3.4.40 \(\int \frac {1}{x^2 (3+2 x^2) \sqrt {1+2 x^2+2 x^4}} \, dx\) [340]

Optimal. Leaf size=399 \[ -\frac {\sqrt {1+2 x^2+2 x^4}}{3 x}+\frac {\sqrt {2} x \sqrt {1+2 x^2+2 x^4}}{3 \left (1+\sqrt {2} x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )}{3 \sqrt {15}}-\frac {\sqrt [4]{2} \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{3 \sqrt {1+2 x^2+2 x^4}}+\frac {\left (5-3 \sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{21\ 2^{3/4} \sqrt {1+2 x^2+2 x^4}}+\frac {\left (3+\sqrt {2}\right )^2 \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{126 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}} \]

[Out]

-1/45*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)-1/3*(2*x^4+2*x^2+1)^(1/2)/x+1/3*x*(2*x^4+2*x^2+1)^
(1/2)*2^(1/2)/(1+x^2*2^(1/2))-1/3*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticE(sin(2*
arctan(2^(1/4)*x)),1/2*(2-2^(1/2))^(1/2))*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*2^(1/4)/(2
*x^4+2*x^2+1)^(1/2)+1/42*2^(1/4)*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticF(sin(2*a
rctan(2^(1/4)*x)),1/2*(2-2^(1/2))^(1/2))*(5-3*2^(1/2))*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/
2)/(2*x^4+2*x^2+1)^(1/2)+1/252*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arctan(2^(1/4)*x))*EllipticPi(sin(2*ar
ctan(2^(1/4)*x)),1/2-11/24*2^(1/2),1/2*(2-2^(1/2))^(1/2))*(3+2^(1/2))^2*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^
2*2^(1/2))^2)^(1/2)*2^(3/4)/(2*x^4+2*x^2+1)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1343, 1728, 1209, 1722, 1117, 1720} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )}{3 \sqrt {15}}+\frac {\left (5-3 \sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} F\left (2 \text {ArcTan}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{21\ 2^{3/4} \sqrt {2 x^4+2 x^2+1}}-\frac {\sqrt [4]{2} \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \text {ArcTan}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{3 \sqrt {2 x^4+2 x^2+1}}+\frac {\left (3+\sqrt {2}\right )^2 \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \text {ArcTan}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{126 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}+\frac {\sqrt {2} \sqrt {2 x^4+2 x^2+1} x}{3 \left (\sqrt {2} x^2+1\right )}-\frac {\sqrt {2 x^4+2 x^2+1}}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*Sqrt[1 + 2*x^2 + 2*x^4]/x + (Sqrt[2]*x*Sqrt[1 + 2*x^2 + 2*x^4])/(3*(1 + Sqrt[2]*x^2)) - ArcTan[(Sqrt[5/3]
*x)/Sqrt[1 + 2*x^2 + 2*x^4]]/(3*Sqrt[15]) - (2^(1/4)*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x
^2)^2]*EllipticE[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(3*Sqrt[1 + 2*x^2 + 2*x^4]) + ((5 - 3*Sqrt[2])*(1 + Sq
rt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(21*
2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + ((3 + Sqrt[2])^2*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^
2)^2]*EllipticPi[(12 - 11*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(126*2^(1/4)*Sqrt[1 + 2*x^2 + 2*
x^4])

Rule 1117

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]

Rule 1209

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)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c))], 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 1343

Int[(x_)^(m_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Simp[x^(m + 1)*(S
qrt[a + b*x^2 + c*x^4]/(a*d*(m + 1))), x] - Dist[1/(a*d*(m + 1)), Int[(x^(m + 2)/((d + e*x^2)*Sqrt[a + b*x^2 +
c*x^4]))*Simp[a*e*(m + 1) + b*d*(m + 2) + (b*e*(m + 2) + c*d*(m + 3))*x^2 + c*e*(m + 3)*x^4, x], x], x] /; Fr
eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && ILtQ[m/2, 0]

Rule 1720

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

Rule 1722

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

Rule 1728

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]
, A = Coeff[P4x, x, 0], B = Coeff[P4x, x, 2], C = Coeff[P4x, x, 4]}, Dist[-C/(e*q), Int[(1 - q*x^2)/Sqrt[a + b
*x^2 + c*x^4], x], x] + Dist[1/(c*e), Int[(A*c*e + a*C*d*q + (B*c*e - C*(c*d - a*e*q))*x^2)/((d + e*x^2)*Sqrt[
a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[P4x, x^2, 2] && NeQ[b^2 - 4*a*c, 0] && NeQ[
c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] &&  !GtQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx &=-\frac {\sqrt {1+2 x^2+2 x^4}}{3 x}+\frac {1}{3} \int \frac {-2+6 x^2+4 x^4}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {\sqrt {1+2 x^2+2 x^4}}{3 x}+\frac {1}{12} \int \frac {-8+12 \sqrt {2}+\left (24-4 \left (6-2 \sqrt {2}\right )\right ) x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx-\frac {1}{3} \sqrt {2} \int \frac {1-\sqrt {2} x^2}{\sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {\sqrt {1+2 x^2+2 x^4}}{3 x}+\frac {\sqrt {2} x \sqrt {1+2 x^2+2 x^4}}{3 \left (1+\sqrt {2} x^2\right )}-\frac {\sqrt [4]{2} \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{3 \sqrt {1+2 x^2+2 x^4}}+\frac {1}{21} \left (2 \left (2+3 \sqrt {2}\right )\right ) \int \frac {1+\sqrt {2} x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx+\frac {1}{21} \left (-6+5 \sqrt {2}\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx\\ &=-\frac {\sqrt {1+2 x^2+2 x^4}}{3 x}+\frac {\sqrt {2} x \sqrt {1+2 x^2+2 x^4}}{3 \left (1+\sqrt {2} x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )}{3 \sqrt {15}}-\frac {\sqrt [4]{2} \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{3 \sqrt {1+2 x^2+2 x^4}}+\frac {\left (5-3 \sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{21\ 2^{3/4} \sqrt {1+2 x^2+2 x^4}}+\frac {\left (3+\sqrt {2}\right )^2 \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{126 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.16, size = 147, normalized size = 0.37 \begin {gather*} -\frac {i \left (-3 i \left (1+2 x^2+2 x^4\right )+\sqrt {1-i} x \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} \left (3 E\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )-3 F\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )-(1+i) \Pi \left (\frac {1}{3}+\frac {i}{3};\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )\right )\right )}{9 x \sqrt {1+2 x^2+2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/9*I)*((-3*I)*(1 + 2*x^2 + 2*x^4) + Sqrt[1 - I]*x*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*(3*EllipticE
[I*ArcSinh[Sqrt[1 - I]*x], I] - 3*EllipticF[I*ArcSinh[Sqrt[1 - I]*x], I] - (1 + I)*EllipticPi[1/3 + I/3, I*Arc
Sinh[Sqrt[1 - I]*x], I])))/(x*Sqrt[1 + 2*x^2 + 2*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.14, size = 178, normalized size = 0.45

method	result	size
default	\(-\frac {\sqrt {2 x^{4}+2 x^{2}+1}}{3 x}+\frac {\left (-\frac {1}{3}+\frac {i}{3}\right ) \sqrt {1+\left (1-i\right ) x^{2}}\, \sqrt {1+\left (1+i\right ) x^{2}}\, \left (\EllipticF \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )-\EllipticE \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {2 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{9 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(178\)
risch	\(-\frac {\sqrt {2 x^{4}+2 x^{2}+1}}{3 x}+\frac {\left (-\frac {1}{3}+\frac {i}{3}\right ) \sqrt {1+\left (1-i\right ) x^{2}}\, \sqrt {1+\left (1+i\right ) x^{2}}\, \left (\EllipticF \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )-\EllipticE \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {2 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{9 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(178\)
elliptic	\(-\frac {\sqrt {2 x^{4}+2 x^{2}+1}}{3 x}-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {i \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticE \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {2 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{9 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(360\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*x^4+2*x^2+1)^(1/2)/x+(-1/3+1/3*I)/(-1+I)^(1/2)*(1+(1-I)*x^2)^(1/2)*(1+(1+I)*x^2)^(1/2)/(2*x^4+2*x^2+1)
^(1/2)*(EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-EllipticE(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2)
))-2/9/(-1+I)^(1/2)*(1+x^2-I*x^2)^(1/2)*(1+x^2+I*x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticPi(x*(-1+I)^(1/2),1/
3+1/3*I,(-1-I)^(1/2)/(-1+I)^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(2*x^4 + 2*x^2 + 1)/(4*x^8 + 10*x^6 + 8*x^4 + 3*x^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \cdot \left (2 x^{2} + 3\right ) \sqrt {2 x^{4} + 2 x^{2} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^2\,\left (2\,x^2+3\right )\,\sqrt {2\,x^4+2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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