\(\int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx\) [96]

Optimal result

Integrand size = 32, antiderivative size = 163 \[ \int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=-\frac {\sqrt {2-\sqrt {3}} (4-b-c x) \sqrt {\frac {16+4 (b+c x)+(b+c x)^2}{\left (4-4 \sqrt {3}-b-c x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {4+4 \sqrt {3}-b-c x}{4-4 \sqrt {3}-b-c x}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} c \sqrt {-\frac {4-b-c x}{\left (4-4 \sqrt {3}-b-c x\right )^2}} \sqrt {-64+(b+c x)^3}} \] Output:

-1/3*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*(-c*x-b+4)*((16+4*c*x+4*b+(c*x+b)^2 
)/(4-4*3^(1/2)-b-c*x)^2)^(1/2)*EllipticF((4+4*3^(1/2)-b-c*x)/(4-4*3^(1/2)- 
b-c*x),2*I-I*3^(1/2))/c/(-(-c*x-b+4)/(4-4*3^(1/2)-b-c*x)^2)^(1/2)/(-64+(c* 
x+b)^3)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.26 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\frac {\sqrt {-4 i+4 \sqrt {3}-2 i b-2 i c x} \sqrt {-\frac {i (-4+b+c x)}{3 i+\sqrt {3}}} \left (2-2 i \sqrt {3}+b+c x\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2 i+2 \sqrt {3}+i b+i c x}}{2 \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )}{c \sqrt {2 i+2 \sqrt {3}+i b+i c x} \sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \] Input:

Integrate[1/Sqrt[-64 + b^3 + 3*b^2*c*x + 3*b*c^2*x^2 + c^3*x^3],x]
 

Output:

(Sqrt[-4*I + 4*Sqrt[3] - (2*I)*b - (2*I)*c*x]*Sqrt[((-I)*(-4 + b + c*x))/( 
3*I + Sqrt[3])]*(2 - (2*I)*Sqrt[3] + b + c*x)*EllipticF[ArcSin[Sqrt[2*I + 
2*Sqrt[3] + I*b + I*c*x]/(2*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])])/(c*Sq 
rt[2*I + 2*Sqrt[3] + I*b + I*c*x]*Sqrt[-64 + b^3 + 3*b^2*c*x + 3*b*c^2*x^2 
 + c^3*x^3])
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.26, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2458, 760}

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.

\(\displaystyle \int \frac {1}{\sqrt {b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3-64}} \, dx\)

\(\Big \downarrow \) 2458

\(\displaystyle \int \frac {1}{\sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}d\left (\frac {b}{c}+x\right )\)

\(\Big \downarrow \) 760

\(\displaystyle -\frac {\sqrt {2-\sqrt {3}} \left (4-c \left (\frac {b}{c}+x\right )\right ) \sqrt {\frac {c^2 \left (\frac {b}{c}+x\right )^2+4 c \left (\frac {b}{c}+x\right )+16}{\left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {4 \left (1+\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}{4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} c \sqrt {-\frac {4-c \left (\frac {b}{c}+x\right )}{\left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )^2}} \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}\)

Input:

Int[1/Sqrt[-64 + b^3 + 3*b^2*c*x + 3*b*c^2*x^2 + c^3*x^3],x]
 

Output:

-((Sqrt[2 - Sqrt[3]]*(4 - c*(b/c + x))*Sqrt[(16 + 4*c*(b/c + x) + c^2*(b/c 
 + x)^2)/(4*(1 - Sqrt[3]) - c*(b/c + x))^2]*EllipticF[ArcSin[(4*(1 + Sqrt[ 
3]) - c*(b/c + x))/(4*(1 - Sqrt[3]) - c*(b/c + x))], -7 + 4*Sqrt[3]])/(3^( 
1/4)*c*Sqrt[-((4 - c*(b/c + x))/(4*(1 - Sqrt[3]) - c*(b/c + x))^2)]*Sqrt[- 
64 + c^3*(b/c + x)^3]))
 

Defintions of rubi rules used

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[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 [A] (verified)

Time = 0.27 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {2 \left (\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}\right ) \sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}\, \sqrt {\frac {x -\frac {-b -2+2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\, \sqrt {\frac {x -\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}, \sqrt {\frac {-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\right )}{\sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}}\)	\(282\)
elliptic	\(\frac {2 \left (\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}\right ) \sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}\, \sqrt {\frac {x -\frac {-b -2+2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\, \sqrt {\frac {x -\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}, \sqrt {\frac {-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\right )}{\sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}}\)	\(282\)

Input:

int(1/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x,method=_RETURNVERBOSE 
)
 

Output:

2*((-b-2-2*I*3^(1/2))/c+(b-4)/c)*((x+(b-4)/c)/((-b-2-2*I*3^(1/2))/c+(b-4)/ 
c))^(1/2)*((x-(-b-2+2*I*3^(1/2))/c)/(-(b-4)/c-(-b-2+2*I*3^(1/2))/c))^(1/2) 
*((x-(-b-2-2*I*3^(1/2))/c)/(-(b-4)/c-(-b-2-2*I*3^(1/2))/c))^(1/2)/(c^3*x^3 
+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2)*EllipticF(((x+(b-4)/c)/((-b-2-2*I*3^( 
1/2))/c+(b-4)/c))^(1/2),((-(b-4)/c-(-b-2-2*I*3^(1/2))/c)/(-(b-4)/c-(-b-2+2 
*I*3^(1/2))/c))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\frac {2 \, \sqrt {c^{3}} {\rm weierstrassPInverse}\left (0, \frac {256}{c^{3}}, \frac {c x + b}{c}\right )}{c^{3}} \] Input:

integrate(1/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x, algorithm="fri 
cas")
 

Output:

2*sqrt(c^3)*weierstrassPInverse(0, 256/c^3, (c*x + b)/c)/c^3
 

Sympy [F]

\[ \int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\int \frac {1}{\sqrt {b^{3} + 3 b^{2} c x + 3 b c^{2} x^{2} + c^{3} x^{3} - 64}}\, dx \] Input:

integrate(1/(c**3*x**3+3*b*c**2*x**2+3*b**2*c*x+b**3-64)**(1/2),x)
 

Output:

Integral(1/sqrt(b**3 + 3*b**2*c*x + 3*b*c**2*x**2 + c**3*x**3 - 64), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\int { \frac {1}{\sqrt {c^{3} x^{3} + 3 \, b c^{2} x^{2} + 3 \, b^{2} c x + b^{3} - 64}} \,d x } \] Input:

integrate(1/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x, algorithm="max 
ima")
 

Output:

integrate(1/sqrt(c^3*x^3 + 3*b*c^2*x^2 + 3*b^2*c*x + b^3 - 64), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\int { \frac {1}{\sqrt {c^{3} x^{3} + 3 \, b c^{2} x^{2} + 3 \, b^{2} c x + b^{3} - 64}} \,d x } \] Input:

integrate(1/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x, algorithm="gia 
c")
 

Output:

integrate(1/sqrt(c^3*x^3 + 3*b*c^2*x^2 + 3*b^2*c*x + b^3 - 64), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\int \frac {1}{\sqrt {b^3+3\,b^2\,c\,x+3\,b\,c^2\,x^2+c^3\,x^3-64}} \,d x \] Input:

int(1/(b^3 + c^3*x^3 + 3*b*c^2*x^2 + 3*b^2*c*x - 64)^(1/2),x)
 

Output:

int(1/(b^3 + c^3*x^3 + 3*b*c^2*x^2 + 3*b^2*c*x - 64)^(1/2), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\int \frac {\sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}}{c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}d x \] Input:

int(1/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x)
 

Output:

int(sqrt(b**3 + 3*b**2*c*x + 3*b*c**2*x**2 + c**3*x**3 - 64)/(b**3 + 3*b** 
2*c*x + 3*b*c**2*x**2 + c**3*x**3 - 64),x)