\(\int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{(d+e x^2) \sqrt {a-c x^4}} \, dx\) [59]

Optimal result

Integrand size = 38, antiderivative size = 113 \[ \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\left (d+e x^2\right ) \sqrt {a-c x^4}} \, dx=\frac {\left (\sqrt {a}-\sqrt {c} x^2\right ) \operatorname {EllipticPi}\left (1-\frac {\sqrt {a} e}{\sqrt {c} d},\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),2\right )}{\sqrt [4]{a} \sqrt [4]{c} d \sqrt {\frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a}+\sqrt {c} x^2}} \sqrt {a-c x^4}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.54 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.32 \[ \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\left (d+e x^2\right ) \sqrt {a-c x^4}} \, dx=\frac {i \sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} \sqrt {1-\frac {c x^4}{a}} \left (\sqrt {c} d \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} x\right ),-1\right )+\left (-\sqrt {c} d+\sqrt {a} e\right ) \operatorname {EllipticPi}\left (-\frac {\sqrt {a} e}{\sqrt {c} d},i \text {arcsinh}\left (\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}} x\right ),-1\right )\right )}{\sqrt {c} d e \sqrt {a-c x^4}} \] Input:

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

Output:

(I*Sqrt[-(Sqrt[c]/Sqrt[a])]*Sqrt[1 - (c*x^4)/a]*(Sqrt[c]*d*EllipticF[I*Arc 
Sinh[Sqrt[-(Sqrt[c]/Sqrt[a])]*x], -1] + (-(Sqrt[c]*d) + Sqrt[a]*e)*Ellipti 
cPi[-((Sqrt[a]*e)/(Sqrt[c]*d)), I*ArcSinh[Sqrt[-(Sqrt[c]/Sqrt[a])]*x], -1] 
))/(Sqrt[c]*d*e*Sqrt[a - c*x^4])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {1787, 414}

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.

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

\(\Big \downarrow \) 1787

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

\(\Big \downarrow \) 414

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

Input:

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

Output:

((a - Sqrt[a]*Sqrt[c]*x^2)*EllipticPi[1 - (Sqrt[a]*e)/(Sqrt[c]*d), ArcTan[ 
(c^(1/4)*x)/a^(1/4)], 2])/(a^(3/4)*c^(1/4)*d*Sqrt[(Sqrt[a] - Sqrt[c]*x^2)/ 
(Sqrt[a] + Sqrt[c]*x^2)]*Sqrt[a - c*x^4])
 

Defintions of rubi rules used

rule 414

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_) 
^2]), x_Symbol] :> Simp[c*(Sqrt[e + f*x^2]/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]* 
Sqrt[c*((e + f*x^2)/(e*(c + d*x^2)))]))*EllipticPi[1 - b*(c/(a*d)), ArcTan[ 
Rt[d/c, 2]*x], 1 - c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ 
[d/c]
 

rule 1787

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + ( 
c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[(a + c*x^(2*n))^FracPart[p]/((d + 
e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPart[p])   Int[(d + e*x^n)^(p + q) 
*(f + g*x^n)^r*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, c, d, e, f, g, n, p 
, q, r}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.65

method	result	size
default	\(\frac {\frac {\sqrt {c}\, \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{e \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}-\frac {\left (\sqrt {c}\, d -\sqrt {a}\, e \right ) \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, -\frac {\sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}}\right )}{e d \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}}{\sqrt {a}}\)	\(187\)
elliptic	\(\frac {\sqrt {\left (-c \,x^{4}+a \right ) a c}\, \left (\sqrt {a}+\sqrt {c}\, x^{2}\right ) \left (\frac {\sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, -\frac {\sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}}\right )}{d \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}+\frac {c \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{e \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-a \,c^{2} x^{4}+a^{2} c}}-\frac {c \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, -\frac {\sqrt {a}\, e}{\sqrt {c}\, d}, \frac {\sqrt {-\frac {\sqrt {c}}{\sqrt {a}}}}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}}\right )}{e \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-a \,c^{2} x^{4}+a^{2} c}}\right )}{\left (c \,x^{2} \sqrt {-c \,x^{4}+a}+\sqrt {\left (-c \,x^{4}+a \right ) a c}\right ) \sqrt {a}}\)	\(336\)

Input:

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

Output:

1/a^(1/2)*(c^(1/2)/e/(c^(1/2)/a^(1/2))^(1/2)*(1-c^(1/2)*x^2/a^(1/2))^(1/2) 
*(1+c^(1/2)*x^2/a^(1/2))^(1/2)/(-c*x^4+a)^(1/2)*EllipticF(x*(c^(1/2)/a^(1/ 
2))^(1/2),I)-(c^(1/2)*d-a^(1/2)*e)/e/d/(c^(1/2)/a^(1/2))^(1/2)*(1-c^(1/2)* 
x^2/a^(1/2))^(1/2)*(1+c^(1/2)*x^2/a^(1/2))^(1/2)/(-c*x^4+a)^(1/2)*Elliptic 
Pi(x*(c^(1/2)/a^(1/2))^(1/2),-a^(1/2)*e/c^(1/2)/d,(-c^(1/2)/a^(1/2))^(1/2) 
/(c^(1/2)/a^(1/2))^(1/2)))
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

(Integral(sqrt(a)/(d*sqrt(a - c*x**4) + e*x**2*sqrt(a - c*x**4)), x) + Int 
egral(sqrt(c)*x**2/(d*sqrt(a - c*x**4) + e*x**2*sqrt(a - c*x**4)), x))/sqr 
t(a)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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