\(\int \frac {1}{(2+5 x^2+5 x^4)^{3/2}} \, dx\) [297]

Optimal result

Integrand size = 16, antiderivative size = 261 \[ \int \frac {1}{\left (2+5 x^2+5 x^4\right )^{3/2}} \, dx=-\frac {x \left (1+5 x^2\right )}{6 \sqrt {2+5 x^2+5 x^4}}+\frac {5 x \sqrt {2+5 x^2+5 x^4}}{6 \left (\sqrt {10}+5 x^2\right )}-\frac {\sqrt [4]{5} \left (2+\sqrt {10} x^2\right ) \sqrt {\frac {2+5 x^2+5 x^4}{\left (2+\sqrt {10} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {5}{2}} x\right )|\frac {1}{8} \left (4-\sqrt {10}\right )\right )}{3\ 2^{3/4} \sqrt {2+5 x^2+5 x^4}}+\frac {\left (5+2 \sqrt {10}\right ) \left (2+\sqrt {10} x^2\right ) \sqrt {\frac {2+5 x^2+5 x^4}{\left (2+\sqrt {10} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {5}{2}} x\right ),\frac {1}{8} \left (4-\sqrt {10}\right )\right )}{6\ 10^{3/4} \sqrt {2+5 x^2+5 x^4}} \] Output:

-1/6*x*(5*x^2+1)/(5*x^4+5*x^2+2)^(1/2)+5*x*(5*x^4+5*x^2+2)^(1/2)/(6*10^(1/ 
2)+30*x^2)-1/6*5^(1/4)*(2+10^(1/2)*x^2)*((5*x^4+5*x^2+2)/(2+10^(1/2)*x^2)^ 
2)^(1/2)*EllipticE(sin(2*arctan(1/2*5^(1/4)*2^(3/4)*x)),1/4*(8-2*10^(1/2)) 
^(1/2))*2^(1/4)/(5*x^4+5*x^2+2)^(1/2)+1/60*(5+2*10^(1/2))*(2+10^(1/2)*x^2) 
*((5*x^4+5*x^2+2)/(2+10^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(1/2*5 
^(1/4)*2^(3/4)*x),1/4*(8-2*10^(1/2))^(1/2))*10^(1/4)/(5*x^4+5*x^2+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.47 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\left (2+5 x^2+5 x^4\right )^{3/2}} \, dx=\frac {-20 \sqrt {-\frac {i}{-5 i+\sqrt {15}}} x \left (1+5 x^2\right )-5 \sqrt {2} \left (\sqrt {3}+i \sqrt {5}\right ) \sqrt {\frac {-5 i+\sqrt {15}-10 i x^2}{-5 i+\sqrt {15}}} \sqrt {\frac {5 i+\sqrt {15}+10 i x^2}{5 i+\sqrt {15}}} E\left (i \text {arcsinh}\left (\sqrt {-\frac {10 i}{-5 i+\sqrt {15}}} x\right )|\frac {5 i-\sqrt {15}}{5 i+\sqrt {15}}\right )+\sqrt {2} \left (5 \sqrt {3}-3 i \sqrt {5}\right ) \sqrt {\frac {-5 i+\sqrt {15}-10 i x^2}{-5 i+\sqrt {15}}} \sqrt {\frac {5 i+\sqrt {15}+10 i x^2}{5 i+\sqrt {15}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {10 i}{-5 i+\sqrt {15}}} x\right ),\frac {5 i-\sqrt {15}}{5 i+\sqrt {15}}\right )}{120 \sqrt {-\frac {i}{-5 i+\sqrt {15}}} \sqrt {2+5 x^2+5 x^4}} \] Input:

Integrate[(2 + 5*x^2 + 5*x^4)^(-3/2),x]
 

Output:

(-20*Sqrt[(-I)/(-5*I + Sqrt[15])]*x*(1 + 5*x^2) - 5*Sqrt[2]*(Sqrt[3] + I*S 
qrt[5])*Sqrt[(-5*I + Sqrt[15] - (10*I)*x^2)/(-5*I + Sqrt[15])]*Sqrt[(5*I + 
 Sqrt[15] + (10*I)*x^2)/(5*I + Sqrt[15])]*EllipticE[I*ArcSinh[Sqrt[(-10*I) 
/(-5*I + Sqrt[15])]*x], (5*I - Sqrt[15])/(5*I + Sqrt[15])] + Sqrt[2]*(5*Sq 
rt[3] - (3*I)*Sqrt[5])*Sqrt[(-5*I + Sqrt[15] - (10*I)*x^2)/(-5*I + Sqrt[15 
])]*Sqrt[(5*I + Sqrt[15] + (10*I)*x^2)/(5*I + Sqrt[15])]*EllipticF[I*ArcSi 
nh[Sqrt[(-10*I)/(-5*I + Sqrt[15])]*x], (5*I - Sqrt[15])/(5*I + Sqrt[15])]) 
/(120*Sqrt[(-I)/(-5*I + Sqrt[15])]*Sqrt[2 + 5*x^2 + 5*x^4])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 27, 1511, 27, 1416, 1509}

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.

\(\displaystyle \int \frac {1}{\left (5 x^4+5 x^2+2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{30} \int \frac {5 \left (5 x^2+4\right )}{\sqrt {5 x^4+5 x^2+2}}dx-\frac {x \left (5 x^2+1\right )}{6 \sqrt {5 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \frac {5 x^2+4}{\sqrt {5 x^4+5 x^2+2}}dx-\frac {x \left (5 x^2+1\right )}{6 \sqrt {5 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {1}{6} \left (\left (4+\sqrt {10}\right ) \int \frac {1}{\sqrt {5 x^4+5 x^2+2}}dx-\sqrt {10} \int \frac {2-\sqrt {10} x^2}{2 \sqrt {5 x^4+5 x^2+2}}dx\right )-\frac {x \left (5 x^2+1\right )}{6 \sqrt {5 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\left (4+\sqrt {10}\right ) \int \frac {1}{\sqrt {5 x^4+5 x^2+2}}dx-\sqrt {\frac {5}{2}} \int \frac {2-\sqrt {10} x^2}{\sqrt {5 x^4+5 x^2+2}}dx\right )-\frac {x \left (5 x^2+1\right )}{6 \sqrt {5 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {1}{6} \left (\frac {\left (4+\sqrt {10}\right ) \left (\sqrt {10} x^2+2\right ) \sqrt {\frac {5 x^4+5 x^2+2}{\left (\sqrt {10} x^2+2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {5}{2}} x\right ),\frac {1}{8} \left (4-\sqrt {10}\right )\right )}{2 \sqrt [4]{10} \sqrt {5 x^4+5 x^2+2}}-\sqrt {\frac {5}{2}} \int \frac {2-\sqrt {10} x^2}{\sqrt {5 x^4+5 x^2+2}}dx\right )-\frac {x \left (5 x^2+1\right )}{6 \sqrt {5 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {1}{6} \left (\frac {\left (4+\sqrt {10}\right ) \left (\sqrt {10} x^2+2\right ) \sqrt {\frac {5 x^4+5 x^2+2}{\left (\sqrt {10} x^2+2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {5}{2}} x\right ),\frac {1}{8} \left (4-\sqrt {10}\right )\right )}{2 \sqrt [4]{10} \sqrt {5 x^4+5 x^2+2}}-\sqrt {\frac {5}{2}} \left (\frac {2^{3/4} \left (\sqrt {10} x^2+2\right ) \sqrt {\frac {5 x^4+5 x^2+2}{\left (\sqrt {10} x^2+2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {5}{2}} x\right )|\frac {1}{8} \left (4-\sqrt {10}\right )\right )}{\sqrt [4]{5} \sqrt {5 x^4+5 x^2+2}}-\frac {2 x \sqrt {5 x^4+5 x^2+2}}{\sqrt {10} x^2+2}\right )\right )-\frac {x \left (5 x^2+1\right )}{6 \sqrt {5 x^4+5 x^2+2}}\)

Input:

Int[(2 + 5*x^2 + 5*x^4)^(-3/2),x]
 

Output:

-1/6*(x*(1 + 5*x^2))/Sqrt[2 + 5*x^2 + 5*x^4] + (-(Sqrt[5/2]*((-2*x*Sqrt[2 
+ 5*x^2 + 5*x^4])/(2 + Sqrt[10]*x^2) + (2^(3/4)*(2 + Sqrt[10]*x^2)*Sqrt[(2 
 + 5*x^2 + 5*x^4)/(2 + Sqrt[10]*x^2)^2]*EllipticE[2*ArcTan[(5/2)^(1/4)*x], 
 (4 - Sqrt[10])/8])/(5^(1/4)*Sqrt[2 + 5*x^2 + 5*x^4]))) + ((4 + Sqrt[10])* 
(2 + Sqrt[10]*x^2)*Sqrt[(2 + 5*x^2 + 5*x^4)/(2 + Sqrt[10]*x^2)^2]*Elliptic 
F[2*ArcTan[(5/2)^(1/4)*x], (4 - Sqrt[10])/8])/(2*10^(1/4)*Sqrt[2 + 5*x^2 + 
 5*x^4]))/6
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1405

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x)*(b^2 
- 2*a*c + b*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c)) 
), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[(b^2 - 2*a*c + 2*(p + 1)*( 
b^2 - 4*a*c) + b*c*(4*p + 7)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; Fr 
eeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]
 

rule 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q 
^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* 
x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 
/(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 
- 4*a*c, 0] && PosQ[c/a]
 

rule 1511

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + b*x^2 + c*x^ 
4], x], x] - Simp[e/q   Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; 
NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos 
Q[c/a]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.36 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {x \left (5 x^{2}+1\right )}{6 \sqrt {5 x^{4}+5 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )}{3 \sqrt {-5+i \sqrt {15}}\, \sqrt {5 x^{4}+5 x^{2}+2}}-\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )\right )}{3 \sqrt {-5+i \sqrt {15}}\, \sqrt {5 x^{4}+5 x^{2}+2}\, \left (5+i \sqrt {15}\right )}\)	\(237\)
default	\(-\frac {10 \left (\frac {1}{60} x +\frac {1}{12} x^{3}\right )}{\sqrt {5 x^{4}+5 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )}{3 \sqrt {-5+i \sqrt {15}}\, \sqrt {5 x^{4}+5 x^{2}+2}}-\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )\right )}{3 \sqrt {-5+i \sqrt {15}}\, \sqrt {5 x^{4}+5 x^{2}+2}\, \left (5+i \sqrt {15}\right )}\)	\(238\)
elliptic	\(-\frac {10 \left (\frac {1}{60} x +\frac {1}{12} x^{3}\right )}{\sqrt {5 x^{4}+5 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )}{3 \sqrt {-5+i \sqrt {15}}\, \sqrt {5 x^{4}+5 x^{2}+2}}-\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+i \sqrt {15}}}{2}, \frac {\sqrt {1+i \sqrt {15}}}{2}\right )\right )}{3 \sqrt {-5+i \sqrt {15}}\, \sqrt {5 x^{4}+5 x^{2}+2}\, \left (5+i \sqrt {15}\right )}\)	\(238\)

Input:

int(1/(5*x^4+5*x^2+2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/6*x*(5*x^2+1)/(5*x^4+5*x^2+2)^(1/2)+4/3/(-5+I*15^(1/2))^(1/2)*(1-(-5/4+ 
1/4*I*15^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*I*15^(1/2))*x^2)^(1/2)/(5*x^4+5*x^ 
2+2)^(1/2)*EllipticF(1/2*x*(-5+I*15^(1/2))^(1/2),1/2*(1+I*15^(1/2))^(1/2)) 
-20/3/(-5+I*15^(1/2))^(1/2)*(1-(-5/4+1/4*I*15^(1/2))*x^2)^(1/2)*(1-(-5/4-1 
/4*I*15^(1/2))*x^2)^(1/2)/(5*x^4+5*x^2+2)^(1/2)/(5+I*15^(1/2))*(EllipticF( 
1/2*x*(-5+I*15^(1/2))^(1/2),1/2*(1+I*15^(1/2))^(1/2))-EllipticE(1/2*x*(-5+ 
I*15^(1/2))^(1/2),1/2*(1+I*15^(1/2))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (2+5 x^2+5 x^4\right )^{3/2}} \, dx=\frac {5 \, \sqrt {2} {\left (25 \, x^{4} + 25 \, x^{2} - \sqrt {-15} {\left (5 \, x^{4} + 5 \, x^{2} + 2\right )} + 10\right )} \sqrt {\sqrt {-15} - 5} E(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {-15} - 5}\right )\,|\,\frac {1}{4} \, \sqrt {-15} + \frac {1}{4}) - \sqrt {2} {\left (225 \, x^{4} + 225 \, x^{2} - \sqrt {-15} {\left (5 \, x^{4} + 5 \, x^{2} + 2\right )} + 90\right )} \sqrt {\sqrt {-15} - 5} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {-15} - 5}\right )\,|\,\frac {1}{4} \, \sqrt {-15} + \frac {1}{4}) - 40 \, \sqrt {5 \, x^{4} + 5 \, x^{2} + 2} {\left (5 \, x^{3} + x\right )}}{240 \, {\left (5 \, x^{4} + 5 \, x^{2} + 2\right )}} \] Input:

integrate(1/(5*x^4+5*x^2+2)^(3/2),x, algorithm="fricas")
 

Output:

1/240*(5*sqrt(2)*(25*x^4 + 25*x^2 - sqrt(-15)*(5*x^4 + 5*x^2 + 2) + 10)*sq 
rt(sqrt(-15) - 5)*elliptic_e(arcsin(1/2*x*sqrt(sqrt(-15) - 5)), 1/4*sqrt(- 
15) + 1/4) - sqrt(2)*(225*x^4 + 225*x^2 - sqrt(-15)*(5*x^4 + 5*x^2 + 2) + 
90)*sqrt(sqrt(-15) - 5)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(-15) - 5)), 1/4* 
sqrt(-15) + 1/4) - 40*sqrt(5*x^4 + 5*x^2 + 2)*(5*x^3 + x))/(5*x^4 + 5*x^2 
+ 2)
 

Sympy [F]

\[ \int \frac {1}{\left (2+5 x^2+5 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (5 x^{4} + 5 x^{2} + 2\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(5*x**4+5*x**2+2)**(3/2),x)
 

Output:

Integral((5*x**4 + 5*x**2 + 2)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (2+5 x^2+5 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (5 \, x^{4} + 5 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(5*x^4+5*x^2+2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((5*x^4 + 5*x^2 + 2)^(-3/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (2+5 x^2+5 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (5 \, x^{4} + 5 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(5*x^4+5*x^2+2)^(3/2),x, algorithm="giac")
 

Output:

integrate((5*x^4 + 5*x^2 + 2)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (2+5 x^2+5 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (5\,x^4+5\,x^2+2\right )}^{3/2}} \,d x \] Input:

int(1/(5*x^2 + 5*x^4 + 2)^(3/2),x)
 

Output:

int(1/(5*x^2 + 5*x^4 + 2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (2+5 x^2+5 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {5 x^{4}+5 x^{2}+2}}{25 x^{8}+50 x^{6}+45 x^{4}+20 x^{2}+4}d x \] Input:

int(1/(5*x^4+5*x^2+2)^(3/2),x)
 

Output:

int(sqrt(5*x**4 + 5*x**2 + 2)/(25*x**8 + 50*x**6 + 45*x**4 + 20*x**2 + 4), 
x)