3.798 \(\int \frac{1}{\sqrt{1+4 x+4 x^2+4 x^4}} \, dx\)

Optimal. Leaf size=108 \[ -\frac{\left (\left (\frac{1}{x}+1\right )^2+\sqrt{5}\right ) \sqrt{\frac{\left (\frac{1}{x}+1\right )^4-2 \left (\frac{1}{x}+1\right )^2+5}{\left (\left (\frac{1}{x}+1\right )^2+\sqrt{5}\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac{1+\frac{1}{x}}{\sqrt [4]{5}}\right )|\frac{1}{10} \left (5+\sqrt{5}\right )\right )}{2 \sqrt [4]{5} \sqrt{4 x^4+4 x^2+4 x+1}} \]

[Out]

-((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*El
lipticF[2*ArcTan[(1 + x^(-1))/5^(1/4)], (5 + Sqrt[5])/10])/(2*5^(1/4)*Sqrt[1 + 4*x + 4*x^2 + 4*x^4])

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Rubi [A]  time = 0.216657, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2069, 6719, 1103} \[ -\frac{\left (\left (\frac{1}{x}+1\right )^2+\sqrt{5}\right ) \sqrt{\frac{\left (\frac{1}{x}+1\right )^4-2 \left (\frac{1}{x}+1\right )^2+5}{\left (\left (\frac{1}{x}+1\right )^2+\sqrt{5}\right )^2}} x^2 F\left (2 \tan ^{-1}\left (\frac{1+\frac{1}{x}}{\sqrt [4]{5}}\right )|\frac{1}{10} \left (5+\sqrt{5}\right )\right )}{2 \sqrt [4]{5} \sqrt{4 x^4+4 x^2+4 x+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + 4*x + 4*x^2 + 4*x^4],x]

[Out]

-((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*El
lipticF[2*ArcTan[(1 + x^(-1))/5^(1/4)], (5 + Sqrt[5])/10])/(2*5^(1/4)*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 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]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{1+4 x+4 x^2+4 x^4}} \, dx &=-\left (16 \operatorname{Subst}\left (\int \frac{1}{(4-4 x)^2 \sqrt{\frac{1280-512 x^2+256 x^4}{(4-4 x)^4}}} \, 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{1}{\sqrt{1280-512 x^2+256 x^4}} \, dx,x,1+\frac{1}{x}\right )}{\sqrt{1+4 x+4 x^2+4 x^4}}\\ &=-\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 F\left (2 \tan ^{-1}\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}}\\ \end{align*}

Mathematica [C]  time = 0.52478, size = 249, normalized size = 2.31 \[ \frac{(2-i) \sqrt{-\frac{1}{10}+\frac{i}{5}} \sqrt{\frac{\left (2 i+\sqrt{-1-2 i}-\sqrt{-1+2 i}\right ) \left (-2 x+\sqrt{-1-2 i}-i\right )}{\left (-2 i+\sqrt{-1-2 i}+\sqrt{-1+2 i}\right ) \left (2 x+\sqrt{-1-2 i}+i\right )}} \left (2 i x^2+2 x+1\right ) F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\left (2 i+\sqrt{-1-2 i}+\sqrt{-1+2 i}\right ) \left (2 x+\sqrt{-1+2 i}-i\right )}{\sqrt{-1+2 i} \left (2 x+\sqrt{-1-2 i}+i\right )}}}{\sqrt{2}}\right )|\frac{1}{2} \left (5-\sqrt{5}\right )\right )}{\sqrt{\frac{(1+2 i) \left ((-1+i)+\sqrt{-1-2 i}\right ) \left (2 i x^2+2 x+1\right )}{\left (2 x+\sqrt{-1-2 i}+i\right )^2}} \sqrt{4 x^4+4 x^2+4 x+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/Sqrt[1 + 4*x + 4*x^2 + 4*x^4],x]

[Out]

((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])

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Maple [C]  time = 0.654, size = 961, normalized size = 8.9 \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)^(1/2),x)

[Out]

(-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,ind
ex=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-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,inde
x=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,in
dex=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(((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)+Ro
otOf(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{4 \, x^{4} + 4 \, x^{2} + 4 \, 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)^(1/2),x, algorithm="fricas")

[Out]

integral(1/sqrt(4*x^4 + 4*x^2 + 4*x + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{4 x^{4} + 4 x^{2} + 4 x + 1}}\, 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)**(1/2),x)

[Out]

Integral(1/sqrt(4*x**4 + 4*x**2 + 4*x + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{4 \, x^{4} + 4 \, x^{2} + 4 \, x + 1}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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