\(\int \frac {x^4}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\) [1211]

Optimal result

Integrand size = 26, antiderivative size = 261 \[ \int \frac {x^4}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {2 (b c+a d) x \sqrt {c+d x^2}}{3 b d^2 \sqrt {a+b x^2}}+\frac {x \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 b d}+\frac {2 \sqrt {a} (b c+a d) \sqrt {c+d x^2} E\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{3 b^{3/2} d^2 \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac {a^{3/2} \sqrt {c+d x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{3 b^{3/2} d \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \] Output:

-2/3*(a*d+b*c)*x*(d*x^2+c)^(1/2)/b/d^2/(b*x^2+a)^(1/2)+1/3*x*(b*x^2+a)^(1/ 
2)*(d*x^2+c)^(1/2)/b/d+2/3*a^(1/2)*(a*d+b*c)*(d*x^2+c)^(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^(3/2)/d^2/(b*x^2+a)^ 
(1/2)/(a*(d*x^2+c)/c/(b*x^2+a))^(1/2)-1/3*a^(3/2)*(d*x^2+c)^(1/2)*InverseJ 
acobiAM(arctan(b^(1/2)*x/a^(1/2)),(1-a*d/b/c)^(1/2))/b^(3/2)/d/(b*x^2+a)^( 
1/2)/(a*(d*x^2+c)/c/(b*x^2+a))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.49 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.77 \[ \int \frac {x^4}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (c+d x^2\right )+2 i c (b c+a d) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} E\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i c (2 b c+a d) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )}{3 b \sqrt {\frac {b}{a}} d^2 \sqrt {a+b x^2} \sqrt {c+d x^2}} \] Input:

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

Output:

(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2) + (2*I)*c*(b*c + a*d)*Sqrt[1 + (b*x 
^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] 
- I*c*(2*b*c + a*d)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*Ar 
cSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(3*b*Sqrt[b/a]*d^2*Sqrt[a + b*x^2]*Sqrt[ 
c + d*x^2])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {381, 406, 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.

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

\(\Big \downarrow \) 381

\(\displaystyle \frac {x \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 b d}-\frac {\int \frac {2 (b c+a d) x^2+a c}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx}{3 b d}\)

\(\Big \downarrow \) 406

\(\displaystyle \frac {x \sqrt {a+b x^2} \sqrt {c+d x^2}}{3 b d}-\frac {a c \int \frac {1}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx+2 (a d+b c) \int \frac {x^2}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx}{3 b d}\)

\(\Big \downarrow \) 320

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

\(\Big \downarrow \) 388

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

\(\Big \downarrow \) 313

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

Input:

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

Output:

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

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 381

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
+ 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) 
Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 
2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q 
}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2 
, p, q, x]
 

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 406

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[e   Int[(a + b*x^2)^p*(c + d*x^2)^q, x], x] + Sim 
p[f   Int[x^2*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, 
f, p, q}, x]
 

Maple [A] (verified)

Time = 5.18 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {x \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}{3 b d}-\frac {\left (\frac {a c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}-\frac {\left (2 a d +2 b c \right ) c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}\, d}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}}{3 b d \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(293\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \left (\frac {x \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}{3 b d}-\frac {a c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{3 d b \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}+\frac {\left (2 a d +2 b c \right ) c \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \left (\operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )-\operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )\right )}{3 b \,d^{2} \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\)	\(301\)
default	\(\frac {\left (\sqrt {-\frac {b}{a}}\, b \,d^{2} x^{5}+\sqrt {-\frac {b}{a}}\, a \,d^{2} x^{3}+\sqrt {-\frac {b}{a}}\, b c d \,x^{3}+a c \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {x^{2} d +c}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) d +2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {x^{2} d +c}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b \,c^{2}-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {x^{2} d +c}{c}}\, \operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) a c d -2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {x^{2} d +c}{c}}\, \operatorname {EllipticE}\left (x \sqrt {-\frac {b}{a}}, \sqrt {\frac {a d}{b c}}\right ) b \,c^{2}+\sqrt {-\frac {b}{a}}\, a c d x \right ) \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}{3 \sqrt {-\frac {b}{a}}\, d^{2} b \left (b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c \right )}\)	\(333\)

Input:

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

Output:

1/3*x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b/d-1/3/b/d*(a*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)*Elliptic 
F(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))-(2*a*d+2*b*c)*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*(E 
llipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))-EllipticE(x*(-b/a)^(1/2) 
,(-1+(a*d+b*c)/c/b)^(1/2))))*((b*x^2+a)*(d*x^2+c))^(1/2)/(b*x^2+a)^(1/2)/( 
d*x^2+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.60 \[ \int \frac {x^4}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {2 \, {\left (b c^{2} + a c d\right )} \sqrt {b d} x \sqrt {-\frac {c}{d}} E(\arcsin \left (\frac {\sqrt {-\frac {c}{d}}}{x}\right )\,|\,\frac {a d}{b c}) - {\left (2 \, b c^{2} + 2 \, a c d + a d^{2}\right )} \sqrt {b d} x \sqrt {-\frac {c}{d}} F(\arcsin \left (\frac {\sqrt {-\frac {c}{d}}}{x}\right )\,|\,\frac {a d}{b c}) + {\left (b d^{2} x^{2} - 2 \, b c d - 2 \, a d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{3 \, b^{2} d^{3} x} \] Input:

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

Output:

1/3*(2*(b*c^2 + a*c*d)*sqrt(b*d)*x*sqrt(-c/d)*elliptic_e(arcsin(sqrt(-c/d) 
/x), a*d/(b*c)) - (2*b*c^2 + 2*a*c*d + a*d^2)*sqrt(b*d)*x*sqrt(-c/d)*ellip 
tic_f(arcsin(sqrt(-c/d)/x), a*d/(b*c)) + (b*d^2*x^2 - 2*b*c*d - 2*a*d^2)*s 
qrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^2*d^3*x)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\int \frac {x^4}{\sqrt {b\,x^2+a}\,\sqrt {d\,x^2+c}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^4}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, x -2 \left (\int \frac {\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, x^{2}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \right ) a d -2 \left (\int \frac {\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, x^{2}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \right ) b c -\left (\int \frac {\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \right ) a c}{3 b d} \] Input:

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

Output:

(sqrt(c + d*x**2)*sqrt(a + b*x**2)*x - 2*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)*a*d - 2*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) 
*b*c - int((sqrt(c + d*x**2)*sqrt(a + b*x**2))/(a*c + a*d*x**2 + b*c*x**2 
+ b*d*x**4),x)*a*c)/(3*b*d)