3.381 \(\int \frac{(d+e x^2)^2}{\sqrt{a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=356 \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} \left (3 c d^2-a e^2\right )}{\sqrt{a}}+2 e (3 c d-b e)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{6 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{2 e x \sqrt{a+b x^2+c x^4} (3 c d-b e)}{3 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{2 \sqrt [4]{a} e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (3 c d-b e) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{e^2 x \sqrt{a+b x^2+c x^4}}{3 c} \]

[Out]

(e^2*x*Sqrt[a + b*x^2 + c*x^4])/(3*c) + (2*e*(3*c*d - b*e)*x*Sqrt[a + b*x^2 + c*x^4])/(3*c^(3/2)*(Sqrt[a] + Sq
rt[c]*x^2)) - (2*a^(1/4)*e*(3*c*d - b*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x
^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(3*c^(7/4)*Sqrt[a + b*x^2 + c*x^
4]) + (a^(1/4)*(2*e*(3*c*d - b*e) + (Sqrt[c]*(3*c*d^2 - a*e^2))/Sqrt[a])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x
^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/
(6*c^(7/4)*Sqrt[a + b*x^2 + c*x^4])

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Rubi [A]  time = 0.191376, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1206, 1197, 1103, 1195} \[ \frac{\sqrt [4]{a} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \left (\frac{\sqrt{c} \left (3 c d^2-a e^2\right )}{\sqrt{a}}+2 e (3 c d-b e)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{6 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{2 e x \sqrt{a+b x^2+c x^4} (3 c d-b e)}{3 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{2 \sqrt [4]{a} e \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} (3 c d-b e) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{e^2 x \sqrt{a+b x^2+c x^4}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*x*Sqrt[a + b*x^2 + c*x^4])/(3*c) + (2*e*(3*c*d - b*e)*x*Sqrt[a + b*x^2 + c*x^4])/(3*c^(3/2)*(Sqrt[a] + Sq
rt[c]*x^2)) - (2*a^(1/4)*e*(3*c*d - b*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x
^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(3*c^(7/4)*Sqrt[a + b*x^2 + c*x^
4]) + (a^(1/4)*(2*e*(3*c*d - b*e) + (Sqrt[c]*(3*c*d^2 - a*e^2))/Sqrt[a])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x
^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/
(6*c^(7/4)*Sqrt[a + b*x^2 + c*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 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 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 1195

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

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^2}{\sqrt{a+b x^2+c x^4}} \, dx &=\frac{e^2 x \sqrt{a+b x^2+c x^4}}{3 c}+\frac{\int \frac{3 c d^2-a e^2+2 e (3 c d-b e) x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{3 c}\\ &=\frac{e^2 x \sqrt{a+b x^2+c x^4}}{3 c}-\frac{\left (2 \sqrt{a} e (3 c d-b e)\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx}{3 c^{3/2}}+\frac{\left (3 c d^2-a e^2+\frac{2 \sqrt{a} e (3 c d-b e)}{\sqrt{c}}\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{3 c}\\ &=\frac{e^2 x \sqrt{a+b x^2+c x^4}}{3 c}+\frac{2 e (3 c d-b e) x \sqrt{a+b x^2+c x^4}}{3 c^{3/2} \left (\sqrt{a}+\sqrt{c} x^2\right )}-\frac{2 \sqrt [4]{a} e (3 c d-b e) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{3 c^{7/4} \sqrt{a+b x^2+c x^4}}+\frac{\left (3 c d^2-a e^2+\frac{2 \sqrt{a} e (3 c d-b e)}{\sqrt{c}}\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{6 \sqrt [4]{a} c^{5/4} \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 1.63391, size = 488, normalized size = 1.37 \[ \frac{i \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \left (c e \left (-3 d \sqrt{b^2-4 a c}+a e+3 b d\right )+b e^2 \left (\sqrt{b^2-4 a c}-b\right )-3 c^2 d^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )-i e \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} (b e-3 c d) E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )+2 c e^2 x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (a+b x^2+c x^4\right )}{6 c^2 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*e^2*x*(a + b*x^2 + c*x^4) - I*(-b + Sqrt[b^2 - 4*a*c])*e*(-3*c*d + b*e)*S
qrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4*a*c] + 4*c*x^2)/(b -
Sqrt[b^2 - 4*a*c])]*EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(
b - Sqrt[b^2 - 4*a*c])] + I*(-3*c^2*d^2 + b*(-b + Sqrt[b^2 - 4*a*c])*e^2 + c*e*(3*b*d - 3*Sqrt[b^2 - 4*a*c]*d
+ a*e))*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 - 4*a*c] + 4*c*
x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 -
4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(6*c^2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4])

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Maple [B]  time = 0.009, size = 756, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

e^2*(1/3/c*x*(c*x^4+b*x^2+a)^(1/2)-1/12/c*a*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)/a)^(1/2)*(4-2*((-4*a*c+b^2)^(1/2)-
b)/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)*EllipticF(1/2*x*2^(1/2)*(((-4*a
*c+b^2)^(1/2)-b)/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))+1/3*b/c*a*2^(1/2)/(((-4*a*c+b^2)^(1/2
)-b)/a)^(1/2)*(4-2*((-4*a*c+b^2)^(1/2)-b)/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a
)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/2*x*2^(1/2)*(((-4*a*c+b^2)^(1/2)-b)/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*
c+b^2)^(1/2))/a/c)^(1/2))-EllipticE(1/2*x*2^(1/2)*(((-4*a*c+b^2)^(1/2)-b)/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)
^(1/2))/a/c)^(1/2))))-d*e*a*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)/a)^(1/2)*(4-2*((-4*a*c+b^2)^(1/2)-b)/a*x^2)^(1/2)*
(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/2*x*2^(1/2)
*(((-4*a*c+b^2)^(1/2)-b)/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))-EllipticE(1/2*x*2^(1/2)*(((-4
*a*c+b^2)^(1/2)-b)/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2)))+1/4*d^2*2^(1/2)/(((-4*a*c+b^2)^(1/
2)-b)/a)^(1/2)*(4-2*((-4*a*c+b^2)^(1/2)-b)/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+
a)^(1/2)*EllipticF(1/2*x*2^(1/2)*(((-4*a*c+b^2)^(1/2)-b)/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(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 )}^{2}}{\sqrt{c x^{4} + b x^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)^2/sqrt(c*x^4 + b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e^2*x^4 + 2*d*e*x^2 + d^2)/sqrt(c*x^4 + b*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral((d + e*x**2)**2/sqrt(a + b*x**2 + c*x**4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)^2/sqrt(c*x^4 + b*x^2 + a), x)