\(\int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx\) [62]

Optimal result

Integrand size = 31, antiderivative size = 217 \[ \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx=\frac {\sqrt [4]{c^3+4 a d^2} \sqrt {\frac {d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (c^3+4 a d^2\right ) \left (\sqrt {c}+\frac {(c+d x)^2}{\sqrt {c^3+4 a d^2}}\right )^2}} \left (\sqrt {c}+\frac {(c+d x)^2}{\sqrt {c^3+4 a d^2}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right ),\frac {1}{2} \left (1+\frac {c^{3/2}}{\sqrt {c^3+4 a d^2}}\right )\right )}{2 \sqrt [4]{c} d \sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \] Output:

1/2*(4*a*d^2+c^3)^(1/4)*(d^2*(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)/(4*a*d^2+ 
c^3)/(c^(1/2)+(d*x+c)^2/(4*a*d^2+c^3)^(1/2))^2)^(1/2)*(c^(1/2)+(d*x+c)^2/( 
4*a*d^2+c^3)^(1/2))*InverseJacobiAM(2*arctan((d*x+c)/c^(1/4)/(4*a*d^2+c^3) 
^(1/4)),1/2*(2+2*c^(3/2)/(4*a*d^2+c^3)^(1/2))^(1/2))/c^(1/4)/d/(d^2*x^4+4* 
c*d*x^3+4*c^2*x^2+4*a*c)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 12.65 (sec) , antiderivative size = 822, normalized size of antiderivative = 3.79 \[ \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx=\frac {2 \left (-c+\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-d x\right ) \left (c+\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}+d x\right ) \sqrt {-\frac {\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d} \left (c-\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}+d x\right )}{\left (\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}+\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}\right ) \left (-c+\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-d x\right )}} \sqrt {-\frac {\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d} \left (c+\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}+d x\right )}{\left (\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}\right ) \left (-c+\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-d x\right )}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {\left (\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}\right ) \left (c+\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}+d x\right )}{\left (\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}+\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}\right ) \left (-c+\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-d x\right )}}\right ),\frac {\left (\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}+\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}\right )^2}{\left (\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}\right )^2}\right )}{d \sqrt {c^2-2 i \sqrt {a} \sqrt {c} d} \sqrt {\frac {\left (\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}\right ) \left (c+\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}+d x\right )}{\left (\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}+\sqrt {c^2+2 i \sqrt {a} \sqrt {c} d}\right ) \left (-c+\sqrt {c^2-2 i \sqrt {a} \sqrt {c} d}-d x\right )}} \sqrt {4 a c+x^2 (2 c+d x)^2}} \] Input:

Integrate[1/Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4],x]
 

Output:

(2*(-c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - d*x)*(c + Sqrt[c^2 - (2*I)* 
Sqrt[a]*Sqrt[c]*d] + d*x)*Sqrt[-((Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d]*(c - 
 Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d] + d*x))/((Sqrt[c^2 - (2*I)*Sqrt[a]*Sq 
rt[c]*d] + Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(-c + Sqrt[c^2 - (2*I)*Sqr 
t[a]*Sqrt[c]*d] - d*x)))]*Sqrt[-((Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d]*(c + 
 Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d] + d*x))/((Sqrt[c^2 - (2*I)*Sqrt[a]*Sq 
rt[c]*d] - Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(-c + Sqrt[c^2 - (2*I)*Sqr 
t[a]*Sqrt[c]*d] - d*x)))]*EllipticF[ArcSin[Sqrt[((Sqrt[c^2 - (2*I)*Sqrt[a] 
*Sqrt[c]*d] - Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(c + Sqrt[c^2 - (2*I)*S 
qrt[a]*Sqrt[c]*d] + d*x))/((Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] + Sqrt[c^2 
 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(-c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - d 
*x))]], (Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] + Sqrt[c^2 + (2*I)*Sqrt[a]*Sq 
rt[c]*d])^2/(Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - Sqrt[c^2 + (2*I)*Sqrt[a 
]*Sqrt[c]*d])^2])/(d*Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d]*Sqrt[((Sqrt[c^2 - 
 (2*I)*Sqrt[a]*Sqrt[c]*d] - Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(c + Sqrt 
[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] + d*x))/((Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c] 
*d] + Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(-c + Sqrt[c^2 - (2*I)*Sqrt[a]* 
Sqrt[c]*d] - d*x))]*Sqrt[4*a*c + x^2*(2*c + d*x)^2])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2458, 1416}

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.

\(\displaystyle \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx\)

\(\Big \downarrow \) 2458

\(\displaystyle \int \frac {1}{\sqrt {c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 \left (\frac {c}{d}+x\right )^2+d^2 \left (\frac {c}{d}+x\right )^4}}d\left (\frac {c}{d}+x\right )\)

\(\Big \downarrow \) 1416

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

Input:

Int[1/Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4],x]
 

Output:

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

Defintions of rubi rules used

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 2458

Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp 
on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x 
- S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp 
on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P 
n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1055\) vs. \(2(192)=384\).

Time = 1.03 (sec) , antiderivative size = 1056, normalized size of antiderivative = 4.87

method	result	size
default	\(\text {Expression too large to display}\)	\(1056\)
elliptic	\(\text {Expression too large to display}\)	\(1056\)

Input:

int(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2*((c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2))/d+(-c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d 
)*((-(c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2))/d+(c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/ 
d)*(x-(-c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)/(-(c+(-2*d*(-a*c)^(1/2)+c^2)^(1 
/2))/d-(-c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)/(x+(c+(2*d*(-a*c)^(1/2)+c^2)^( 
1/2))/d))^(1/2)*(x+(c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)^2*((-(c+(2*d*(-a*c) 
^(1/2)+c^2)^(1/2))/d-(-c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)*(x-(-c+(-2*d*(-a 
*c)^(1/2)+c^2)^(1/2))/d)/((-c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2))/d-(-c+(2*d*(- 
a*c)^(1/2)+c^2)^(1/2))/d)/(x+(c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d))^(1/2)*(( 
-(c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d-(-c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)*( 
x+(c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)/(-(c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2)) 
/d-(-c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)/(x+(c+(2*d*(-a*c)^(1/2)+c^2)^(1/2) 
)/d))^(1/2)/(-(c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2))/d+(c+(2*d*(-a*c)^(1/2)+c^2 
)^(1/2))/d)/(-(c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d-(-c+(2*d*(-a*c)^(1/2)+c^2 
)^(1/2))/d)/(d^2*(x-(-c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)*(x+(c+(2*d*(-a*c) 
^(1/2)+c^2)^(1/2))/d)*(x-(-c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)*(x+(c+(-2*d 
*(-a*c)^(1/2)+c^2)^(1/2))/d))^(1/2)*EllipticF(((-(c+(-2*d*(-a*c)^(1/2)+c^2 
)^(1/2))/d+(c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)*(x-(-c+(2*d*(-a*c)^(1/2)+c^ 
2)^(1/2))/d)/(-(c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2))/d-(-c+(2*d*(-a*c)^(1/2)+c 
^2)^(1/2))/d)/(x+(c+(2*d*(-a*c)^(1/2)+c^2)^(1/2))/d))^(1/2),((-(c+(2*d*(-a 
*c)^(1/2)+c^2)^(1/2))/d-(-c+(-2*d*(-a*c)^(1/2)+c^2)^(1/2))/d)*((c+(-2*d...
 

Fricas [F]

\[ \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx=\int { \frac {1}{\sqrt {d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c}} \,d x } \] Input:

integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2),x, algorithm="fricas 
")
 

Output:

integral(1/sqrt(d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c), x)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx=\int \frac {1}{\sqrt {4 a c + 4 c^{2} x^{2} + 4 c d x^{3} + d^{2} x^{4}}}\, dx \] Input:

integrate(1/(d**2*x**4+4*c*d*x**3+4*c**2*x**2+4*a*c)**(1/2),x)
 

Output:

Integral(1/sqrt(4*a*c + 4*c**2*x**2 + 4*c*d*x**3 + d**2*x**4), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx=\int { \frac {1}{\sqrt {d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c}} \,d x } \] Input:

integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2),x, algorithm="maxima 
")
 

Output:

integrate(1/sqrt(d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx=\int { \frac {1}{\sqrt {d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c}} \,d x } \] Input:

integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/sqrt(d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx=\int \frac {1}{\sqrt {4\,c^2\,x^2+4\,c\,d\,x^3+4\,a\,c+d^2\,x^4}} \,d x \] Input:

int(1/(4*a*c + 4*c^2*x^2 + d^2*x^4 + 4*c*d*x^3)^(1/2),x)
 

Output:

int(1/(4*a*c + 4*c^2*x^2 + d^2*x^4 + 4*c*d*x^3)^(1/2), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx=\int \frac {\sqrt {d^{2} x^{4}+4 c d \,x^{3}+4 c^{2} x^{2}+4 a c}}{d^{2} x^{4}+4 c d \,x^{3}+4 c^{2} x^{2}+4 a c}d x \] Input:

int(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c)^(1/2),x)
 

Output:

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