Integrand size = 31, antiderivative size = 106 \[ \int \frac {946+315 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx=\frac {631 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {2+3 x^2+x^4}}-\frac {2525 \left (2+x^2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{14 \sqrt {2} \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}} \]
-2525/28*(x^2+2)*(1/(x^2+1))^(1/2)*(x^2+1)^(1/2)*EllipticPi(x/(x^2+1)^(1/2 ),2/7,1/2*2^(1/2))*2^(1/2)/((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+631 /4*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))* ((x^2+2)/(x^2+1))^(1/2)*2^(1/2)/(x^4+3*x^2+2)^(1/2)
Result contains complex when optimal does not.
Time = 10.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70 \[ \int \frac {946+315 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx=-\frac {i \sqrt {1+x^2} \sqrt {2+x^2} \left (441 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )+505 \operatorname {EllipticPi}\left (\frac {10}{7},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )\right )}{7 \sqrt {2+3 x^2+x^4}} \]
((-1/7*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(441*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] + 505*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2]))/Sqrt[2 + 3*x^2 + x^4 ]
Time = 0.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2218, 27, 1412, 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.
\(\displaystyle \int \frac {315 x^2+946}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}} \, dx\) |
\(\Big \downarrow \) 2218 |
\(\displaystyle \frac {631}{2} \int \frac {1}{\sqrt {x^4+3 x^2+2}}dx-\frac {2525}{8} \int \frac {4 \left (x^2+1\right )}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {631}{2} \int \frac {1}{\sqrt {x^4+3 x^2+2}}dx-\frac {2525}{2} \int \frac {x^2+1}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}}dx\) |
\(\Big \downarrow \) 1412 |
\(\displaystyle \frac {631 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {2525}{2} \int \frac {x^2+1}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}}dx\) |
\(\Big \downarrow \) 1786 |
\(\displaystyle \frac {631 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {2525 \sqrt {x^2+1} \sqrt {x^2+2} \int \frac {\sqrt {x^2+1}}{\sqrt {x^2+2} \left (5 x^2+7\right )}dx}{2 \sqrt {x^4+3 x^2+2}}\) |
\(\Big \downarrow \) 414 |
\(\displaystyle \frac {631 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {2525 \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{14 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\) |
(631*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(2*Sqr t[2]*Sqrt[2 + 3*x^2 + x^4]) - (2525*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1 /2])/(14*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])
3.1.32.3.1 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[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]
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[((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]
Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Sqrt[b^2 - 4*a*c]}, Simp[(2*a*B - A *(b + q))/(2*a*e - d*(b + q)) Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Sim p[(B*d - A*e)/(2*a*e - d*(b + q)) Int[(2*a + (b + q)*x^2)/((d + e*x^2)*Sq rt[a + b*x^2 + c*x^4]), x], x] /; RationalQ[q]] /; FreeQ[{a, b, c, d, e, A, B}, x] && GtQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*A^ 2 - b*A*B + a*B^2, 0]
Result contains complex when optimal does not.
Time = 1.54 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {63 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}-\frac {505 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{7 \sqrt {x^{4}+3 x^{2}+2}}\) | \(93\) |
elliptic | \(-\frac {63 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2 \sqrt {x^{4}+3 x^{2}+2}}-\frac {505 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{7 \sqrt {x^{4}+3 x^{2}+2}}\) | \(93\) |
-63/2*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*Elliptic F(1/2*I*2^(1/2)*x,2^(1/2))-505/7*I*2^(1/2)*(1+1/2*x^2)^(1/2)*(x^2+1)^(1/2) /(x^4+3*x^2+2)^(1/2)*EllipticPi(1/2*I*2^(1/2)*x,10/7,2^(1/2))
\[ \int \frac {946+315 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx=\int { \frac {315 \, x^{2} + 946}{\sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}} \,d x } \]
\[ \int \frac {946+315 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx=\int \frac {315 x^{2} + 946}{\sqrt {\left (x^{2} + 1\right ) \left (x^{2} + 2\right )} \left (5 x^{2} + 7\right )}\, dx \]
\[ \int \frac {946+315 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx=\int { \frac {315 \, x^{2} + 946}{\sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}} \,d x } \]
\[ \int \frac {946+315 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx=\int { \frac {315 \, x^{2} + 946}{\sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}} \,d x } \]
Timed out. \[ \int \frac {946+315 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx=\int \frac {315\,x^2+946}{\left (5\,x^2+7\right )\,\sqrt {x^4+3\,x^2+2}} \,d x \]