\(\int \frac {1}{\sqrt {\frac {e (a+b x^2)}{c+d x^2}}} \, dx\) [83]

Optimal result

Integrand size = 22, antiderivative size = 230 \[ \int \frac {1}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\frac {x}{\sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}-\frac {\sqrt {a} E\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{\sqrt {b} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}}+\frac {\sqrt {a} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{\sqrt {b} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \sqrt {\frac {b e}{d}-\frac {(b c-a d) e}{d \left (c+d x^2\right )}}} \] Output:

x/(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)-a^(1/2)*EllipticE(b^(1/2)*x/a^(1/ 
2)/(1+b*x^2/a)^(1/2),(1-a*d/b/c)^(1/2))/b^(1/2)/(a*(d*x^2+c)/c/(b*x^2+a))^ 
(1/2)/(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)+a^(1/2)*InverseJacobiAM(arcta 
n(b^(1/2)*x/a^(1/2)),(1-a*d/b/c)^(1/2))/b^(1/2)/(a*(d*x^2+c)/c/(b*x^2+a))^ 
(1/2)/(b*e/d-(-a*d+b*c)*e/d/(d*x^2+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.82 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.37 \[ \int \frac {1}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=\frac {\sqrt {\frac {a+b x^2}{a}} E\left (\arcsin \left (\sqrt {-\frac {b}{a}} x\right )|\frac {a d}{b c}\right )}{\sqrt {-\frac {b}{a}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {\frac {c+d x^2}{c}}} \] Input:

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

Output:

(Sqrt[(a + b*x^2)/a]*EllipticE[ArcSin[Sqrt[-(b/a)]*x], (a*d)/(b*c)])/(Sqrt 
[-(b/a)]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[(c + d*x^2)/c])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2058, 324, 320, 388, 313}

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.

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

\(\Big \downarrow \) 2058

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

\(\Big \downarrow \) 324

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

\(\Big \downarrow \) 320

\(\displaystyle \frac {\sqrt {a+b x^2} \left (d \int \frac {x^2}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx+\frac {c^{3/2} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\right )}{\sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}\)

\(\Big \downarrow \) 388

\(\displaystyle \frac {\sqrt {a+b x^2} \left (d \left (\frac {x \sqrt {a+b x^2}}{b \sqrt {c+d x^2}}-\frac {c \int \frac {\sqrt {b x^2+a}}{\left (d x^2+c\right )^{3/2}}dx}{b}\right )+\frac {c^{3/2} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\right )}{\sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}\)

\(\Big \downarrow \) 313

\(\displaystyle \frac {\sqrt {a+b x^2} \left (\frac {c^{3/2} \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+d \left (\frac {x \sqrt {a+b x^2}}{b \sqrt {c+d x^2}}-\frac {\sqrt {c} \sqrt {a+b x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{b \sqrt {d} \sqrt {c+d x^2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\right )\right )}{\sqrt {c+d x^2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}\)

Input:

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

Output:

(Sqrt[a + b*x^2]*(d*((x*Sqrt[a + b*x^2])/(b*Sqrt[c + d*x^2]) - (Sqrt[c]*Sq 
rt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(b* 
Sqrt[d]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])) + (c^(3/2) 
*Sqrt[a + b*x^2]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/ 
(a*Sqrt[d]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2])))/(Sqrt[ 
(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[c + d*x^2])
 

Defintions of rubi rules used

rule 313

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Sim 
p[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c 
+ d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; FreeQ 
[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]
 

rule 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( 
c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 324

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
a   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] + Simp[b   Int[x^2/(Sqr 
t[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[d/c 
] && PosQ[b/a]
 

rule 388

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] 
 :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt[c + d*x^2])), x] - Simp[c/b   Int[Sqrt[ 
a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 2058

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ 
(r_.))^(p_), x_Symbol] :> Simp[Simp[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + 
 b*x^n)^(p*q)*(c + d*x^n)^(p*r))]   Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)^(p* 
r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]
 

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.55

method	result	size
default	\(\frac {\left (b \,x^{2}+a \right ) c \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, \operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right )}{\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, \sqrt {d b \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}}\)	\(127\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx=-\frac {c \sqrt {\frac {b e}{d}} x \sqrt {-\frac {c}{d}} E(\arcsin \left (\frac {\sqrt {-\frac {c}{d}}}{x}\right )\,|\,\frac {a d}{b c}) - {\left (c + d\right )} \sqrt {\frac {b e}{d}} x \sqrt {-\frac {c}{d}} F(\arcsin \left (\frac {\sqrt {-\frac {c}{d}}}{x}\right )\,|\,\frac {a d}{b c}) - {\left (d x^{2} + c\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{b e x} \] Input:

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

Output:

-(c*sqrt(b*e/d)*x*sqrt(-c/d)*elliptic_e(arcsin(sqrt(-c/d)/x), a*d/(b*c)) - 
 (c + d)*sqrt(b*e/d)*x*sqrt(-c/d)*elliptic_f(arcsin(sqrt(-c/d)/x), a*d/(b* 
c)) - (d*x^2 + c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b*e*x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*int((sqrt(c + d*x**2)*sqrt(a + b*x**2))/(a + b*x**2),x))/e