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

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

[Out]

(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])

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Rubi [A]  time = 0.0820125, antiderivative size = 168, normalized size of antiderivative = 1., 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 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 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 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*}

Mathematica [C]  time = 0.135387, 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]

((-I/7)*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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Maple [C]  time = 0.016, size = 107, normalized size = 0.6 \begin{align*}{\frac{1}{7\,\sqrt{-3/8+i/8\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticPi} \left ( \sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}x,-{\frac{5}{-{\frac{21}{8}}+{\frac{7\,i}{8}}\sqrt{7}}},{\frac{\sqrt{-{\frac{3}{8}}-{\frac{i}{8}}\sqrt{7}}}{\sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}} \end{align*}

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)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{4} + 3 \, x^{2} + 4}{\left (5 \, x^{2} + 7\right )}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{4} + 3 \, x^{2} + 4}}{5 \, x^{6} + 22 \, x^{4} + 41 \, x^{2} + 28}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\left (x^{2} - x + 2\right ) \left (x^{2} + x + 2\right )} \left (5 x^{2} + 7\right )}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{4} + 3 \, x^{2} + 4}{\left (5 \, x^{2} + 7\right )}}\,{d x} \end{align*}

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)