Integrand size = 20, antiderivative size = 199 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^2} \, dx=\frac {24 x \sqrt {5+x^4}}{\sqrt {5}+x^2}+\frac {6}{35} x \left (25+14 x^2\right ) \sqrt {5+x^4}-\frac {\left (14-3 x^2\right ) \left (5+x^4\right )^{3/2}}{7 x}-\frac {24 \sqrt [4]{5} \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {5+x^4}}+\frac {6 \sqrt [4]{5} \left (14+5 \sqrt {5}\right ) \left (\sqrt {5}+x^2\right ) \sqrt {\frac {5+x^4}{\left (\sqrt {5}+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{7 \sqrt {5+x^4}} \]
-1/7*(-3*x^2+14)*(x^4+5)^(3/2)/x+6/35*x*(14*x^2+25)*(x^4+5)^(1/2)+24*x*(x^ 4+5)^(1/2)/(x^2+5^(1/2))-24*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)+6/7*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))*(14+5*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 = 4.85 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.27 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^2} \, dx=-\frac {10 \sqrt {5} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},\frac {3}{4},-\frac {x^4}{5}\right )}{x}+15 \sqrt {5} x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {x^4}{5}\right ) \]
(-10*Sqrt[5]*Hypergeometric2F1[-3/2, -1/4, 3/4, -1/5*x^4])/x + 15*Sqrt[5]* x*Hypergeometric2F1[-3/2, 1/4, 5/4, -1/5*x^4]
Time = 0.33 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1595, 25, 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.
| \(\displaystyle \int \frac {\left (3 x^2+2\right ) \left (x^4+5\right )^{3/2}}{x^2} \, dx\) |
| \(\Big \downarrow \) 1595 |
| \(\displaystyle -\frac {6}{7} \int -\left (\left (14 x^2+15\right ) \sqrt {x^4+5}\right )dx-\frac {\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {6}{7} \int \left (14 x^2+15\right ) \sqrt {x^4+5}dx-\frac {\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}\) |
| \(\Big \downarrow \) 1491 |
| \(\displaystyle \frac {6}{7} \left (\frac {1}{15} \int \frac {30 \left (14 x^2+25\right )}{\sqrt {x^4+5}}dx+\frac {1}{5} x \sqrt {x^4+5} \left (14 x^2+25\right )\right )-\frac {\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {6}{7} \left (2 \int \frac {14 x^2+25}{\sqrt {x^4+5}}dx+\frac {1}{5} x \sqrt {x^4+5} \left (14 x^2+25\right )\right )-\frac {\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {6}{7} \left (2 \left (\left (25+14 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-14 \sqrt {5} \int \frac {\sqrt {5}-x^2}{\sqrt {5} \sqrt {x^4+5}}dx\right )+\frac {1}{5} x \sqrt {x^4+5} \left (14 x^2+25\right )\right )-\frac {\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {6}{7} \left (2 \left (\left (25+14 \sqrt {5}\right ) \int \frac {1}{\sqrt {x^4+5}}dx-14 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\right )+\frac {1}{5} x \sqrt {x^4+5} \left (14 x^2+25\right )\right )-\frac {\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {6}{7} \left (2 \left (\frac {\left (25+14 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-14 \int \frac {\sqrt {5}-x^2}{\sqrt {x^4+5}}dx\right )+\frac {1}{5} x \sqrt {x^4+5} \left (14 x^2+25\right )\right )-\frac {\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {6}{7} \left (2 \left (\frac {\left (25+14 \sqrt {5}\right ) \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{5} \sqrt {x^4+5}}-14 \left (\frac {\sqrt [4]{5} \left (x^2+\sqrt {5}\right ) \sqrt {\frac {x^4+5}{\left (x^2+\sqrt {5}\right )^2}} E\left (2 \arctan \left (\frac {x}{\sqrt [4]{5}}\right )|\frac {1}{2}\right )}{\sqrt {x^4+5}}-\frac {x \sqrt {x^4+5}}{x^2+\sqrt {5}}\right )\right )+\frac {1}{5} x \sqrt {x^4+5} \left (14 x^2+25\right )\right )-\frac {\left (14-3 x^2\right ) \left (x^4+5\right )^{3/2}}{7 x}\) |
-1/7*((14 - 3*x^2)*(5 + x^4)^(3/2))/x + (6*((x*(25 + 14*x^2)*Sqrt[5 + x^4] )/5 + 2*(-14*(-((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]) + ((25 + 14*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] ))))/7
3.1.30.3.1 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]
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])
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 1.36 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.19
| method | result | size |
| meijerg | \(-\frac {10 \sqrt {5}\, {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{2},-\frac {1}{4};\frac {3}{4};-\frac {x^{4}}{5}\right )}{x}+15 \sqrt {5}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {x^{4}}{5}\right )\) | \(38\) |
| risch | \(\frac {15 x^{10}+14 x^{8}+300 x^{6}-280 x^{4}+1125 x^{2}-1750}{35 x \sqrt {x^{4}+5}}+\frac {12 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{7 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}+\frac {24 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(185\) |
| default | \(\frac {3 x^{5} \sqrt {x^{4}+5}}{7}+\frac {45 x \sqrt {x^{4}+5}}{7}+\frac {12 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{7 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {10 \sqrt {x^{4}+5}}{x}+\frac {2 x^{3} \sqrt {x^{4}+5}}{5}+\frac {24 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(192\) |
| elliptic | \(\frac {3 x^{5} \sqrt {x^{4}+5}}{7}+\frac {45 x \sqrt {x^{4}+5}}{7}+\frac {12 \sqrt {5}\, \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )}{7 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}-\frac {10 \sqrt {x^{4}+5}}{x}+\frac {2 x^{3} \sqrt {x^{4}+5}}{5}+\frac {24 i \sqrt {25-5 i \sqrt {5}\, x^{2}}\, \sqrt {25+5 i \sqrt {5}\, x^{2}}\, \left (F\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )-E\left (\frac {x \sqrt {5}\, \sqrt {i \sqrt {5}}}{5}, i\right )\right )}{5 \sqrt {i \sqrt {5}}\, \sqrt {x^{4}+5}}\) | \(192\) |
-10*5^(1/2)/x*hypergeom([-3/2,-1/4],[3/4],-1/5*x^4)+15*5^(1/2)*x*hypergeom ([-3/2,1/4],[5/4],-1/5*x^4)
\[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^2} \, dx=\int { \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}} {\left (3 \, x^{2} + 2\right )}}{x^{2}} \,d x } \]
Result contains complex when optimal does not.
Time = 1.84 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.80 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^2} \, dx=\frac {3 \sqrt {5} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {\sqrt {5} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} + \frac {15 \sqrt {5} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {5 \sqrt {5} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {x^{4} e^{i \pi }}{5}} \right )}}{2 x \Gamma \left (\frac {3}{4}\right )} \]
3*sqrt(5)*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), x**4*exp_polar(I*pi)/ 5)/(4*gamma(9/4)) + sqrt(5)*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), x** 4*exp_polar(I*pi)/5)/(2*gamma(7/4)) + 15*sqrt(5)*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), x**4*exp_polar(I*pi)/5)/(4*gamma(5/4)) + 5*sqrt(5)*gamma(-1 /4)*hyper((-1/2, -1/4), (3/4,), x**4*exp_polar(I*pi)/5)/(2*x*gamma(3/4))
\[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^2} \, dx=\int { \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}} {\left (3 \, x^{2} + 2\right )}}{x^{2}} \,d x } \]
\[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^2} \, dx=\int { \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}} {\left (3 \, x^{2} + 2\right )}}{x^{2}} \,d x } \]
Time = 7.77 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.24 \[ \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^2} \, dx=15\,\sqrt {5}\,x\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{4};\ \frac {5}{4};\ -\frac {x^4}{5}\right )+\frac {2\,{\left (x^4+5\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {5}{4};\ -\frac {1}{4};\ -\frac {5}{x^4}\right )}{5\,x\,{\left (\frac {5}{x^4}+1\right )}^{3/2}} \]