\(\int \frac {1}{(d+e x^2) \sqrt {-a-c x^4}} \, dx\) [431]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 345 \[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a-c x^4}} \, dx=\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {c d^2+a e^2} x}{\sqrt {d} \sqrt {e} \sqrt {-a-c x^4}}\right )}{2 \sqrt {d} \sqrt {c d^2+a e^2}}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d-\sqrt {a} e\right ) \sqrt {-a-c x^4}}-\frac {\left (\sqrt {c} d+\sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} c^{3/4} d \left (d-\frac {\sqrt {a} e}{\sqrt {c}}\right ) \sqrt {-a-c x^4}} \] Output:

1/2*e^(1/2)*arctanh((a*e^2+c*d^2)^(1/2)*x/d^(1/2)/e^(1/2)/(-c*x^4-a)^(1/2) 
)/d^(1/2)/(a*e^2+c*d^2)^(1/2)+1/2*c^(1/4)*(a^(1/2)+c^(1/2)*x^2)*((c*x^4+a) 
/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x/a^(1/4) 
),1/2*2^(1/2))/a^(1/4)/(c^(1/2)*d-a^(1/2)*e)/(-c*x^4-a)^(1/2)-1/4*(c^(1/2) 
*d+a^(1/2)*e)*(a^(1/2)+c^(1/2)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1 
/2)*EllipticPi(sin(2*arctan(c^(1/4)*x/a^(1/4))),-1/4*(c^(1/2)*d-a^(1/2)*e) 
^2/a^(1/2)/c^(1/2)/d/e,1/2*2^(1/2))/a^(1/4)/c^(3/4)/d/(d-a^(1/2)*e/c^(1/2) 
)/(-c*x^4-a)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.15 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.28 \[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a-c x^4}} \, dx=-\frac {i \sqrt {1+\frac {c x^4}{a}} \operatorname {EllipticPi}\left (-\frac {i \sqrt {a} e}{\sqrt {c} d},i \text {arcsinh}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right ),-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} d \sqrt {-a-c x^4}} \] Input:

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

Output:

((-I)*Sqrt[1 + (c*x^4)/a]*EllipticPi[((-I)*Sqrt[a]*e)/(Sqrt[c]*d), I*ArcSi 
nh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1])/(Sqrt[(I*Sqrt[c])/Sqrt[a]]*d*Sqrt[-a 
 - c*x^4])
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1541, 27, 761, 2223}

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.

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

\(\Big \downarrow \) 1541

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 2223

\(\displaystyle \frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt {-a-c x^4} \left (\sqrt {c} d-\sqrt {a} e\right )}-\frac {e \left (\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {a} e+\sqrt {c} d\right ) \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \left (\frac {\sqrt {c} d}{\sqrt {a}}-e\right )^2}{4 \sqrt {c} d e},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e \sqrt {-a-c x^4}}-\frac {\left (\sqrt {c} d-\sqrt {a} e\right ) \text {arctanh}\left (\frac {x \sqrt {a e^2+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {-a-c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {a e^2+c d^2}}\right )}{\sqrt {c} d-\sqrt {a} e}\)

Input:

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

Output:

(c^(1/4)*(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])/(2*a^(1/4)*(Sqrt[c]*d - 
Sqrt[a]*e)*Sqrt[-a - c*x^4]) - (e*(-1/2*((Sqrt[c]*d - Sqrt[a]*e)*ArcTanh[( 
Sqrt[c*d^2 + a*e^2]*x)/(Sqrt[d]*Sqrt[e]*Sqrt[-a - c*x^4])])/(Sqrt[d]*Sqrt[ 
e]*Sqrt[c*d^2 + a*e^2]) + ((Sqrt[c]*d + Sqrt[a]*e)*(Sqrt[a] + Sqrt[c]*x^2) 
*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[-1/4*(Sqrt[a]*((Sq 
rt[c]*d)/Sqrt[a] - e)^2)/(Sqrt[c]*d*e), 2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2 
])/(4*a^(1/4)*c^(1/4)*d*e*Sqrt[-a - c*x^4])))/(Sqrt[c]*d - Sqrt[a]*e)
 

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 2223
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))*(ArcTanh[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*Arc 
Tan[q*x], 1/2], x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0 
] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && NegQ[c*(d/e) + a*(e 
/d)]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.47 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.32

method result size
default \(\frac {\sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}, -\frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{d \sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}-a}}\) \(110\)
elliptic \(\frac {\sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}, -\frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{d \sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}-a}}\) \(110\)

Input:

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

Output:

1/d/(-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)*EllipticPi(x*(-I/a^(1/2)*c^(1/2))^(1 
/2),-I*a^(1/2)/c^(1/2)/d*e,(I/a^(1/2)*c^(1/2))^(1/2)/(-I/a^(1/2)*c^(1/2))^ 
(1/2))
 

Fricas [F]

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

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

Output:

integral(-sqrt(-c*x^4 - a)/(c*e*x^6 + c*d*x^4 + a*e*x^2 + a*d), x)
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt(-a - c*x**4)*(d + e*x**2)), x)
 

Maxima [F]

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

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

Output:

integrate(1/(sqrt(-c*x^4 - a)*(e*x^2 + d)), x)
 

Giac [F]

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

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

Output:

integrate(1/(sqrt(-c*x^4 - a)*(e*x^2 + d)), x)
 

Mupad [F(-1)]

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

int(1/((- a - c*x^4)^(1/2)*(d + e*x^2)),x)
 

Output:

int(1/((- a - c*x^4)^(1/2)*(d + e*x^2)), x)
 

Reduce [F]

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

int(1/(e*x^2+d)/(-c*x^4-a)^(1/2),x)
 

Output:

 - int(sqrt(a + c*x**4)/(a*d + a*e*x**2 + c*d*x**4 + c*e*x**6),x)*i