\(\int \frac {1}{(a+b x^2) \sqrt {4+5 x^4}} \, dx\) [171]

Optimal result

Integrand size = 21, antiderivative size = 310 \[ \int \frac {1}{\left (a+b x^2\right ) \sqrt {4+5 x^4}} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt {5 a^2+4 b^2} x}{\sqrt {a} \sqrt {b} \sqrt {4+5 x^4}}\right )}{2 \sqrt {a} \sqrt {5 a^2+4 b^2}}+\frac {\sqrt [4]{5} \left (\sqrt {5} a+2 b\right ) \left (2+\sqrt {5} x^2\right ) \sqrt {\frac {4+5 x^4}{\left (2+\sqrt {5} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right ),\frac {1}{2}\right )}{2 \sqrt {2} \left (5 a^2-4 b^2\right ) \sqrt {4+5 x^4}}-\frac {\left (\sqrt {5} a+2 b\right )^2 \left (2+\sqrt {5} x^2\right ) \sqrt {\frac {4+5 x^4}{\left (2+\sqrt {5} x^2\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {5} a-2 b\right )^2}{8 \sqrt {5} a b},2 \arctan \left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} \sqrt [4]{5} a \left (5 a^2-4 b^2\right ) \sqrt {4+5 x^4}} \]

[Out]

1/2*arctan(x*(5*a^2+4*b^2)^(1/2)/a^(1/2)/b^(1/2)/(5*x^4+4)^(1/2))*b^(1/2)/a^(1/2)/(5*a^2+4*b^2)^(1/2)+1/4*5^(1
/4)*(cos(2*arctan(1/2*5^(1/4)*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*5^(1/4)*x*2^(1/2)))*EllipticF(sin(2*arctan
(1/2*5^(1/4)*x*2^(1/2))),1/2*2^(1/2))*(2*b+a*5^(1/2))*(2+x^2*5^(1/2))*((5*x^4+4)/(2+x^2*5^(1/2))^2)^(1/2)/(5*a
^2-4*b^2)*2^(1/2)/(5*x^4+4)^(1/2)-1/40*(cos(2*arctan(1/2*5^(1/4)*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*5^(1/4)
*x*2^(1/2)))*EllipticPi(sin(2*arctan(1/2*5^(1/4)*x*2^(1/2))),-1/40*(-2*b+a*5^(1/2))^2/a/b*5^(1/2),1/2*2^(1/2))
*(2*b+a*5^(1/2))^2*(2+x^2*5^(1/2))*((5*x^4+4)/(2+x^2*5^(1/2))^2)^(1/2)*5^(3/4)/a/(5*a^2-4*b^2)*2^(1/2)/(5*x^4+
4)^(1/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1231, 226, 1721} \[ \int \frac {1}{\left (a+b x^2\right ) \sqrt {4+5 x^4}} \, dx=\frac {\sqrt {b} \arctan \left (\frac {x \sqrt {5 a^2+4 b^2}}{\sqrt {a} \sqrt {b} \sqrt {5 x^4+4}}\right )}{2 \sqrt {a} \sqrt {5 a^2+4 b^2}}+\frac {\sqrt [4]{5} \left (\sqrt {5} x^2+2\right ) \sqrt {\frac {5 x^4+4}{\left (\sqrt {5} x^2+2\right )^2}} \left (\sqrt {5} a+2 b\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right ),\frac {1}{2}\right )}{2 \sqrt {2} \sqrt {5 x^4+4} \left (5 a^2-4 b^2\right )}-\frac {\left (\sqrt {5} x^2+2\right ) \sqrt {\frac {5 x^4+4}{\left (\sqrt {5} x^2+2\right )^2}} \left (\sqrt {5} a+2 b\right )^2 \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {5} a-2 b\right )^2}{8 \sqrt {5} a b},2 \arctan \left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} \sqrt [4]{5} a \sqrt {5 x^4+4} \left (5 a^2-4 b^2\right )} \]

[In]

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

[Out]

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

Rule 226

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 1231

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(c*d + a*e*q
)/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] - Dist[(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 1721

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)*(A + B*x^2)*(Sqrt[A^2*((a + c*x^4)/(a*(A + B*x^2)^2))]/(4*d*e*A*q*Sqrt[a + c*x^4]))*El
lipticPi[Cancel[-(B*d - A*e)^2/(4*d*e*A*B)], 2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (2 b \left (\sqrt {5} a+2 b\right )\right ) \int \frac {1+\frac {\sqrt {5} x^2}{2}}{\left (a+b x^2\right ) \sqrt {4+5 x^4}} \, dx}{5 a^2-4 b^2}+\frac {\left (5 a+2 \sqrt {5} b\right ) \int \frac {1}{\sqrt {4+5 x^4}} \, dx}{5 a^2-4 b^2} \\ & = \frac {\sqrt {b} \tan ^{-1}\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 {\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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right )|\frac {1}{2}\right )}{2 \sqrt {2} \left (5 a^2-4 b^2\right ) \sqrt {4+5 x^4}}-\frac {\left (\sqrt {5} a+2 b\right )^2 \left (2+\sqrt {5} x^2\right ) \sqrt {\frac {4+5 x^4}{\left (2+\sqrt {5} x^2\right )^2}} \Pi \left (-\frac {\left (\sqrt {5} a-2 b\right )^2}{8 \sqrt {5} a b};2 \tan ^{-1}\left (\frac {\sqrt [4]{5} x}{\sqrt {2}}\right )|\frac {1}{2}\right )}{4 \sqrt {2} \sqrt [4]{5} a \left (5 a^2-4 b^2\right ) \sqrt {4+5 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.14 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\left (a+b x^2\right ) \sqrt {4+5 x^4}} \, dx=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \operatorname {EllipticPi}\left (-\frac {2 i b}{\sqrt {5} a},i \text {arcsinh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{5} x\right ),-1\right )}{\sqrt [4]{5} a} \]

[In]

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

[Out]

((-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)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.28

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

[In]

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

[Out]

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)*b/a,(-1/2*I*5^(1/2))^(1/2)/(1/2*I*5^(1/2))^(1/2))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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