\(\int \frac {(1+2 x^2)^2 (4-7 x^2+x^4)}{\sqrt {2+5 x^2+3 x^4}} \, dx\) [205]

Optimal result

Integrand size = 36, antiderivative size = 206 \[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\sqrt {2+5 x^2+3 x^4}} \, dx=-\frac {2771 x \left (2+3 x^2\right )}{945 \sqrt {2+5 x^2+3 x^4}}+\frac {187}{63} x \sqrt {2+5 x^2+3 x^4}-\frac {208}{105} x^3 \sqrt {2+5 x^2+3 x^4}+\frac {4}{21} x^5 \sqrt {2+5 x^2+3 x^4}+\frac {2771 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{945 \sqrt {2+5 x^2+3 x^4}}-\frac {61 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{63 \sqrt {2+5 x^2+3 x^4}} \] Output:

-2771/945*x*(3*x^2+2)/(3*x^4+5*x^2+2)^(1/2)+187/63*x*(3*x^4+5*x^2+2)^(1/2) 
-208/105*x^3*(3*x^4+5*x^2+2)^(1/2)+4/21*x^5*(3*x^4+5*x^2+2)^(1/2)+2771/945 
*2^(1/2)*(x^2+1)*((3*x^2+2)/(x^2+1))^(1/2)*EllipticE(x/(x^2+1)^(1/2),1/2*I 
*2^(1/2))/(3*x^4+5*x^2+2)^(1/2)-61/63*2^(1/2)*(x^2+1)*((3*x^2+2)/(x^2+1))^ 
(1/2)*InverseJacobiAM(arctan(x),1/2*I*2^(1/2))/(3*x^4+5*x^2+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.67 \[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\sqrt {2+5 x^2+3 x^4}} \, dx=\frac {5610 x+10281 x^3-585 x^5-4716 x^7+540 x^9+2771 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )-2161 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{945 \sqrt {2+5 x^2+3 x^4}} \] Input:

Integrate[((1 + 2*x^2)^2*(4 - 7*x^2 + x^4))/Sqrt[2 + 5*x^2 + 3*x^4],x]
 

Output:

(5610*x + 10281*x^3 - 585*x^5 - 4716*x^7 + 540*x^9 + (2771*I)*Sqrt[3]*Sqrt 
[1 + x^2]*Sqrt[2 + 3*x^2]*EllipticE[I*ArcSinh[Sqrt[3/2]*x], 2/3] - (2161*I 
)*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*EllipticF[I*ArcSinh[Sqrt[3/2]*x], 
2/3])/(945*Sqrt[2 + 5*x^2 + 3*x^4])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2207, 2207, 27, 2207, 25, 1503, 1413, 1456}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (2 x^2+1\right )^2 \left (x^4-7 x^2+4\right )}{\sqrt {3 x^4+5 x^2+2}} \, dx\)

\(\Big \downarrow \) 2207

\(\displaystyle \frac {1}{21} \int \frac {-624 x^6-271 x^4+189 x^2+84}{\sqrt {3 x^4+5 x^2+2}}dx+\frac {4}{21} \sqrt {3 x^4+5 x^2+2} x^5\)

\(\Big \downarrow \) 2207

\(\displaystyle \frac {1}{21} \left (\frac {1}{15} \int \frac {9 \left (935 x^4+731 x^2+140\right )}{\sqrt {3 x^4+5 x^2+2}}dx-\frac {208}{5} x^3 \sqrt {3 x^4+5 x^2+2}\right )+\frac {4}{21} \sqrt {3 x^4+5 x^2+2} x^5\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{21} \left (\frac {3}{5} \int \frac {935 x^4+731 x^2+140}{\sqrt {3 x^4+5 x^2+2}}dx-\frac {208}{5} x^3 \sqrt {3 x^4+5 x^2+2}\right )+\frac {4}{21} \sqrt {3 x^4+5 x^2+2} x^5\)

\(\Big \downarrow \) 2207

\(\displaystyle \frac {1}{21} \left (\frac {3}{5} \left (\frac {1}{9} \int -\frac {2771 x^2+610}{\sqrt {3 x^4+5 x^2+2}}dx+\frac {935}{9} \sqrt {3 x^4+5 x^2+2} x\right )-\frac {208}{5} x^3 \sqrt {3 x^4+5 x^2+2}\right )+\frac {4}{21} \sqrt {3 x^4+5 x^2+2} x^5\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{21} \left (\frac {3}{5} \left (\frac {935}{9} x \sqrt {3 x^4+5 x^2+2}-\frac {1}{9} \int \frac {2771 x^2+610}{\sqrt {3 x^4+5 x^2+2}}dx\right )-\frac {208}{5} x^3 \sqrt {3 x^4+5 x^2+2}\right )+\frac {4}{21} \sqrt {3 x^4+5 x^2+2} x^5\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{21} \left (\frac {3}{5} \left (\frac {1}{9} \left (-610 \int \frac {1}{\sqrt {3 x^4+5 x^2+2}}dx-2771 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx\right )+\frac {935}{9} \sqrt {3 x^4+5 x^2+2} x\right )-\frac {208}{5} x^3 \sqrt {3 x^4+5 x^2+2}\right )+\frac {4}{21} \sqrt {3 x^4+5 x^2+2} x^5\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {1}{21} \left (\frac {3}{5} \left (\frac {1}{9} \left (-2771 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx-\frac {305 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {3 x^4+5 x^2+2}}\right )+\frac {935}{9} \sqrt {3 x^4+5 x^2+2} x\right )-\frac {208}{5} x^3 \sqrt {3 x^4+5 x^2+2}\right )+\frac {4}{21} \sqrt {3 x^4+5 x^2+2} x^5\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {1}{21} \left (\frac {3}{5} \left (\frac {1}{9} \left (-\frac {305 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {3 x^4+5 x^2+2}}-2771 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )\right )+\frac {935}{9} \sqrt {3 x^4+5 x^2+2} x\right )-\frac {208}{5} x^3 \sqrt {3 x^4+5 x^2+2}\right )+\frac {4}{21} \sqrt {3 x^4+5 x^2+2} x^5\)

Input:

Int[((1 + 2*x^2)^2*(4 - 7*x^2 + x^4))/Sqrt[2 + 5*x^2 + 3*x^4],x]
 

Output:

(4*x^5*Sqrt[2 + 5*x^2 + 3*x^4])/21 + ((-208*x^3*Sqrt[2 + 5*x^2 + 3*x^4])/5 
 + (3*((935*x*Sqrt[2 + 5*x^2 + 3*x^4])/9 + (-2771*((x*(2 + 3*x^2))/(3*Sqrt 
[2 + 5*x^2 + 3*x^4]) - (Sqrt[2]*(1 + x^2)*Sqrt[(2 + 3*x^2)/(1 + x^2)]*Elli 
pticE[ArcTan[x], -1/2])/(3*Sqrt[2 + 5*x^2 + 3*x^4])) - (305*Sqrt[2]*(1 + x 
^2)*Sqrt[(2 + 3*x^2)/(1 + x^2)]*EllipticF[ArcTan[x], -1/2])/Sqrt[2 + 5*x^2 
 + 3*x^4])/9))/5)/21
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1413

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]))*EllipticF 
[ArcTan[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; 
 FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1456

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]] /; FreeQ[ 
{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1503

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d   Int[1/Sqrt[a + b*x^2 + c*x^4] 
, x], x] + Simp[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]
 

rule 2207

Int[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{n = 
 Expon[Px, x^2], e = Coeff[Px, x^2, Expon[Px, x^2]]}, Simp[e*x^(2*n - 3)*(( 
a + b*x^2 + c*x^4)^(p + 1)/(c*(2*n + 4*p + 1))), x] + Simp[1/(c*(2*n + 4*p 
+ 1))   Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*n + 4*p + 1)*Px - a*e*(2 
*n - 3)*x^(2*n - 4) - b*e*(2*n + 2*p - 1)*x^(2*n - 2) - c*e*(2*n + 4*p + 1) 
*x^(2*n), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Px, x^2] && Expon[ 
Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]
 

Maple [A] (verified)

Time = 11.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.63

method	result	size
risch	\(\frac {x \left (60 x^{4}-624 x^{2}+935\right ) \sqrt {3 x^{4}+5 x^{2}+2}}{315}+\frac {61 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{63 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {2771 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{945 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(130\)
default	\(\frac {61 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{63 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {2771 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{945 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {187 x \sqrt {3 x^{4}+5 x^{2}+2}}{63}-\frac {208 x^{3} \sqrt {3 x^{4}+5 x^{2}+2}}{105}+\frac {4 x^{5} \sqrt {3 x^{4}+5 x^{2}+2}}{21}\)	\(156\)
elliptic	\(\frac {61 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{63 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {2771 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{945 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {187 x \sqrt {3 x^{4}+5 x^{2}+2}}{63}-\frac {208 x^{3} \sqrt {3 x^{4}+5 x^{2}+2}}{105}+\frac {4 x^{5} \sqrt {3 x^{4}+5 x^{2}+2}}{21}\)	\(156\)

Input:

int((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(1/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/315*x*(60*x^4-624*x^2+935)*(3*x^4+5*x^2+2)^(1/2)+61/63*I*(x^2+1)^(1/2)*( 
6*x^2+4)^(1/2)/(3*x^4+5*x^2+2)^(1/2)*EllipticF(I*x,1/2*6^(1/2))-2771/945*I 
*(x^2+1)^(1/2)*(6*x^2+4)^(1/2)/(3*x^4+5*x^2+2)^(1/2)*(EllipticF(I*x,1/2*6^ 
(1/2))-EllipticE(I*x,1/2*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.37 \[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\sqrt {2+5 x^2+3 x^4}} \, dx=\frac {5542 \, \sqrt {3} \sqrt {-\frac {2}{3}} x E(\arcsin \left (\frac {\sqrt {-\frac {2}{3}}}{x}\right )\,|\,\frac {3}{2}) - 8287 \, \sqrt {3} \sqrt {-\frac {2}{3}} x F(\arcsin \left (\frac {\sqrt {-\frac {2}{3}}}{x}\right )\,|\,\frac {3}{2}) + 3 \, {\left (180 \, x^{6} - 1872 \, x^{4} + 2805 \, x^{2} - 2771\right )} \sqrt {3 \, x^{4} + 5 \, x^{2} + 2}}{2835 \, x} \] Input:

integrate((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(1/2),x, algorithm="fr 
icas")
 

Output:

1/2835*(5542*sqrt(3)*sqrt(-2/3)*x*elliptic_e(arcsin(sqrt(-2/3)/x), 3/2) - 
8287*sqrt(3)*sqrt(-2/3)*x*elliptic_f(arcsin(sqrt(-2/3)/x), 3/2) + 3*(180*x 
^6 - 1872*x^4 + 2805*x^2 - 2771)*sqrt(3*x^4 + 5*x^2 + 2))/x
 

Sympy [F]

\[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\sqrt {2+5 x^2+3 x^4}} \, dx=\int \frac {\left (2 x^{2} + 1\right )^{2} \left (x^{4} - 7 x^{2} + 4\right )}{\sqrt {\left (x^{2} + 1\right ) \left (3 x^{2} + 2\right )}}\, dx \] Input:

integrate((2*x**2+1)**2*(x**4-7*x**2+4)/(3*x**4+5*x**2+2)**(1/2),x)
 

Output:

Integral((2*x**2 + 1)**2*(x**4 - 7*x**2 + 4)/sqrt((x**2 + 1)*(3*x**2 + 2)) 
, x)
 

Maxima [F]

\[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\sqrt {2+5 x^2+3 x^4}} \, dx=\int { \frac {{\left (x^{4} - 7 \, x^{2} + 4\right )} {\left (2 \, x^{2} + 1\right )}^{2}}{\sqrt {3 \, x^{4} + 5 \, x^{2} + 2}} \,d x } \] Input:

integrate((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(1/2),x, algorithm="ma 
xima")
 

Output:

integrate((x^4 - 7*x^2 + 4)*(2*x^2 + 1)^2/sqrt(3*x^4 + 5*x^2 + 2), x)
 

Giac [F]

\[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\sqrt {2+5 x^2+3 x^4}} \, dx=\int { \frac {{\left (x^{4} - 7 \, x^{2} + 4\right )} {\left (2 \, x^{2} + 1\right )}^{2}}{\sqrt {3 \, x^{4} + 5 \, x^{2} + 2}} \,d x } \] Input:

integrate((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(1/2),x, algorithm="gi 
ac")
 

Output:

integrate((x^4 - 7*x^2 + 4)*(2*x^2 + 1)^2/sqrt(3*x^4 + 5*x^2 + 2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\sqrt {2+5 x^2+3 x^4}} \, dx=\int \frac {{\left (2\,x^2+1\right )}^2\,\left (x^4-7\,x^2+4\right )}{\sqrt {3\,x^4+5\,x^2+2}} \,d x \] Input:

int(((2*x^2 + 1)^2*(x^4 - 7*x^2 + 4))/(5*x^2 + 3*x^4 + 2)^(1/2),x)
 

Output:

int(((2*x^2 + 1)^2*(x^4 - 7*x^2 + 4))/(5*x^2 + 3*x^4 + 2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right )}{\sqrt {2+5 x^2+3 x^4}} \, dx=\frac {4 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{5}}{21}-\frac {208 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{3}}{105}+\frac {187 \sqrt {3 x^{4}+5 x^{2}+2}\, x}{63}-\frac {122 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{3 x^{4}+5 x^{2}+2}d x \right )}{63}-\frac {2771 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{3 x^{4}+5 x^{2}+2}d x \right )}{315} \] Input:

int((2*x^2+1)^2*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(1/2),x)
 

Output:

(60*sqrt(3*x**4 + 5*x**2 + 2)*x**5 - 624*sqrt(3*x**4 + 5*x**2 + 2)*x**3 + 
935*sqrt(3*x**4 + 5*x**2 + 2)*x - 610*int(sqrt(3*x**4 + 5*x**2 + 2)/(3*x** 
4 + 5*x**2 + 2),x) - 2771*int((sqrt(3*x**4 + 5*x**2 + 2)*x**2)/(3*x**4 + 5 
*x**2 + 2),x))/315