\(\int \frac {1}{(d+e x^2) \sqrt {a+c x^4}} \, dx\) [425]

Optimal result

Integrand size = 21, antiderivative size = 336 \[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {a+c x^4}} \, dx=\frac {\sqrt {e} \arctan \left (\frac {\sqrt {c d^2+a e^2} x}{\sqrt {d} \sqrt {e} \sqrt {a+c x^4}}\right )}{2 \sqrt {d} \sqrt {c d^2+a e^2}}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d-\sqrt {a} e\right ) \sqrt {a+c x^4}}-\frac {\left (\sqrt {c} d+\sqrt {a} e\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} c^{3/4} d \left (d-\frac {\sqrt {a} e}{\sqrt {c}}\right ) \sqrt {a+c x^4}} \] Output:

1/2*e^(1/2)*arctan((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 = 95, 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 = 0.87 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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.

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

\(\Big \downarrow \) 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

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

\(\Big \downarrow \) 2221

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

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)*ArcTan[(Sq 
rt[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)*Sq 
rt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[-1/4*(Sqrt[a]*((Sqrt[ 
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 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 = 0.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.32

method	result	size
default	\(\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, \frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{d \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(107\)
elliptic	\(\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, \frac {i \sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {i \sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}\right )}{d \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(107\)

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]

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