3.11.11 \(\int \frac {1}{x^4 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx\) [1011]

3.11.11.1 Optimal result
3.11.11.2 Mathematica [C] (verified)
3.11.11.3 Rubi [A] (verified)
3.11.11.4 Maple [C] (verified)
3.11.11.5 Fricas [A] (verification not implemented)
3.11.11.6 Sympy [C] (verification not implemented)
3.11.11.7 Maxima [F]
3.11.11.8 Giac [F]
3.11.11.9 Mupad [F(-1)]

3.11.11.1 Optimal result

Integrand size = 29, antiderivative size = 110 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {a+c x^4}}{3 a x^3}-\frac {c^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+c x^4}} \]

output
-1/3*(c*x^4+a)^(1/2)/a/x^3-1/6*c^(3/4)*(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^(5/4)/(c*x^4+a)^(1/2)
 
3.11.11.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.46 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=-\frac {\sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},-\frac {c x^4}{a}\right )}{3 x^3 \sqrt {a+c x^4}} \]

input
Integrate[1/(x^4*Sqrt[a + (2 + 2*b - 2*(1 + b))*x^2 + c*x^4]),x]
 
output
-1/3*(Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[-3/4, 1/2, 1/4, -((c*x^4)/a)]) 
/(x^3*Sqrt[a + c*x^4])
 
3.11.11.3 Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4, 847, 761}

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.

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

\(\Big \downarrow \) 4

\(\displaystyle \int \frac {1}{x^4 \sqrt {a+c x^4}}dx\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {c \int \frac {1}{\sqrt {c x^4+a}}dx}{3 a}-\frac {\sqrt {a+c x^4}}{3 a x^3}\)

\(\Big \downarrow \) 761

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

input
Int[1/(x^4*Sqrt[a + (2 + 2*b - 2*(1 + b))*x^2 + c*x^4]),x]
 
output
-1/3*Sqrt[a + c*x^4]/(a*x^3) - (c^(3/4)*(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])/(6*a^(5/4)*Sqrt[a + c*x^4])
 

3.11.11.3.1 Defintions of rubi rules used

rule 4
Int[(u_.)*((a_.) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> 
 Int[u*(a + c*x^(2*n))^p, x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && 
 EqQ[b, 0]
 

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 847
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 
3.11.11.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85

method result size
default \(-\frac {\sqrt {c \,x^{4}+a}}{3 a \,x^{3}}-\frac {c \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) \(93\)
risch \(-\frac {\sqrt {c \,x^{4}+a}}{3 a \,x^{3}}-\frac {c \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) \(93\)
elliptic \(-\frac {\sqrt {c \,x^{4}+a}}{3 a \,x^{3}}-\frac {c \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) \(93\)

input
int(1/x^4/(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/3*(c*x^4+a)^(1/2)/a/x^3-1/3*c/a/(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)*Ellipti 
cF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)
 
3.11.11.5 Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {\sqrt {a} x^{3} \left (-\frac {c}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - \sqrt {c x^{4} + a}}{3 \, a x^{3}} \]

input
integrate(1/x^4/(c*x^4+a)^(1/2),x, algorithm="fricas")
 
output
1/3*(sqrt(a)*x^3*(-c/a)^(3/4)*elliptic_f(arcsin(x*(-c/a)^(1/4)), -1) - sqr 
t(c*x^4 + a))/(a*x^3)
 
3.11.11.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.37 \[ \int \frac {1}{x^4 \sqrt {a+(2+2 b-2 (1+b)) x^2+c x^4}} \, dx=\frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} x^{3} \Gamma \left (\frac {1}{4}\right )} \]

input
integrate(1/x**4/(c*x**4+a)**(1/2),x)
 
output
gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a 
)*x**3*gamma(1/4))
 
3.11.11.7 Maxima [F]

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

input
integrate(1/x^4/(c*x^4+a)^(1/2),x, algorithm="maxima")
 
output
integrate(1/(sqrt(c*x^4 + a)*x^4), x)
 
3.11.11.8 Giac [F]

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

input
integrate(1/x^4/(c*x^4+a)^(1/2),x, algorithm="giac")
 
output
integrate(1/(sqrt(c*x^4 + a)*x^4), x)
 
3.11.11.9 Mupad [F(-1)]

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

input
int(1/(x^4*(a + c*x^4)^(1/2)),x)
 
output
int(1/(x^4*(a + c*x^4)^(1/2)), x)