Integrand size = 34, antiderivative size = 235 \[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\frac {\sqrt [4]{5 d^4+256 a e^3} \sqrt {\frac {e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (5 d^4+256 a e^3\right ) \left (1+\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}\right )^2}} \left (1+\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {5 d^4+256 a e^3}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right ),\frac {1}{2} \left (1+\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}\right )\right )}{\sqrt {2} e \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]
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(si n(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)
Leaf count is larger than twice the leaf count of optimal. \(1065\) vs. \(2(235)=470\).
Time = 11.41 (sec) , antiderivative size = 1065, normalized size of antiderivative = 4.53 \[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=-\frac {\left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right ) \left (d-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}+4 e x\right ) \sqrt {-\frac {\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}} \left (d+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}+4 e x\right )}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}} \sqrt {\frac {3 d^2-2 \sqrt {d^4-64 a e^3}-\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}} \sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}+d \left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right )+4 e \left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) x}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+4 e x\right )}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}}\right ),\frac {\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right )^2}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right )^2}\right )}{2 e \left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \sqrt {\frac {\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}} \left (-d+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}-4 e x\right )}{\left (\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}+\sqrt {3 d^2+2 \sqrt {d^4-64 a e^3}}\right ) \left (-d+\sqrt {3 d^2-2 \sqrt {d^4-64 a e^3}}-4 e x\right )}} \sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]
-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...
Time = 0.30 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.19, 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 |
| \(\displaystyle \frac {\sqrt [4]{256 a e^3+5 d^4} \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right ) \sqrt {\frac {256 a e^3+5 d^4-96 d^2 e^2 \left (\frac {d}{4 e}+x\right )^2+256 e^4 \left (\frac {d}{4 e}+x\right )^4}{\left (256 a e^3+5 d^4\right ) \left (\frac {16 e^2 \left (\frac {d}{4 e}+x\right )^2}{\sqrt {256 a e^3+5 d^4}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {4 e \left (\frac {d}{4 e}+x\right )}{\sqrt [4]{5 d^4+256 a e^3}}\right ),\frac {1}{2} \left (\frac {3 d^2}{\sqrt {5 d^4+256 a e^3}}+1\right )\right )}{\sqrt {2} e \sqrt {256 a e^2+\frac {5 d^4}{e}-96 d^2 e \left (\frac {d}{4 e}+x\right )^2+256 e^3 \left (\frac {d}{4 e}+x\right )^4}}\) |
((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])
3.8.79.3.1 Defintions of rubi rules used
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1703\) vs. \(2(273)=546\).
Time = 1.23 (sec) , antiderivative size = 1704, normalized size of antiderivative = 7.25
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1704\) |
| elliptic | \(\text {Expression too large to display}\) | \(1704\) |
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*(-6 4*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)*(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...
\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}} \,d x } \]
\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int \frac {1}{\sqrt {8 a e^{2} - d^{3} x + 8 d e^{2} x^{3} + 8 e^{3} x^{4}}}\, dx \]
\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}} \,d x } \]
\[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int { \frac {1}{\sqrt {8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \, dx=\int \frac {1}{\sqrt {-d^3\,x+8\,d\,e^2\,x^3+8\,e^3\,x^4+8\,a\,e^2}} \,d x \]