\(\int (2+3 x^2+x^4)^{3/2} \, dx\) [296]

Optimal result

Integrand size = 14, antiderivative size = 172 \[ \int \left (2+3 x^2+x^4\right )^{3/2} \, dx=\frac {6 x \left (2+x^2\right )}{5 \sqrt {2+3 x^2+x^4}}+\frac {1}{35} x \left (29+9 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {1}{7} x \left (2+3 x^2+x^4\right )^{3/2}-\frac {6 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{5 \sqrt {2+3 x^2+x^4}}+\frac {31 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{35 \sqrt {2+3 x^2+x^4}} \]

[Out]

1/7*x*(x^4+3*x^2+2)^(3/2)+6/5*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-6/5*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticE(x/(x
^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+31/35*(x^2+1)^(3/2)*(1/(x^2+1))^(
1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+1/35*x*(9*x^2+
29)*(x^4+3*x^2+2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1105, 1190, 1203, 1113, 1149} \[ \int \left (2+3 x^2+x^4\right )^{3/2} \, dx=\frac {31 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{35 \sqrt {x^4+3 x^2+2}}-\frac {6 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{5 \sqrt {x^4+3 x^2+2}}+\frac {1}{7} x \left (x^4+3 x^2+2\right )^{3/2}+\frac {1}{35} x \left (9 x^2+29\right ) \sqrt {x^4+3 x^2+2}+\frac {6 x \left (x^2+2\right )}{5 \sqrt {x^4+3 x^2+2}} \]

[In]

Int[(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

(6*x*(2 + x^2))/(5*Sqrt[2 + 3*x^2 + x^4]) + (x*(29 + 9*x^2)*Sqrt[2 + 3*x^2 + x^4])/35 + (x*(2 + 3*x^2 + x^4)^(
3/2))/7 - (6*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(5*Sqrt[2 + 3*x^2 + x^4])
+ (31*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(35*Sqrt[2 + 3*x^2 + x^4])

Rule 1105

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*((a + b*x^2 + c*x^4)^p/(4*p + 1)), x] + Dis
t[2*(p/(4*p + 1)), Int[(2*a + b*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4
*a*c, 0] && GtQ[p, 0] && IntegerQ[2*p]

Rule 1113

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(2*a + (b + q
)*x^2)*(Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*Elli
pticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1149

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[x*((b +
q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4])), x] - Simp[Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*(Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q
/(b + q))], x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1190

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*(2*b*e*p + c*d*(4*p +
3) + c*e*(4*p + 1)*x^2)*((a + b*x^2 + c*x^4)^p/(c*(4*p + 1)*(4*p + 3))), x] + Dist[2*(p/(c*(4*p + 1)*(4*p + 3
))), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*x^2, x]*(a +
b*x^2 + c*x^4)^(p - 1), 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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

Rule 1203

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{7} x \left (2+3 x^2+x^4\right )^{3/2}+\frac {3}{7} \int \left (4+3 x^2\right ) \sqrt {2+3 x^2+x^4} \, dx \\ & = \frac {1}{35} x \left (29+9 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {1}{7} x \left (2+3 x^2+x^4\right )^{3/2}+\frac {1}{35} \int \frac {62+42 x^2}{\sqrt {2+3 x^2+x^4}} \, dx \\ & = \frac {1}{35} x \left (29+9 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {1}{7} x \left (2+3 x^2+x^4\right )^{3/2}+\frac {6}{5} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {62}{35} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx \\ & = \frac {6 x \left (2+x^2\right )}{5 \sqrt {2+3 x^2+x^4}}+\frac {1}{35} x \left (29+9 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {1}{7} x \left (2+3 x^2+x^4\right )^{3/2}-\frac {6 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {2+3 x^2+x^4}}+\frac {31 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {2+3 x^2+x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.66 \[ \int \left (2+3 x^2+x^4\right )^{3/2} \, dx=\frac {78 x+165 x^3+121 x^5+39 x^7+5 x^9-42 i \sqrt {1+x^2} \sqrt {2+x^2} E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-20 i \sqrt {1+x^2} \sqrt {2+x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{35 \sqrt {2+3 x^2+x^4}} \]

[In]

Integrate[(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

(78*x + 165*x^3 + 121*x^5 + 39*x^7 + 5*x^9 - (42*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]]
, 2] - (20*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2])/(35*Sqrt[2 + 3*x^2 + x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.68 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {x \left (5 x^{4}+24 x^{2}+39\right ) \sqrt {x^{4}+3 x^{2}+2}}{35}-\frac {31 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{35 \sqrt {x^{4}+3 x^{2}+2}}+\frac {3 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{5 \sqrt {x^{4}+3 x^{2}+2}}\)	\(133\)
default	\(\frac {x^{5} \sqrt {x^{4}+3 x^{2}+2}}{7}+\frac {24 x^{3} \sqrt {x^{4}+3 x^{2}+2}}{35}+\frac {39 x \sqrt {x^{4}+3 x^{2}+2}}{35}-\frac {31 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{35 \sqrt {x^{4}+3 x^{2}+2}}+\frac {3 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{5 \sqrt {x^{4}+3 x^{2}+2}}\)	\(155\)
elliptic	\(\frac {x^{5} \sqrt {x^{4}+3 x^{2}+2}}{7}+\frac {24 x^{3} \sqrt {x^{4}+3 x^{2}+2}}{35}+\frac {39 x \sqrt {x^{4}+3 x^{2}+2}}{35}-\frac {31 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{35 \sqrt {x^{4}+3 x^{2}+2}}+\frac {3 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{5 \sqrt {x^{4}+3 x^{2}+2}}\)	\(155\)

[In]

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

[Out]

1/35*x*(5*x^4+24*x^2+39)*(x^4+3*x^2+2)^(1/2)-31/35*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)
*EllipticF(1/2*I*2^(1/2)*x,2^(1/2))+3/5*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*(EllipticF
(1/2*I*2^(1/2)*x,2^(1/2))-EllipticE(1/2*I*2^(1/2)*x,2^(1/2)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.34 \[ \int \left (2+3 x^2+x^4\right )^{3/2} \, dx=\frac {-42 i \, x E(\arcsin \left (\frac {i}{x}\right )\,|\,2) + 104 i \, x F(\arcsin \left (\frac {i}{x}\right )\,|\,2) + {\left (5 \, x^{6} + 24 \, x^{4} + 39 \, x^{2} + 42\right )} \sqrt {x^{4} + 3 \, x^{2} + 2}}{35 \, x} \]

[In]

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

[Out]

1/35*(-42*I*x*elliptic_e(arcsin(I/x), 2) + 104*I*x*elliptic_f(arcsin(I/x), 2) + (5*x^6 + 24*x^4 + 39*x^2 + 42)
*sqrt(x^4 + 3*x^2 + 2))/x

Sympy [F]

\[ \int \left (2+3 x^2+x^4\right )^{3/2} \, dx=\int \left (x^{4} + 3 x^{2} + 2\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \left (2+3 x^2+x^4\right )^{3/2} \, dx=\int { {\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

integrate((x^4 + 3*x^2 + 2)^(3/2), x)

Giac [F]

\[ \int \left (2+3 x^2+x^4\right )^{3/2} \, dx=\int { {\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

integrate((x^4 + 3*x^2 + 2)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \left (2+3 x^2+x^4\right )^{3/2} \, dx=\int {\left (x^4+3\,x^2+2\right )}^{3/2} \,d x \]

[In]

int((3*x^2 + x^4 + 2)^(3/2),x)

[Out]

int((3*x^2 + x^4 + 2)^(3/2), x)