Integrand size = 17, antiderivative size = 155 \[ \int \frac {2+3 x^2}{\sqrt {5+x^4}} \, dx=\frac {3 x \sqrt {5+x^4}}{\sqrt {5}+x^2}-\frac {3 \sqrt [4]{5} \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {5+x^4}}+\frac {\left (2+3 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {5+x^4}} \]
3*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-3*5^(1/4)*(cos(2*arctan(1/5*x*5^(3/4)))^2) ^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*EllipticE(sin(2*arctan(1/5*x*5^(3/4))) ,1/2*2^(1/2))*(x^2+5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2)/(x^4+5)^(1/2)+ 1/10*(cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*E llipticF(sin(2*arctan(1/5*x*5^(3/4))),1/2*2^(1/2))*(x^2+5^(1/2))*(2+3*5^(1 /2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2)*5^(3/4)/(x^4+5)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.31 \[ \int \frac {2+3 x^2}{\sqrt {5+x^4}} \, dx=\frac {x \left (2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {x^4}{5}\right )+x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {x^4}{5}\right )\right )}{\sqrt {5}} \]
(x*(2*Hypergeometric2F1[1/4, 1/2, 5/4, -1/5*x^4] + x^2*Hypergeometric2F1[1 /2, 3/4, 7/4, -1/5*x^4]))/Sqrt[5]
Time = 0.25 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1512, 27, 761, 1510}
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.
\(\displaystyle \int \frac {3 x^2+2}{\sqrt {x^4+5}} \, dx\) |
\(\Big \downarrow \) 1512 |
\(\displaystyle \left (2+3 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-3 \sqrt {5} \int \frac {\sqrt {5}-x^2}{\sqrt {5} \sqrt {x^4+5}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (2+3 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-3 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\left (2+3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-3 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {\left (2+3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-3 \left (\frac {\sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {x \sqrt {x^4+5}}{x^2+\sqrt {5}}\right )\) |
-3*(-((x*Sqrt[5 + x^4])/(Sqrt[5] + x^2)) + (5^(1/4)*(Sqrt[5] + x^2)*Sqrt[( 5 + x^4)/(Sqrt[5] + x^2)^2]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt[5 + x^4]) + ((2 + 3*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2] *EllipticF[2*ArcTan[x/5^(1/4)], 1/2])/(2*5^(1/4)*Sqrt[5 + x^4])
3.1.41.3.1 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_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c , d, e}, x] && PosQ[c/a]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.81 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.25
method | result | size |
meijerg | \(\frac {2 \sqrt {5}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {x^{4}}{5}\right )}{5}+\frac {\sqrt {5}\, x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {x^{4}}{5}\right )}{5}\) | \(38\) |
default | \(\frac {2 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(146\) |
elliptic | \(\frac {2 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(146\) |
2/5*5^(1/2)*x*hypergeom([1/4,1/2],[5/4],-1/5*x^4)+1/5*5^(1/2)*x^3*hypergeo m([1/2,3/4],[7/4],-1/5*x^4)
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.30 \[ \int \frac {2+3 x^2}{\sqrt {5+x^4}} \, dx=\frac {15 \, \left (-5\right )^{\frac {3}{4}} x E(\arcsin \left (\frac {\left (-5\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 13 \, \left (-5\right )^{\frac {3}{4}} x F(\arcsin \left (\frac {\left (-5\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 15 \, \sqrt {x^{4} + 5}}{5 \, x} \]
1/5*(15*(-5)^(3/4)*x*elliptic_e(arcsin((-5)^(1/4)/x), -1) - 13*(-5)^(3/4)* x*elliptic_f(arcsin((-5)^(1/4)/x), -1) + 15*sqrt(x^4 + 5))/x
Result contains complex when optimal does not.
Time = 0.78 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.47 \[ \int \frac {2+3 x^2}{\sqrt {5+x^4}} \, dx=\frac {3 \sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{20 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{10 \Gamma \left (\frac {5}{4}\right )} \]
3*sqrt(5)*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), x**4*exp_polar(I*pi)/5 )/(20*gamma(7/4)) + sqrt(5)*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), x**4*ex p_polar(I*pi)/5)/(10*gamma(5/4))
\[ \int \frac {2+3 x^2}{\sqrt {5+x^4}} \, dx=\int { \frac {3 \, x^{2} + 2}{\sqrt {x^{4} + 5}} \,d x } \]
\[ \int \frac {2+3 x^2}{\sqrt {5+x^4}} \, dx=\int { \frac {3 \, x^{2} + 2}{\sqrt {x^{4} + 5}} \,d x } \]
Timed out. \[ \int \frac {2+3 x^2}{\sqrt {5+x^4}} \, dx=\int \frac {3\,x^2+2}{\sqrt {x^4+5}} \,d x \]