Integrand size = 24, antiderivative size = 157 \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\frac {135 x \left (2+x^2\right )}{\sqrt {2+3 x^2+x^4}}+75 x \sqrt {2+3 x^2+x^4}+25 x^3 \sqrt {2+3 x^2+x^4}-\frac {135 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {2+3 x^2+x^4}}+\frac {193 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+3 x^2+x^4}} \]
135*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)+193/2*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*El lipticF(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)-135*(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)+75*x*( x^4+3*x^2+2)^(1/2)+25*x^3*(x^4+3*x^2+2)^(1/2)
Result contains complex when optimal does not.
Time = 10.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.68 \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\frac {25 x \left (6+11 x^2+6 x^4+x^6\right )-135 i \sqrt {1+x^2} \sqrt {2+x^2} E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-58 i \sqrt {1+x^2} \sqrt {2+x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{\sqrt {2+3 x^2+x^4}} \]
(25*x*(6 + 11*x^2 + 6*x^4 + x^6) - (135*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*Ell ipticE[I*ArcSinh[x/Sqrt[2]], 2] - (58*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*Ellip ticF[I*ArcSinh[x/Sqrt[2]], 2])/Sqrt[2 + 3*x^2 + x^4]
Time = 0.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1518, 27, 2207, 27, 1503, 1412, 1455}
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.
\(\displaystyle \int \frac {\left (5 x^2+7\right )^3}{\sqrt {x^4+3 x^2+2}} \, dx\) |
\(\Big \downarrow \) 1518 |
\(\displaystyle \frac {1}{5} \int \frac {5 \left (225 x^4+585 x^2+343\right )}{\sqrt {x^4+3 x^2+2}}dx+25 \sqrt {x^4+3 x^2+2} x^3\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {225 x^4+585 x^2+343}{\sqrt {x^4+3 x^2+2}}dx+25 \sqrt {x^4+3 x^2+2} x^3\) |
\(\Big \downarrow \) 2207 |
\(\displaystyle \frac {1}{3} \int \frac {3 \left (135 x^2+193\right )}{\sqrt {x^4+3 x^2+2}}dx+75 \sqrt {x^4+3 x^2+2} x+25 \sqrt {x^4+3 x^2+2} x^3\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {135 x^2+193}{\sqrt {x^4+3 x^2+2}}dx+75 \sqrt {x^4+3 x^2+2} x+25 \sqrt {x^4+3 x^2+2} x^3\) |
\(\Big \downarrow \) 1503 |
\(\displaystyle 193 \int \frac {1}{\sqrt {x^4+3 x^2+2}}dx+135 \int \frac {x^2}{\sqrt {x^4+3 x^2+2}}dx+75 \sqrt {x^4+3 x^2+2} x+25 \sqrt {x^4+3 x^2+2} x^3\) |
\(\Big \downarrow \) 1412 |
\(\displaystyle 135 \int \frac {x^2}{\sqrt {x^4+3 x^2+2}}dx+\frac {193 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}}+75 \sqrt {x^4+3 x^2+2} x+25 \sqrt {x^4+3 x^2+2} x^3\) |
\(\Big \downarrow \) 1455 |
\(\displaystyle \frac {193 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}}+135 \left (\frac {x \left (x^2+2\right )}{\sqrt {x^4+3 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+3 x^2+2}}\right )+75 \sqrt {x^4+3 x^2+2} x+25 \sqrt {x^4+3 x^2+2} x^3\) |
75*x*Sqrt[2 + 3*x^2 + x^4] + 25*x^3*Sqrt[2 + 3*x^2 + x^4] + 135*((x*(2 + x ^2))/Sqrt[2 + 3*x^2 + x^4] - (Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]* EllipticE[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4]) + (193*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^ 4])
3.3.100.3.1 Defintions of rubi rules used
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] && !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*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] && !(PosQ[ (b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*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[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x _Symbol] :> Simp[e^q*x^(2*q - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Simp[1/(c*(4*p + 2*q + 1)) Int[(a + b*x^2 + c*x^4)^p*Expand ToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2* p + 2*q - 1)*e^q*x^(2*q - 2) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1]
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]
Result contains complex when optimal does not.
Time = 4.61 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.80
method | result | size |
risch | \(25 x \left (x^{2}+3\right ) \sqrt {x^{4}+3 x^{2}+2}-\frac {193 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+\frac {135 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}\) | \(126\) |
default | \(-\frac {193 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+25 x^{3} \sqrt {x^{4}+3 x^{2}+2}+75 x \sqrt {x^{4}+3 x^{2}+2}+\frac {135 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}\) | \(138\) |
elliptic | \(-\frac {193 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}+25 x^{3} \sqrt {x^{4}+3 x^{2}+2}+75 x \sqrt {x^{4}+3 x^{2}+2}+\frac {135 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}\) | \(138\) |
25*x*(x^2+3)*(x^4+3*x^2+2)^(1/2)-193/2*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))+135/2*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)))
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.34 \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\frac {-135 i \, x E(\arcsin \left (\frac {i}{x}\right )\,|\,2) + 328 i \, x F(\arcsin \left (\frac {i}{x}\right )\,|\,2) + 5 \, {\left (5 \, x^{4} + 15 \, x^{2} + 27\right )} \sqrt {x^{4} + 3 \, x^{2} + 2}}{x} \]
(-135*I*x*elliptic_e(arcsin(I/x), 2) + 328*I*x*elliptic_f(arcsin(I/x), 2) + 5*(5*x^4 + 15*x^2 + 27)*sqrt(x^4 + 3*x^2 + 2))/x
\[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\int \frac {\left (5 x^{2} + 7\right )^{3}}{\sqrt {\left (x^{2} + 1\right ) \left (x^{2} + 2\right )}}\, dx \]
\[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\int { \frac {{\left (5 \, x^{2} + 7\right )}^{3}}{\sqrt {x^{4} + 3 \, x^{2} + 2}} \,d x } \]
\[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\int { \frac {{\left (5 \, x^{2} + 7\right )}^{3}}{\sqrt {x^{4} + 3 \, x^{2} + 2}} \,d x } \]
Timed out. \[ \int \frac {\left (7+5 x^2\right )^3}{\sqrt {2+3 x^2+x^4}} \, dx=\int \frac {{\left (5\,x^2+7\right )}^3}{\sqrt {x^4+3\,x^2+2}} \,d x \]