Integrand size = 17, antiderivative size = 176 \[ \int \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx=\frac {6 x \sqrt {5+x^4}}{\sqrt {5}+x^2}+\frac {1}{15} x \left (10+9 x^2\right ) \sqrt {5+x^4}-\frac {6 \sqrt [4]{5} \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {5+x^4}}+\frac {\sqrt [4]{5} \left (9+2 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{3 \sqrt {5+x^4}} \]
1/15*x*(9*x^2+10)*(x^4+5)^(1/2)+6*x*(x^4+5)^(1/2)/(x^2+5^(1/2))-6*5^(1/4)* (cos(2*arctan(1/5*x*5^(3/4)))^2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*Ellipt icE(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)+1/3*5^(1/4)*(cos(2*arctan(1/5*x*5^(3/4)))^ 2)^(1/2)/cos(2*arctan(1/5*x*5^(3/4)))*EllipticF(sin(2*arctan(1/5*x*5^(3/4) )),1/2*2^(1/2))*(x^2+5^(1/2))*(9+2*5^(1/2))*((x^4+5)/(x^2+5^(1/2))^2)^(1/2 )/(x^4+5)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.57 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.27 \[ \int \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx=\sqrt {5} x \left (2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {x^4}{5}\right )+x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {x^4}{5}\right )\right ) \]
Sqrt[5]*x*(2*Hypergeometric2F1[-1/2, 1/4, 5/4, -1/5*x^4] + x^2*Hypergeomet ric2F1[-1/2, 3/4, 7/4, -1/5*x^4])
Time = 0.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1491, 27, 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.
\(\displaystyle \int \left (3 x^2+2\right ) \sqrt {x^4+5} \, dx\) |
\(\Big \downarrow \) 1491 |
\(\displaystyle \frac {1}{15} \int \frac {10 \left (9 x^2+10\right )}{\sqrt {x^4+5}}dx+\frac {1}{15} x \sqrt {x^4+5} \left (9 x^2+10\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \int \frac {9 x^2+10}{\sqrt {x^4+5}}dx+\frac {1}{15} x \sqrt {x^4+5} \left (9 x^2+10\right )\) |
\(\Big \downarrow \) 1512 |
\(\displaystyle \frac {2}{3} \left (\left (10+9 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-9 \sqrt {5} \int \frac {\sqrt {5}-x^2}{\sqrt {5} \sqrt {x^4+5}}dx\right )+\frac {1}{15} x \sqrt {x^4+5} \left (9 x^2+10\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \left (\left (10+9 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-9 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\right )+\frac {1}{15} x \sqrt {x^4+5} \left (9 x^2+10\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2}{3} \left (\frac {\left (10+9 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-9 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\right )+\frac {1}{15} x \sqrt {x^4+5} \left (9 x^2+10\right )\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {2}{3} \left (\frac {\left (10+9 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-9 \left (\frac {\sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {x \sqrt {x^4+5}}{x^2+\sqrt {5}}\right )\right )+\frac {1}{15} x \sqrt {x^4+5} \left (9 x^2+10\right )\) |
(x*(10 + 9*x^2)*Sqrt[5 + x^4])/15 + (2*(-9*(-((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]*Ellipt icE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt[5 + x^4]) + ((10 + 9*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
3.1.17.3.1 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*(4*p + 3) + e*(4*p + 1)*x^2)*((a + c*x^4)^p/((4*p + 1)*(4*p + 3))), x] + Simp[2*(p/((4*p + 1)*(4*p + 3))) Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4*p + 1))*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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[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 = 1.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.21
method | result | size |
meijerg | \(2 \sqrt {5}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {x^{4}}{5}\right )+\sqrt {5}\, x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {x^{4}}{5}\right )\) | \(37\) |
risch | \(\frac {x \left (9 x^{2}+10\right ) \sqrt {x^{4}+5}}{15}+\frac {4 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{15 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {6 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(163\) |
default | \(\frac {2 x \sqrt {x^{4}+5}}{3}+\frac {4 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{15 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 x^{3} \sqrt {x^{4}+5}}{5}+\frac {6 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(168\) |
elliptic | \(\frac {2 x \sqrt {x^{4}+5}}{3}+\frac {4 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{15 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {3 x^{3} \sqrt {x^{4}+5}}{5}+\frac {6 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(168\) |
2*5^(1/2)*x*hypergeom([-1/2,1/4],[5/4],-1/5*x^4)+5^(1/2)*x^3*hypergeom([-1 /2,3/4],[7/4],-1/5*x^4)
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.33 \[ \int \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx=\frac {90 \, \left (-5\right )^{\frac {3}{4}} x E(\arcsin \left (\frac {\left (-5\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 70 \, \left (-5\right )^{\frac {3}{4}} x F(\arcsin \left (\frac {\left (-5\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (9 \, x^{4} + 10 \, x^{2} + 90\right )} \sqrt {x^{4} + 5}}{15 \, x} \]
1/15*(90*(-5)^(3/4)*x*elliptic_e(arcsin((-5)^(1/4)/x), -1) - 70*(-5)^(3/4) *x*elliptic_f(arcsin((-5)^(1/4)/x), -1) + (9*x^4 + 10*x^2 + 90)*sqrt(x^4 + 5))/x
Result contains complex when optimal does not.
Time = 0.82 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.43 \[ \int \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx=\frac {3 \sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac {5}{4}\right )} \]
3*sqrt(5)*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), x**4*exp_polar(I*pi)/ 5)/(4*gamma(7/4)) + sqrt(5)*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), x**4*e xp_polar(I*pi)/5)/(2*gamma(5/4))
\[ \int \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx=\int { \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )} \,d x } \]
\[ \int \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx=\int { \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )} \,d x } \]
Timed out. \[ \int \left (2+3 x^2\right ) \sqrt {5+x^4} \, dx=\int \sqrt {x^4+5}\,\left (3\,x^2+2\right ) \,d x \]