\(\int (1+2 x^2)^2 (4-7 x^2+x^4) \sqrt {2+5 x^2+3 x^4} \, dx\) [191]

Optimal result

Integrand size = 36, antiderivative size = 234 \[ \int \left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right ) \sqrt {2+5 x^2+3 x^4} \, dx=\frac {1291552 x \left (2+3 x^2\right )}{841995 \sqrt {2+5 x^2+3 x^4}}+\frac {x \left (88715-195687 x^2\right ) \sqrt {2+5 x^2+3 x^4}}{280665}+\frac {5893 x \left (2+5 x^2+3 x^4\right )^{3/2}}{6237}-\frac {952}{891} x^3 \left (2+5 x^2+3 x^4\right )^{3/2}+\frac {4}{33} x^5 \left (2+5 x^2+3 x^4\right )^{3/2}-\frac {1291552 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{841995 \sqrt {2+5 x^2+3 x^4}}+\frac {100715 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{56133 \sqrt {2+5 x^2+3 x^4}} \] Output:

1291552/841995*x*(3*x^2+2)/(3*x^4+5*x^2+2)^(1/2)+1/280665*x*(-195687*x^2+8 
8715)*(3*x^4+5*x^2+2)^(1/2)+5893/6237*x*(3*x^4+5*x^2+2)^(3/2)-952/891*x^3* 
(3*x^4+5*x^2+2)^(3/2)+4/33*x^5*(3*x^4+5*x^2+2)^(3/2)-1291552/841995*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)+100715/56133*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 = 150, normalized size of antiderivative = 0.64 \[ \int \left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right ) \sqrt {2+5 x^2+3 x^4} \, dx=\frac {3 x \left (1238170+4156381 x^2+3238035 x^4-3046671 x^6-5350995 x^8-1678320 x^{10}+306180 x^{12}\right )-1291552 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )+284402 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{841995 \sqrt {2+5 x^2+3 x^4}} \] Input:

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

Output:

(3*x*(1238170 + 4156381*x^2 + 3238035*x^4 - 3046671*x^6 - 5350995*x^8 - 16 
78320*x^10 + 306180*x^12) - (1291552*I)*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x 
^2]*EllipticE[I*ArcSinh[Sqrt[3/2]*x], 2/3] + (284402*I)*Sqrt[3]*Sqrt[1 + x 
^2]*Sqrt[2 + 3*x^2]*EllipticF[I*ArcSinh[Sqrt[3/2]*x], 2/3])/(841995*Sqrt[2 
 + 5*x^2 + 3*x^4])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2207, 2207, 27, 2207, 1490, 27, 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.

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

\(\Big \downarrow \) 2207

\(\displaystyle \frac {1}{33} \int \sqrt {3 x^4+5 x^2+2} \left (-952 x^6-403 x^4+297 x^2+132\right )dx+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

\(\Big \downarrow \) 2207

\(\displaystyle \frac {1}{33} \left (\frac {1}{27} \int 3 \sqrt {3 x^4+5 x^2+2} \left (5893 x^4+4577 x^2+1188\right )dx-\frac {952}{27} x^3 \left (3 x^4+5 x^2+2\right )^{3/2}\right )+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{33} \left (\frac {1}{9} \int \sqrt {3 x^4+5 x^2+2} \left (5893 x^4+4577 x^2+1188\right )dx-\frac {952}{27} x^3 \left (3 x^4+5 x^2+2\right )^{3/2}\right )+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

\(\Big \downarrow \) 2207

\(\displaystyle \frac {1}{33} \left (\frac {1}{9} \left (\frac {1}{21} \int \left (13162-21743 x^2\right ) \sqrt {3 x^4+5 x^2+2}dx+\frac {5893}{21} x \left (3 x^4+5 x^2+2\right )^{3/2}\right )-\frac {952}{27} x^3 \left (3 x^4+5 x^2+2\right )^{3/2}\right )+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

\(\Big \downarrow \) 1490

\(\displaystyle \frac {1}{33} \left (\frac {1}{9} \left (\frac {1}{21} \left (\frac {1}{45} \int \frac {2 \left (645776 x^2+503575\right )}{\sqrt {3 x^4+5 x^2+2}}dx+\frac {1}{45} x \sqrt {3 x^4+5 x^2+2} \left (88715-195687 x^2\right )\right )+\frac {5893}{21} x \left (3 x^4+5 x^2+2\right )^{3/2}\right )-\frac {952}{27} x^3 \left (3 x^4+5 x^2+2\right )^{3/2}\right )+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{33} \left (\frac {1}{9} \left (\frac {1}{21} \left (\frac {2}{45} \int \frac {645776 x^2+503575}{\sqrt {3 x^4+5 x^2+2}}dx+\frac {1}{45} x \sqrt {3 x^4+5 x^2+2} \left (88715-195687 x^2\right )\right )+\frac {5893}{21} x \left (3 x^4+5 x^2+2\right )^{3/2}\right )-\frac {952}{27} x^3 \left (3 x^4+5 x^2+2\right )^{3/2}\right )+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{33} \left (\frac {1}{9} \left (\frac {1}{21} \left (\frac {2}{45} \left (503575 \int \frac {1}{\sqrt {3 x^4+5 x^2+2}}dx+645776 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx\right )+\frac {1}{45} x \sqrt {3 x^4+5 x^2+2} \left (88715-195687 x^2\right )\right )+\frac {5893}{21} x \left (3 x^4+5 x^2+2\right )^{3/2}\right )-\frac {952}{27} x^3 \left (3 x^4+5 x^2+2\right )^{3/2}\right )+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {1}{33} \left (\frac {1}{9} \left (\frac {1}{21} \left (\frac {2}{45} \left (645776 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx+\frac {503575 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )+\frac {1}{45} x \sqrt {3 x^4+5 x^2+2} \left (88715-195687 x^2\right )\right )+\frac {5893}{21} x \left (3 x^4+5 x^2+2\right )^{3/2}\right )-\frac {952}{27} x^3 \left (3 x^4+5 x^2+2\right )^{3/2}\right )+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {1}{33} \left (\frac {1}{9} \left (\frac {1}{21} \left (\frac {2}{45} \left (\frac {503575 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}+645776 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )\right )+\frac {1}{45} x \sqrt {3 x^4+5 x^2+2} \left (88715-195687 x^2\right )\right )+\frac {5893}{21} x \left (3 x^4+5 x^2+2\right )^{3/2}\right )-\frac {952}{27} x^3 \left (3 x^4+5 x^2+2\right )^{3/2}\right )+\frac {4}{33} \left (3 x^4+5 x^2+2\right )^{3/2} x^5\)

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*(2 + 5*x^2 + 3*x^4)^(3/2))/33 + ((-952*x^3*(2 + 5*x^2 + 3*x^4)^(3/2 
))/27 + ((5893*x*(2 + 5*x^2 + 3*x^4)^(3/2))/21 + ((x*(88715 - 195687*x^2)* 
Sqrt[2 + 5*x^2 + 3*x^4])/45 + (2*(645776*((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[Arc 
Tan[x], -1/2])/(3*Sqrt[2 + 5*x^2 + 3*x^4])) + (503575*(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])))/45)/21)/9)/33
 

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 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 1490

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb 
ol] :> 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] + Simp[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 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 = 10.84 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.60

method	result	size
risch	\(\frac {x \left (102060 x^{8}-729540 x^{6}-635805 x^{4}+530478 x^{2}+619085\right ) \sqrt {3 x^{4}+5 x^{2}+2}}{280665}-\frac {100715 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{56133 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {1291552 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{841995 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(140\)
default	\(\frac {123817 x \sqrt {3 x^{4}+5 x^{2}+2}}{56133}-\frac {100715 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{56133 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {1291552 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{841995 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {58942 x^{3} \sqrt {3 x^{4}+5 x^{2}+2}}{31185}-\frac {14129 x^{5} \sqrt {3 x^{4}+5 x^{2}+2}}{6237}-\frac {772 x^{7} \sqrt {3 x^{4}+5 x^{2}+2}}{297}+\frac {4 x^{9} \sqrt {3 x^{4}+5 x^{2}+2}}{11}\)	\(194\)
elliptic	\(\frac {123817 x \sqrt {3 x^{4}+5 x^{2}+2}}{56133}-\frac {100715 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{56133 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {1291552 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{841995 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {58942 x^{3} \sqrt {3 x^{4}+5 x^{2}+2}}{31185}-\frac {14129 x^{5} \sqrt {3 x^{4}+5 x^{2}+2}}{6237}-\frac {772 x^{7} \sqrt {3 x^{4}+5 x^{2}+2}}{297}+\frac {4 x^{9} \sqrt {3 x^{4}+5 x^{2}+2}}{11}\)	\(194\)

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:

1/280665*x*(102060*x^8-729540*x^6-635805*x^4+530478*x^2+619085)*(3*x^4+5*x 
^2+2)^(1/2)-100715/56133*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))+1291552/841995*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 = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.37 \[ \int \left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right ) \sqrt {2+5 x^2+3 x^4} \, dx=-\frac {2583104 \, \sqrt {3} \sqrt {-\frac {2}{3}} x E(\arcsin \left (\frac {\sqrt {-\frac {2}{3}}}{x}\right )\,|\,\frac {3}{2}) - 7115279 \, \sqrt {3} \sqrt {-\frac {2}{3}} x F(\arcsin \left (\frac {\sqrt {-\frac {2}{3}}}{x}\right )\,|\,\frac {3}{2}) - 3 \, {\left (306180 \, x^{10} - 2188620 \, x^{8} - 1907415 \, x^{6} + 1591434 \, x^{4} + 1857255 \, x^{2} + 1291552\right )} \sqrt {3 \, x^{4} + 5 \, x^{2} + 2}}{2525985 \, x} \] 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:

-1/2525985*(2583104*sqrt(3)*sqrt(-2/3)*x*elliptic_e(arcsin(sqrt(-2/3)/x), 
3/2) - 7115279*sqrt(3)*sqrt(-2/3)*x*elliptic_f(arcsin(sqrt(-2/3)/x), 3/2) 
- 3*(306180*x^10 - 2188620*x^8 - 1907415*x^6 + 1591434*x^4 + 1857255*x^2 + 
 1291552)*sqrt(3*x^4 + 5*x^2 + 2))/x
 

Sympy [F]

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

Maxima [F]

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

Giac [F]

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

Mupad [F(-1)]

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

\[ \int \left (1+2 x^2\right )^2 \left (4-7 x^2+x^4\right ) \sqrt {2+5 x^2+3 x^4} \, dx=\frac {4 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{9}}{11}-\frac {772 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{7}}{297}-\frac {14129 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{5}}{6237}+\frac {58942 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{3}}{31185}+\frac {123817 \sqrt {3 x^{4}+5 x^{2}+2}\, x}{56133}+\frac {201430 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{3 x^{4}+5 x^{2}+2}d x \right )}{56133}+\frac {1291552 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{3 x^{4}+5 x^{2}+2}d x \right )}{280665} \] Input:

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

Output:

(102060*sqrt(3*x**4 + 5*x**2 + 2)*x**9 - 729540*sqrt(3*x**4 + 5*x**2 + 2)* 
x**7 - 635805*sqrt(3*x**4 + 5*x**2 + 2)*x**5 + 530478*sqrt(3*x**4 + 5*x**2 
 + 2)*x**3 + 619085*sqrt(3*x**4 + 5*x**2 + 2)*x + 1007150*int(sqrt(3*x**4 
+ 5*x**2 + 2)/(3*x**4 + 5*x**2 + 2),x) + 1291552*int((sqrt(3*x**4 + 5*x**2 
 + 2)*x**2)/(3*x**4 + 5*x**2 + 2),x))/280665