Integrand size = 27, antiderivative size = 66 \[ \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (-\frac {5}{3},2,\frac {1}{2},-\frac {2}{3},\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{320 c^2 x^5 \sqrt {c+d x^3}} \] Output:
-1/320*(1+d*x^3/c)^(1/2)*AppellF1(-5/3,1/2,2,-2/3,-d*x^3/c,1/8*d*x^3/c)/c^ 2/x^5/(d*x^3+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(279\) vs. \(2(66)=132\).
Time = 10.31 (sec) , antiderivative size = 279, normalized size of antiderivative = 4.23 \[ \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\frac {64 \left (c+d x^3\right ) \left (864 c^2-1080 c d x^3+119 d^2 x^6\right )}{c^4 x^5 \left (-8 c+d x^3\right )}-\frac {119 d^3 x^4 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{c^5}+\frac {1404928 d^2 x \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{c^2 \left (8 c-d x^3\right ) \left (32 c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+3 d x^3 \left (\operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-4 \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )\right )}}{2211840 \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^6*(8*c - d*x^3)^2*Sqrt[c + d*x^3]),x]
Output:
((64*(c + d*x^3)*(864*c^2 - 1080*c*d*x^3 + 119*d^2*x^6))/(c^4*x^5*(-8*c + d*x^3)) - (119*d^3*x^4*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d *x^3)/c), (d*x^3)/(8*c)])/c^5 + (1404928*d^2*x*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), (d*x^3)/(8*c)])/(c^2*(8*c - d*x^3)*(32*c*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), (d*x^3)/(8*c)] + 3*d*x^3*(AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), (d*x^3)/(8*c)] - 4*AppellF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), (d*x^3)/(8*c)]))))/(2211840*Sqrt[c + d*x^3])
Time = 0.36 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {\frac {d x^3}{c}+1}}dx}{\sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (-\frac {5}{3},2,\frac {1}{2},-\frac {2}{3},\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{320 c^2 x^5 \sqrt {c+d x^3}}\) |
Input:
Int[1/(x^6*(8*c - d*x^3)^2*Sqrt[c + d*x^3]),x]
Output:
-1/320*(Sqrt[1 + (d*x^3)/c]*AppellF1[-5/3, 2, 1/2, -2/3, (d*x^3)/(8*c), -( (d*x^3)/c)])/(c^2*x^5*Sqrt[c + d*x^3])
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])
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])
Result contains higher order function than in optimal. Order 9 vs. order 6.
Time = 4.14 (sec) , antiderivative size = 765, normalized size of antiderivative = 11.59
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(765\) |
risch | \(\text {Expression too large to display}\) | \(1464\) |
default | \(\text {Expression too large to display}\) | \(1783\) |
Input:
int(1/x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/320*(d*x^3+c)^(1/2)/c^3/x^5+9/2560*d*(d*x^3+c)^(1/2)/c^4/x^2+1/13824*x* d^2/c^4*(d*x^3+c)^(1/2)/(-d*x^3+8*c)-119/103680*I*d/c^4*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)*Ellipt icF(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))-7/31104*I/d/c^4*2^(1/2)* sum(1/_alpha^2*(-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3) +(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c* d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)* (-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*( -c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c *d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(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),-1/18/d*(2*I*(-c*d^2)^(1/3)*_alpha^2*3^(1/2)*d-I*(-c*d^2)^(2/3)*_alpha* 3^(1/2)+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/c,(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)),_a lpha=RootOf(_Z^3*d-8*c))
Leaf count of result is larger than twice the leaf count of optimal. 2561 vs. \(2 (52) = 104\).
Time = 3.55 (sec) , antiderivative size = 2561, normalized size of antiderivative = 38.80 \[ \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\text {Too large to display} \] Input:
integrate(1/x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="fricas")
Output:
1/2488320*(11088*(d^2*x^8 - 8*c*d*x^5)*sqrt(d)*weierstrassPInverse(0, -4*c /d, x) + 35*(c^4*d*x^8 - 8*c^5*x^5 + sqrt(-3)*(c^4*d*x^8 - 8*c^5*x^5))*(d^ 10/c^25)^(1/6)*log(16807*(d^11*x^9 + 318*c*d^10*x^6 + 1200*c^2*d^9*x^3 + 6 40*c^3*d^8 - 9*(c^17*d^4*x^8 + 38*c^18*d^3*x^5 + 64*c^19*d^2*x^2 + sqrt(-3 )*(c^17*d^4*x^8 + 38*c^18*d^3*x^5 + 64*c^19*d^2*x^2))*(d^10/c^25)^(2/3) + 3*sqrt(d*x^3 + c)*((c^21*d^2*x^7 + 80*c^22*d*x^4 + 160*c^23*x - sqrt(-3)*( c^21*d^2*x^7 + 80*c^22*d*x^4 + 160*c^23*x))*(d^10/c^25)^(5/6) - 2*(7*c^13* d^5*x^6 + 152*c^14*d^4*x^3 + 64*c^15*d^3)*sqrt(d^10/c^25) + 6*(5*c^5*d^8*x ^5 + 32*c^6*d^7*x^2 + sqrt(-3)*(5*c^5*d^8*x^5 + 32*c^6*d^7*x^2))*(d^10/c^2 5)^(1/6)) - 9*(5*c^9*d^7*x^7 + 64*c^10*d^6*x^4 + 32*c^11*d^5*x - sqrt(-3)* (5*c^9*d^7*x^7 + 64*c^10*d^6*x^4 + 32*c^11*d^5*x))*(d^10/c^25)^(1/3))/(d^3 *x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 35*(c^4*d*x^8 - 8*c^5*x^ 5 + sqrt(-3)*(c^4*d*x^8 - 8*c^5*x^5))*(d^10/c^25)^(1/6)*log(16807*(d^11*x^ 9 + 318*c*d^10*x^6 + 1200*c^2*d^9*x^3 + 640*c^3*d^8 - 9*(c^17*d^4*x^8 + 38 *c^18*d^3*x^5 + 64*c^19*d^2*x^2 + sqrt(-3)*(c^17*d^4*x^8 + 38*c^18*d^3*x^5 + 64*c^19*d^2*x^2))*(d^10/c^25)^(2/3) - 3*sqrt(d*x^3 + c)*((c^21*d^2*x^7 + 80*c^22*d*x^4 + 160*c^23*x - sqrt(-3)*(c^21*d^2*x^7 + 80*c^22*d*x^4 + 16 0*c^23*x))*(d^10/c^25)^(5/6) - 2*(7*c^13*d^5*x^6 + 152*c^14*d^4*x^3 + 64*c ^15*d^3)*sqrt(d^10/c^25) + 6*(5*c^5*d^8*x^5 + 32*c^6*d^7*x^2 + sqrt(-3)*(5 *c^5*d^8*x^5 + 32*c^6*d^7*x^2))*(d^10/c^25)^(1/6)) - 9*(5*c^9*d^7*x^7 +...
\[ \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^{6} \left (- 8 c + d x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(1/x**6/(-d*x**3+8*c)**2/(d*x**3+c)**(1/2),x)
Output:
Integral(1/(x**6*(-8*c + d*x**3)**2*sqrt(c + d*x**3)), x)
\[ \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )}^{2} x^{6}} \,d x } \] Input:
integrate(1/x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)^2*x^6), x)
\[ \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )}^{2} x^{6}} \,d x } \] Input:
integrate(1/x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="giac")
Output:
integrate(1/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)^2*x^6), x)
Timed out. \[ \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^6\,\sqrt {d\,x^3+c}\,{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:
int(1/(x^6*(c + d*x^3)^(1/2)*(8*c - d*x^3)^2),x)
Output:
int(1/(x^6*(c + d*x^3)^(1/2)*(8*c - d*x^3)^2), x)
\[ \int \frac {1}{x^6 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {-8 \sqrt {d \,x^{3}+c}\, c +10 \sqrt {d \,x^{3}+c}\, d \,x^{3}+336 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{2} d^{2} x^{5}-42 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{3} x^{8}-280 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{3} x^{5}+35 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) d^{4} x^{8}}{320 c^{3} x^{5} \left (-d \,x^{3}+8 c \right )} \] Input:
int(1/x^6/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x)
Output:
( - 8*sqrt(c + d*x**3)*c + 10*sqrt(c + d*x**3)*d*x**3 + 336*int(sqrt(c + d *x**3)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c**2*d** 2*x**5 - 42*int(sqrt(c + d*x**3)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x** 6 + d**3*x**9),x)*c*d**3*x**8 - 280*int((sqrt(c + d*x**3)*x**3)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c*d**3*x**5 + 35*int((sqr t(c + d*x**3)*x**3)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9 ),x)*d**4*x**8)/(320*c**3*x**5*(8*c - d*x**3))