Integrand size = 34, antiderivative size = 167 \[ \int \frac {\left (1+2 x^2\right ) \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\frac {151 x \left (2+3 x^2\right )}{9 \sqrt {2+5 x^2+3 x^4}}-\frac {x \left (113+149 x^2\right )}{3 \sqrt {2+5 x^2+3 x^4}}-\frac {151 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{9 \sqrt {2+5 x^2+3 x^4}}+\frac {119 \left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{3 \sqrt {2} \sqrt {2+5 x^2+3 x^4}} \] Output:
151/9*x*(3*x^2+2)/(3*x^4+5*x^2+2)^(1/2)-1/3*x*(149*x^2+113)/(3*x^4+5*x^2+2 )^(1/2)-151/9*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)+119/6*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)
Result contains complex when optimal does not.
Time = 10.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.74 \[ \int \frac {\left (1+2 x^2\right ) \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=-\frac {339 x+447 x^3+151 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )-32 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{9 \sqrt {2+5 x^2+3 x^4}} \] Input:
Integrate[((1 + 2*x^2)*(4 - 7*x^2 + x^4))/(2 + 5*x^2 + 3*x^4)^(3/2),x]
Output:
-1/9*(339*x + 447*x^3 + (151*I)*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*Elli pticE[I*ArcSinh[Sqrt[3/2]*x], 2/3] - (32*I)*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*EllipticF[I*ArcSinh[Sqrt[3/2]*x], 2/3])/Sqrt[2 + 5*x^2 + 3*x^4]
Time = 0.35 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {2206, 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.
| \(\displaystyle \int \frac {\left (2 x^2+1\right ) \left (x^4-7 x^2+4\right )}{\left (3 x^4+5 x^2+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2206 |
| \(\displaystyle -\frac {1}{2} \int -\frac {2 \left (151 x^2+119\right )}{3 \sqrt {3 x^4+5 x^2+2}}dx-\frac {x \left (149 x^2+113\right )}{3 \sqrt {3 x^4+5 x^2+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {151 x^2+119}{\sqrt {3 x^4+5 x^2+2}}dx-\frac {x \left (149 x^2+113\right )}{3 \sqrt {3 x^4+5 x^2+2}}\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle \frac {1}{3} \left (119 \int \frac {1}{\sqrt {3 x^4+5 x^2+2}}dx+151 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx\right )-\frac {x \left (149 x^2+113\right )}{3 \sqrt {3 x^4+5 x^2+2}}\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle \frac {1}{3} \left (151 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx+\frac {119 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}\right )-\frac {x \left (149 x^2+113\right )}{3 \sqrt {3 x^4+5 x^2+2}}\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle \frac {1}{3} \left (\frac {119 \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}}+151 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )\right )-\frac {x \left (149 x^2+113\right )}{3 \sqrt {3 x^4+5 x^2+2}}\) |
Input:
Int[((1 + 2*x^2)*(4 - 7*x^2 + x^4))/(2 + 5*x^2 + 3*x^4)^(3/2),x]
Output:
-1/3*(x*(113 + 149*x^2))/Sqrt[2 + 5*x^2 + 3*x^4] + (151*((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])) + (119*(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]))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
Int[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[Poly nomialRemainder[Px, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^ 4)^(p + 1)*((a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)/(2*a*(p + 1)*(b ^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c *x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[Px, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4* p + 7)*(b*d - 2*a*e)*x^2, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Px, x ^2] && Expon[Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1]
Time = 6.91 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.75
| method | result | size |
| risch | \(-\frac {x \left (149 x^{2}+113\right )}{3 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {119 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{6 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {151 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {3 x^{4}+5 x^{2}+2}}\) | \(125\) |
| elliptic | \(-\frac {6 \left (\frac {149}{18} x^{3}+\frac {113}{18} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}-\frac {119 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{6 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {151 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {3 x^{4}+5 x^{2}+2}}\) | \(126\) |
| default | \(-\frac {6 \left (x^{3}+\frac {5}{6} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}-\frac {119 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{6 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {151 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {3 x^{4}+5 x^{2}+2}}-\frac {24 \left (-\frac {5}{4} x^{3}-\frac {13}{12} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}+\frac {-65 x^{3}-52 x}{\sqrt {3 x^{4}+5 x^{2}+2}}-\frac {12 \left (\frac {13}{18} x^{3}+\frac {5}{9} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}\) | \(199\) |
Input:
int((2*x^2+1)*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/3*x*(149*x^2+113)/(3*x^4+5*x^2+2)^(1/2)-119/6*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))+151/9*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))-Elli pticE(I*x,1/2*6^(1/2)))
Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.66 \[ \int \frac {\left (1+2 x^2\right ) \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=-\frac {604 \, \sqrt {3} \sqrt {-\frac {2}{3}} {\left (3 \, x^{5} + 5 \, x^{3} + 2 \, x\right )} E(\arcsin \left (\frac {\sqrt {-\frac {2}{3}}}{x}\right )\,|\,\frac {3}{2}) - 1675 \, \sqrt {3} \sqrt {-\frac {2}{3}} {\left (3 \, x^{5} + 5 \, x^{3} + 2 \, x\right )} F(\arcsin \left (\frac {\sqrt {-\frac {2}{3}}}{x}\right )\,|\,\frac {3}{2}) - 12 \, {\left (3 \, x^{4} + 208 \, x^{2} + 151\right )} \sqrt {3 \, x^{4} + 5 \, x^{2} + 2}}{54 \, {\left (3 \, x^{5} + 5 \, x^{3} + 2 \, x\right )}} \] Input:
integrate((2*x^2+1)*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x, algorithm="fric as")
Output:
-1/54*(604*sqrt(3)*sqrt(-2/3)*(3*x^5 + 5*x^3 + 2*x)*elliptic_e(arcsin(sqrt (-2/3)/x), 3/2) - 1675*sqrt(3)*sqrt(-2/3)*(3*x^5 + 5*x^3 + 2*x)*elliptic_f (arcsin(sqrt(-2/3)/x), 3/2) - 12*(3*x^4 + 208*x^2 + 151)*sqrt(3*x^4 + 5*x^ 2 + 2))/(3*x^5 + 5*x^3 + 2*x)
\[ \int \frac {\left (1+2 x^2\right ) \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {\left (2 x^{2} + 1\right ) \left (x^{4} - 7 x^{2} + 4\right )}{\left (\left (x^{2} + 1\right ) \left (3 x^{2} + 2\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((2*x**2+1)*(x**4-7*x**2+4)/(3*x**4+5*x**2+2)**(3/2),x)
Output:
Integral((2*x**2 + 1)*(x**4 - 7*x**2 + 4)/((x**2 + 1)*(3*x**2 + 2))**(3/2) , x)
\[ \int \frac {\left (1+2 x^2\right ) \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int { \frac {{\left (x^{4} - 7 \, x^{2} + 4\right )} {\left (2 \, x^{2} + 1\right )}}{{\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((2*x^2+1)*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x, algorithm="maxi ma")
Output:
integrate((x^4 - 7*x^2 + 4)*(2*x^2 + 1)/(3*x^4 + 5*x^2 + 2)^(3/2), x)
\[ \int \frac {\left (1+2 x^2\right ) \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int { \frac {{\left (x^{4} - 7 \, x^{2} + 4\right )} {\left (2 \, x^{2} + 1\right )}}{{\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((2*x^2+1)*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x, algorithm="giac ")
Output:
integrate((x^4 - 7*x^2 + 4)*(2*x^2 + 1)/(3*x^4 + 5*x^2 + 2)^(3/2), x)
Timed out. \[ \int \frac {\left (1+2 x^2\right ) \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {\left (2\,x^2+1\right )\,\left (x^4-7\,x^2+4\right )}{{\left (3\,x^4+5\,x^2+2\right )}^{3/2}} \,d x \] Input:
int(((2*x^2 + 1)*(x^4 - 7*x^2 + 4))/(5*x^2 + 3*x^4 + 2)^(3/2),x)
Output:
int(((2*x^2 + 1)*(x^4 - 7*x^2 + 4))/(5*x^2 + 3*x^4 + 2)^(3/2), x)
\[ \int \frac {\left (1+2 x^2\right ) \left (4-7 x^2+x^4\right )}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\frac {6 \sqrt {3 x^{4}+5 x^{2}+2}\, x^{3}+59 \sqrt {3 x^{4}+5 x^{2}+2}\, x -246 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right ) x^{4}-410 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right ) x^{2}-164 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right )-81 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right ) x^{4}-135 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right ) x^{2}-54 \left (\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}\, x^{2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \right )}{27 x^{4}+45 x^{2}+18} \] Input:
int((2*x^2+1)*(x^4-7*x^2+4)/(3*x^4+5*x^2+2)^(3/2),x)
Output:
(6*sqrt(3*x**4 + 5*x**2 + 2)*x**3 + 59*sqrt(3*x**4 + 5*x**2 + 2)*x - 246*i nt(sqrt(3*x**4 + 5*x**2 + 2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x) *x**4 - 410*int(sqrt(3*x**4 + 5*x**2 + 2)/(9*x**8 + 30*x**6 + 37*x**4 + 20 *x**2 + 4),x)*x**2 - 164*int(sqrt(3*x**4 + 5*x**2 + 2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x) - 81*int((sqrt(3*x**4 + 5*x**2 + 2)*x**2)/(9*x* *8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x)*x**4 - 135*int((sqrt(3*x**4 + 5*x **2 + 2)*x**2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x)*x**2 - 54*int ((sqrt(3*x**4 + 5*x**2 + 2)*x**2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x))/(9*(3*x**4 + 5*x**2 + 2))