3.8.79 \(\int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx\) [779]

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

[Out]

1/2*(256*a*e^3+5*d^4)^(1/4)*(cos(2*arctan((4*e*x+d)/(256*a*e^3+5*d^4)^(1/4)))^2)^(1/2)/cos(2*arctan((4*e*x+d)/
(256*a*e^3+5*d^4)^(1/4)))*EllipticF(sin(2*arctan((4*e*x+d)/(256*a*e^3+5*d^4)^(1/4))),1/2*(2+6*d^2/(256*a*e^3+5
*d^4)^(1/2))^(1/2))*(1+16*e^2*(1/4*d/e+x)^2/(256*a*e^3+5*d^4)^(1/2))*(e*(8*e^3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)/
(256*a*e^3+5*d^4)/(1+16*e^2*(1/4*d/e+x)^2/(256*a*e^3+5*d^4)^(1/2))^2)^(1/2)/e*2^(1/2)/(8*e^3*x^4+8*d*e^2*x^3-d
^3*x+8*a*e^2)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 235, normalized size of antiderivative = 1.00, 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 = {1120, 1117} \begin {gather*} \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 \text {ArcTan}\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}} \end {gather*}

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 1117

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)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1120

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]

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 &=\text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1065\) vs. \(2(235)=470\).
time = 11.55, size = 1065, normalized size = 4.53 \begin {gather*} -\frac {\left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right ) \left (d-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}+4 e x\right ) \sqrt {-\frac {\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}} \left (d+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}+4 e x\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+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}} \sqrt {\frac {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 \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 )+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}{\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+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\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+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+4 e x\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+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\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+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}-4 e x\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+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \end {gather*}

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]

-1/2*((-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)*S
qrt[-((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[(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]]
+ 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))]*EllipticF[ArcSi
n[Sqrt[((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
+ 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*Sq
rt[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])/(e*(S
qrt[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] Leaf count of result is larger than twice the leaf count of optimal. \(1703\) vs. \(2(273)=546\).
time = 0.05, size = 1704, normalized size = 7.25

method	result	size
default	\(\text {Expression too large to display}\)	\(1704\)
elliptic	\(\text {Expression too large to display}\)	\(1704\)

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,method=_RETURNVERBOSE)

[Out]

1/2*(1/4*(-d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*
e^2)^(1/2))/e^2)*((-1/4*(d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*d^2*e^2+2*(-64*a*e
^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x-1/4*(-d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1/4*(d*e+(3*
d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/
(x+1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2))^(1/2)*(x+1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4
)^(1/2)*e^2)^(1/2))/e^2)^2*((-1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*d^2*e^2
+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x-1/4*(-d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(1/
4*(-d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(
1/2))/e^2)/(x+1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2))^(1/2)*((-1/4*(d*e+(3*d^2*e^2+2*(-6
4*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x+1/4*(d*e+(
3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1/4*(d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e
^2-1/4*(-d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(x+1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2
)*e^2)^(1/2))/e^2))^(1/2)/(-1/4*(d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*d^2*e^2+2*
(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e
+(3*d^2*e^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*d^2*e^2+2*(-64*a*e^3+d^4)^(1
/2)*e^2)^(1/2))/e^2)*(x+1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x-1/4*(-d*e+(3*d^2*e^2-2
*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(x+1/4*(d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2))^(1/2)
*EllipticF(((-1/4*(d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4
)^(1/2)*e^2)^(1/2))/e^2)*(x-1/4*(-d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1/4*(d*e+(3*d^2*e^
2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(x+1/4
*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2))^(1/2),((-1/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2
)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)*(1/4*(-d*e+(3*d^2*e^2+2*(-64*a
*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(1/4*(-d*e+(3*d^2
*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2-1/4*(-d*e+(3*d^2*e^2-2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2)/(-1
/4*(d*e+(3*d^2*e^2+2*(-64*a*e^3+d^4)^(1/2)*e^2)^(1/2))/e^2+1/4*(d*e+(3*d^2*e^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*x^4*e^3 + 8*d*x^3*e^2 - d^3*x + 8*a*e^2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {8 a e^{2} - d^{3} x + 8 d e^{2} x^{3} + 8 e^{3} x^{4}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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*x^4*e^3 + 8*d*x^3*e^2 - d^3*x + 8*a*e^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{\sqrt {-d^3\,x+8\,d\,e^2\,x^3+8\,e^3\,x^4+8\,a\,e^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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