\(\int \frac {1}{(2+5 x^2+4 x^4)^{3/2}} \, dx\) [298]

Optimal result

Integrand size = 16, antiderivative size = 257 \[ \int \frac {1}{\left (2+5 x^2+4 x^4\right )^{3/2}} \, dx=-\frac {x \left (9+20 x^2\right )}{14 \sqrt {2+5 x^2+4 x^4}}+\frac {5 x \sqrt {2+5 x^2+4 x^4}}{7 \sqrt {2} \left (1+\sqrt {2} x^2\right )}-\frac {5 \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {2+5 x^2+4 x^4}{\left (1+\sqrt {2} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{2} x\right )|\frac {1}{16} \left (8-5 \sqrt {2}\right )\right )}{7 \sqrt [4]{2} \sqrt {2+5 x^2+4 x^4}}+\frac {\left (5+4 \sqrt {2}\right ) \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {2+5 x^2+4 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{16} \left (8-5 \sqrt {2}\right )\right )}{14 \sqrt [4]{2} \sqrt {2+5 x^2+4 x^4}} \] Output:

-1/14*x*(20*x^2+9)/(4*x^4+5*x^2+2)^(1/2)+5/14*x*(4*x^4+5*x^2+2)^(1/2)*2^(1 
/2)/(1+x^2*2^(1/2))-5/14*(1+x^2*2^(1/2))*((4*x^4+5*x^2+2)/(1+x^2*2^(1/2))^ 
2)^(1/2)*EllipticE(sin(2*arctan(2^(1/4)*x)),1/4*(8-5*2^(1/2))^(1/2))*2^(3/ 
4)/(4*x^4+5*x^2+2)^(1/2)+1/28*(5+4*2^(1/2))*(1+x^2*2^(1/2))*((4*x^4+5*x^2+ 
2)/(1+x^2*2^(1/2))^2)^(1/2)*InverseJacobiAM(2*arctan(2^(1/4)*x),1/4*(8-5*2 
^(1/2))^(1/2))*2^(3/4)/(4*x^4+5*x^2+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.46 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\left (2+5 x^2+4 x^4\right )^{3/2}} \, dx=\frac {-8 \sqrt {-\frac {i}{-5 i+\sqrt {7}}} x \left (9+20 x^2\right )-5 \sqrt {2} \left (5 i+\sqrt {7}\right ) \sqrt {\frac {-5 i+\sqrt {7}-8 i x^2}{-5 i+\sqrt {7}}} \sqrt {\frac {5 i+\sqrt {7}+8 i x^2}{5 i+\sqrt {7}}} E\left (i \text {arcsinh}\left (2 \sqrt {-\frac {2 i}{-5 i+\sqrt {7}}} x\right )|\frac {5 i-\sqrt {7}}{5 i+\sqrt {7}}\right )+\sqrt {2} \left (-7 i+5 \sqrt {7}\right ) \sqrt {\frac {-5 i+\sqrt {7}-8 i x^2}{-5 i+\sqrt {7}}} \sqrt {\frac {5 i+\sqrt {7}+8 i x^2}{5 i+\sqrt {7}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (2 \sqrt {-\frac {2 i}{-5 i+\sqrt {7}}} x\right ),\frac {5 i-\sqrt {7}}{5 i+\sqrt {7}}\right )}{112 \sqrt {-\frac {i}{-5 i+\sqrt {7}}} \sqrt {2+5 x^2+4 x^4}} \] Input:

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

Output:

(-8*Sqrt[(-I)/(-5*I + Sqrt[7])]*x*(9 + 20*x^2) - 5*Sqrt[2]*(5*I + Sqrt[7]) 
*Sqrt[(-5*I + Sqrt[7] - (8*I)*x^2)/(-5*I + Sqrt[7])]*Sqrt[(5*I + Sqrt[7] + 
 (8*I)*x^2)/(5*I + Sqrt[7])]*EllipticE[I*ArcSinh[2*Sqrt[(-2*I)/(-5*I + Sqr 
t[7])]*x], (5*I - Sqrt[7])/(5*I + Sqrt[7])] + Sqrt[2]*(-7*I + 5*Sqrt[7])*S 
qrt[(-5*I + Sqrt[7] - (8*I)*x^2)/(-5*I + Sqrt[7])]*Sqrt[(5*I + Sqrt[7] + ( 
8*I)*x^2)/(5*I + Sqrt[7])]*EllipticF[I*ArcSinh[2*Sqrt[(-2*I)/(-5*I + Sqrt[ 
7])]*x], (5*I - Sqrt[7])/(5*I + Sqrt[7])])/(112*Sqrt[(-I)/(-5*I + Sqrt[7]) 
]*Sqrt[2 + 5*x^2 + 4*x^4])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1405, 27, 1511, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{14} \int \frac {4 \left (5 x^2+4\right )}{\sqrt {4 x^4+5 x^2+2}}dx-\frac {x \left (20 x^2+9\right )}{14 \sqrt {4 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{7} \int \frac {5 x^2+4}{\sqrt {4 x^4+5 x^2+2}}dx-\frac {x \left (20 x^2+9\right )}{14 \sqrt {4 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {2}{7} \left (\frac {1}{2} \left (8+5 \sqrt {2}\right ) \int \frac {1}{\sqrt {4 x^4+5 x^2+2}}dx-\frac {5 \int \frac {1-\sqrt {2} x^2}{\sqrt {4 x^4+5 x^2+2}}dx}{\sqrt {2}}\right )-\frac {x \left (20 x^2+9\right )}{14 \sqrt {4 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {2}{7} \left (\frac {\left (8+5 \sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {4 x^4+5 x^2+2}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{16} \left (8-5 \sqrt {2}\right )\right )}{4\ 2^{3/4} \sqrt {4 x^4+5 x^2+2}}-\frac {5 \int \frac {1-\sqrt {2} x^2}{\sqrt {4 x^4+5 x^2+2}}dx}{\sqrt {2}}\right )-\frac {x \left (20 x^2+9\right )}{14 \sqrt {4 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {2}{7} \left (\frac {\left (8+5 \sqrt {2}\right ) \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {4 x^4+5 x^2+2}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{16} \left (8-5 \sqrt {2}\right )\right )}{4\ 2^{3/4} \sqrt {4 x^4+5 x^2+2}}-\frac {5 \left (\frac {\left (\sqrt {2} x^2+1\right ) \sqrt {\frac {4 x^4+5 x^2+2}{\left (\sqrt {2} x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt [4]{2} x\right )|\frac {1}{16} \left (8-5 \sqrt {2}\right )\right )}{2^{3/4} \sqrt {4 x^4+5 x^2+2}}-\frac {x \sqrt {4 x^4+5 x^2+2}}{2 \left (\sqrt {2} x^2+1\right )}\right )}{\sqrt {2}}\right )-\frac {x \left (20 x^2+9\right )}{14 \sqrt {4 x^4+5 x^2+2}}\)

Input:

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

Output:

-1/14*(x*(9 + 20*x^2))/Sqrt[2 + 5*x^2 + 4*x^4] + (2*((-5*(-1/2*(x*Sqrt[2 + 
 5*x^2 + 4*x^4])/(1 + Sqrt[2]*x^2) + ((1 + Sqrt[2]*x^2)*Sqrt[(2 + 5*x^2 + 
4*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticE[2*ArcTan[2^(1/4)*x], (8 - 5*Sqrt[2]) 
/16])/(2^(3/4)*Sqrt[2 + 5*x^2 + 4*x^4])))/Sqrt[2] + ((8 + 5*Sqrt[2])*(1 + 
Sqrt[2]*x^2)*Sqrt[(2 + 5*x^2 + 4*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*Arc 
Tan[2^(1/4)*x], (8 - 5*Sqrt[2])/16])/(4*2^(3/4)*Sqrt[2 + 5*x^2 + 4*x^4]))) 
/7
 

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 = 1.93 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.92

method	result	size
risch	\(-\frac {x \left (20 x^{2}+9\right )}{14 \sqrt {4 x^{4}+5 x^{2}+2}}+\frac {16 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )}{7 \sqrt {-5+i \sqrt {7}}\, \sqrt {4 x^{4}+5 x^{2}+2}}-\frac {80 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )\right )}{7 \sqrt {-5+i \sqrt {7}}\, \sqrt {4 x^{4}+5 x^{2}+2}\, \left (i \sqrt {7}+5\right )}\)	\(237\)
default	\(-\frac {8 \left (\frac {9}{112} x +\frac {5}{28} x^{3}\right )}{\sqrt {4 x^{4}+5 x^{2}+2}}+\frac {16 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )}{7 \sqrt {-5+i \sqrt {7}}\, \sqrt {4 x^{4}+5 x^{2}+2}}-\frac {80 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )\right )}{7 \sqrt {-5+i \sqrt {7}}\, \sqrt {4 x^{4}+5 x^{2}+2}\, \left (i \sqrt {7}+5\right )}\)	\(238\)
elliptic	\(-\frac {8 \left (\frac {9}{112} x +\frac {5}{28} x^{3}\right )}{\sqrt {4 x^{4}+5 x^{2}+2}}+\frac {16 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )}{7 \sqrt {-5+i \sqrt {7}}\, \sqrt {4 x^{4}+5 x^{2}+2}}-\frac {80 \sqrt {1-\left (-\frac {5}{4}+\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {i \sqrt {7}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+i \sqrt {7}}}{2}, \frac {\sqrt {9+5 i \sqrt {7}}}{4}\right )\right )}{7 \sqrt {-5+i \sqrt {7}}\, \sqrt {4 x^{4}+5 x^{2}+2}\, \left (i \sqrt {7}+5\right )}\)	\(238\)

Input:

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

Output:

-1/14*x*(20*x^2+9)/(4*x^4+5*x^2+2)^(1/2)+16/7/(-5+I*7^(1/2))^(1/2)*(1-(-5/ 
4+1/4*I*7^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*I*7^(1/2))*x^2)^(1/2)/(4*x^4+5*x^ 
2+2)^(1/2)*EllipticF(1/2*x*(-5+I*7^(1/2))^(1/2),1/4*(9+5*I*7^(1/2))^(1/2)) 
-80/7/(-5+I*7^(1/2))^(1/2)*(1-(-5/4+1/4*I*7^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4 
*I*7^(1/2))*x^2)^(1/2)/(4*x^4+5*x^2+2)^(1/2)/(I*7^(1/2)+5)*(EllipticF(1/2* 
x*(-5+I*7^(1/2))^(1/2),1/4*(9+5*I*7^(1/2))^(1/2))-EllipticE(1/2*x*(-5+I*7^ 
(1/2))^(1/2),1/4*(9+5*I*7^(1/2))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\left (2+5 x^2+4 x^4\right )^{3/2}} \, dx=\frac {5 \, \sqrt {2} {\left (20 \, x^{4} + 25 \, x^{2} - \sqrt {-7} {\left (4 \, x^{4} + 5 \, x^{2} + 2\right )} + 10\right )} \sqrt {\sqrt {-7} - 5} E(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {-7} - 5}\right )\,|\,\frac {5}{16} \, \sqrt {-7} + \frac {9}{16}) - \sqrt {2} {\left (180 \, x^{4} + 225 \, x^{2} - \sqrt {-7} {\left (4 \, x^{4} + 5 \, x^{2} + 2\right )} + 90\right )} \sqrt {\sqrt {-7} - 5} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {-7} - 5}\right )\,|\,\frac {5}{16} \, \sqrt {-7} + \frac {9}{16}) - 8 \, \sqrt {4 \, x^{4} + 5 \, x^{2} + 2} {\left (20 \, x^{3} + 9 \, x\right )}}{112 \, {\left (4 \, x^{4} + 5 \, x^{2} + 2\right )}} \] Input:

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

Output:

1/112*(5*sqrt(2)*(20*x^4 + 25*x^2 - sqrt(-7)*(4*x^4 + 5*x^2 + 2) + 10)*sqr 
t(sqrt(-7) - 5)*elliptic_e(arcsin(1/2*x*sqrt(sqrt(-7) - 5)), 5/16*sqrt(-7) 
 + 9/16) - sqrt(2)*(180*x^4 + 225*x^2 - sqrt(-7)*(4*x^4 + 5*x^2 + 2) + 90) 
*sqrt(sqrt(-7) - 5)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(-7) - 5)), 5/16*sqrt 
(-7) + 9/16) - 8*sqrt(4*x^4 + 5*x^2 + 2)*(20*x^3 + 9*x))/(4*x^4 + 5*x^2 + 
2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (2+5 x^2+4 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {4 x^{4}+5 x^{2}+2}}{16 x^{8}+40 x^{6}+41 x^{4}+20 x^{2}+4}d x \] Input:

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

Output:

int(sqrt(4*x**4 + 5*x**2 + 2)/(16*x**8 + 40*x**6 + 41*x**4 + 20*x**2 + 4), 
x)