3.368 \(\int \frac {1}{(7+5 x^2) \sqrt {4+3 x^2+x^4}} \, dx\)

Optimal. Leaf size=168 \[ \frac {1}{4} \sqrt {\frac {5}{77}} \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {x^4+3 x^2+4}}\right )-\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {17 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac {9}{280};2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{84 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

[Out]

1/308*arctan(2/35*x*385^(1/2)/(x^4+3*x^2+4)^(1/2))*385^(1/2)-1/12*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/
2)/cos(2*arctan(1/2*x*2^(1/2)))*EllipticF(sin(2*arctan(1/2*x*2^(1/2))),1/4*2^(1/2))*((x^4+3*x^2+4)/(x^2+2)^2)^
(1/2)*2^(1/2)/(x^4+3*x^2+4)^(1/2)+17/168*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(
1/2)))*EllipticPi(sin(2*arctan(1/2*x*2^(1/2))),-9/280,1/4*2^(1/2))*((x^4+3*x^2+4)/(x^2+2)^2)^(1/2)*2^(1/2)/(x^
4+3*x^2+4)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 168, 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 = {1216, 1103, 1706} \[ \frac {1}{4} \sqrt {\frac {5}{77}} \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {x^4+3 x^2+4}}\right )-\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {17 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac {9}{280};2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{84 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Int[1/((7 + 5*x^2)*Sqrt[4 + 3*x^2 + x^4]),x]

[Out]

(Sqrt[5/77]*ArcTan[(2*Sqrt[11/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/4 - ((2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2
]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(6*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) + (17*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4
)/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(84*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

Rule 1103

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

Rule 1216

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Di
st[(c*d + a*e*q)/(c*d^2 - a*e^2), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2)
, Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a
*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a]

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + c*x^4))/(a*(A + B*x
^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^
2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx &=-\left (\frac {1}{3} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx\right )+\frac {10}{3} \int \frac {1+\frac {x^2}{2}}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {1}{4} \sqrt {\frac {5}{77}} \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {4+3 x^2+x^4}}\right )-\frac {\left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {4+3 x^2+x^4}}+\frac {17 \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} \Pi \left (-\frac {9}{280};2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{84 \sqrt {2} \sqrt {4+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 159, normalized size = 0.95 \[ -\frac {i \sqrt {1-\frac {2 x^2}{-3-i \sqrt {7}}} \sqrt {1-\frac {2 x^2}{-3+i \sqrt {7}}} \Pi \left (-\frac {5}{14} \left (-3-i \sqrt {7}\right );i \sinh ^{-1}\left (\sqrt {-\frac {2}{-3-i \sqrt {7}}} x\right )|\frac {-3-i \sqrt {7}}{-3+i \sqrt {7}}\right )}{7 \sqrt {2} \sqrt {-\frac {1}{-3-i \sqrt {7}}} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((7 + 5*x^2)*Sqrt[4 + 3*x^2 + x^4]),x]

[Out]

((-1/7*I)*Sqrt[1 - (2*x^2)/(-3 - I*Sqrt[7])]*Sqrt[1 - (2*x^2)/(-3 + I*Sqrt[7])]*EllipticPi[(-5*(-3 - I*Sqrt[7]
))/14, I*ArcSinh[Sqrt[-2/(-3 - I*Sqrt[7])]*x], (-3 - I*Sqrt[7])/(-3 + I*Sqrt[7])])/(Sqrt[2]*Sqrt[-(-3 - I*Sqrt
[7])^(-1)]*Sqrt[4 + 3*x^2 + x^4])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 3 \, x^{2} + 4}}{5 \, x^{6} + 22 \, x^{4} + 41 \, x^{2} + 28}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+7)/(x^4+3*x^2+4)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 3*x^2 + 4)/(5*x^6 + 22*x^4 + 41*x^2 + 28), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 4} {\left (5 \, x^{2} + 7\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+7)/(x^4+3*x^2+4)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(x^4 + 3*x^2 + 4)*(5*x^2 + 7)), x)

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maple [C]  time = 0.02, size = 107, normalized size = 0.64 \[ \frac {\sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticPi \left (\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, x , -\frac {5}{7 \left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right )}, \frac {\sqrt {-\frac {3}{8}-\frac {i \sqrt {7}}{8}}}{\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}}\right )}{7 \sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, \sqrt {x^{4}+3 x^{2}+4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x^2+7)/(x^4+3*x^2+4)^(1/2),x)

[Out]

1/7/(-3/8+1/8*I*7^(1/2))^(1/2)*(3/8*x^2-1/8*I*7^(1/2)*x^2+1)^(1/2)*(3/8*x^2+1/8*I*7^(1/2)*x^2+1)^(1/2)/(x^4+3*
x^2+4)^(1/2)*EllipticPi((-3/8+1/8*I*7^(1/2))^(1/2)*x,-5/7/(-3/8+1/8*I*7^(1/2)),(-3/8-1/8*I*7^(1/2))^(1/2)/(-3/
8+1/8*I*7^(1/2))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 4} {\left (5 \, x^{2} + 7\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+7)/(x^4+3*x^2+4)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(x^4 + 3*x^2 + 4)*(5*x^2 + 7)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (5\,x^2+7\right )\,\sqrt {x^4+3\,x^2+4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((5*x^2 + 7)*(3*x^2 + x^4 + 4)^(1/2)),x)

[Out]

int(1/((5*x^2 + 7)*(3*x^2 + x^4 + 4)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\left (x^{2} - x + 2\right ) \left (x^{2} + x + 2\right )} \left (5 x^{2} + 7\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x**2+7)/(x**4+3*x**2+4)**(1/2),x)

[Out]

Integral(1/(sqrt((x**2 - x + 2)*(x**2 + x + 2))*(5*x**2 + 7)), x)

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