\, \sqrt {2}}{2 x \sqrt {a b}}\right ) \sqrt {2}}{2 \sqrt {a b}}\)	\(38\)
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/(I*a/b)^(1/2)*(1-I*a/b*x^2)^(1/2)*(1+I*a/b*x^2)^(1/2)/(a^2*x^4+b^2)^(1/2)*EllipticF(x*(I*a/b)^(1/2),I)-2/(I*
a/b)^(1/2)*(1-I*a/b*x^2)^(1/2)*(1+I*a/b*x^2)^(1/2)/(a^2*x^4+b^2)^(1/2)*EllipticPi(x*(I*a/b)^(1/2),-I,(-I*a/b)^
(1/2)/(I*a/b)^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [A]
time = 0.40, size = 132, normalized size = 0.99 \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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {a\,x^2+b}{\sqrt {a^2\,x^4+b^2}\,\left (b-a\,x^2\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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