\(\int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx\) [1007]

Optimal result

Integrand size = 25, antiderivative size = 88 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}} \]

[Out]

1/2*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)
*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/c^(1/4)/(c*x
^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4, 226} \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}} \]

[In]

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

[Out]

((Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)],
1/2])/(2*a^(1/4)*c^(1/4)*Sqrt[a + c*x^4])

Rule 4

Int[(u_.)*((a_.) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(a + c*x^(2*n))^p, x] /; Fre
eQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[b, 0]

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]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {i \sqrt {1+\frac {c x^4}{a}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right ),-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} \sqrt {a+c x^4}} \]

[In]

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

[Out]

((-I)*Sqrt[1 + (c*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1])/(Sqrt[(I*Sqrt[c])/Sqrt[a]]*Sq
rt[a + c*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(70\)
elliptic	\(\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(70\)

[In]

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

[Out]

1/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*El
lipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {a} \left (-\frac {c}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1)}{c} \]

[In]

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

[Out]

-sqrt(a)*(-c/a)^(3/4)*elliptic_f(arcsin(x*(-c/a)^(1/4)), -1)/c

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} \]

[In]

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

[Out]

x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4))

Maxima [F]

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

[In]

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

[Out]

integrate(1/sqrt(c*x^4 + a), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/sqrt(c*x^4 + a), x)

Mupad [B] (verification not implemented)

Time = 12.90 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {x\,\sqrt {\frac {c\,x^4}{a}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {5}{4};\ -\frac {c\,x^4}{a}\right )}{\sqrt {c\,x^4+a}} \]

[In]

int(1/(a + c*x^4)^(1/2),x)

[Out]

(x*((c*x^4)/a + 1)^(1/2)*hypergeom([1/4, 1/2], 5/4, -(c*x^4)/a))/(a + c*x^4)^(1/2)