Integrand size = 17, antiderivative size = 180 \[ \int \frac {2+3 x^2}{\left (5+x^4\right )^{3/2}} \, dx=\frac {x \left (2+3 x^2\right )}{10 \sqrt {5+x^4}}-\frac {3 x \sqrt {5+x^4}}{10 \left (\sqrt {5}+x^2\right )}+\frac {3 \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{2\ 5^{3/4} \sqrt {5+x^4}}+\frac {\left (2-3 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{20 \sqrt [4]{5} \sqrt {5+x^4}} \]
1/10*x*(3*x^2+2)/(x^4+5)^(1/2)-3/10*x*(x^4+5)^(1/2)/(x^2+5^(1/2))+3/10*5^( 1/4)*(cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*E llipticE(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)+1/100*(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))*(2-3*5^(1/2))*(x^2+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 = 66, normalized size of antiderivative = 0.37 \[ \int \frac {2+3 x^2}{\left (5+x^4\right )^{3/2}} \, dx=\frac {1}{25} x \left (\frac {5}{\sqrt {5+x^4}}+\sqrt {5} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {x^4}{5}\right )+\sqrt {5} x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {x^4}{5}\right )\right ) \]
(x*(5/Sqrt[5 + x^4] + Sqrt[5]*Hypergeometric2F1[1/4, 1/2, 5/4, -1/5*x^4] + Sqrt[5]*x^2*Hypergeometric2F1[3/4, 3/2, 7/4, -1/5*x^4]))/25
Time = 0.28 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1493, 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.
| \(\displaystyle \int \frac {3 x^2+2}{\left (x^4+5\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1493 |
| \(\displaystyle \frac {x \left (3 x^2+2\right )}{10 \sqrt {x^4+5}}-\frac {1}{10} \int -\frac {2-3 x^2}{\sqrt {x^4+5}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{10} \int \frac {2-3 x^2}{\sqrt {x^4+5}}dx+\frac {x \left (3 x^2+2\right )}{10 \sqrt {x^4+5}}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {1}{10} \left (\left (2-3 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx+3 \sqrt {5} \int \frac {\sqrt {5}-x^2}{\sqrt {5} \sqrt {x^4+5}}dx\right )+\frac {x \left (3 x^2+2\right )}{10 \sqrt {x^4+5}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{10} \left (\left (2-3 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx+3 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\right )+\frac {x \left (3 x^2+2\right )}{10 \sqrt {x^4+5}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {1}{10} \left (3 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx+\frac {\left (2-3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}\right )+\frac {x \left (3 x^2+2\right )}{10 \sqrt {x^4+5}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {1}{10} \left (\frac {\left (2-3 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}+3 \left (\frac {\sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {x \sqrt {x^4+5}}{x^2+\sqrt {5}}\right )\right )+\frac {x \left (3 x^2+2\right )}{10 \sqrt {x^4+5}}\) |
(x*(2 + 3*x^2))/(10*Sqrt[5 + x^4]) + (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]))/10
3.1.52.3.1 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)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x )*(d + e*x^2)*((a + c*x^4)^(p + 1)/(4*a*(p + 1))), x] + Simp[1/(4*a*(p + 1) ) Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && Integer Q[2*p]
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 = 1.57 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.21
| method | result | size |
| meijerg | \(\frac {2 \sqrt {5}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {3}{2};\frac {5}{4};-\frac {x^{4}}{5}\right )}{25}+\frac {\sqrt {5}\, x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {x^{4}}{5}\right )}{25}\) | \(38\) |
| risch | \(\frac {x \left (3 x^{2}+2\right )}{10 \sqrt {x^{4}+5}}+\frac {\sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{125 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {3 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{50 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(163\) |
| elliptic | \(-\frac {2 \left (-\frac {3}{20} x^{3}-\frac {1}{10} x \right )}{\sqrt {x^{4}+5}}+\frac {\sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{125 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {3 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{50 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(164\) |
| default | \(\frac {x}{5 \sqrt {x^{4}+5}}+\frac {\sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{125 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 x^{3}}{10 \sqrt {x^{4}+5}}-\frac {3 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{50 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(168\) |
2/25*5^(1/2)*x*hypergeom([1/4,3/2],[5/4],-1/5*x^4)+1/25*5^(1/2)*x^3*hyperg eom([3/4,3/2],[7/4],-1/5*x^4)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.54 \[ \int \frac {2+3 x^2}{\left (5+x^4\right )^{3/2}} \, dx=-\frac {3 \, \sqrt {5} {\left (-i \, x^{4} - 5 i\right )} \sqrt {i \, \sqrt {5}} E(\arcsin \left (\frac {1}{5} \, \sqrt {5} x \sqrt {i \, \sqrt {5}}\right )\,|\,-1) + 5 \, \sqrt {5} {\left (i \, x^{4} + 5 i\right )} \sqrt {i \, \sqrt {5}} F(\arcsin \left (\frac {1}{5} \, \sqrt {5} x \sqrt {i \, \sqrt {5}}\right )\,|\,-1) - 5 \, \sqrt {x^{4} + 5} {\left (3 \, x^{3} + 2 \, x\right )}}{50 \, {\left (x^{4} + 5\right )}} \]
-1/50*(3*sqrt(5)*(-I*x^4 - 5*I)*sqrt(I*sqrt(5))*elliptic_e(arcsin(1/5*sqrt (5)*x*sqrt(I*sqrt(5))), -1) + 5*sqrt(5)*(I*x^4 + 5*I)*sqrt(I*sqrt(5))*elli ptic_f(arcsin(1/5*sqrt(5)*x*sqrt(I*sqrt(5))), -1) - 5*sqrt(x^4 + 5)*(3*x^3 + 2*x))/(x^4 + 5)
Result contains complex when optimal does not.
Time = 1.97 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.41 \[ \int \frac {2+3 x^2}{\left (5+x^4\right )^{3/2}} \, dx=\frac {3 \sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{100 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{50 \Gamma \left (\frac {5}{4}\right )} \]
3*sqrt(5)*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), x**4*exp_polar(I*pi)/5 )/(100*gamma(7/4)) + sqrt(5)*x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), x**4*e xp_polar(I*pi)/5)/(50*gamma(5/4))
\[ \int \frac {2+3 x^2}{\left (5+x^4\right )^{3/2}} \, dx=\int { \frac {3 \, x^{2} + 2}{{\left (x^{4} + 5\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {2+3 x^2}{\left (5+x^4\right )^{3/2}} \, dx=\int { \frac {3 \, x^{2} + 2}{{\left (x^{4} + 5\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {2+3 x^2}{\left (5+x^4\right )^{3/2}} \, dx=\int \frac {3\,x^2+2}{{\left (x^4+5\right )}^{3/2}} \,d x \]