3.1.53 \(\int \frac {1}{\sqrt {-3+7 x^2+2 x^4}} \, dx\) [53]

Optimal. Leaf size=148 \[ \frac {\sqrt {\frac {6-\left (7-\sqrt {73}\right ) x^2}{6-\left (7+\sqrt {73}\right ) x^2}} \sqrt {-6+\left (7+\sqrt {73}\right ) x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{73} x}{\sqrt {-6+\left (7+\sqrt {73}\right ) x^2}}\right )|\frac {1}{146} \left (73+7 \sqrt {73}\right )\right )}{2 \sqrt {3} \sqrt [4]{73} \sqrt {\frac {1}{6-\left (7+\sqrt {73}\right ) x^2}} \sqrt {-3+7 x^2+2 x^4}} \]

[Out]

1/438*EllipticF(73^(1/4)*x*2^(1/2)/(-6+x^2*(7+73^(1/2)))^(1/2),1/146*(10658+1022*73^(1/2))^(1/2))*((6-x^2*(7-7
3^(1/2)))/(6-x^2*(7+73^(1/2))))^(1/2)*(-6+x^2*(7+73^(1/2)))^(1/2)*73^(3/4)*3^(1/2)/(2*x^4+7*x^2-3)^(1/2)/(1/(6
-x^2*(7+73^(1/2))))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1112} \begin {gather*} \frac {\sqrt {\frac {6-\left (7-\sqrt {73}\right ) x^2}{6-\left (7+\sqrt {73}\right ) x^2}} \sqrt {\left (7+\sqrt {73}\right ) x^2-6} F\left (\text {ArcSin}\left (\frac {\sqrt {2} \sqrt [4]{73} x}{\sqrt {\left (7+\sqrt {73}\right ) x^2-6}}\right )|\frac {1}{146} \left (73+7 \sqrt {73}\right )\right )}{2 \sqrt {3} \sqrt [4]{73} \sqrt {\frac {1}{6-\left (7+\sqrt {73}\right ) x^2}} \sqrt {2 x^4+7 x^2-3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-3 + 7*x^2 + 2*x^4],x]

[Out]

(Sqrt[(6 - (7 - Sqrt[73])*x^2)/(6 - (7 + Sqrt[73])*x^2)]*Sqrt[-6 + (7 + Sqrt[73])*x^2]*EllipticF[ArcSin[(Sqrt[
2]*73^(1/4)*x)/Sqrt[-6 + (7 + Sqrt[73])*x^2]], (73 + 7*Sqrt[73])/146])/(2*Sqrt[3]*73^(1/4)*Sqrt[(6 - (7 + Sqrt
[73])*x^2)^(-1)]*Sqrt[-3 + 7*x^2 + 2*x^4])

Rule 1112

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2*a + (b + q)
*x^2)]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b + q)/(2*q)], x]] /; FreeQ[{a, b, c}, x] && Gt
Q[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {-3+7 x^2+2 x^4}} \, dx &=\frac {\sqrt {\frac {6-\left (7-\sqrt {73}\right ) x^2}{6-\left (7+\sqrt {73}\right ) x^2}} \sqrt {-6+\left (7+\sqrt {73}\right ) x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{73} x}{\sqrt {-6+\left (7+\sqrt {73}\right ) x^2}}\right )|\frac {1}{146} \left (73+7 \sqrt {73}\right )\right )}{2 \sqrt {3} \sqrt [4]{73} \sqrt {\frac {1}{6-\left (7+\sqrt {73}\right ) x^2}} \sqrt {-3+7 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.04, size = 80, normalized size = 0.54 \begin {gather*} -\frac {i \sqrt {6-14 x^2-4 x^4} F\left (i \sinh ^{-1}\left (\frac {2 x}{\sqrt {7+\sqrt {73}}}\right )|\frac {1}{12} \left (-61-7 \sqrt {73}\right )\right )}{\sqrt {-7+\sqrt {73}} \sqrt {-3+7 x^2+2 x^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/Sqrt[-3 + 7*x^2 + 2*x^4],x]

[Out]

((-I)*Sqrt[6 - 14*x^2 - 4*x^4]*EllipticF[I*ArcSinh[(2*x)/Sqrt[7 + Sqrt[73]]], (-61 - 7*Sqrt[73])/12])/(Sqrt[-7
+ Sqrt[73]]*Sqrt[-3 + 7*x^2 + 2*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.04, size = 84, normalized size = 0.57

method	result	size
default	\(\frac {6 \sqrt {1-\left (\frac {7}{6}-\frac {\sqrt {73}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {7}{6}+\frac {\sqrt {73}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {\sqrt {42-6 \sqrt {73}}\, x}{6}, \frac {7 i \sqrt {6}}{12}+\frac {i \sqrt {438}}{12}\right )}{\sqrt {42-6 \sqrt {73}}\, \sqrt {2 x^{4}+7 x^{2}-3}}\)	\(84\)
elliptic	\(\frac {6 \sqrt {1-\left (\frac {7}{6}-\frac {\sqrt {73}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {7}{6}+\frac {\sqrt {73}}{6}\right ) x^{2}}\, \EllipticF \left (\frac {\sqrt {42-6 \sqrt {73}}\, x}{6}, \frac {7 i \sqrt {6}}{12}+\frac {i \sqrt {438}}{12}\right )}{\sqrt {42-6 \sqrt {73}}\, \sqrt {2 x^{4}+7 x^{2}-3}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^4+7*x^2-3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

6/(42-6*73^(1/2))^(1/2)*(1-(7/6-1/6*73^(1/2))*x^2)^(1/2)*(1-(7/6+1/6*73^(1/2))*x^2)^(1/2)/(2*x^4+7*x^2-3)^(1/2
)*EllipticF(1/6*(42-6*73^(1/2))^(1/2)*x,7/12*I*6^(1/2)+1/12*I*438^(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(1/(2*x^4+7*x^2-3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(2*x^4 + 7*x^2 - 3), x)

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Fricas [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(1/(2*x^4+7*x^2-3)^(1/2),x, algorithm="fricas")

[Out]

0

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {2 x^{4} + 7 x^{2} - 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x**4+7*x**2-3)**(1/2),x)

[Out]

Integral(1/sqrt(2*x**4 + 7*x**2 - 3), 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(1/(2*x^4+7*x^2-3)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(2*x^4 + 7*x^2 - 3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {2\,x^4+7\,x^2-3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(7*x^2 + 2*x^4 - 3)^(1/2),x)

[Out]

int(1/(7*x^2 + 2*x^4 - 3)^(1/2), x)

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