3.779 \(\int \frac{1}{\sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx\)

Optimal. Leaf size=235 \[ \frac{\sqrt [4]{256 a e^3+5 d^4} \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right )^2}} \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right ) F\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}+1\right )\right )}{\sqrt{2} e \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]

[Out]

((5*d^4 + 256*a*e^3)^(1/4)*Sqrt[(e*(8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4))/((5*d^4 + 256*a*e^3)*(1 + (16*
e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])^2)]*(1 + (16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])*Ellip
ticF[2*ArcTan[(d + 4*e*x)/(5*d^4 + 256*a*e^3)^(1/4)], (1 + (3*d^2)/Sqrt[5*d^4 + 256*a*e^3])/2])/(Sqrt[2]*e*Sqr
t[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4])

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Rubi [A]  time = 0.177037, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {1106, 1103} \[ \frac{\sqrt [4]{256 a e^3+5 d^4} \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right )^2}} \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right ) F\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}+1\right )\right )}{\sqrt{2} e \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4],x]

[Out]

((5*d^4 + 256*a*e^3)^(1/4)*Sqrt[(e*(8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4))/((5*d^4 + 256*a*e^3)*(1 + (16*
e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])^2)]*(1 + (16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256*a*e^3])*Ellip
ticF[2*ArcTan[(d + 4*e*x)/(5*d^4 + 256*a*e^3)^(1/4)], (1 + (3*d^2)/Sqrt[5*d^4 + 256*a*e^3])/2])/(Sqrt[2]*e*Sqr
t[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4])

Rule 1106

Int[(P4_)^(p_), x_Symbol] :> With[{a = Coeff[P4, x, 0], b = Coeff[P4, x, 1], c = Coeff[P4, x, 2], d = Coeff[P4
, x, 3], e = Coeff[P4, x, 4]}, Subst[Int[SimplifyIntegrand[(a + d^4/(256*e^3) - (b*d)/(8*e) + (c - (3*d^2)/(8*
e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0] && NeQ[d, 0]] /; FreeQ[p, x] &&
PolyQ[P4, x, 4] && NeQ[p, 2] && NeQ[p, 3]

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]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{1}{32} \left (\frac{5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4}} \, dx,x,\frac{d}{4 e}+x\right )\\ &=\frac{\sqrt [4]{5 d^4+256 a e^3} \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (5 d^4+256 a e^3\right ) \left (1+\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{5 d^4+256 a e^3}}\right )^2}} \left (1+\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{5 d^4+256 a e^3}}\right ) F\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (1+\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}\right )\right )}{\sqrt{2} e \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}\\ \end{align*}

Mathematica [B]  time = 2.2346, size = 1065, normalized size = 4.53 \[ -\frac{\left (-d-4 e x+\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}\right ) \left (d+4 e x-\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) \sqrt{-\frac{\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}} \left (d+4 e x+\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right )}{\left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}-\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) \left (-d-4 e x+\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}\right )}} \sqrt{\frac{3 d^2+\left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}-\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) d+4 e \left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}-\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) x-2 \sqrt{d^4-64 a e^3}-\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}} \sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}}{\left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}+\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) \left (-d-4 e x+\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}\right )}} F\left (\sin ^{-1}\left (\sqrt{\frac{\left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}-\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) \left (d+4 e x+\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}\right )}{\left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}+\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) \left (-d-4 e x+\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}\right )}}\right )|\frac{\left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}+\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right )^2}{\left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}-\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right )^2}\right )}{2 e \left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}-\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) \sqrt{\frac{\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}} \left (-d-4 e x+\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right )}{\left (\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}+\sqrt{3 d^2+2 \sqrt{d^4-64 a e^3}}\right ) \left (-d-4 e x+\sqrt{3 d^2-2 \sqrt{d^4-64 a e^3}}\right )}} \sqrt{8 e^3 x^4+8 d e^2 x^3-d^3 x+8 a e^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[8*a*e^2 - d^3*x + 8*d*e^2*x^3 + 8*e^3*x^4],x]

[Out]

-((-d + Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - 4*e*x)*(d - Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]] + 4*e*x)*Sqrt[
-((Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]]*(d + Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]] + 4*e*x))/((Sqrt[3*d^2 - 2*S
qrt[d^4 - 64*a*e^3]] - Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]])*(-d + Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - 4*e*
x)))]*Sqrt[(3*d^2 - 2*Sqrt[d^4 - 64*a*e^3] - Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]]*Sqrt[3*d^2 + 2*Sqrt[d^4 - 64
*a*e^3]] + d*(Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]]) + 4*e*(Sqrt[3*d^2 -
2*Sqrt[d^4 - 64*a*e^3]] - Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]])*x)/((Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] + S
qrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]])*(-d + Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - 4*e*x))]*EllipticF[ArcSin[Sq
rt[((Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]])*(d + Sqrt[3*d^2 - 2*Sqrt[d^4
- 64*a*e^3]] + 4*e*x))/((Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] + Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]])*(-d + S
qrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - 4*e*x))]], (Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] + Sqrt[3*d^2 + 2*Sqrt[d
^4 - 64*a*e^3]])^2/(Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]])^2])/(2*e*(Sqr
t[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]])*Sqrt[(Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a
*e^3]]*(-d + Sqrt[3*d^2 + 2*Sqrt[d^4 - 64*a*e^3]] - 4*e*x))/((Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] + Sqrt[3*d^
2 + 2*Sqrt[d^4 - 64*a*e^3]])*(-d + Sqrt[3*d^2 - 2*Sqrt[d^4 - 64*a*e^3]] - 4*e*x))]*Sqrt[8*a*e^2 - d^3*x + 8*d*
e^2*x^3 + 8*e^3*x^4])

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Maple [B]  time = 0.024, size = 1704, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2),x)

[Out]

1/2*(1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*
e^2)^(1/2))/e^2)*((-1/4*(d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*e^2*d^2+2*(-64*a*e
^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x-1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1/4*(d*e+(3*
e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/
(x+1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2))^(1/2)*(x+1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4
)^(1/2)*e^2)^(1/2))/e^2)^2*((-1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*e^2*d^2
+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x-1/4*(-d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(1/
4*(-d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(
1/2))/e^2)/(x+1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2))^(1/2)*((-1/4*(d*e+(3*e^2*d^2+2*(-6
4*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x+1/4*(d*e+(
3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1/4*(d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e
^2-1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(x+1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2
)*e^2)^(1/2))/e^2))^(1/2)/(-1/4*(d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*e^2*d^2+2*
(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e
+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*2^(1/2)/(e^3*(x-1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1
/2)*e^2)^(1/2))/e^2)*(x+1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x-1/4*(-d*e+(3*e^2*d^2-2
*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x+1/4*(d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2))^(1/2)
*EllipticF(((-1/4*(d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4
)^(1/2)*e^2)^(1/2))/e^2)*(x-1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1/4*(d*e+(3*e^2*d^
2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(x+1/4
*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2))^(1/2),((-1/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2
)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(1/4*(-d*e+(3*e^2*d^2+2*(-64*a
*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(1/4*(-d*e+(3*e^2
*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1
/4*(d*e+(3*e^2*d^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*e^2*d^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1
/2))/e^2))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2),x, algorithm="fricas")

[Out]

integral(1/sqrt(8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 a e^{2} - d^{3} x + 8 d e^{2} x^{3} + 8 e^{3} x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(8*e**3*x**4+8*d*e**2*x**3-d**3*x+8*a*e**2)**(1/2),x)

[Out]

Integral(1/sqrt(8*a*e**2 - d**3*x + 8*d*e**2*x**3 + 8*e**3*x**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(8*e^3*x^4 + 8*d*e^2*x^3 - d^3*x + 8*a*e^2), x)