\(\int \frac {4-7 x^2+x^4}{(1+2 x^2) \sqrt {2+5 x^2+3 x^4}} \, dx\) [208]

Optimal result

Integrand size = 36, antiderivative size = 201 \[ \int \frac {4-7 x^2+x^4}{\left (1+2 x^2\right ) \sqrt {2+5 x^2+3 x^4}} \, dx=\frac {x \left (2+3 x^2\right )}{6 \sqrt {2+5 x^2+3 x^4}}-\frac {\left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {2} \sqrt {2+5 x^2+3 x^4}}-\frac {27 \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+5 x^2+3 x^4}}+\frac {31 \left (1+x^2\right ) \operatorname {EllipticPi}\left (-\frac {1}{3},\arctan \left (\sqrt {\frac {3}{2}} x\right ),\frac {1}{3}\right )}{\sqrt {3} \sqrt {\frac {1+x^2}{2+3 x^2}} \sqrt {2+5 x^2+3 x^4}} \] Output:

1/6*x*(3*x^2+2)/(3*x^4+5*x^2+2)^(1/2)-1/6*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)-2 
7/2*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)+31/3*(x^2+1)*EllipticPi(x*6^(1/2)/(6*x^ 
2+4)^(1/2),-1/3,1/3*3^(1/2))*3^(1/2)/((x^2+1)/(3*x^2+2))^(1/2)/(3*x^4+5*x^ 
2+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.53 \[ \int \frac {4-7 x^2+x^4}{\left (1+2 x^2\right ) \sqrt {2+5 x^2+3 x^4}} \, dx=-\frac {i \sqrt {1+x^2} \sqrt {2+3 x^2} \left (2 E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )-17 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )+31 \operatorname {EllipticPi}\left (\frac {4}{3},i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )\right )}{4 \sqrt {6+15 x^2+9 x^4}} \] Input:

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

Output:

((-1/4*I)*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*(2*EllipticE[I*ArcSinh[Sqrt[3/2]*x 
], 2/3] - 17*EllipticF[I*ArcSinh[Sqrt[3/2]*x], 2/3] + 31*EllipticPi[4/3, I 
*ArcSinh[Sqrt[3/2]*x], 2/3]))/Sqrt[6 + 15*x^2 + 9*x^4]
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.33, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2234, 1503, 1413, 1456, 1538, 27, 1413, 1786, 414}

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.

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

\(\Big \downarrow \) 2234

\(\displaystyle \frac {31}{4} \int \frac {1}{\left (2 x^2+1\right ) \sqrt {3 x^4+5 x^2+2}}dx-\frac {1}{4} \int \frac {15-2 x^2}{\sqrt {3 x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{4} \left (2 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx-15 \int \frac {1}{\sqrt {3 x^4+5 x^2+2}}dx\right )+\frac {31}{4} \int \frac {1}{\left (2 x^2+1\right ) \sqrt {3 x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {1}{4} \left (2 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx-\frac {15 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )+\frac {31}{4} \int \frac {1}{\left (2 x^2+1\right ) \sqrt {3 x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {31}{4} \int \frac {1}{\left (2 x^2+1\right ) \sqrt {3 x^4+5 x^2+2}}dx+\frac {1}{4} \left (2 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )-\frac {15 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )\)

\(\Big \downarrow \) 1538

\(\displaystyle \frac {31}{4} \left (\int \frac {2 \left (3 x^2+2\right )}{\left (2 x^2+1\right ) \sqrt {3 x^4+5 x^2+2}}dx-3 \int \frac {1}{\sqrt {3 x^4+5 x^2+2}}dx\right )+\frac {1}{4} \left (2 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )-\frac {15 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {31}{4} \left (2 \int \frac {3 x^2+2}{\left (2 x^2+1\right ) \sqrt {3 x^4+5 x^2+2}}dx-3 \int \frac {1}{\sqrt {3 x^4+5 x^2+2}}dx\right )+\frac {1}{4} \left (2 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )-\frac {15 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {31}{4} \left (2 \int \frac {3 x^2+2}{\left (2 x^2+1\right ) \sqrt {3 x^4+5 x^2+2}}dx-\frac {3 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )+\frac {1}{4} \left (2 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )-\frac {15 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )\)

\(\Big \downarrow \) 1786

\(\displaystyle \frac {31}{4} \left (\frac {2 \sqrt {x^2+1} \sqrt {3 x^2+2} \int \frac {\sqrt {3 x^2+2}}{\sqrt {x^2+1} \left (2 x^2+1\right )}dx}{\sqrt {3 x^4+5 x^2+2}}-\frac {3 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )+\frac {1}{4} \left (2 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )-\frac {15 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )\)

\(\Big \downarrow \) 414

\(\displaystyle \frac {31}{4} \left (\frac {4 \left (x^2+1\right ) \operatorname {EllipticPi}\left (-\frac {1}{3},\arctan \left (\sqrt {\frac {3}{2}} x\right ),\frac {1}{3}\right )}{\sqrt {3} \sqrt {\frac {x^2+1}{3 x^2+2}} \sqrt {3 x^4+5 x^2+2}}-\frac {3 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )+\frac {1}{4} \left (2 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )-\frac {15 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )\)

Input:

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

Output:

(2*((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)]*EllipticE[ArcTan[x], -1/2])/(3*Sqrt[2 + 5*x^2 + 3*x 
^4])) - (15*(1 + x^2)*Sqrt[(2 + 3*x^2)/(1 + x^2)]*EllipticF[ArcTan[x], -1/ 
2])/(Sqrt[2]*Sqrt[2 + 5*x^2 + 3*x^4]))/4 + (31*((-3*(1 + x^2)*Sqrt[(2 + 3* 
x^2)/(1 + x^2)]*EllipticF[ArcTan[x], -1/2])/(Sqrt[2]*Sqrt[2 + 5*x^2 + 3*x^ 
4]) + (4*(1 + x^2)*EllipticPi[-1/3, ArcTan[Sqrt[3/2]*x], 1/3])/(Sqrt[3]*Sq 
rt[(1 + x^2)/(2 + 3*x^2)]*Sqrt[2 + 5*x^2 + 3*x^4])))/4
 

Defintions of rubi rules used

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 414

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_) 
^2]), x_Symbol] :> Simp[c*(Sqrt[e + f*x^2]/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]* 
Sqrt[c*((e + f*x^2)/(e*(c + d*x^2)))]))*EllipticPi[1 - b*(c/(a*d)), ArcTan[ 
Rt[d/c, 2]*x], 1 - c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ 
[d/c]
 

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 1538

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_S 
ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/(2*c*d - e*(b - q)))   I 
nt[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[e/(2*c*d - e*(b - q))   Int[(b 
- q + 2*c*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, 
b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && 
 !LtQ[c, 0]
 

rule 1786

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + ( 
b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[(a + b*x^n + c*x 
^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPart[p]) 
   Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c/e)*x^n)^p, x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, n, p, q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c 
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p]
 

rule 2234

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]) 
, x_Symbol] :> With[{A = Coeff[P4x, x, 0], B = Coeff[P4x, x, 2], C = Coeff[ 
P4x, x, 4]}, Simp[-(e^2)^(-1)   Int[(C*d - B*e - C*e*x^2)/Sqrt[a + b*x^2 + 
c*x^4], x], x] + Simp[(C*d^2 - B*d*e + A*e^2)/e^2   Int[1/((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[c*d^2 - a*e^2, 0]
 

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.68

method	result	size
elliptic	\(\frac {49 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{24 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )}{6 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {31 i \sqrt {x^{2}+1}\, \sqrt {1+\frac {3 x^{2}}{2}}\, \operatorname {EllipticPi}\left (i x , 2, \frac {i \sqrt {-3}\, \sqrt {2}}{2}\right )}{4 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(136\)
default	\(\frac {15 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{8 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{6 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {31 i \sqrt {x^{2}+1}\, \sqrt {1+\frac {3 x^{2}}{2}}\, \operatorname {EllipticPi}\left (i x , 2, \frac {i \sqrt {-3}\, \sqrt {2}}{2}\right )}{4 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(149\)

Input:

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

Output:

49/24*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))-1/6*I*(x^2+1)^(1/2)*(6*x^2+4)^(1/2)/(3*x^4+5*x^2+2)^(1/2)*Ell 
ipticE(I*x,1/2*6^(1/2))-31/4*I*(x^2+1)^(1/2)*(1+3/2*x^2)^(1/2)/(3*x^4+5*x^ 
2+2)^(1/2)*EllipticPi(I*x,2,1/2*I*(-3)^(1/2)*2^(1/2))
 

Fricas [F]

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

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

Output:

integral(sqrt(3*x^4 + 5*x^2 + 2)*(x^4 - 7*x^2 + 4)/(6*x^6 + 13*x^4 + 9*x^2 
 + 2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {4-7 x^2+x^4}{\left (1+2 x^2\right ) \sqrt {2+5 x^2+3 x^4}} \, dx=4 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{6 x^{6}+13 x^{4}+9 x^{2}+2}d x \right )+\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{4}}{6 x^{6}+13 x^{4}+9 x^{2}+2}d x -7 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{6 x^{6}+13 x^{4}+9 x^{2}+2}d x \right ) \] Input:

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

Output:

4*int(sqrt(3*x**4 + 5*x**2 + 2)/(6*x**6 + 13*x**4 + 9*x**2 + 2),x) + int(( 
sqrt(3*x**4 + 5*x**2 + 2)*x**4)/(6*x**6 + 13*x**4 + 9*x**2 + 2),x) - 7*int 
((sqrt(3*x**4 + 5*x**2 + 2)*x**2)/(6*x**6 + 13*x**4 + 9*x**2 + 2),x)