\(\int \frac {1}{(a+b x^2) \sqrt {4+5 x^4}} \, dx\) [433]

Optimal result

Integrand size = 21, antiderivative size = 299 \[ \int \frac {1}{\left (a+b x^2\right ) \sqrt {4+5 x^4}} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt {5 a^2+4 b^2} x}{\sqrt {a} \sqrt {b} \sqrt {4+5 x^4}}\right )}{2 \sqrt {a} \sqrt {5 a^2+4 b^2}}+\frac {5^{3/4} \left (2+\sqrt {5} x^2\right ) \sqrt {\frac {4+5 x^4}{\left (2+\sqrt {5} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right ),\frac {1}{2}\right )}{2 \sqrt {2} \left (5 a-2 \sqrt {5} b\right ) \sqrt {4+5 x^4}}-\frac {\sqrt [4]{5} \left (\sqrt {5} a+2 b\right ) \left (2+\sqrt {5} x^2\right ) \sqrt {\frac {4+5 x^4}{\left (2+\sqrt {5} x^2\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {5} a-2 b\right )^2}{8 \sqrt {5} a b},2 \arctan \left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} a \left (5 a-2 \sqrt {5} b\right ) \sqrt {4+5 x^4}} \] Output:

1/2*b^(1/2)*arctan((5*a^2+4*b^2)^(1/2)*x/a^(1/2)/b^(1/2)/(5*x^4+4)^(1/2))/ 
a^(1/2)/(5*a^2+4*b^2)^(1/2)+1/4*5^(3/4)*(2+5^(1/2)*x^2)*((5*x^4+4)/(2+5^(1 
/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(1/2*5^(1/4)*x*2^(1/2)),1/2*2^(1 
/2))*2^(1/2)/(5*a-2*b*5^(1/2))/(5*x^4+4)^(1/2)-1/8*5^(1/4)*(5^(1/2)*a+2*b) 
*(2+5^(1/2)*x^2)*((5*x^4+4)/(2+5^(1/2)*x^2)^2)^(1/2)*EllipticPi(sin(2*arct 
an(1/2*5^(1/4)*x*2^(1/2))),-1/40*(5^(1/2)*a-2*b)^2*5^(1/2)/a/b,1/2*2^(1/2) 
)*2^(1/2)/a/(5*a-2*b*5^(1/2))/(5*x^4+4)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.17 \[ \int \frac {1}{\left (a+b x^2\right ) \sqrt {4+5 x^4}} \, dx=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \operatorname {EllipticPi}\left (-\frac {2 i b}{\sqrt {5} a},i \text {arcsinh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{5} x\right ),-1\right )}{\sqrt [4]{5} a} \] Input:

Integrate[1/((a + b*x^2)*Sqrt[4 + 5*x^4]),x]
 

Output:

((-1/2 - I/2)*EllipticPi[((-2*I)*b)/(Sqrt[5]*a), I*ArcSinh[(1/2 + I/2)*5^( 
1/4)*x], -1])/(5^(1/4)*a)
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1541, 27, 761, 2221}

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.

\(\displaystyle \int \frac {1}{\sqrt {5 x^4+4} \left (a+b x^2\right )} \, dx\)

\(\Big \downarrow \) 1541

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 2221

\(\displaystyle \frac {\left (\sqrt {5} x^2+2\right ) \sqrt {\frac {5 x^4+4}{\left (\sqrt {5} x^2+2\right )^2}} \left (5 a+2 \sqrt {5} b\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right ),\frac {1}{2}\right )}{2 \sqrt {2} \sqrt [4]{5} \sqrt {5 x^4+4} \left (5 a^2-4 b^2\right )}-\frac {b \left (\sqrt {5} a+2 b\right ) \left (\frac {\left (\sqrt {5} x^2+2\right ) \sqrt {\frac {5 x^4+4}{\left (\sqrt {5} x^2+2\right )^2}} \left (\sqrt {5} a+2 b\right ) \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {5} a-2 b\right )^2}{8 \sqrt {5} a b},2 \arctan \left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} \sqrt [4]{5} a b \sqrt {5 x^4+4}}-\frac {\left (\sqrt {5} a-2 b\right ) \arctan \left (\frac {x \sqrt {5 a^2+4 b^2}}{\sqrt {a} \sqrt {b} \sqrt {5 x^4+4}}\right )}{2 \sqrt {a} \sqrt {b} \sqrt {5 a^2+4 b^2}}\right )}{5 a^2-4 b^2}\)

Input:

Int[1/((a + b*x^2)*Sqrt[4 + 5*x^4]),x]
 

Output:

((5*a + 2*Sqrt[5]*b)*(2 + Sqrt[5]*x^2)*Sqrt[(4 + 5*x^4)/(2 + Sqrt[5]*x^2)^ 
2]*EllipticF[2*ArcTan[(5^(1/4)*x)/Sqrt[2]], 1/2])/(2*Sqrt[2]*5^(1/4)*(5*a^ 
2 - 4*b^2)*Sqrt[4 + 5*x^4]) - (b*(Sqrt[5]*a + 2*b)*(-1/2*((Sqrt[5]*a - 2*b 
)*ArcTan[(Sqrt[5*a^2 + 4*b^2]*x)/(Sqrt[a]*Sqrt[b]*Sqrt[4 + 5*x^4])])/(Sqrt 
[a]*Sqrt[b]*Sqrt[5*a^2 + 4*b^2]) + ((Sqrt[5]*a + 2*b)*(2 + Sqrt[5]*x^2)*Sq 
rt[(4 + 5*x^4)/(2 + Sqrt[5]*x^2)^2]*EllipticPi[-1/8*(Sqrt[5]*a - 2*b)^2/(S 
qrt[5]*a*b), 2*ArcTan[(5^(1/4)*x)/Sqrt[2]], 1/2])/(4*Sqrt[2]*5^(1/4)*a*b*S 
qrt[4 + 5*x^4])))/(5*a^2 - 4*b^2)
 

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 761

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]
 

rule 1541

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

rule 2221

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]) 
, x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(ArcTan[Rt[c*(d/e 
) + a*(e/d), 2]*(x/Sqrt[a + c*x^4])]/(2*d*e*Rt[c*(d/e) + a*(e/d), 2])), x] 
+ Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4* 
d*e*q*Sqrt[a + c*x^4]))*EllipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x 
], 1/2], x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && Po 
sQ[c/a] && EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && PosQ[c*(d/e) + a*(e/d)]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.72 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.29

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {1-\frac {i \sqrt {5}\, x^{2}}{2}}\, \sqrt {1+\frac {i \sqrt {5}\, x^{2}}{2}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {2}\, \sqrt {i \sqrt {5}}\, x}{2}, \frac {2 i \sqrt {5}\, b}{5 a}, \frac {\sqrt {-\frac {i \sqrt {5}}{2}}\, \sqrt {2}}{\sqrt {i \sqrt {5}}}\right )}{a \sqrt {i \sqrt {5}}\, \sqrt {5 x^{4}+4}}\)	\(86\)
elliptic	\(\frac {\sqrt {2}\, \sqrt {1-\frac {i \sqrt {5}\, x^{2}}{2}}\, \sqrt {1+\frac {i \sqrt {5}\, x^{2}}{2}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {2}\, \sqrt {i \sqrt {5}}\, x}{2}, \frac {2 i \sqrt {5}\, b}{5 a}, \frac {\sqrt {-\frac {i \sqrt {5}}{2}}\, \sqrt {2}}{\sqrt {i \sqrt {5}}}\right )}{a \sqrt {i \sqrt {5}}\, \sqrt {5 x^{4}+4}}\)	\(86\)

Input:

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

Output:

1/a/(1/2*I*5^(1/2))^(1/2)*(1-1/2*I*5^(1/2)*x^2)^(1/2)*(1+1/2*I*5^(1/2)*x^2 
)^(1/2)/(5*x^4+4)^(1/2)*EllipticPi((1/2*I*5^(1/2))^(1/2)*x,2/5*I*5^(1/2)/a 
*b,(-1/2*I*5^(1/2))^(1/2)/(1/2*I*5^(1/2))^(1/2))
 

Fricas [F]

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

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

Output:

integral(sqrt(5*x^4 + 4)/(5*b*x^6 + 5*a*x^4 + 4*b*x^2 + 4*a), x)
 

Sympy [F]

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

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

Output:

Integral(1/((a + b*x**2)*sqrt(5*x**4 + 4)), x)
 

Maxima [F]

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

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

Output:

integrate(1/(sqrt(5*x^4 + 4)*(b*x^2 + a)), x)
 

Giac [F]

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

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

Output:

integrate(1/(sqrt(5*x^4 + 4)*(b*x^2 + a)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b x^2\right ) \sqrt {4+5 x^4}} \, dx=\int \frac {\sqrt {5 x^{4}+4}}{5 b \,x^{6}+5 a \,x^{4}+4 b \,x^{2}+4 a}d x \] Input:

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

Output:

int(sqrt(5*x**4 + 4)/(5*a*x**4 + 4*a + 5*b*x**6 + 4*b*x**2),x)