\(\int \frac {x^2}{\sqrt {(a+b x^2) (c+d x^2)}} \, dx\) [703]

Optimal result
Mathematica [C] (verified)
Rubi [B] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \frac {x^2}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=\frac {x \left (c+d x^2\right )}{d \sqrt {a c+(b c+a d) x^2+b d x^4}}-\frac {c \left (a+b x^2\right ) \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} E\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{\sqrt {a} \sqrt {b} d \sqrt {a c+(b c+a d) x^2+b d x^4}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.71 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=-\frac {i c \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \left (E\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )\right )}{\sqrt {\frac {b}{a}} d \sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \] Input:

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

Output:

((-I)*c*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*(EllipticE[I*ArcSinh[Sqrt[ 
b/a]*x], (a*d)/(b*c)] - EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/( 
Sqrt[b/a]*d*Sqrt[(a + b*x^2)*(c + d*x^2)])
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(447\) vs. \(2(137)=274\).

Time = 0.88 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.26, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2048, 1459, 27, 1416, 1509}

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.

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

\(\Big \downarrow \) 2048

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

\(\Big \downarrow \) 1459

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 1509

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

Input:

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

Output:

-((-((x*Sqrt[a*c + (b*c + a*d)*x^2 + b*d*x^4])/(Sqrt[a]*Sqrt[c] + Sqrt[b]* 
Sqrt[d]*x^2)) + (a^(1/4)*c^(1/4)*(Sqrt[a]*Sqrt[c] + Sqrt[b]*Sqrt[d]*x^2)*S 
qrt[(a*c + (b*c + a*d)*x^2 + b*d*x^4)/(Sqrt[a]*Sqrt[c] + Sqrt[b]*Sqrt[d]*x 
^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*d^(1/4)*x)/(a^(1/4)*c^(1/4))], (2 - (b* 
c + a*d)/(Sqrt[a]*Sqrt[b]*Sqrt[c]*Sqrt[d]))/4])/(b^(1/4)*d^(1/4)*Sqrt[a*c 
+ (b*c + a*d)*x^2 + b*d*x^4]))/(Sqrt[b]*Sqrt[d])) + (a^(1/4)*c^(1/4)*(Sqrt 
[a]*Sqrt[c] + Sqrt[b]*Sqrt[d]*x^2)*Sqrt[(a*c + (b*c + a*d)*x^2 + b*d*x^4)/ 
(Sqrt[a]*Sqrt[c] + Sqrt[b]*Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*d^( 
1/4)*x)/(a^(1/4)*c^(1/4))], (2 - (b*c + a*d)/(Sqrt[a]*Sqrt[b]*Sqrt[c]*Sqrt 
[d]))/4])/(2*b^(3/4)*d^(3/4)*Sqrt[a*c + (b*c + a*d)*x^2 + b*d*x^4])
 

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 1416
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1459
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[1/q   Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[1/q 
 Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]] /; FreeQ[{a, b, c}, x] && 
 NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q 
^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* 
x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 
/(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 
- 4*a*c, 0] && PosQ[c/a]
 

rule 2048
Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_) 
, x_Symbol] :> Int[u*(a*c*e + (b*c + a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; F 
reeQ[{a, b, c, d, e, n, p}, x]
 
Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91

method result size
default \(-\frac {c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+x^{2} d a +b c \,x^{2}+a c}\, d}\) \(124\)
elliptic \(-\frac {c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+x^{2} d a +b c \,x^{2}+a c}\, d}\) \(124\)

Input:

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

Output:

-c/(-b/a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x 
^2+a*c)^(1/2)/d*(EllipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))-Ellipt 
icE(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {x^2}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=-\frac {\sqrt {b d} c x \sqrt {-\frac {c}{d}} E(\arcsin \left (\frac {\sqrt {-\frac {c}{d}}}{x}\right )\,|\,\frac {a d}{b c}) - \sqrt {b d} c x \sqrt {-\frac {c}{d}} F(\arcsin \left (\frac {\sqrt {-\frac {c}{d}}}{x}\right )\,|\,\frac {a d}{b c}) - \sqrt {b d x^{4} + {\left (b c + a d\right )} x^{2} + a c} d}{b d^{2} x} \] Input:

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

Output:

-(sqrt(b*d)*c*x*sqrt(-c/d)*elliptic_e(arcsin(sqrt(-c/d)/x), a*d/(b*c)) - s 
qrt(b*d)*c*x*sqrt(-c/d)*elliptic_f(arcsin(sqrt(-c/d)/x), a*d/(b*c)) - sqrt 
(b*d*x^4 + (b*c + a*d)*x^2 + a*c)*d)/(b*d^2*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**2)/(a*c + a*d*x**2 + b*c*x**2 + 
b*d*x**4),x)