Integrand size = 27, antiderivative size = 66 \[ \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {3}{2},\frac {1}{3},\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{16 c^2 x^2 \sqrt {c+d x^3}} \] Output:
-1/16*(1+d*x^3/c)^(1/2)*AppellF1(-2/3,3/2,1,1/3,-d*x^3/c,1/8*d*x^3/c)/c^2/ x^2/(d*x^3+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(248\) vs. \(2(66)=132\).
Time = 11.20 (sec) , antiderivative size = 248, normalized size of antiderivative = 3.76 \[ \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {59 d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+64 c \left (-27 c-59 d x^3-\frac {7360 c^2 d x^3 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{\left (8 c-d x^3\right ) \left (32 c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+3 d x^3 \left (\operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-4 \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )\right )}\right )}{27648 c^4 x^2 \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^3*(8*c - d*x^3)*(c + d*x^3)^(3/2)),x]
Output:
(59*d^2*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), ( d*x^3)/(8*c)] + 64*c*(-27*c - 59*d*x^3 - (7360*c^2*d*x^3*AppellF1[1/3, 1/2 , 1, 4/3, -((d*x^3)/c), (d*x^3)/(8*c)])/((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)])))))/(27648*c^4*x^2*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.
\(\displaystyle \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (\frac {d x^3}{c}+1\right )^{3/2}}dx}{c \sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {3}{2},\frac {1}{3},\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{16 c^2 x^2 \sqrt {c+d x^3}}\) |
Input:
Int[1/(x^3*(8*c - d*x^3)*(c + d*x^3)^(3/2)),x]
Output:
-1/16*(Sqrt[1 + (d*x^3)/c]*AppellF1[-2/3, 1, 3/2, 1/3, (d*x^3)/(8*c), -((d *x^3)/c)])/(c^2*x^2*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 = 2.39 (sec) , antiderivative size = 736, normalized size of antiderivative = 11.15
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(736\) |
risch | \(\text {Expression too large to display}\) | \(1028\) |
default | \(\text {Expression too large to display}\) | \(1053\) |
Input:
int(1/x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/16*(d*x^3+c)^(1/2)/c^3/x^2-2/27*d/c^3*x/((x^3+c/d)*d)^(1/2)+59/1296*I/c ^3*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))-1/1 944*I/c^3/d^2*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)),_alpha=RootOf(_Z^3*d-8*c))
Leaf count of result is larger than twice the leaf count of optimal. 2495 vs. \(2 (52) = 104\).
Time = 2.00 (sec) , antiderivative size = 2495, normalized size of antiderivative = 37.80 \[ \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
-1/31104*(4176*(d*x^5 + c*x^2)*sqrt(d)*weierstrassPInverse(0, -4*c/d, x) - (c^3*d*x^5 + c^4*x^2 + sqrt(-3)*(c^3*d*x^5 + c^4*x^2))*(d^4/c^19)^(1/6)*l og((d^6*x^9 + 318*c*d^5*x^6 + 1200*c^2*d^4*x^3 + 640*c^3*d^3 - 9*(c^13*d^3 *x^8 + 38*c^14*d^2*x^5 + 64*c^15*d*x^2 + sqrt(-3)*(c^13*d^3*x^8 + 38*c^14* d^2*x^5 + 64*c^15*d*x^2))*(d^4/c^19)^(2/3) + 3*sqrt(d*x^3 + c)*((c^16*d^2* x^7 + 80*c^17*d*x^4 + 160*c^18*x - sqrt(-3)*(c^16*d^2*x^7 + 80*c^17*d*x^4 + 160*c^18*x))*(d^4/c^19)^(5/6) - 2*(7*c^10*d^3*x^6 + 152*c^11*d^2*x^3 + 6 4*c^12*d)*sqrt(d^4/c^19) + 6*(5*c^4*d^4*x^5 + 32*c^5*d^3*x^2 + sqrt(-3)*(5 *c^4*d^4*x^5 + 32*c^5*d^3*x^2))*(d^4/c^19)^(1/6)) - 9*(5*c^7*d^4*x^7 + 64* c^8*d^3*x^4 + 32*c^9*d^2*x - sqrt(-3)*(5*c^7*d^4*x^7 + 64*c^8*d^3*x^4 + 32 *c^9*d^2*x))*(d^4/c^19)^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 5 12*c^3)) + (c^3*d*x^5 + c^4*x^2 + sqrt(-3)*(c^3*d*x^5 + c^4*x^2))*(d^4/c^1 9)^(1/6)*log((d^6*x^9 + 318*c*d^5*x^6 + 1200*c^2*d^4*x^3 + 640*c^3*d^3 - 9 *(c^13*d^3*x^8 + 38*c^14*d^2*x^5 + 64*c^15*d*x^2 + sqrt(-3)*(c^13*d^3*x^8 + 38*c^14*d^2*x^5 + 64*c^15*d*x^2))*(d^4/c^19)^(2/3) - 3*sqrt(d*x^3 + c)*( (c^16*d^2*x^7 + 80*c^17*d*x^4 + 160*c^18*x - sqrt(-3)*(c^16*d^2*x^7 + 80*c ^17*d*x^4 + 160*c^18*x))*(d^4/c^19)^(5/6) - 2*(7*c^10*d^3*x^6 + 152*c^11*d ^2*x^3 + 64*c^12*d)*sqrt(d^4/c^19) + 6*(5*c^4*d^4*x^5 + 32*c^5*d^3*x^2 + s qrt(-3)*(5*c^4*d^4*x^5 + 32*c^5*d^3*x^2))*(d^4/c^19)^(1/6)) - 9*(5*c^7*d^4 *x^7 + 64*c^8*d^3*x^4 + 32*c^9*d^2*x - sqrt(-3)*(5*c^7*d^4*x^7 + 64*c^8...
\[ \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=- \int \frac {1}{- 8 c^{2} x^{3} \sqrt {c + d x^{3}} - 7 c d x^{6} \sqrt {c + d x^{3}} + d^{2} x^{9} \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(1/x**3/(-d*x**3+8*c)/(d*x**3+c)**(3/2),x)
Output:
-Integral(1/(-8*c**2*x**3*sqrt(c + d*x**3) - 7*c*d*x**6*sqrt(c + d*x**3) + d**2*x**9*sqrt(c + d*x**3)), x)
\[ \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { -\frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )} x^{3}} \,d x } \] Input:
integrate(1/x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
-integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)*x^3), x)
\[ \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { -\frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )} x^{3}} \,d x } \] Input:
integrate(1/x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
integrate(-1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{x^3\,{\left (d\,x^3+c\right )}^{3/2}\,\left (8\,c-d\,x^3\right )} \,d x \] Input:
int(1/(x^3*(c + d*x^3)^(3/2)*(8*c - d*x^3)),x)
Output:
int(1/(x^3*(c + d*x^3)^(3/2)*(8*c - d*x^3)), x)
\[ \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}-52 \left (\int \frac {\sqrt {d \,x^{3}+c}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c^{2} d \,x^{2}-52 \left (\int \frac {\sqrt {d \,x^{3}+c}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c \,d^{2} x^{5}+7 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c \,d^{2} x^{2}+7 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) d^{3} x^{5}}{32 c^{2} x^{2} \left (d \,x^{3}+c \right )} \] Input:
int(1/x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x)
Output:
( - 2*sqrt(c + d*x**3) - 52*int(sqrt(c + d*x**3)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x**9),x)*c**2*d*x**2 - 52*int(sqrt(c + d*x**3)/(8*c **3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x**9),x)*c*d**2*x**5 + 7*int(( sqrt(c + d*x**3)*x**3)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x** 9),x)*c*d**2*x**2 + 7*int((sqrt(c + d*x**3)*x**3)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x**9),x)*d**3*x**5)/(32*c**2*x**2*(c + d*x**3))