Integrand size = 20, antiderivative size = 173 \[ \int \frac {2+3 x^2}{x^2 \sqrt {5+x^4}} \, dx=-\frac {2 \sqrt {5+x^4}}{5 x}+\frac {2 x \sqrt {5+x^4}}{5 \left (\sqrt {5}+x^2\right )}-\frac {2 \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{5^{3/4} \sqrt {5+x^4}}+\frac {\left (2+3 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2\ 5^{3/4} \sqrt {5+x^4}} \]
-2/5*(x^4+5)^(1/2)/x+2/5*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-2/5*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)))*EllipticF(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^(1/4)/(x^4 +5)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.31 \[ \int \frac {2+3 x^2}{x^2 \sqrt {5+x^4}} \, dx=-\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-\frac {x^4}{5}\right )}{\sqrt {5} x}+\frac {3 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {x^4}{5}\right )}{\sqrt {5}} \]
(-2*Hypergeometric2F1[-1/4, 1/2, 3/4, -1/5*x^4])/(Sqrt[5]*x) + (3*x*Hyperg eometric2F1[1/4, 1/2, 5/4, -1/5*x^4])/Sqrt[5]
Time = 0.28 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1605, 25, 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.
\(\displaystyle \int \frac {3 x^2+2}{x^2 \sqrt {x^4+5}} \, dx\) |
\(\Big \downarrow \) 1605 |
\(\displaystyle -\frac {1}{5} \int -\frac {2 x^2+15}{\sqrt {x^4+5}}dx-\frac {2 \sqrt {x^4+5}}{5 x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \int \frac {2 x^2+15}{\sqrt {x^4+5}}dx-\frac {2 \sqrt {x^4+5}}{5 x}\) |
\(\Big \downarrow \) 1512 |
\(\displaystyle \frac {1}{5} \left (\left (15+2 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-2 \sqrt {5} \int \frac {\sqrt {5}-x^2}{\sqrt {5} \sqrt {x^4+5}}dx\right )-\frac {2 \sqrt {x^4+5}}{5 x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\left (15+2 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-2 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\right )-\frac {2 \sqrt {x^4+5}}{5 x}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {1}{5} \left (\frac {\left (15+2 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-2 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\right )-\frac {2 \sqrt {x^4+5}}{5 x}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {1}{5} \left (\frac {\left (15+2 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-2 \left (\frac {\sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {x \sqrt {x^4+5}}{x^2+\sqrt {5}}\right )\right )-\frac {2 \sqrt {x^4+5}}{5 x}\) |
(-2*Sqrt[5 + x^4])/(5*x) + (-2*(-((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]) + ((15 + 2*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]))/5
3.1.42.3.1 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]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_ Symbol] :> Simp[d*(f*x)^(m + 1)*((a + c*x^4)^(p + 1)/(a*f*(m + 1))), x] + S imp[1/(a*f^2*(m + 1)) Int[(f*x)^(m + 2)*(a + c*x^4)^p*(a*e*(m + 1) - c*d* (m + 4*p + 5)*x^2), x], x] /; FreeQ[{a, c, d, e, f, p}, x] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 1.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.22
method | result | size |
meijerg | \(-\frac {2 \sqrt {5}\, {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {x^{4}}{5}\right )}{5 x}+\frac {3 \sqrt {5}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {x^{4}}{5}\right )}{5}\) | \(38\) |
default | \(\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{5 x}+\frac {2 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(158\) |
risch | \(\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{5 x}+\frac {2 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(158\) |
elliptic | \(\frac {3 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{5 x}+\frac {2 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{25 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(158\) |
-2/5*5^(1/2)/x*hypergeom([-1/4,1/2],[3/4],-1/5*x^4)+3/5*5^(1/2)*x*hypergeo m([1/4,1/2],[5/4],-1/5*x^4)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.42 \[ \int \frac {2+3 x^2}{x^2 \sqrt {5+x^4}} \, dx=\frac {-2 i \, \sqrt {5} x \sqrt {i \, \sqrt {5}} E(\arcsin \left (\frac {1}{5} \, \sqrt {5} x \sqrt {i \, \sqrt {5}}\right )\,|\,-1) - 13 i \, \sqrt {5} x \sqrt {i \, \sqrt {5}} F(\arcsin \left (\frac {1}{5} \, \sqrt {5} x \sqrt {i \, \sqrt {5}}\right )\,|\,-1) - 10 \, \sqrt {x^{4} + 5}}{25 \, x} \]
1/25*(-2*I*sqrt(5)*x*sqrt(I*sqrt(5))*elliptic_e(arcsin(1/5*sqrt(5)*x*sqrt( I*sqrt(5))), -1) - 13*I*sqrt(5)*x*sqrt(I*sqrt(5))*elliptic_f(arcsin(1/5*sq rt(5)*x*sqrt(I*sqrt(5))), -1) - 10*sqrt(x^4 + 5))/x
Result contains complex when optimal does not.
Time = 0.81 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.43 \[ \int \frac {2+3 x^2}{x^2 \sqrt {5+x^4}} \, dx=\frac {3 \sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{20 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {5} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{10 x \Gamma \left (\frac {3}{4}\right )} \]
3*sqrt(5)*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), x**4*exp_polar(I*pi)/5)/( 20*gamma(5/4)) + sqrt(5)*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), x**4*exp_p olar(I*pi)/5)/(10*x*gamma(3/4))
\[ \int \frac {2+3 x^2}{x^2 \sqrt {5+x^4}} \, dx=\int { \frac {3 \, x^{2} + 2}{\sqrt {x^{4} + 5} x^{2}} \,d x } \]
\[ \int \frac {2+3 x^2}{x^2 \sqrt {5+x^4}} \, dx=\int { \frac {3 \, x^{2} + 2}{\sqrt {x^{4} + 5} x^{2}} \,d x } \]
Time = 8.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.28 \[ \int \frac {2+3 x^2}{x^2 \sqrt {5+x^4}} \, dx=\frac {3\,\sqrt {5}\,x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {5}{4};\ -\frac {x^4}{5}\right )}{5}-\frac {2\,\sqrt {\frac {5}{x^4}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {5}{x^4}\right )}{3\,x\,\sqrt {x^4+5}} \]