\(\int \frac {d+e x}{\sqrt {a+b x^2+c x^4}} \, dx\) [262]

Optimal result

Integrand size = 22, antiderivative size = 160 \[ \int \frac {d+e x}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {e \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c}}+\frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b x^2+c x^4}} \] Output:

1/2*e*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(1/2)+1/2*d 
*(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)*Inv 
erseJacobiAM(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^(1/4)/(c*x^4+b*x^2+a)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.40 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.44 \[ \int \frac {d+e x}{\sqrt {a+b x^2+c x^4}} \, dx=-\frac {i d \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {a+b x^2+c x^4}}-\frac {e \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )}{2 \sqrt {c}} \] Input:

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

Output:

((-I)*d*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sq 
rt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[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]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4]) 
- (e*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]])/(2*Sqrt[c])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2202, 27, 1416, 1432, 1092, 219}

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.

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

\(\Big \downarrow \) 2202

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 1432

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+b x^2+c x^4}}+\frac {e \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {c}}\)

Input:

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

Output:

(e*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(2*Sqrt[c]) 
 + (d*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]* 
x^2)^2]*EllipticF[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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, 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 1432

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 
 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
 

rule 2202

Int[(Pn_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{n 
 = Expon[Pn, x], k}, Int[Sum[Coeff[Pn, x, 2*k]*x^(2*k), {k, 0, n/2}]*(a + b 
*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pn, x, 2*k + 1]*x^(2*k), {k, 0, (n - 
1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pn, x] 
 &&  !PolyQ[Pn, x^2]
 

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {d \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {e \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {c}}\)	\(181\)
elliptic	\(\frac {d \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}+\frac {e \ln \left (\frac {2 c \,x^{2}+b}{\sqrt {c}}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {c}}\)	\(182\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.18 \[ \int \frac {d+e x}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {2 \, \sqrt {\frac {1}{2}} {\left (c d \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b d\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} F(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) + a \sqrt {c} e \log \left (8 \, c^{2} x^{4} + 8 \, b c x^{2} + b^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} + 4 \, a c\right )}{4 \, a c} \] Input:

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

Output:

1/4*(2*sqrt(1/2)*(c*d*sqrt((b^2 - 4*a*c)/c^2) + b*d)*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*sqrt(c)*e*log(8*c^2*x^4 + 8*b*c*x^2 + b^2 + 4*sqrt(c*x^4 + b*x^2 
 + a)*(2*c*x^2 + b)*sqrt(c) + 4*a*c))/(a*c)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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