\(\int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+c x^4}} \, dx\) [120]

Optimal result

Integrand size = 30, antiderivative size = 660 \[ \int \frac {A+B x^2}{\sqrt {d+e x^2} \sqrt {a+c x^4}} \, dx=\frac {B \sqrt {c+\frac {a}{x^4}} \sqrt {e+\frac {d}{x^2}} x^3 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {e+\frac {d}{x^2}}}{\sqrt {e} \sqrt {c+\frac {a}{x^4}}}\right )}{2 \sqrt {c} \sqrt {e} \sqrt {d+e x^2} \sqrt {a+c x^4}}+\frac {\left (\sqrt {a} B c d^2+a^{3/2} B e^2-(A c d+a B e) \sqrt {c d^2+a e^2}\right ) \left (1+\frac {\sqrt {a} \left (e+\frac {d}{x^2}\right )}{\sqrt {c d^2+a e^2}}\right ) \sqrt {\frac {c+\frac {a}{x^4}}{\left (c+\frac {a e^2}{d^2}\right ) \left (1+\frac {\sqrt {a} \left (e+\frac {d}{x^2}\right )}{\sqrt {c d^2+a e^2}}\right )^2}} \sqrt {e+\frac {d}{x^2}} x^3 \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt {e+\frac {d}{x^2}}}{\sqrt [4]{c d^2+a e^2}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {a} e}{\sqrt {c d^2+a e^2}}\right )\right )}{2 \sqrt [4]{a} c d^2 \sqrt [4]{c d^2+a e^2} \sqrt {d+e x^2} \sqrt {a+c x^4}}+\frac {B \sqrt [4]{c d^2+a e^2} \left (\sqrt {a} e-\sqrt {c d^2+a e^2}\right )^2 \left (1+\frac {\sqrt {a} \left (e+\frac {d}{x^2}\right )}{\sqrt {c d^2+a e^2}}\right ) \sqrt {\frac {c+\frac {a}{x^4}}{\left (c+\frac {a e^2}{d^2}\right ) \left (1+\frac {\sqrt {a} \left (e+\frac {d}{x^2}\right )}{\sqrt {c d^2+a e^2}}\right )^2}} \sqrt {e+\frac {d}{x^2}} x^3 \operatorname {EllipticPi}\left (\frac {\left (\sqrt {a} e+\sqrt {c d^2+a e^2}\right )^2}{4 \sqrt {a} e \sqrt {c d^2+a e^2}},2 \arctan \left (\frac {\sqrt [4]{a} \sqrt {e+\frac {d}{x^2}}}{\sqrt [4]{c d^2+a e^2}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {a} e}{\sqrt {c d^2+a e^2}}\right )\right )}{4 \sqrt [4]{a} c d^2 e \sqrt {d+e x^2} \sqrt {a+c x^4}} \] Output:

1/2*B*(c+a/x^4)^(1/2)*(e+d/x^2)^(1/2)*x^3*arctanh(c^(1/2)*(e+d/x^2)^(1/2)/ 
e^(1/2)/(c+a/x^4)^(1/2))/c^(1/2)/e^(1/2)/(e*x^2+d)^(1/2)/(c*x^4+a)^(1/2)+1 
/2*(a^(1/2)*B*c*d^2+a^(3/2)*B*e^2-(A*c*d+B*a*e)*(a*e^2+c*d^2)^(1/2))*(1+a^ 
(1/2)*(e+d/x^2)/(a*e^2+c*d^2)^(1/2))*((c+a/x^4)/(c+a*e^2/d^2)/(1+a^(1/2)*( 
e+d/x^2)/(a*e^2+c*d^2)^(1/2))^2)^(1/2)*(e+d/x^2)^(1/2)*x^3*InverseJacobiAM 
(2*arctan(a^(1/4)*(e+d/x^2)^(1/2)/(a*e^2+c*d^2)^(1/4)),1/2*(2+2*a^(1/2)/(a 
*e^2+c*d^2)^(1/2)*e)^(1/2))/a^(1/4)/c/d^2/(a*e^2+c*d^2)^(1/4)/(e*x^2+d)^(1 
/2)/(c*x^4+a)^(1/2)+1/4*B*(a*e^2+c*d^2)^(1/4)*(a^(1/2)*e-(a*e^2+c*d^2)^(1/ 
2))^2*(1+a^(1/2)*(e+d/x^2)/(a*e^2+c*d^2)^(1/2))*((c+a/x^4)/(c+a*e^2/d^2)/( 
1+a^(1/2)*(e+d/x^2)/(a*e^2+c*d^2)^(1/2))^2)^(1/2)*(e+d/x^2)^(1/2)*x^3*Elli 
pticPi(sin(2*arctan(a^(1/4)*(e+d/x^2)^(1/2)/(a*e^2+c*d^2)^(1/4))),1/4*(a^( 
1/2)*e+(a*e^2+c*d^2)^(1/2))^2/a^(1/2)/e/(a*e^2+c*d^2)^(1/2),1/2*(2+2*a^(1/ 
2)/(a*e^2+c*d^2)^(1/2)*e)^(1/2))/a^(1/4)/c/d^2/e/(e*x^2+d)^(1/2)/(c*x^4+a) 
^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [F]

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.

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

\(\Big \downarrow \) 2261

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 2261

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Unintegrable[Px*(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, 
p, q}, x] && PolyQ[Px, x]
 

Maple [F]

\[\int \frac {B \,x^{2}+A}{\sqrt {e \,x^{2}+d}\, \sqrt {c \,x^{4}+a}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(c*x^4 + a)*(B*x^2 + A)*sqrt(e*x^2 + d)/(c*e*x^6 + c*d*x^4 + 
a*e*x^2 + a*d), x)
 

Sympy [F]

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

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

Output:

Integral((A + B*x**2)/(sqrt(a + c*x**4)*sqrt(d + e*x**2)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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