3.394 \(\int \frac{(d+e x^2)^3}{\sqrt{2+3 x^2+x^4}} \, dx\)

Optimal. Leaf size=229 \[ \frac{\left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \left (5 d^3-10 d e^2+8 e^3\right ) \text{EllipticF}\left (\tan ^{-1}(x),\frac{1}{2}\right )}{5 \sqrt{2} \sqrt{x^4+3 x^2+2}}+\frac{3 e \left (x^2+2\right ) x \left (5 d^2-10 d e+6 e^2\right )}{5 \sqrt{x^4+3 x^2+2}}-\frac{3 \sqrt{2} e \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \left (5 d^2-10 d e+6 e^2\right ) E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{x^4+3 x^2+2}}+\frac{1}{5} e^2 \sqrt{x^4+3 x^2+2} x (5 d-4 e)+\frac{1}{5} e^3 \sqrt{x^4+3 x^2+2} x^3 \]

[Out]

(3*e*(5*d^2 - 10*d*e + 6*e^2)*x*(2 + x^2))/(5*Sqrt[2 + 3*x^2 + x^4]) + ((5*d - 4*e)*e^2*x*Sqrt[2 + 3*x^2 + x^4
])/5 + (e^3*x^3*Sqrt[2 + 3*x^2 + x^4])/5 - (3*Sqrt[2]*e*(5*d^2 - 10*d*e + 6*e^2)*(1 + x^2)*Sqrt[(2 + x^2)/(1 +
x^2)]*EllipticE[ArcTan[x], 1/2])/(5*Sqrt[2 + 3*x^2 + x^4]) + ((5*d^3 - 10*d*e^2 + 8*e^3)*(1 + x^2)*Sqrt[(2 +
x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(5*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4])

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Rubi [A]  time = 0.153778, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {1206, 1679, 1189, 1099, 1135} \[ \frac{3 e \left (x^2+2\right ) x \left (5 d^2-10 d e+6 e^2\right )}{5 \sqrt{x^4+3 x^2+2}}+\frac{\left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \left (5 d^3-10 d e^2+8 e^3\right ) F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{2} \sqrt{x^4+3 x^2+2}}-\frac{3 \sqrt{2} e \left (x^2+1\right ) \sqrt{\frac{x^2+2}{x^2+1}} \left (5 d^2-10 d e+6 e^2\right ) E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{x^4+3 x^2+2}}+\frac{1}{5} e^2 \sqrt{x^4+3 x^2+2} x (5 d-4 e)+\frac{1}{5} e^3 \sqrt{x^4+3 x^2+2} x^3 \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^3/Sqrt[2 + 3*x^2 + x^4],x]

[Out]

(3*e*(5*d^2 - 10*d*e + 6*e^2)*x*(2 + x^2))/(5*Sqrt[2 + 3*x^2 + x^4]) + ((5*d - 4*e)*e^2*x*Sqrt[2 + 3*x^2 + x^4
])/5 + (e^3*x^3*Sqrt[2 + 3*x^2 + x^4])/5 - (3*Sqrt[2]*e*(5*d^2 - 10*d*e + 6*e^2)*(1 + x^2)*Sqrt[(2 + x^2)/(1 +
x^2)]*EllipticE[ArcTan[x], 1/2])/(5*Sqrt[2 + 3*x^2 + x^4]) + ((5*d^3 - 10*d*e^2 + 8*e^3)*(1 + x^2)*Sqrt[(2 +
x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(5*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4])

Rule 1206

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e^q*x^(2*q - 3)*(
a + b*x^2 + c*x^4)^(p + 1))/(c*(4*p + 2*q + 1)), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2*p + 2*q - 1)*e^q*x^(2*q - 2) -
c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && IGtQ[q, 1]

Rule 1679

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[(e*x^(2*q - 3)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(2*q + 4*p + 1)), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]

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]

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]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^3}{\sqrt{2+3 x^2+x^4}} \, dx &=\frac{1}{5} e^3 x^3 \sqrt{2+3 x^2+x^4}+\frac{1}{5} \int \frac{5 d^3+3 e \left (5 d^2-2 e^2\right ) x^2+3 (5 d-4 e) e^2 x^4}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{1}{5} (5 d-4 e) e^2 x \sqrt{2+3 x^2+x^4}+\frac{1}{5} e^3 x^3 \sqrt{2+3 x^2+x^4}+\frac{1}{15} \int \frac{3 \left (5 d^3-10 d e^2+8 e^3\right )+9 e \left (5 d^2-10 d e+6 e^2\right ) x^2}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{1}{5} (5 d-4 e) e^2 x \sqrt{2+3 x^2+x^4}+\frac{1}{5} e^3 x^3 \sqrt{2+3 x^2+x^4}+\frac{1}{5} \left (3 e \left (5 d^2-10 d e+6 e^2\right )\right ) \int \frac{x^2}{\sqrt{2+3 x^2+x^4}} \, dx+\frac{1}{5} \left (5 d^3-10 d e^2+8 e^3\right ) \int \frac{1}{\sqrt{2+3 x^2+x^4}} \, dx\\ &=\frac{3 e \left (5 d^2-10 d e+6 e^2\right ) x \left (2+x^2\right )}{5 \sqrt{2+3 x^2+x^4}}+\frac{1}{5} (5 d-4 e) e^2 x \sqrt{2+3 x^2+x^4}+\frac{1}{5} e^3 x^3 \sqrt{2+3 x^2+x^4}-\frac{3 \sqrt{2} e \left (5 d^2-10 d e+6 e^2\right ) \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{2+3 x^2+x^4}}+\frac{\left (5 d^3-10 d e^2+8 e^3\right ) \left (1+x^2\right ) \sqrt{\frac{2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac{1}{2}\right )}{5 \sqrt{2} \sqrt{2+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.229417, size = 154, normalized size = 0.67 \[ \frac{-5 i \sqrt{x^2+1} \sqrt{x^2+2} \left (-3 d^2 e+d^3+4 d e^2-2 e^3\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right ),2\right )-3 i e \sqrt{x^2+1} \sqrt{x^2+2} \left (5 d^2-10 d e+6 e^2\right ) E\left (\left .i \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |2\right )+e^2 x \left (x^4+3 x^2+2\right ) \left (5 d+e \left (x^2-4\right )\right )}{5 \sqrt{x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)^3/Sqrt[2 + 3*x^2 + x^4],x]

[Out]

(e^2*x*(2 + 3*x^2 + x^4)*(5*d + e*(-4 + x^2)) - (3*I)*e*(5*d^2 - 10*d*e + 6*e^2)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*E
llipticE[I*ArcSinh[x/Sqrt[2]], 2] - (5*I)*(d^3 - 3*d^2*e + 4*d*e^2 - 2*e^3)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*Ellipt
icF[I*ArcSinh[x/Sqrt[2]], 2])/(5*Sqrt[2 + 3*x^2 + x^4])

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Maple [C]  time = 0.01, size = 380, normalized size = 1.7 \begin{align*}{e}^{3} \left ({\frac{{x}^{3}}{5}\sqrt{{x}^{4}+3\,{x}^{2}+2}}-{\frac{4\,x}{5}\sqrt{{x}^{4}+3\,{x}^{2}+2}}-{{\frac{4\,i}{5}}\sqrt{2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}+{{\frac{9\,i}{5}}\sqrt{2} \left ({\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) -{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}} \right ) +3\,d{e}^{2} \left ( 1/3\,x\sqrt{{x}^{4}+3\,{x}^{2}+2}+{\frac{i/3\sqrt{2}\sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\it EllipticF} \left ( i/2x\sqrt{2},\sqrt{2} \right ) }{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}-{\frac{i\sqrt{2}\sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1} \left ({\it EllipticF} \left ( i/2x\sqrt{2},\sqrt{2} \right ) -{\it EllipticE} \left ( i/2x\sqrt{2},\sqrt{2} \right ) \right ) }{\sqrt{{x}^{4}+3\,{x}^{2}+2}}} \right ) +{{\frac{3\,i}{2}}{d}^{2}e\sqrt{2} \left ({\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) -{\it EllipticE} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}}-{{\frac{i}{2}}{d}^{3}\sqrt{2}{\it EllipticF} \left ({\frac{i}{2}}x\sqrt{2},\sqrt{2} \right ) \sqrt{2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3/(x^4+3*x^2+2)^(1/2),x)

[Out]

e^3*(1/5*x^3*(x^4+3*x^2+2)^(1/2)-4/5*x*(x^4+3*x^2+2)^(1/2)-4/5*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*x*2^(1/2),2^(1/2))+9/5*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*x*2^(1/2),2^(1/2))-EllipticE(1/2*I*x*2^(1/2),2^(1/2))))+3*d*e^2*(1/3*x*(x^4+3*x^2+2)^(1/2)+
1/3*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*x*2^(1/2),2^(1/2))-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*x*2^(1/2),2^(1/2))-EllipticE(1/2*I*x*2^(1/2)
,2^(1/2))))+3/2*I*d^2*e*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*x*2^(1/2),2
^(1/2))-EllipticE(1/2*I*x*2^(1/2),2^(1/2)))-1/2*I*d^3*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*x*2^(1/2),2^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)^3/sqrt(x^4 + 3*x^2 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3)/sqrt(x^4 + 3*x^2 + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3/(x**4+3*x**2+2)**(1/2),x)

[Out]

Integral((d + e*x**2)**3/sqrt((x**2 + 1)*(x**2 + 2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)^3/sqrt(x^4 + 3*x^2 + 2), x)