3.4.38 \(\int \frac {x^2}{(3+2 x^2) \sqrt {1+2 x^2+2 x^4}} \, dx\) [338]

Optimal. Leaf size=247 \[ -\frac {1}{4} \sqrt {\frac {3}{5}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )-\frac {\left (3+\sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{14\ 2^{3/4} \sqrt {1+2 x^2+2 x^4}}+\frac {\left (3+\sqrt {2}\right )^2 \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \Pi \left (\frac {1}{24} \left (12-11 \sqrt {2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{56 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}} \]

[Out]

-1/20*arctan(1/3*x*15^(1/2)/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)-1/28*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*arct
an(2^(1/4)*x))*EllipticF(sin(2*arctan(2^(1/4)*x)),1/2*(2-2^(1/2))^(1/2))*(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^(1/4)/(2*x^4+2*x^2+1)^(1/2)+1/112*(cos(2*arctan(2^(1/4)*x))^2)^(1/2)/cos(2*
arctan(2^(1/4)*x))*EllipticPi(sin(2*arctan(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 = 0.08, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1333, 1117, 1720} \begin {gather*} -\frac {1}{4} \sqrt {\frac {3}{5}} \text {ArcTan}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )-\frac {\left (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 )}{14\ 2^{3/4} \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 )}{56 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[3/5]*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]]) - ((3 + Sqrt[2])*(1 + Sqrt[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])/(14*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])/(56*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 1333

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

Rubi steps

\begin {align*} \int \frac {x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx &=-\left (\frac {1}{14} \left (2+3 \sqrt {2}\right ) \int \frac {1}{\sqrt {1+2 x^2+2 x^4}} \, dx\right )+\frac {1}{14} \left (3 \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}{4} \sqrt {\frac {3}{5}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )-\frac {\left (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 )}{14\ 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 )}{56 \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 = 10.11, size = 99, normalized size = 0.40 \begin {gather*} \frac {(1-i)^{3/2} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} \left (F\left (\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )-\Pi \left (\frac {1}{3}+\frac {i}{3};\left .i \sinh ^{-1}\left (\sqrt {1-i} x\right )\right |i\right )\right )}{4 \sqrt {1+2 x^2+2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - I)^(3/2)*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*(EllipticF[I*ArcSinh[Sqrt[1 - I]*x], I] - EllipticP
i[1/3 + I/3, I*ArcSinh[Sqrt[1 - I]*x], I]))/(4*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 = 134, normalized size = 0.54

method	result	size
default	\(\frac {\sqrt {1+\left (1-i\right ) x^{2}}\, \sqrt {1+\left (1+i\right ) x^{2}}\, \EllipticF \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(134\)
elliptic	\(\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticF \left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \EllipticPi \left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{2 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

[Out]

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

[Out]

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

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

[Out]

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

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