Integrand size = 24, antiderivative size = 64 \[ \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {x \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {1}{3},2,\frac {3}{2},\frac {4}{3},\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{64 c^3 \sqrt {c+d x^3}} \] Output:
1/64*x*(1+d*x^3/c)^(1/2)*AppellF1(1/3,3/2,2,4/3,-d*x^3/c,1/8*d*x^3/c)/c^3/ (d*x^3+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(253\) vs. \(2(64)=128\).
Time = 10.28 (sec) , antiderivative size = 253, normalized size of antiderivative = 3.95 \[ \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {x \left (-15 d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+192 c \left (\frac {-43 c+5 d x^3}{-8 c+d x^3}+\frac {1216 c^2 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{\left (8 c-d x^3\right ) \left (32 c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+3 d x^3 \left (\operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-4 \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )\right )}\right )\right )}{124416 c^4 \sqrt {c+d x^3}} \] Input:
Integrate[1/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
(x*(-15*d*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), (d*x^3)/(8*c)] + 192*c*((-43*c + 5*d*x^3)/(-8*c + d*x^3) + (1216*c^2*Appe llF1[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*(AppellF 1[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)]))))))/(124416*c^4*Sqrt[c + d*x^3])
Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {937, 936}
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.
\(\displaystyle \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{\left (8 c-d x^3\right )^2 \left (\frac {d x^3}{c}+1\right )^{3/2}}dx}{c \sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (\frac {1}{3},2,\frac {3}{2},\frac {4}{3},\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{64 c^3 \sqrt {c+d x^3}}\) |
Input:
Int[1/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
(x*Sqrt[1 + (d*x^3)/c]*AppellF1[1/3, 2, 3/2, 4/3, (d*x^3)/(8*c), -((d*x^3) /c)])/(64*c^3*Sqrt[c + d*x^3])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
Result contains higher order function than in optimal. Order 9 vs. order 6.
Time = 1.10 (sec) , antiderivative size = 748, normalized size of antiderivative = 11.69
method | result | size |
default | \(\text {Expression too large to display}\) | \(748\) |
elliptic | \(\text {Expression too large to display}\) | \(748\) |
Input:
int(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/243*x/c^3/((x^3+c/d)*d)^(1/2)+1/1944/c^3*x*(d*x^3+c)^(1/2)/(-d*x^3+8*c)- 5/1944*I/c^3*3^(1/2)/d*(-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))-1/972*I/c^3/d^3*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))*Elli pticPi(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)*_a lpha-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. 2640 vs. \(2 (50) = 100\).
Time = 0.78 (sec) , antiderivative size = 2640, normalized size of antiderivative = 41.25 \[ \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
1/15552*(192*(d^2*x^6 - 7*c*d*x^3 - 8*c^2)*sqrt(d)*weierstrassPInverse(0, -4*c/d, x) + (c^3*d^3*x^6 - 7*c^4*d^2*x^3 - 8*c^5*d + sqrt(-3)*(c^3*d^3*x^ 6 - 7*c^4*d^2*x^3 - 8*c^5*d))*(1/(c^19*d^2))^(1/6)*log((d^3*x^9 + 318*c*d^ 2*x^6 + 1200*c^2*d*x^3 + 640*c^3 - 9*(c^13*d^4*x^8 + 38*c^14*d^3*x^5 + 64* c^15*d^2*x^2 + sqrt(-3)*(c^13*d^4*x^8 + 38*c^14*d^3*x^5 + 64*c^15*d^2*x^2) )*(1/(c^19*d^2))^(2/3) + 3*sqrt(d*x^3 + c)*((c^16*d^4*x^7 + 80*c^17*d^3*x^ 4 + 160*c^18*d^2*x - sqrt(-3)*(c^16*d^4*x^7 + 80*c^17*d^3*x^4 + 160*c^18*d ^2*x))*(1/(c^19*d^2))^(5/6) - 2*(7*c^10*d^3*x^6 + 152*c^11*d^2*x^3 + 64*c^ 12*d)*sqrt(1/(c^19*d^2)) + 6*(5*c^4*d^2*x^5 + 32*c^5*d*x^2 + sqrt(-3)*(5*c ^4*d^2*x^5 + 32*c^5*d*x^2))*(1/(c^19*d^2))^(1/6)) - 9*(5*c^7*d^3*x^7 + 64* c^8*d^2*x^4 + 32*c^9*d*x - sqrt(-3)*(5*c^7*d^3*x^7 + 64*c^8*d^2*x^4 + 32*c ^9*d*x))*(1/(c^19*d^2))^(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^3*x^6 - 7*c^4*d^2*x^3 - 8*c^5*d + sqrt(-3)*(c^3*d^3*x^6 - 7*c^4*d^2*x^3 - 8*c^5*d))*(1/(c^19*d^2))^(1/6)*log((d^3*x^9 + 318*c*d^2* x^6 + 1200*c^2*d*x^3 + 640*c^3 - 9*(c^13*d^4*x^8 + 38*c^14*d^3*x^5 + 64*c^ 15*d^2*x^2 + sqrt(-3)*(c^13*d^4*x^8 + 38*c^14*d^3*x^5 + 64*c^15*d^2*x^2))* (1/(c^19*d^2))^(2/3) - 3*sqrt(d*x^3 + c)*((c^16*d^4*x^7 + 80*c^17*d^3*x^4 + 160*c^18*d^2*x - sqrt(-3)*(c^16*d^4*x^7 + 80*c^17*d^3*x^4 + 160*c^18*d^2 *x))*(1/(c^19*d^2))^(5/6) - 2*(7*c^10*d^3*x^6 + 152*c^11*d^2*x^3 + 64*c^12 *d)*sqrt(1/(c^19*d^2)) + 6*(5*c^4*d^2*x^5 + 32*c^5*d*x^2 + sqrt(-3)*(5*...
\[ \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{\left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)
Output:
Integral(1/((-8*c + d*x**3)**2*(c + d*x**3)**(3/2)), x)
\[ \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)
\[ \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)
Timed out. \[ \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{{\left (d\,x^3+c\right )}^{3/2}\,{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:
int(1/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2),x)
Output:
int(1/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2), x)
\[ \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{d^{4} x^{12}-14 c \,d^{3} x^{9}+33 c^{2} d^{2} x^{6}+112 c^{3} d \,x^{3}+64 c^{4}}d x \] Input:
int(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)
Output:
int(sqrt(c + d*x**3)/(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)