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 + 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])
________________________________________________________________________________________
Rubi [A] time = 0.376027, antiderivative size = 367, normalized size of antiderivative = 1.,
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 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]
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 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 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 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 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]
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*}
Mathematica [C] time = 4.31841, 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])
________________________________________________________________________________________
Maple [C] time = 0.016, size = 2564, normalized size = 7. \begin{align*} \text{result too large to display} \end{align*}
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=4)+Root
Of(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))
*(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,in
dex=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))/(Ro
otOf(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))/(
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,ind
ex=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=4)+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=3))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=2)-Roo
tOf(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=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=2))*(x-Ro
otOf(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)-R
ootOf(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=4)+RootOf(4*_Z^4+4*_Z^2+4*_Z+1,index=1))/(RootOf(4*_Z^4+4*_Z^2+4*_Z+1,in
dex=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))+(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(((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,
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=4)+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=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))
/(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=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)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}
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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (4 x^{4} + 4 x^{2} + 4 x + 1\right )^{\frac{3}{2}}}\, dx \end{align*}
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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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
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)