Integrand size = 19, antiderivative size = 108 \[ \int \frac {1}{\sqrt {1+4 x+4 x^2+4 x^4}} \, dx=-\frac {\left (\sqrt {5}+\left (1+\frac {1}{x}\right )^2\right ) \sqrt {\frac {5-2 \left (1+\frac {1}{x}\right )^2+\left (1+\frac {1}{x}\right )^4}{\left (\sqrt {5}+\left (1+\frac {1}{x}\right )^2\right )^2}} x^2 \operatorname {EllipticF}\left (2 \arctan \left (\frac {1+\frac {1}{x}}{\sqrt [4]{5}}\right ),\frac {1}{10} \left (5+\sqrt {5}\right )\right )}{2 \sqrt [4]{5} \sqrt {1+4 x+4 x^2+4 x^4}} \]
-1/10*x^2*(cos(2*arctan(1/5*(1+1/x)*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*(1 +1/x)*5^(3/4)))*EllipticF(sin(2*arctan(1/5*(1+1/x)*5^(3/4))),1/10*(50+10*5 ^(1/2))^(1/2))*((1+1/x)^2+5^(1/2))*((5-2*(1+1/x)^2+(1+1/x)^4)/((1+1/x)^2+5 ^(1/2))^2)^(1/2)*5^(3/4)/(4*x^4+4*x^2+4*x+1)^(1/2)
Result contains complex when optimal does not.
Time = 10.38 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.31 \[ \int \frac {1}{\sqrt {1+4 x+4 x^2+4 x^4}} \, dx=\frac {(2-i) \sqrt {-\frac {1}{10}+\frac {i}{5}} \sqrt {\frac {\left (2 i+\sqrt {-1-2 i}-\sqrt {-1+2 i}\right ) \left (-i+\sqrt {-1-2 i}-2 x\right )}{\left (-2 i+\sqrt {-1-2 i}+\sqrt {-1+2 i}\right ) \left (i+\sqrt {-1-2 i}+2 x\right )}} \left (1+2 x+2 i x^2\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {\left (2 i+\sqrt {-1-2 i}+\sqrt {-1+2 i}\right ) \left (-i+\sqrt {-1+2 i}+2 x\right )}{\sqrt {-1+2 i} \left (i+\sqrt {-1-2 i}+2 x\right )}}}{\sqrt {2}}\right ),\frac {1}{2} \left (5-\sqrt {5}\right )\right )}{\sqrt {\frac {(1+2 i) \left ((-1+i)+\sqrt {-1-2 i}\right ) \left (1+2 x+2 i x^2\right )}{\left (i+\sqrt {-1-2 i}+2 x\right )^2}} \sqrt {1+4 x+4 x^2+4 x^4}} \]
((2 - I)*Sqrt[-1/10 + I/5]*Sqrt[((2*I + Sqrt[-1 - 2*I] - Sqrt[-1 + 2*I])*( -I + Sqrt[-1 - 2*I] - 2*x))/((-2*I + Sqrt[-1 - 2*I] + Sqrt[-1 + 2*I])*(I + Sqrt[-1 - 2*I] + 2*x))]*(1 + 2*x + (2*I)*x^2)*EllipticF[ArcSin[Sqrt[((2*I + Sqrt[-1 - 2*I] + Sqrt[-1 + 2*I])*(-I + Sqrt[-1 + 2*I] + 2*x))/(Sqrt[-1 + 2*I]*(I + Sqrt[-1 - 2*I] + 2*x))]/Sqrt[2]], (5 - Sqrt[5])/2])/(Sqrt[((1 + 2*I)*((-1 + I) + Sqrt[-1 - 2*I])*(1 + 2*x + (2*I)*x^2))/(I + Sqrt[-1 - 2 *I] + 2*x)^2]*Sqrt[1 + 4*x + 4*x^2 + 4*x^4])
Time = 0.37 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2504, 27, 7270, 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.
\(\displaystyle \int \frac {1}{\sqrt {4 x^4+4 x^2+4 x+1}} \, dx\) |
\(\Big \downarrow \) 2504 |
\(\displaystyle -16 \int \frac {x^2}{16 \sqrt {\left (\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5\right ) x^4}}d\left (1+\frac {1}{x}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {x^2}{\sqrt {\left (\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5\right ) x^4}}d\left (1+\frac {1}{x}\right )\) |
\(\Big \downarrow \) 7270 |
\(\displaystyle -\frac {\sqrt {\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5} x^2 \int \frac {1}{\sqrt {\left (1+\frac {1}{x}\right )^4-2 \left (1+\frac {1}{x}\right )^2+5}}d\left (1+\frac {1}{x}\right )}{\sqrt {\left (\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5\right ) x^4}}\) |
\(\Big \downarrow \) 1416 |
\(\displaystyle -\frac {\left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right ) \sqrt {\frac {\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5}{\left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right )^2}} x^2 \operatorname {EllipticF}\left (2 \arctan \left (\frac {1+\frac {1}{x}}{\sqrt [4]{5}}\right ),\frac {1}{10} \left (5+\sqrt {5}\right )\right )}{2 \sqrt [4]{5} \sqrt {\left (\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5\right ) x^4}}\) |
-1/2*((Sqrt[5] + (1 + x^(-1))^2)*Sqrt[(5 - 2*(1 + x^(-1))^2 + (1 + x^(-1)) ^4)/(Sqrt[5] + (1 + x^(-1))^2)^2]*x^2*EllipticF[2*ArcTan[(1 + x^(-1))/5^(1 /4)], (5 + Sqrt[5])/10])/(5^(1/4)*Sqrt[(5 - 2*(1 + x^(-1))^2 + (1 + x^(-1) )^4)*x^4])
3.8.98.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(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]}, Simp[-16* a^2 Subst[Int[(1/(b - 4*a*x)^2)*(a*((-3*b^4 + 16*a*b^2*c - 64*a^2*b*d + 2 56*a^3*e - 32*a^2*(3*b^2 - 8*a*c)*x^2 + 256*a^4*x^4)/(b - 4*a*x)^4))^p, x], x, b/(4*a) + 1/x], x] /; NeQ[a, 0] && NeQ[b, 0] && EqQ[b^3 - 4*a*b*c + 8*a ^2*d, 0]] /; FreeQ[p, x] && PolyQ[P4, x, 4] && IntegerQ[2*p] && !IGtQ[p, 0 ]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p ]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p]))) Int[u*v ^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] && !IntegerQ[p] && !Free Q[v, x] && !FreeQ[w, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.83 (sec) , antiderivative size = 961, normalized size of antiderivative = 8.90
method | result | size |
default | \(\text {Expression too large to display}\) | \(961\) |
elliptic | \(\text {Expression too large to display}\) | \(961\) |
(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4) )*((RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index =2))*(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1 ,index=4)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))/(x-RootOf(4*_Z^4+4*_Z^2+4* _Z+1,index=2)))^(1/2)*(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2))^2*((RootOf( 4*_Z^4+4*_Z^2+4*_Z+1,index=2)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))*(x-Roo tOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3)-R ootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))/(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index= 2)))^(1/2)*((RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-RootOf(4*_Z^4+4*_Z^2+4*_ Z+1,index=1))*(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4))/(RootOf(4*_Z^4+4*_Z ^2+4*_Z+1,index=4)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))/(x-RootOf(4*_Z^4+ 4*_Z^2+4*_Z+1,index=2)))^(1/2)/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4)-RootO f(4*_Z^4+4*_Z^2+4*_Z+1,index=2))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-Roo tOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))/((x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1 ))*(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2))*(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1 ,index=3))*(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4)))^(1/2)*EllipticF(((Roo tOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2))*(x -RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index= 4)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))/(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,in dex=2)))^(1/2),((RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-RootOf(4*_Z^4+4*_...
\[ \int \frac {1}{\sqrt {1+4 x+4 x^2+4 x^4}} \, dx=\int { \frac {1}{\sqrt {4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1}} \,d x } \]
\[ \int \frac {1}{\sqrt {1+4 x+4 x^2+4 x^4}} \, dx=\int \frac {1}{\sqrt {4 x^{4} + 4 x^{2} + 4 x + 1}}\, dx \]
\[ \int \frac {1}{\sqrt {1+4 x+4 x^2+4 x^4}} \, dx=\int { \frac {1}{\sqrt {4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1}} \,d x } \]
\[ \int \frac {1}{\sqrt {1+4 x+4 x^2+4 x^4}} \, dx=\int { \frac {1}{\sqrt {4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {1+4 x+4 x^2+4 x^4}} \, dx=\int \frac {1}{\sqrt {4\,x^4+4\,x^2+4\,x+1}} \,d x \]