3.799 \(\int \frac {1}{(1+4 x+4 x^2+4 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=367 \[ -\frac {\left (3-\left (\frac {1}{x}+1\right )^2\right ) x^2}{\sqrt {4 x^4+4 x^2+4 x+1}}+\frac {\left (13-9 \left (\frac {1}{x}+1\right )^2\right ) \left (\frac {1}{x}+1\right ) x^2}{10 \sqrt {4 x^4+4 x^2+4 x+1}}+\frac {9 \left (\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5\right ) \left (\frac {1}{x}+1\right ) x^2}{10 \left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right ) \sqrt {4 x^4+4 x^2+4 x+1}}+\frac {3 \left (3-\sqrt {5}\right ) \left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right ) \sqrt {\frac {\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5}{\left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac {1+\frac {1}{x}}{\sqrt [4]{5}}\right )|\frac {1}{10} \left (5+\sqrt {5}\right )\right )}{4\ 5^{3/4} \sqrt {4 x^4+4 x^2+4 x+1}}-\frac {9 \left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right ) \sqrt {\frac {\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5}{\left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right )^2}} x^2 E\left (2 \tan ^{-1}\left (\frac {1+\frac {1}{x}}{\sqrt [4]{5}}\right )|\frac {1}{10} \left (5+\sqrt {5}\right )\right )}{2\ 5^{3/4} \sqrt {4 x^4+4 x^2+4 x+1}} \]

[Out]

-(3-(1+1/x)^2)*x^2/(4*x^4+4*x^2+4*x+1)^(1/2)+1/10*(13-9*(1+1/x)^2)*(1+1/x)*x^2/(4*x^4+4*x^2+4*x+1)^(1/2)+9/10*
(5-2*(1+1/x)^2+(1+1/x)^4)*(1+1/x)*x^2/((1+1/x)^2+5^(1/2))/(4*x^4+4*x^2+4*x+1)^(1/2)-9/10*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)))*EllipticE(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^(1/4)/
(4*x^4+4*x^2+4*x+1)^(1/2)+3/20*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))*(3-5^(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^(1/4)/(4*x^4+4*x^2+4*x+1)^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {2069, 6719, 1673, 1678, 1197, 1103, 1195, 1247, 636} \[ -\frac {\left (3-\left (\frac {1}{x}+1\right )^2\right ) x^2}{\sqrt {4 x^4+4 x^2+4 x+1}}+\frac {\left (13-9 \left (\frac {1}{x}+1\right )^2\right ) \left (\frac {1}{x}+1\right ) x^2}{10 \sqrt {4 x^4+4 x^2+4 x+1}}+\frac {9 \left (\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5\right ) \left (\frac {1}{x}+1\right ) x^2}{10 \left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right ) \sqrt {4 x^4+4 x^2+4 x+1}}+\frac {3 \left (3-\sqrt {5}\right ) \left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right ) \sqrt {\frac {\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5}{\left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac {1+\frac {1}{x}}{\sqrt [4]{5}}\right )|\frac {1}{10} \left (5+\sqrt {5}\right )\right )}{4\ 5^{3/4} \sqrt {4 x^4+4 x^2+4 x+1}}-\frac {9 \left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right ) \sqrt {\frac {\left (\frac {1}{x}+1\right )^4-2 \left (\frac {1}{x}+1\right )^2+5}{\left (\left (\frac {1}{x}+1\right )^2+\sqrt {5}\right )^2}} x^2 E\left (2 \tan ^{-1}\left (\frac {1+\frac {1}{x}}{\sqrt [4]{5}}\right )|\frac {1}{10} \left (5+\sqrt {5}\right )\right )}{2\ 5^{3/4} \sqrt {4 x^4+4 x^2+4 x+1}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 4*x + 4*x^2 + 4*x^4)^(-3/2),x]

[Out]

-(((3 - (1 + x^(-1))^2)*x^2)/Sqrt[1 + 4*x + 4*x^2 + 4*x^4]) + ((13 - 9*(1 + x^(-1))^2)*(1 + x^(-1))*x^2)/(10*S
qrt[1 + 4*x + 4*x^2 + 4*x^4]) + (9*(5 - 2*(1 + x^(-1))^2 + (1 + x^(-1))^4)*(1 + x^(-1))*x^2)/(10*(Sqrt[5] + (1
+ x^(-1))^2)*Sqrt[1 + 4*x + 4*x^2 + 4*x^4]) - (9*(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*EllipticE[2*ArcTan[(1 + x^(-1))/5^(1/4)], (5 + Sqrt[5])/10])/(2*
5^(3/4)*Sqrt[1 + 4*x + 4*x^2 + 4*x^4]) + (3*(3 - Sqrt[5])*(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])/(4*5^(3/4)*Sqrt[1 + 4*x + 4*x^2 + 4*x^4])

Rule 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

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]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 1673

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rule 1678

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2
+ c*x^4)^(p + 1)*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*
a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuot
ient[Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p + 7)*(b*d - 2*a*e)*x^2,
x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 2069

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]}, Dist[-16*a^2, Subst[Int[(1*((a*(-3*b^4 + 16*a*b^2*c - 64*a^2*b*d + 256*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)/(b - 4*a*x)^2, 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]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

\begin {align*} \int \frac {1}{\left (1+4 x+4 x^2+4 x^4\right )^{3/2}} \, dx &=-\left (16 \operatorname {Subst}\left (\int \frac {1}{(4-4 x)^2 \left (\frac {1280-512 x^2+256 x^4}{(4-4 x)^4}\right )^{3/2}} \, dx,x,1+\frac {1}{x}\right )\right )\\ &=-\frac {\left (\sqrt {1280-512 \left (1+\frac {1}{x}\right )^2+256 \left (1+\frac {1}{x}\right )^4} x^2\right ) \operatorname {Subst}\left (\int \frac {(4-4 x)^4}{\left (1280-512 x^2+256 x^4\right )^{3/2}} \, dx,x,1+\frac {1}{x}\right )}{\sqrt {1+4 x+4 x^2+4 x^4}}\\ &=-\frac {\left (\sqrt {1280-512 \left (1+\frac {1}{x}\right )^2+256 \left (1+\frac {1}{x}\right )^4} x^2\right ) \operatorname {Subst}\left (\int \frac {x \left (-1024-1024 x^2\right )}{\left (1280-512 x^2+256 x^4\right )^{3/2}} \, dx,x,1+\frac {1}{x}\right )}{\sqrt {1+4 x+4 x^2+4 x^4}}-\frac {\left (\sqrt {1280-512 \left (1+\frac {1}{x}\right )^2+256 \left (1+\frac {1}{x}\right )^4} x^2\right ) \operatorname {Subst}\left (\int \frac {256+1536 x^2+256 x^4}{\left (1280-512 x^2+256 x^4\right )^{3/2}} \, dx,x,1+\frac {1}{x}\right )}{\sqrt {1+4 x+4 x^2+4 x^4}}\\ &=\frac {\left (13-9 \left (1+\frac {1}{x}\right )^2\right ) \left (1+\frac {1}{x}\right ) x^2}{10 \sqrt {1+4 x+4 x^2+4 x^4}}-\frac {\left (\sqrt {1280-512 \left (1+\frac {1}{x}\right )^2+256 \left (1+\frac {1}{x}\right )^4} x^2\right ) \operatorname {Subst}\left (\int \frac {2013265920-1207959552 x^2}{\sqrt {1280-512 x^2+256 x^4}} \, dx,x,1+\frac {1}{x}\right )}{1342177280 \sqrt {1+4 x+4 x^2+4 x^4}}-\frac {\left (\sqrt {1280-512 \left (1+\frac {1}{x}\right )^2+256 \left (1+\frac {1}{x}\right )^4} x^2\right ) \operatorname {Subst}\left (\int \frac {-1024-1024 x}{\left (1280-512 x+256 x^2\right )^{3/2}} \, dx,x,\left (1+\frac {1}{x}\right )^2\right )}{2 \sqrt {1+4 x+4 x^2+4 x^4}}\\ &=-\frac {\left (3-\left (1+\frac {1}{x}\right )^2\right ) x^2}{\sqrt {1+4 x+4 x^2+4 x^4}}+\frac {\left (13-9 \left (1+\frac {1}{x}\right )^2\right ) \left (1+\frac {1}{x}\right ) x^2}{10 \sqrt {1+4 x+4 x^2+4 x^4}}-\frac {\left (9 \sqrt {1280-512 \left (1+\frac {1}{x}\right )^2+256 \left (1+\frac {1}{x}\right )^4} x^2\right ) \operatorname {Subst}\left (\int \frac {1-\frac {x^2}{\sqrt {5}}}{\sqrt {1280-512 x^2+256 x^4}} \, dx,x,1+\frac {1}{x}\right )}{2 \sqrt {5} \sqrt {1+4 x+4 x^2+4 x^4}}-\frac {\left (3 \left (5-3 \sqrt {5}\right ) \sqrt {1280-512 \left (1+\frac {1}{x}\right )^2+256 \left (1+\frac {1}{x}\right )^4} x^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1280-512 x^2+256 x^4}} \, dx,x,1+\frac {1}{x}\right )}{10 \sqrt {1+4 x+4 x^2+4 x^4}}\\ &=-\frac {\left (3-\left (1+\frac {1}{x}\right )^2\right ) x^2}{\sqrt {1+4 x+4 x^2+4 x^4}}+\frac {\left (13-9 \left (1+\frac {1}{x}\right )^2\right ) \left (1+\frac {1}{x}\right ) x^2}{10 \sqrt {1+4 x+4 x^2+4 x^4}}+\frac {9 \left (5-2 \left (1+\frac {1}{x}\right )^2+\left (1+\frac {1}{x}\right )^4\right ) \left (1+\frac {1}{x}\right ) x^2}{10 \left (\sqrt {5}+\left (1+\frac {1}{x}\right )^2\right ) \sqrt {1+4 x+4 x^2+4 x^4}}-\frac {9 \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 E\left (2 \tan ^{-1}\left (\frac {1+\frac {1}{x}}{\sqrt [4]{5}}\right )|\frac {1}{10} \left (5+\sqrt {5}\right )\right )}{2\ 5^{3/4} \sqrt {1+4 x+4 x^2+4 x^4}}+\frac {3 \left (3-\sqrt {5}\right ) \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 F\left (2 \tan ^{-1}\left (\frac {1+\frac {1}{x}}{\sqrt [4]{5}}\right )|\frac {1}{10} \left (5+\sqrt {5}\right )\right )}{4\ 5^{3/4} \sqrt {1+4 x+4 x^2+4 x^4}}\\ \end {align*}

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Mathematica [C]  time = 4.76, size = 602, normalized size = 1.64 \[ \frac {36 x^3-16 x^2+\frac {(6-3 i) \sqrt {-\frac {2}{5}+\frac {4 i}{5}} \sqrt {\frac {\left (2 i+\sqrt {-1-2 i}-\sqrt {-1+2 i}\right ) \left (-2 x+\sqrt {-1-2 i}-i\right )}{\left (-2 i+\sqrt {-1-2 i}+\sqrt {-1+2 i}\right ) \left (2 x+\sqrt {-1-2 i}+i\right )}} \left (2 i x^2+2 x+1\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {\left (2 i+\sqrt {-1-2 i}+\sqrt {-1+2 i}\right ) \left (2 x+\sqrt {-1+2 i}-i\right )}{\sqrt {-1+2 i} \left (2 x+\sqrt {-1-2 i}+i\right )}}}{\sqrt {2}}\right )|\frac {1}{2} \left (5-\sqrt {5}\right )\right )}{\sqrt {\frac {(1+2 i) \left ((-1+i)+\sqrt {-1-2 i}\right ) \left (2 i x^2+2 x+1\right )}{\left (2 x+\sqrt {-1-2 i}+i\right )^2}}}-\frac {9 i \sqrt {-\frac {2}{5}+\frac {4 i}{5}} \left (-2 i+\sqrt {-1-2 i}+\sqrt {-1+2 i}\right ) \left (2 i+\sqrt {-1-2 i}+\sqrt {-1+2 i}\right ) \left (x+\frac {1}{2} \left (i+\sqrt {-1-2 i}\right )\right )^2 \sqrt {\frac {\left (2 i+\sqrt {-1-2 i}-\sqrt {-1+2 i}\right ) \left (-2 x+\sqrt {-1-2 i}-i\right )}{\left (-2 i+\sqrt {-1-2 i}+\sqrt {-1+2 i}\right ) \left (2 x+\sqrt {-1-2 i}+i\right )}} \sqrt {\frac {(1+2 i) \left ((-1+i)+\sqrt {-1-2 i}\right ) \left (2 i x^2+2 x+1\right )}{\left (2 x+\sqrt {-1-2 i}+i\right )^2}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {\left (2 i+\sqrt {-1-2 i}+\sqrt {-1+2 i}\right ) \left (2 x+\sqrt {-1+2 i}-i\right )}{\sqrt {-1+2 i} \left (2 x+\sqrt {-1-2 i}+i\right )}}}{\sqrt {2}}\right )|\frac {1}{2} \left (5-\sqrt {5}\right )\right )}{(-1+i)+\sqrt {-1-2 i}}+42 x+\frac {9}{2} \left (-2 x+\sqrt {-1-2 i}-i\right ) \left (2 x-\sqrt {-1+2 i}-i\right ) \left (2 x+\sqrt {-1+2 i}-i\right )+19}{10 \sqrt {4 x^4+4 x^2+4 x+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + 4*x + 4*x^2 + 4*x^4)^(-3/2),x]

[Out]

(19 + 42*x - 16*x^2 + 36*x^3 + (9*(-I + Sqrt[-1 - 2*I] - 2*x)*(-I - Sqrt[-1 + 2*I] + 2*x)*(-I + Sqrt[-1 + 2*I]
+ 2*x))/2 - ((9*I)*Sqrt[-2/5 + (4*I)/5]*(-2*I + Sqrt[-1 - 2*I] + Sqrt[-1 + 2*I])*(2*I + Sqrt[-1 - 2*I] + Sqrt
[-1 + 2*I])*((I + Sqrt[-1 - 2*I])/2 + x)^2*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))]*Sqrt[((1 + 2*I)*((-1 + I) + Sqr
t[-1 - 2*I])*(1 + 2*x + (2*I)*x^2))/(I + Sqrt[-1 - 2*I] + 2*x)^2]*EllipticE[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 - Sq
rt[5])/2])/((-1 + I) + Sqrt[-1 - 2*I]) + ((6 - 3*I)*Sqrt[-2/5 + (4*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])/(10*Sqrt[1 + 4*x + 4*x^2 + 4*x^4])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1}}{16 \, x^{8} + 32 \, x^{6} + 32 \, x^{5} + 24 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 8 \, x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x^4+4*x^2+4*x+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(4*x^4 + 4*x^2 + 4*x + 1)/(16*x^8 + 32*x^6 + 32*x^5 + 24*x^4 + 32*x^3 + 24*x^2 + 8*x + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x^4+4*x^2+4*x+1)^(3/2),x, algorithm="giac")

[Out]

integrate((4*x^4 + 4*x^2 + 4*x + 1)^(-3/2), x)

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maple [C]  time = 0.11, size = 2564, normalized size = 6.99 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*x^4+4*x^2+4*x+1)^(3/2),x)

[Out]

-8*(-9/20*x^3+1/5*x^2-21/40*x-19/80)/(4*x^4+4*x^2+4*x+1)^(1/2)+3/5*(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1)-RootO
f(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,ind
ex=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-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3))/(Roo
tOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3)-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=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)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2))/(R
ootOf(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,inde
x=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(((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),((RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-RootOf(4*_Z^4
+4*_Z^2+4*_Z+1,index=3))*(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)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-RootO
f(4*_Z^4+4*_Z^2+4*_Z+1,index=4)))^(1/2))-9/5*((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=3))*(x-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4))+(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-RootO
f(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-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3))/(RootOf(4*_Z
^4+4*_Z^2+4*_Z+1,index=3)-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=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=2)*RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1)-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=2)*RootOf(4*
_Z^4+4*_Z^2+4*_Z+1,index=4)+RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)^2)/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4)-Root
Of(4*_Z^4+4*_Z^2+4*_Z+1,index=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))*
EllipticF(((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),((RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3))*(
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)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,ind
ex=4)))^(1/2))+(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=3))*EllipticE(((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,inde
x=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=3))*(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)-RootOf(4*_Z^4+4*
_Z^2+4*_Z+1,index=3))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=4)))^(1/2))/(Roo
tOf(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,inde
x=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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x^4+4*x^2+4*x+1)^(3/2),x, algorithm="maxima")

[Out]

integrate((4*x^4 + 4*x^2 + 4*x + 1)^(-3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (4\,x^4+4\,x^2+4\,x+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(4*x + 4*x^2 + 4*x^4 + 1)^(3/2),x)

[Out]

int(1/(4*x + 4*x^2 + 4*x^4 + 1)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (4 x^{4} + 4 x^{2} + 4 x + 1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(4*x**4+4*x**2+4*x+1)**(3/2),x)

[Out]

Integral((4*x**4 + 4*x**2 + 4*x + 1)**(-3/2), x)

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