Integrand size = 24, antiderivative size = 168 \[ \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx=\frac {1}{4} \sqrt {\frac {5}{77}} \arctan \left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {4+3 x^2+x^4}}\right )-\frac {\left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt {2}}\right ),\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {4+3 x^2+x^4}}+\frac {17 \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} \operatorname {EllipticPi}\left (-\frac {9}{280},2 \arctan \left (\frac {x}{\sqrt {2}}\right ),\frac {1}{8}\right )}{84 \sqrt {2} \sqrt {4+3 x^2+x^4}} \]
1/308*arctan(2/35*x*385^(1/2)/(x^4+3*x^2+4)^(1/2))*385^(1/2)-1/12*(x^2+2)* (cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2)))*Ellipt icF(sin(2*arctan(1/2*x*2^(1/2))),1/4*2^(1/2))*((x^4+3*x^2+4)/(x^2+2)^2)^(1 /2)*2^(1/2)/(x^4+3*x^2+4)^(1/2)+17/168*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2) ))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2)))*EllipticPi(sin(2*arctan(1/2*x*2^( 1/2))),-9/280,1/4*2^(1/2))*((x^4+3*x^2+4)/(x^2+2)^2)^(1/2)*2^(1/2)/(x^4+3* x^2+4)^(1/2)
Result contains complex when optimal does not.
Time = 10.16 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx=-\frac {i \sqrt {1-\frac {2 x^2}{-3-i \sqrt {7}}} \sqrt {1-\frac {2 x^2}{-3+i \sqrt {7}}} \operatorname {EllipticPi}\left (-\frac {5}{14} \left (-3-i \sqrt {7}\right ),i \text {arcsinh}\left (\sqrt {-\frac {2}{-3-i \sqrt {7}}} x\right ),\frac {-3-i \sqrt {7}}{-3+i \sqrt {7}}\right )}{7 \sqrt {2} \sqrt {-\frac {1}{-3-i \sqrt {7}}} \sqrt {4+3 x^2+x^4}} \]
((-1/7*I)*Sqrt[1 - (2*x^2)/(-3 - I*Sqrt[7])]*Sqrt[1 - (2*x^2)/(-3 + I*Sqrt [7])]*EllipticPi[(-5*(-3 - I*Sqrt[7]))/14, I*ArcSinh[Sqrt[-2/(-3 - I*Sqrt[ 7])]*x], (-3 - I*Sqrt[7])/(-3 + I*Sqrt[7])])/(Sqrt[2]*Sqrt[-(-3 - I*Sqrt[7 ])^(-1)]*Sqrt[4 + 3*x^2 + x^4])
Time = 0.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1540, 27, 1416, 2220}
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.
\(\displaystyle \int \frac {1}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+4}} \, dx\) |
\(\Big \downarrow \) 1540 |
\(\displaystyle \frac {10}{3} \int \frac {x^2+2}{2 \left (5 x^2+7\right ) \sqrt {x^4+3 x^2+4}}dx-\frac {1}{3} \int \frac {1}{\sqrt {x^4+3 x^2+4}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{3} \int \frac {x^2+2}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+4}}dx-\frac {1}{3} \int \frac {1}{\sqrt {x^4+3 x^2+4}}dx\) |
\(\Big \downarrow \) 1416 |
\(\displaystyle \frac {5}{3} \int \frac {x^2+2}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+4}}dx-\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt {2}}\right ),\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {x^4+3 x^2+4}}\) |
\(\Big \downarrow \) 2220 |
\(\displaystyle \frac {5}{3} \left (\frac {3 \arctan \left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {x^4+3 x^2+4}}\right )}{4 \sqrt {385}}+\frac {17 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \operatorname {EllipticPi}\left (-\frac {9}{280},2 \arctan \left (\frac {x}{\sqrt {2}}\right ),\frac {1}{8}\right )}{140 \sqrt {2} \sqrt {x^4+3 x^2+4}}\right )-\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt {2}}\right ),\frac {1}{8}\right )}{6 \sqrt {2} \sqrt {x^4+3 x^2+4}}\) |
-1/6*((2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/S qrt[2]], 1/8])/(Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) + (5*((3*ArcTan[(2*Sqrt[11/ 35]*x)/Sqrt[4 + 3*x^2 + x^4]])/(4*Sqrt[385]) + (17*(2 + x^2)*Sqrt[(4 + 3*x ^2 + x^4)/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(140* Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])))/3
3.4.68.3.1 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[c /a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ (2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) ], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_S ymbol] :> With[{q = Rt[c/a, 2]}, Simp[(c*d + a*e*q)/(c*d^2 - a*e^2) Int[1 /Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[(a*e*(e + d*q))/(c*d^2 - a*e^2) I nt[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), 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] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a]
Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(A rcTan[Rt[-b + c*(d/e) + a*(e/d), 2]*(x/Sqrt[a + b*x^2 + c*x^4])]/(2*d*e*Rt[ -b + c*(d/e) + a*(e/d), 2])), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4*d*e*q*Sqrt[a + b*x^2 + c*x^4]))*El lipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x], 1/2 - b/(4*a*q^2)], x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] & & EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && PosQ[-b + c*(d/e) + a*(e/d)]
Result contains complex when optimal does not.
Time = 0.45 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.64
method | result | size |
default | \(\frac {\sqrt {1+\frac {3 x^{2}}{8}-\frac {i x^{2} \sqrt {7}}{8}}\, \sqrt {1+\frac {3 x^{2}}{8}+\frac {i x^{2} \sqrt {7}}{8}}\, \Pi \left (\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, x , -\frac {5}{7 \left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right )}, \frac {\sqrt {-\frac {3}{8}-\frac {i \sqrt {7}}{8}}}{\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}}\right )}{7 \sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, \sqrt {x^{4}+3 x^{2}+4}}\) | \(107\) |
elliptic | \(\frac {\sqrt {1+\frac {3 x^{2}}{8}-\frac {i x^{2} \sqrt {7}}{8}}\, \sqrt {1+\frac {3 x^{2}}{8}+\frac {i x^{2} \sqrt {7}}{8}}\, \Pi \left (\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, x , -\frac {5}{7 \left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right )}, \frac {\sqrt {-\frac {3}{8}-\frac {i \sqrt {7}}{8}}}{\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}}\right )}{7 \sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, \sqrt {x^{4}+3 x^{2}+4}}\) | \(107\) |
1/7/(-3/8+1/8*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8* x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticPi((-3/8+1/8*I*7^ (1/2))^(1/2)*x,-5/7/(-3/8+1/8*I*7^(1/2)),(-3/8-1/8*I*7^(1/2))^(1/2)/(-3/8+ 1/8*I*7^(1/2))^(1/2))
\[ \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 4} {\left (5 \, x^{2} + 7\right )}} \,d x } \]
\[ \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx=\int \frac {1}{\sqrt {\left (x^{2} - x + 2\right ) \left (x^{2} + x + 2\right )} \left (5 x^{2} + 7\right )}\, dx \]
\[ \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 4} {\left (5 \, x^{2} + 7\right )}} \,d x } \]
\[ \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx=\int { \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 4} {\left (5 \, x^{2} + 7\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx=\int \frac {1}{\left (5\,x^2+7\right )\,\sqrt {x^4+3\,x^2+4}} \,d x \]