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

Optimal. Leaf size=149 \[ \frac {5 x \left (x^2+2\right )}{\sqrt {x^4+3 x^2+2}}+\frac {1}{3} x \left (3 x^2+10\right ) \sqrt {x^4+3 x^2+2}+\frac {11 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {x^4+3 x^2+2}}-\frac {5 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}} \]

[Out]

5*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-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)+11/3*(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)+1/3*x*(3*x^2+10)*(x^4+3*x^2+2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1176, 1189, 1099, 1135} \[ \frac {5 x \left (x^2+2\right )}{\sqrt {x^4+3 x^2+2}}+\frac {1}{3} x \left (3 x^2+10\right ) \sqrt {x^4+3 x^2+2}+\frac {11 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {x^4+3 x^2+2}}-\frac {5 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}} \]

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] + (x*(10 + 3*x^2)*Sqrt[2 + 3*x^2 + x^4])/3 - (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] + (11*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1
+ x^2)]*EllipticF[ArcTan[x], 1/2])/(3*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 1176

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(2*b*e*p + c*d*(4*p
+ 3) + c*e*(4*p + 1)*x^2)*(a + b*x^2 + c*x^4)^p)/(c*(4*p + 1)*(4*p + 3)), x] + Dist[(2*p)/(c*(4*p + 1)*(4*p +
3)), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*x^2, x]*(a +
b*x^2 + c*x^4)^(p - 1), 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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

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 \left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4} \, dx &=\frac {1}{3} x \left (10+3 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {1}{15} \int \frac {110+75 x^2}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {1}{3} x \left (10+3 x^2\right ) \sqrt {2+3 x^2+x^4}+5 \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {22}{3} \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 {1}{3} x \left (10+3 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 {11 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {2+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 109, normalized size = 0.73 \[ \frac {3 x^7+19 x^5+36 x^3-7 i \sqrt {x^2+1} \sqrt {x^2+2} F\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-15 i \sqrt {x^2+1} \sqrt {x^2+2} E\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+20 x}{3 \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(20*x + 36*x^3 + 19*x^5 + 3*x^7 - (15*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] - (7*I
)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2])/(3*Sqrt[2 + 3*x^2 + x^4])

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}, x\right ) \]

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

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

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

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maple [C]  time = 0.01, size = 137, normalized size = 0.92 \[ \sqrt {x^{4}+3 x^{2}+2}\, x^{3}+\frac {10 \sqrt {x^{4}+3 x^{2}+2}\, x}{3}-\frac {11 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{3 \sqrt {x^{4}+3 x^{2}+2}}+\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )+\EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}} \]

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]

(x^4+3*x^2+2)^(1/2)*x^3+10/3*(x^4+3*x^2+2)^(1/2)*x-11/3*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))+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)*(Elli
pticF(1/2*I*2^(1/2)*x,2^(1/2))-EllipticE(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 \[ \int \sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}\,{d x} \]

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

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

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 \[ \int \sqrt {\left (x^{2} + 1\right ) \left (x^{2} + 2\right )} \left (5 x^{2} + 7\right )\, dx \]

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

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