\(\int \frac {A+B x+C x^2}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx\) [76]

Optimal result

Integrand size = 43, antiderivative size = 431 \[ \int \frac {A+B x+C x^2}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\frac {2 C \sqrt {-64+(b+c x)^3}}{3 c^3}-\frac {2 (B c-2 b C) \sqrt {-64+(b+c x)^3}}{c^3 \left (4-4 \sqrt {3}-b-c x\right )}+\frac {2 \sqrt [4]{3} \sqrt {2+\sqrt {3}} (B c-2 b C) (4-b-c x) \sqrt {\frac {16+4 (b+c x)+(b+c x)^2}{\left (4-4 \sqrt {3}-b-c x\right )^2}} E\left (\arcsin \left (\frac {4+4 \sqrt {3}-b-c x}{4-4 \sqrt {3}-b-c x}\right )|-7+4 \sqrt {3}\right )}{c^3 \sqrt {-\frac {4-b-c x}{\left (4-4 \sqrt {3}-b-c x\right )^2}} \sqrt {-64+(b+c x)^3}}+\frac {\sqrt {2-\sqrt {3}} \left (b B c-A c^2-b^2 C-4 \left (1+\sqrt {3}\right ) (B c-2 b C)\right ) (4-b-c x) \sqrt {\frac {16+4 (b+c x)+(b+c x)^2}{\left (4-4 \sqrt {3}-b-c x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {4+4 \sqrt {3}-b-c x}{4-4 \sqrt {3}-b-c x}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} c^3 \sqrt {-\frac {4-b-c x}{\left (4-4 \sqrt {3}-b-c x\right )^2}} \sqrt {-64+(b+c x)^3}} \] Output:

2/3*C*(-64+(c*x+b)^3)^(1/2)/c^3-2*(B*c-2*C*b)*(-64+(c*x+b)^3)^(1/2)/c^3/(4 
-4*3^(1/2)-b-c*x)+2*3^(1/4)*(1/2*6^(1/2)+1/2*2^(1/2))*(B*c-2*C*b)*(-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)*EllipticE((4+4*3 
^(1/2)-b-c*x)/(4-4*3^(1/2)-b-c*x),2*I-I*3^(1/2))/c^3/(-(-c*x-b+4)/(4-4*3^( 
1/2)-b-c*x)^2)^(1/2)/(-64+(c*x+b)^3)^(1/2)+1/3*(1/2*6^(1/2)-1/2*2^(1/2))*( 
B*b*c-A*c^2-C*b^2-4*(1+3^(1/2))*(B*c-2*C*b))*(-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))*3^(3/4)/c^3/(-(-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 = 10.66 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x+C x^2}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\frac {2 C \left (-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3\right )-3 \sqrt {2} \sqrt {-\frac {i (-4+b+c x)}{3 i+\sqrt {3}}} \sqrt {16+b^2+4 c x+c^2 x^2+2 b (2+c x)} \left (-2 \left (3 i+\sqrt {3}\right ) (B c-2 b C) E\left (\arcsin \left (\frac {\sqrt {2 i+2 \sqrt {3}+i b+i c x}}{2 \sqrt [4]{3}}\right )|\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )+i \left (-((-4+b) B c)+A c^2+(-8+b) b C\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2 i+2 \sqrt {3}+i b+i c x}}{2 \sqrt [4]{3}}\right ),\frac {2 \sqrt {3}}{3 i+\sqrt {3}}\right )\right )}{3 c^3 \sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \] Input:

Integrate[(A + B*x + C*x^2)/Sqrt[-64 + b^3 + 3*b^2*c*x + 3*b*c^2*x^2 + c^3 
*x^3],x]
 

Output:

(2*C*(-64 + b^3 + 3*b^2*c*x + 3*b*c^2*x^2 + c^3*x^3) - 3*Sqrt[2]*Sqrt[((-I 
)*(-4 + b + c*x))/(3*I + Sqrt[3])]*Sqrt[16 + b^2 + 4*c*x + c^2*x^2 + 2*b*( 
2 + c*x)]*(-2*(3*I + Sqrt[3])*(B*c - 2*b*C)*EllipticE[ArcSin[Sqrt[2*I + 2* 
Sqrt[3] + I*b + I*c*x]/(2*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])] + I*(-(( 
-4 + b)*B*c) + A*c^2 + (-8 + b)*b*C)*EllipticF[ArcSin[Sqrt[2*I + 2*Sqrt[3] 
 + I*b + I*c*x]/(2*3^(1/4))], (2*Sqrt[3])/(3*I + Sqrt[3])]))/(3*c^3*Sqrt[- 
64 + b^3 + 3*b^2*c*x + 3*b*c^2*x^2 + c^3*x^3])
 

Rubi [A] (warning: unable to verify)

Time = 1.21 (sec) , antiderivative size = 537, normalized size of antiderivative = 1.25, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {2459, 2425, 793, 2419, 760, 2418}

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.

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

\(\Big \downarrow \) 2459

\(\displaystyle \int \frac {A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )+C \left (\frac {b}{c}+x\right )^2}{\sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}d\left (\frac {b}{c}+x\right )\)

\(\Big \downarrow \) 2425

\(\displaystyle \int \frac {A-\frac {b (B c-b C)}{c^2}+\left (B-\frac {2 b C}{c}\right ) \left (\frac {b}{c}+x\right )}{\sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}d\left (\frac {b}{c}+x\right )+C \int \frac {\left (\frac {b}{c}+x\right )^2}{\sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}d\left (\frac {b}{c}+x\right )\)

\(\Big \downarrow \) 793

\(\displaystyle \int \frac {A-\frac {b (B c-b C)}{c^2}+\left (B-\frac {2 b C}{c}\right ) \left (\frac {b}{c}+x\right )}{\sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}d\left (\frac {b}{c}+x\right )+\frac {2 C \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}{3 c^3}\)

\(\Big \downarrow \) 2419

\(\displaystyle \frac {\left (A c^2+\left (-b+4 \sqrt {3}+4\right ) B c-\left (-b+8 \sqrt {3}+8\right ) b C\right ) \int \frac {1}{\sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}d\left (\frac {b}{c}+x\right )}{c^2}-\frac {(B c-2 b C) \int \frac {4 \left (1+\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}{\sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}d\left (\frac {b}{c}+x\right )}{c^2}+\frac {2 C \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}{3 c^3}\)

\(\Big \downarrow \) 760

\(\displaystyle -\frac {(B c-2 b C) \int \frac {4 \left (1+\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}{\sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}d\left (\frac {b}{c}+x\right )}{c^2}-\frac {\sqrt {2-\sqrt {3}} \left (4-c \left (\frac {b}{c}+x\right )\right ) \sqrt {\frac {c^2 \left (\frac {b}{c}+x\right )^2+4 c \left (\frac {b}{c}+x\right )+16}{\left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )^2}} \left (A c^2+\left (-b+4 \sqrt {3}+4\right ) B c-\left (-b+8 \sqrt {3}+8\right ) b C\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {4 \left (1+\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}{4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} c^3 \sqrt {-\frac {4-c \left (\frac {b}{c}+x\right )}{\left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )^2}} \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}+\frac {2 C \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}{3 c^3}\)

\(\Big \downarrow \) 2418

\(\displaystyle -\frac {\sqrt {2-\sqrt {3}} \left (4-c \left (\frac {b}{c}+x\right )\right ) \sqrt {\frac {c^2 \left (\frac {b}{c}+x\right )^2+4 c \left (\frac {b}{c}+x\right )+16}{\left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )^2}} \left (A c^2+\left (-b+4 \sqrt {3}+4\right ) B c-\left (-b+8 \sqrt {3}+8\right ) b C\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {4 \left (1+\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}{4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} c^3 \sqrt {-\frac {4-c \left (\frac {b}{c}+x\right )}{\left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )^2}} \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}-\frac {(B c-2 b C) \left (\frac {2 \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}{c \left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )}-\frac {2 \sqrt [4]{3} \sqrt {2+\sqrt {3}} \left (4-c \left (\frac {b}{c}+x\right )\right ) \sqrt {\frac {c^2 \left (\frac {b}{c}+x\right )^2+4 c \left (\frac {b}{c}+x\right )+16}{\left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )^2}} E\left (\arcsin \left (\frac {4 \left (1+\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}{4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )}\right )|-7+4 \sqrt {3}\right )}{c \sqrt {-\frac {4-c \left (\frac {b}{c}+x\right )}{\left (4 \left (1-\sqrt {3}\right )-c \left (\frac {b}{c}+x\right )\right )^2}} \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}\right )}{c^2}+\frac {2 C \sqrt {c^3 \left (\frac {b}{c}+x\right )^3-64}}{3 c^3}\)

Input:

Int[(A + B*x + C*x^2)/Sqrt[-64 + b^3 + 3*b^2*c*x + 3*b*c^2*x^2 + c^3*x^3], 
x]
 

Output:

(2*C*Sqrt[-64 + c^3*(b/c + x)^3])/(3*c^3) - ((B*c - 2*b*C)*((2*Sqrt[-64 + 
c^3*(b/c + x)^3])/(c*(4*(1 - Sqrt[3]) - c*(b/c + x))) - (2*3^(1/4)*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]*EllipticE[ArcSin[(4*(1 + Sqrt[3]) - c*(b 
/c + x))/(4*(1 - Sqrt[3]) - c*(b/c + x))], -7 + 4*Sqrt[3]])/(c*Sqrt[-((4 - 
 c*(b/c + x))/(4*(1 - Sqrt[3]) - c*(b/c + x))^2)]*Sqrt[-64 + c^3*(b/c + x) 
^3])))/c^2 - (Sqrt[2 - Sqrt[3]]*((4 + 4*Sqrt[3] - b)*B*c + A*c^2 - (8 + 8* 
Sqrt[3] - b)*b*C)*(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^3*Sqrt[-((4 - c*(b/c + x))/(4*(1 - Sqrt[3]) - c*(b/c + x))^2)]*Sqrt[-6 
4 + 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 793

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n) 
^(p + 1)/(b*n*(p + 1)), x] /; FreeQ[{a, b, m, n, p}, x] && EqQ[m, n - 1] && 
 NeQ[p, -1]
 

rule 2418

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 + Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 + Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 - Sqrt[3])*s + r*x))), x] + S 
imp[3^(1/4)*Sqrt[2 + Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 - Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[(-s)*((s + r*x)/((1 - S 
qrt[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 + Sqrt[3])*s + r*x)/((1 - Sqrt[ 
3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && NegQ[a] && 
 EqQ[b*c^3 - 2*(5 + 3*Sqrt[3])*a*d^3, 0]
 

rule 2419

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(c*r - (1 + Sqrt[3])*d*s)/r 
  Int[1/Sqrt[a + b*x^3], x], x] + Simp[d/r   Int[((1 + Sqrt[3])*s + r*x)/Sq 
rt[a + b*x^3], x], x]] /; FreeQ[{a, b, c, d}, x] && NegQ[a] && NeQ[b*c^3 - 
2*(5 + 3*Sqrt[3])*a*d^3, 0]
 

rule 2425

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Pq, x, n - 
 1]   Int[x^(n - 1)*(a + b*x^n)^p, x], x] + Int[ExpandToSum[Pq - Coeff[Pq, 
x, n - 1]*x^(n - 1), x]*(a + b*x^n)^p, x] /; FreeQ[{a, b, p}, x] && PolyQ[P 
q, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1
 

rule 2459

Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 
]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x 
 -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial 
Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - 
> x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ 
[Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] &&  !(MonomialQ[Qx, x] 
&& IGtQ[p, 0])
 

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 755, normalized size of antiderivative = 1.75

method	result	size
elliptic	\(\frac {2 C \sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}}{3 c^{3}}+\frac {2 \left (A -\frac {b^{2} C}{c^{2}}\right ) \left (\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}\right ) \sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}\, \sqrt {\frac {x -\frac {-b -2+2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\, \sqrt {\frac {x -\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}, \sqrt {\frac {-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\right )}{\sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}}+\frac {2 \left (B -\frac {2 b C}{c}\right ) \left (\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}\right ) \sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}\, \sqrt {\frac {x -\frac {-b -2+2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\, \sqrt {\frac {x -\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}}\, \left (\left (-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}, \sqrt {\frac {-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\right )+\frac {\left (-b -2+2 i \sqrt {3}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b -4}{c}}{\frac {-b -2-2 i \sqrt {3}}{c}+\frac {b -4}{c}}}, \sqrt {\frac {-\frac {b -4}{c}-\frac {-b -2-2 i \sqrt {3}}{c}}{-\frac {b -4}{c}-\frac {-b -2+2 i \sqrt {3}}{c}}}\right )}{c}\right )}{\sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}}\)	\(755\)
risch	\(\text {Expression too large to display}\)	\(1038\)
default	\(\text {Expression too large to display}\)	\(1446\)

Input:

int((C*x^2+B*x+A)/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

2/3*C/c^3*(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2)+2*(A-b^2/c^2*C)*((- 
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))+2*(B-2*b/c*C)*((-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)*((-(b-4 
)/c-(-b-2+2*I*3^(1/2))/c)*EllipticE(((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))+(-b-2+2*I*3^(1/2))/c*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 = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.28 \[ \int \frac {A+B x+C x^2}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\frac {2 \, {\left (\sqrt {c^{3} x^{3} + 3 \, b c^{2} x^{2} + 3 \, b^{2} c x + b^{3} - 64} C c^{2} + 3 \, {\left (C b^{2} - B b c + A c^{2}\right )} \sqrt {c^{3}} {\rm weierstrassPInverse}\left (0, \frac {256}{c^{3}}, \frac {c x + b}{c}\right ) + 3 \, {\left (2 \, C b c - B c^{2}\right )} \sqrt {c^{3}} {\rm weierstrassZeta}\left (0, \frac {256}{c^{3}}, {\rm weierstrassPInverse}\left (0, \frac {256}{c^{3}}, \frac {c x + b}{c}\right )\right )\right )}}{3 \, c^{5}} \] Input:

integrate((C*x^2+B*x+A)/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x, al 
gorithm="fricas")
 

Output:

2/3*(sqrt(c^3*x^3 + 3*b*c^2*x^2 + 3*b^2*c*x + b^3 - 64)*C*c^2 + 3*(C*b^2 - 
 B*b*c + A*c^2)*sqrt(c^3)*weierstrassPInverse(0, 256/c^3, (c*x + b)/c) + 3 
*(2*C*b*c - B*c^2)*sqrt(c^3)*weierstrassZeta(0, 256/c^3, weierstrassPInver 
se(0, 256/c^3, (c*x + b)/c)))/c^5
 

Sympy [F]

\[ \int \frac {A+B x+C x^2}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\int \frac {A + B x + C x^{2}}{\sqrt {\left (b + c x - 4\right ) \left (b^{2} + 2 b c x + 4 b + c^{2} x^{2} + 4 c x + 16\right )}}\, dx \] Input:

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

Output:

Integral((A + B*x + C*x**2)/sqrt((b + c*x - 4)*(b**2 + 2*b*c*x + 4*b + c** 
2*x**2 + 4*c*x + 16)), x)
 

Maxima [F]

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

integrate((C*x^2+B*x+A)/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x, al 
gorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((C*x^2+B*x+A)/(c^3*x^3+3*b*c^2*x^2+3*b^2*c*x+b^3-64)^(1/2),x, al 
gorithm="giac")
 

Output:

integrate((C*x^2 + B*x + A)/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. \[ \int \frac {A+B x+C x^2}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\int \frac {C\,x^2+B\,x+A}{\sqrt {b^3+3\,b^2\,c\,x+3\,b\,c^2\,x^2+c^3\,x^3-64}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {A+B x+C x^2}{\sqrt {-64+b^3+3 b^2 c x+3 b c^2 x^2+c^3 x^3}} \, dx=\frac {2 \sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}+6 \left (\int \frac {\sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}}{c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}d x \right ) a \,c^{2}-3 \left (\int \frac {\sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}}{c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}d x \right ) b^{2} c +3 \left (\int \frac {\sqrt {c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}\, x^{2}}{c^{3} x^{3}+3 b \,c^{2} x^{2}+3 b^{2} c x +b^{3}-64}d x \right ) c^{3}}{6 c^{2}} \] Input:

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

Output:

(2*sqrt(b**3 + 3*b**2*c*x + 3*b*c**2*x**2 + c**3*x**3 - 64) + 6*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)*a*c**2 - 3*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)*b**2*c + 3*int((sqrt(b**3 + 3*b**2*c*x + 3*b*c**2*x**2 + c**3 
*x**3 - 64)*x**2)/(b**3 + 3*b**2*c*x + 3*b*c**2*x**2 + c**3*x**3 - 64),x)* 
c**3)/(6*c**2)