\(\int \frac {x^6}{(8 c-d x^3)^2 (c+d x^3)^{3/2}} \, dx\) [628]

Optimal result

Integrand size = 27, antiderivative size = 256 \[ \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 x \left (4 c+d x^3\right )}{81 c d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{81 \sqrt [4]{3} c d^{7/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:

2/81*x*(d*x^3+4*c)/c/d^2/(-d*x^3+8*c)/(d*x^3+c)^(1/2)-2/243*(1/2*6^(1/2)+1 
/2*2^(1/2))*(c^(1/3)+d^(1/3)*x)*((c^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/( 
(1+3^(1/2))*c^(1/3)+d^(1/3)*x)^2)^(1/2)*EllipticF(((1-3^(1/2))*c^(1/3)+d^( 
1/3)*x)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x),I*3^(1/2)+2*I)*3^(3/4)/c/d^(7/3)/( 
c^(1/3)*(c^(1/3)+d^(1/3)*x)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)^2)^(1/2)/(d*x^ 
3+c)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.65 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.74 \[ \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {6 \sqrt [3]{-d} x \left (4 c+d x^3\right )+2 i 3^{3/4} \sqrt [3]{c} \sqrt {\frac {(-1)^{5/6} \left (-\sqrt [3]{c}+\sqrt [3]{-d} x\right )}{\sqrt [3]{c}}} \sqrt {1+\frac {\sqrt [3]{-d} x}{\sqrt [3]{c}}+\frac {(-d)^{2/3} x^2}{c^{2/3}}} \left (-8 c+d x^3\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {-(-1)^{5/6}-\frac {i \sqrt [3]{-d} x}{\sqrt [3]{c}}}}{\sqrt [4]{3}}\right ),\sqrt [3]{-1}\right )}{243 c (-d)^{7/3} \left (-8 c+d x^3\right ) \sqrt {c+d x^3}} \] Input:

Integrate[x^6/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
 

Output:

-1/243*(6*(-d)^(1/3)*x*(4*c + d*x^3) + (2*I)*3^(3/4)*c^(1/3)*Sqrt[((-1)^(5 
/6)*(-c^(1/3) + (-d)^(1/3)*x))/c^(1/3)]*Sqrt[1 + ((-d)^(1/3)*x)/c^(1/3) + 
((-d)^(2/3)*x^2)/c^(2/3)]*(-8*c + d*x^3)*EllipticF[ArcSin[Sqrt[-(-1)^(5/6) 
 - (I*(-d)^(1/3)*x)/c^(1/3)]/3^(1/4)], (-1)^(1/3)])/(c*(-d)^(7/3)*(-8*c + 
d*x^3)*Sqrt[c + d*x^3])
 

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 0.36 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1013, 1012}

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.

\(\displaystyle \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (\frac {d x^3}{c}+1\right )^{3/2}}dx}{c \sqrt {c+d x^3}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {x^7 \sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (\frac {7}{3},2,\frac {3}{2},\frac {10}{3},\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{448 c^3 \sqrt {c+d x^3}}\)

Input:

Int[x^6/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
 

Output:

(x^7*Sqrt[1 + (d*x^3)/c]*AppellF1[7/3, 2, 3/2, 10/3, (d*x^3)/(8*c), -((d*x 
^3)/c)])/(448*c^3*Sqrt[c + d*x^3])
 

Defintions of rubi rules used

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ 
n/a))^FracPart[p])   Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [A] (verified)

Time = 1.95 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.32

method	result	size
elliptic	\(\frac {2 x}{243 d^{2} c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {8 x \sqrt {d \,x^{3}+c}}{243 c \,d^{2} \left (-d \,x^{3}+8 c \right )}+\frac {2 i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}}{-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{243 d^{3} c \sqrt {d \,x^{3}+c}}\)	\(339\)
default	\(\text {Expression too large to display}\)	\(1792\)

Input:

int(x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/243/d^2*x/c/((x^3+c/d)*d)^(1/2)+8/243*x/c/d^2*(d*x^3+c)^(1/2)/(-d*x^3+8* 
c)+2/243*I/d^3/c*3^(1/2)*(-c*d^2)^(1/3)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3 
^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)*((x-1/d*(-c*d^2)^ 
(1/3))/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)*(-I*( 
x+1/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^ 
(1/3))^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1 
/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),(I*3^( 
1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3 
)))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.35 \[ \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left ({\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {d} {\rm weierstrassPInverse}\left (0, -\frac {4 \, c}{d}, x\right ) + {\left (d^{2} x^{4} + 4 \, c d x\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (c d^{5} x^{6} - 7 \, c^{2} d^{4} x^{3} - 8 \, c^{3} d^{3}\right )}} \] Input:

integrate(x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
 

Output:

-2/81*((d^2*x^6 - 7*c*d*x^3 - 8*c^2)*sqrt(d)*weierstrassPInverse(0, -4*c/d 
, x) + (d^2*x^4 + 4*c*d*x)*sqrt(d*x^3 + c))/(c*d^5*x^6 - 7*c^2*d^4*x^3 - 8 
*c^3*d^3)
 

Sympy [F]

\[ \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^{6}}{\left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**6/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)
 

Output:

Integral(x**6/((-8*c + d*x**3)**2*(c + d*x**3)**(3/2)), x)
 

Maxima [F]

\[ \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x^{6}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:

integrate(x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^6/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)
 

Giac [F]

\[ \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x^{6}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:

integrate(x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
 

Output:

integrate(x^6/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^6}{{\left (d\,x^3+c\right )}^{3/2}\,{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:

int(x^6/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2),x)
 

Output:

int(x^6/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2), x)
 

Reduce [F]

\[ \int \frac {x^6}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{6}}{d^{4} x^{12}-14 c \,d^{3} x^{9}+33 c^{2} d^{2} x^{6}+112 c^{3} d \,x^{3}+64 c^{4}}d x \] Input:

int(x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)
 

Output:

int((sqrt(c + d*x**3)*x**6)/(64*c**4 + 112*c**3*d*x**3 + 33*c**2*d**2*x**6 
 - 14*c*d**3*x**9 + d**4*x**12),x)