\(\int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx\) [66]

Optimal result

Integrand size = 34, antiderivative size = 219 \[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\frac {\sqrt [4]{5 d^4+256 a e^3} \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right ) \sqrt {\frac {5 d^4+256 a e^3-6 d^2 (d+4 e x)^2+(d+4 e x)^4}{\left (5 d^4+256 a e^3\right ) \left (1+\frac {(d+4 e x)^2}{\sqrt {5 d^4+256 a e^3}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right ),\frac {1}{2} \left (1+\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}\right )\right )}{8 e \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \] Output:

1/8*(256*a*e^3+5*d^4)^(1/4)*(1+(4*e*x+d)^2/(256*a*e^3+5*d^4)^(1/2))*((5*d^ 
4+256*a*e^3-6*d^2*(4*e*x+d)^2+(4*e*x+d)^4)/(256*a*e^3+5*d^4)/(1+(4*e*x+d)^ 
2/(256*a*e^3+5*d^4)^(1/2))^2)^(1/2)*InverseJacobiAM(2*arctan((4*e*x+d)/(25 
6*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))/e/(8*e^ 
3*x^4+8*d*e^2*x^3-d^3*x+8*a*e^2)^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1065\) vs. \(2(219)=438\).

Time = 12.39 (sec) , antiderivative size = 1065, normalized size of antiderivative = 4.86 \[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=-\frac {\left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right ) \left (d-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}+4 e x\right ) \sqrt {-\frac {\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}} \left (d+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}+4 e x\right )}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}} \sqrt {\frac {3 d^2-2 \sqrt {d^4-64 a e^3}-\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}} \sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}+d \left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right )+4 e \left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) x}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+4 e x\right )}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}}\right ),\frac {\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right )^2}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right )^2}\right )}{2 e \left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \sqrt {\frac {\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}} \left (-d+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}-4 e x\right )}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \] Input:

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

Output:

-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)*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[(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]] + Sqr 
t[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[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 - 6 
4*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 - 6 
4*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*(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[(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...
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.27, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2458, 1416}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2458

\(\displaystyle \int \frac {1}{\sqrt {\frac {1}{32} \left (256 a e^2+\frac {5 d^4}{e}\right )-3 d^2 e \left (\frac {d}{4 e}+x\right )^2+8 e^3 \left (\frac {d}{4 e}+x\right )^4}}d\left (\frac {d}{4 e}+x\right )\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {\sqrt [4]{256 a e^3+5 d^4} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) \sqrt {\frac {256 a e^3+5 d^4-96 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2+256 e^4 \left (\frac {d}{4 e}+x\right )^4}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {4 e \left (\frac {d}{4 e}+x\right )}{\sqrt [4]{5 d^4+256 a e^3}}\right ),\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{\sqrt {2} e \sqrt {256 a e^2+\frac {5 d^4}{e}-96 d^2 e \left (\frac {d}{4 e}+x\right )^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4}}\)

Input:

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

Output:

((5*d^4 + 256*a*e^3)^(1/4)*(1 + (16*e^2*(d/(4*e) + x)^2)/Sqrt[5*d^4 + 256* 
a*e^3])*Sqrt[(5*d^4 + 256*a*e^3 - 96*d^2*e^2*(d/(4*e) + x)^2 + 256*e^4*(d/ 
(4*e) + 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)]*EllipticF[2*ArcTan[(4*e*(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*Sqrt[(5* 
d^4)/e + 256*a*e^2 - 96*d^2*e*(d/(4*e) + x)^2 + 256*e^3*(d/(4*e) + x)^4])
 

Defintions of rubi rules used

rule 1416

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 2458

Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp 
on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x 
- S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp 
on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P 
n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1703\) vs. \(2(203)=406\).

Time = 0.69 (sec) , antiderivative size = 1704, normalized size of antiderivative = 7.78

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

Input:

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)
 

Output:

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*(-6 
4*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)*(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...
 

Fricas [F]

\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int \frac {1}{\sqrt {8 a e^{2} - d^{3} x + 8 d e^{2} x^{3} + 8 e^{3} x^{4}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}} \,d x } \] Input:

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 
")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int \frac {1}{\sqrt {-d^3\,x+8\,d\,e^2\,x^3+8\,e^3\,x^4+8\,a\,e^2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int \frac {\sqrt {8 e^{3} x^{4}+8 d \,e^{2} x^{3}-d^{3} x +8 a \,e^{2}}}{8 e^{3} x^{4}+8 d \,e^{2} x^{3}-d^{3} x +8 a \,e^{2}}d x \] Input:

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

Output:

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