\(\int \frac {A+B x^2}{\sqrt {a+b x^2+c x^4}} \, dx\) [170]

Optimal result

Integrand size = 24, antiderivative size = 284 \[ \int \frac {A+B x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {B x \sqrt {a+b x^2+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} B \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{c^{3/4} \sqrt {a+b x^2+c x^4}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+b x^2+c x^4}} \] Output:

B*x*(c*x^4+b*x^2+a)^(1/2)/c^(1/2)/(a^(1/2)+c^(1/2)*x^2)-a^(1/4)*B*(a^(1/2) 
+c^(1/2)*x^2)*((c*x^4+b*x^2+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticE(si 
n(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))/c^(3/4)/(c 
*x^4+b*x^2+a)^(1/2)+1/2*(a^(1/2)*B+A*c^(1/2))*(a^(1/2)+c^(1/2)*x^2)*((c*x^ 
4+b*x^2+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4) 
*x/a^(1/4)),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))/a^(1/4)/c^(3/4)/(c*x^4+b*x^2+ 
a)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {i \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \left (B \left (-b+\sqrt {b^2-4 a c}\right ) E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+\left (b B-2 A c-B \sqrt {b^2-4 a c}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{2 \sqrt {2} c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {a+b x^2+c x^4}} \] Input:

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

Output:

((I/2)*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqr 
t[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*(B*(-b + Sqrt[b^2 - 4*a*c])*Ellip 
ticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 
 4*a*c])/(b - Sqrt[b^2 - 4*a*c])] + (b*B - 2*A*c - B*Sqrt[b^2 - 4*a*c])*El 
lipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^ 
2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])]))/(Sqrt[2]*c*Sqrt[c/(b + Sqrt[b^2 - 4 
*a*c])]*Sqrt[a + b*x^2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1511, 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.

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

\(\Big \downarrow \) 1511

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 1509

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

Input:

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

Output:

-((B*(-((x*Sqrt[a + b*x^2 + c*x^4])/(Sqrt[a] + Sqrt[c]*x^2)) + (a^(1/4)*(S 
qrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]* 
EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(c^ 
(1/4)*Sqrt[a + b*x^2 + c*x^4])))/Sqrt[c]) + ((A + (Sqrt[a]*B)/Sqrt[c])*(Sq 
rt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*E 
llipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*a 
^(1/4)*c^(1/4)*Sqrt[a + b*x^2 + c*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 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 1511

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

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.27

method	result	size
default	\(\frac {A \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {B a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(362\)
elliptic	\(\frac {A \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {B a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(362\)

Input:

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

Output:

1/4*A*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b+(-4*a*c+b^2)^(1/2 
))/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^( 
1/2)*EllipticF(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b 
*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))-1/2*B*a*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2 
))/a)^(1/2)*(4-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2) 
^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(Ellipti 
cF(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+ 
b^2)^(1/2))/a/c)^(1/2))-EllipticE(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a 
)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {2 \, \sqrt {c x^{4} + b x^{2} + a} B a c + \sqrt {\frac {1}{2}} {\left (B a c x \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - B a b x\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} E(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) - \sqrt {\frac {1}{2}} {\left ({\left (B a c - A c^{2}\right )} x \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - {\left (B a b + A b c\right )} x\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} F(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c})}{2 \, a c^{2} x} \] Input:

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

Output:

1/2*(2*sqrt(c*x^4 + b*x^2 + a)*B*a*c + sqrt(1/2)*(B*a*c*x*sqrt((b^2 - 4*a* 
c)/c^2) - B*a*b*x)*sqrt(c)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)*ellipti 
c_e(arcsin(sqrt(1/2)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)/x), 1/2*(b*c* 
sqrt((b^2 - 4*a*c)/c^2) + b^2 - 2*a*c)/(a*c)) - sqrt(1/2)*((B*a*c - A*c^2) 
*x*sqrt((b^2 - 4*a*c)/c^2) - (B*a*b + A*b*c)*x)*sqrt(c)*sqrt((c*sqrt((b^2 
- 4*a*c)/c^2) - b)/c)*elliptic_f(arcsin(sqrt(1/2)*sqrt((c*sqrt((b^2 - 4*a* 
c)/c^2) - b)/c)/x), 1/2*(b*c*sqrt((b^2 - 4*a*c)/c^2) + b^2 - 2*a*c)/(a*c)) 
)/(a*c^2*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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