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

Optimal. Leaf size=121 \[ \frac {5 x \left (x^2+2\right )}{\sqrt {x^4+3 x^2+2}}+\frac {7 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {5 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}} \]

[Out]

5*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)+7/2*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(
1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)-5*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticE(x/(x^2+1)^(1/2),1
/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1189, 1099, 1135} \[ \frac {5 x \left (x^2+2\right )}{\sqrt {x^4+3 x^2+2}}+\frac {7 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {5 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x*(2 + x^2))/Sqrt[2 + 3*x^2 + x^4] - (5*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/
2])/Sqrt[2 + 3*x^2 + x^4] + (7*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2
+ 3*x^2 + x^4])

Rule 1099

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

Rule 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1189

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

Rubi steps

\begin {align*} \int \frac {7+5 x^2}{\sqrt {2+3 x^2+x^4}} \, dx &=5 \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+7 \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {5 x \left (2+x^2\right )}{\sqrt {2+3 x^2+x^4}}-\frac {5 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2+3 x^2+x^4}}+\frac {7 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 69, normalized size = 0.57 \[ -\frac {i \sqrt {x^2+1} \sqrt {x^2+2} \left (2 F\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+5 E\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )\right )}{\sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(5*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] + 2*EllipticF[I*ArcSinh[x/Sqrt[2]], 2]
))/Sqrt[2 + 3*x^2 + x^4]

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {5 \, x^{2} + 7}{\sqrt {x^{4} + 3 \, x^{2} + 2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.00, size = 106, normalized size = 0.88 \[ -\frac {7 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )+\EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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