Integrand size = 20, antiderivative size = 192 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x^4} \, dx=-\frac {6 \sqrt {5+x^4}}{x}-\frac {\left (2-9 x^2\right ) \sqrt {5+x^4}}{3 x^3}+\frac {6 x \sqrt {5+x^4}}{\sqrt {5}+x^2}-\frac {6 \sqrt [4]{5} \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {5+x^4}}+\frac {\left (2+9 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{5} \sqrt {5+x^4}} \]
-6*(x^4+5)^(1/2)/x-1/3*(-9*x^2+2)*(x^4+5)^(1/2)/x^3+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)))*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)+1/15*(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+9*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 = 7.72 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.28 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x^4} \, dx=-\frac {\sqrt {5} \left (2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{4},-\frac {x^4}{5}\right )+9 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},-\frac {x^4}{5}\right )\right )}{3 x^3} \]
-1/3*(Sqrt[5]*(2*Hypergeometric2F1[-3/4, -1/2, 1/4, -1/5*x^4] + 9*x^2*Hype rgeometric2F1[-1/2, -1/4, 3/4, -1/5*x^4]))/x^3
Time = 0.32 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1595, 25, 1605, 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.
| \(\displaystyle \int \frac {\left (3 x^2+2\right ) \sqrt {x^4+5}}{x^4} \, dx\) |
| \(\Big \downarrow \) 1595 |
| \(\displaystyle -\frac {2}{3} \int -\frac {2 x^2+45}{x^2 \sqrt {x^4+5}}dx-\frac {\sqrt {x^4+5} \left (2-9 x^2\right )}{3 x^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2}{3} \int \frac {2 x^2+45}{x^2 \sqrt {x^4+5}}dx-\frac {\left (2-9 x^2\right ) \sqrt {x^4+5}}{3 x^3}\) |
| \(\Big \downarrow \) 1605 |
| \(\displaystyle \frac {2}{3} \left (-\frac {1}{5} \int -\frac {5 \left (9 x^2+2\right )}{\sqrt {x^4+5}}dx-\frac {9 \sqrt {x^4+5}}{x}\right )-\frac {\left (2-9 x^2\right ) \sqrt {x^4+5}}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{3} \left (\int \frac {9 x^2+2}{\sqrt {x^4+5}}dx-\frac {9 \sqrt {x^4+5}}{x}\right )-\frac {\left (2-9 x^2\right ) \sqrt {x^4+5}}{3 x^3}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {2}{3} \left (\left (2+9 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-9 \sqrt {5} \int \frac {\sqrt {5}-x^2}{\sqrt {5} \sqrt {x^4+5}}dx-\frac {9 \sqrt {x^4+5}}{x}\right )-\frac {\left (2-9 x^2\right ) \sqrt {x^4+5}}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{3} \left (\left (2+9 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-9 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx-\frac {9 \sqrt {x^4+5}}{x}\right )-\frac {\left (2-9 x^2\right ) \sqrt {x^4+5}}{3 x^3}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2}{3} \left (-9 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx+\frac {\left (2+9 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-\frac {9 \sqrt {x^4+5}}{x}\right )-\frac {\left (2-9 x^2\right ) \sqrt {x^4+5}}{3 x^3}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {2}{3} \left (\frac {\left (2+9 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-9 \left (\frac {\sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {x \sqrt {x^4+5}}{x^2+\sqrt {5}}\right )-\frac {9 \sqrt {x^4+5}}{x}\right )-\frac {\left (2-9 x^2\right ) \sqrt {x^4+5}}{3 x^3}\) |
-1/3*((2 - 9*x^2)*Sqrt[5 + x^4])/x^3 + (2*((-9*Sqrt[5 + x^4])/x - 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]*EllipticE[2*ArcTan[x/5^(1/4)], 1/2])/Sqrt[5 + x^4]) + ( (2 + 9*Sqrt[5])*(Sqrt[5] + x^2)*Sqrt[(5 + x^4)/(Sqrt[5] + x^2)^2]*Elliptic F[2*ArcTan[x/5^(1/4)], 1/2])/(2*5^(1/4)*Sqrt[5 + x^4])))/3
3.1.19.3.1 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[(f*x)^(m + 1)*(a + c*x^4)^p*((d*(m + 4*p + 3) + e*(m + 1)* x^2)/(f*(m + 1)*(m + 4*p + 3))), x] + Simp[4*(p/(f^2*(m + 1)*(m + 4*p + 3)) ) Int[(f*x)^(m + 2)*(a + c*x^4)^(p - 1)*(a*e*(m + 1) - c*d*(m + 4*p + 3)* x^2), x], x] /; FreeQ[{a, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1] && m + 4*p + 3 != 0 && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
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 = 2.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.21
| method | result | size |
| meijerg | \(-\frac {2 \sqrt {5}\, {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {x^{4}}{5}\right )}{3 x^{3}}-\frac {3 \sqrt {5}\, {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};-\frac {x^{4}}{5}\right )}{x}\) | \(40\) |
| default | \(-\frac {3 \sqrt {x^{4}+5}}{x}+\frac {6 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{3 x^{3}}+\frac {4 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{75 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(170\) |
| elliptic | \(-\frac {3 \sqrt {x^{4}+5}}{x}+\frac {6 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {2 \sqrt {x^{4}+5}}{3 x^{3}}+\frac {4 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{75 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(170\) |
| risch | \(-\frac {9 x^{6}+2 x^{4}+45 x^{2}+10}{3 x^{3} \sqrt {x^{4}+5}}+\frac {4 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{75 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {6 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(175\) |
-2/3*5^(1/2)/x^3*hypergeom([-3/4,-1/2],[1/4],-1/5*x^4)-3*5^(1/2)/x*hyperge om([-1/2,-1/4],[3/4],-1/5*x^4)
\[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x^4} \, dx=\int { \frac {\sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )}}{x^{4}} \,d x } \]
Result contains complex when optimal does not.
Time = 1.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.43 \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x^4} \, dx=\frac {3 \sqrt {5} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} + \frac {\sqrt {5} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 x^{3} \Gamma \left (\frac {1}{4}\right )} \]
3*sqrt(5)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), x**4*exp_polar(I*pi)/5)/ (4*x*gamma(3/4)) + sqrt(5)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), x**4*ex p_polar(I*pi)/5)/(2*x**3*gamma(1/4))
\[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x^4} \, dx=\int { \frac {\sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )}}{x^{4}} \,d x } \]
\[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x^4} \, dx=\int { \frac {\sqrt {x^{4} + 5} {\left (3 \, x^{2} + 2\right )}}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (2+3 x^2\right ) \sqrt {5+x^4}}{x^4} \, dx=\int \frac {\sqrt {x^4+5}\,\left (3\,x^2+2\right )}{x^4} \,d x \]