Integrand size = 21, antiderivative size = 59 \[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx=\frac {x \sqrt {1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},3,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \sqrt {a+b x^3}} \] Output:
x*(1+b*x^3/a)^(1/2)*AppellF1(1/3,1/2,3,4/3,-b*x^3/a,-d*x^3/c)/c^3/(b*x^3+a )^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(420\) vs. \(2(59)=118\).
Time = 10.53 (sec) , antiderivative size = 420, normalized size of antiderivative = 7.12 \[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx=\frac {x \left (b d (-19 b c+10 a d) x^3 \sqrt {1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {8 c \left (-8 a c \left (6 a^2 d^2 \left (6 c+5 d x^3\right )+b^2 c \left (36 c^2+11 c d x^3-19 d^2 x^6\right )+2 a b d \left (-36 c^2-25 c d x^3+5 d^2 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+3 d x^3 \left (a+b x^3\right ) \left (2 a d \left (8 c+5 d x^3\right )-b c \left (25 c+19 d x^3\right )\right ) \left (2 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{\left (c+d x^3\right )^2 \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+3 x^3 \left (2 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )}{288 c^3 (b c-a d)^2 \sqrt {a+b x^3}} \] Input:
Integrate[1/(Sqrt[a + b*x^3]*(c + d*x^3)^3),x]
Output:
(x*(b*d*(-19*b*c + 10*a*d)*x^3*Sqrt[1 + (b*x^3)/a]*AppellF1[4/3, 1/2, 1, 7 /3, -((b*x^3)/a), -((d*x^3)/c)] + (8*c*(-8*a*c*(6*a^2*d^2*(6*c + 5*d*x^3) + b^2*c*(36*c^2 + 11*c*d*x^3 - 19*d^2*x^6) + 2*a*b*d*(-36*c^2 - 25*c*d*x^3 + 5*d^2*x^6))*AppellF1[1/3, 1/2, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + 3* d*x^3*(a + b*x^3)*(2*a*d*(8*c + 5*d*x^3) - b*c*(25*c + 19*d*x^3))*(2*a*d*A ppellF1[4/3, 1/2, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + b*c*AppellF1[4/3, 3/2, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])))/((c + d*x^3)^2*(-8*a*c*AppellF 1[1/3, 1/2, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + 3*x^3*(2*a*d*AppellF1[4/ 3, 1/2, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + b*c*AppellF1[4/3, 3/2, 1, 7/ 3, -((b*x^3)/a), -((d*x^3)/c)])))))/(288*c^3*(b*c - a*d)^2*Sqrt[a + b*x^3] )
Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
\(\displaystyle \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\sqrt {\frac {b x^3}{a}+1} \int \frac {1}{\sqrt {\frac {b x^3}{a}+1} \left (d x^3+c\right )^3}dx}{\sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \sqrt {\frac {b x^3}{a}+1} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},3,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \sqrt {a+b x^3}}\) |
Input:
Int[1/(Sqrt[a + b*x^3]*(c + d*x^3)^3),x]
Output:
(x*Sqrt[1 + (b*x^3)/a]*AppellF1[1/3, 1/2, 3, 4/3, -((b*x^3)/a), -((d*x^3)/ c)])/(c^3*Sqrt[a + b*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 = 1.68 (sec) , antiderivative size = 836, normalized size of antiderivative = 14.17
method | result | size |
default | \(\text {Expression too large to display}\) | \(836\) |
elliptic | \(\text {Expression too large to display}\) | \(836\) |
Input:
int(1/(b*x^3+a)^(1/2)/(d*x^3+c)^3,x,method=_RETURNVERBOSE)
Output:
1/6/c*d/(a*d-b*c)*x*(b*x^3+a)^(1/2)/(d*x^3+c)^2+1/36*d*(10*a*d-19*b*c)/(a* d-b*c)^2/c^2*x*(b*x^3+a)^(1/2)/(d*x^3+c)-1/108*I*(10*a*d-19*b*c)/(a*d-b*c) ^2/c^2*3^(1/2)*(-a*b^2)^(1/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*( -a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3 /2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(- a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/ 2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I* 3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a *b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)) -1/216*I/b^2/c^2*2^(1/2)*sum((-40*a^2*d^2+104*a*b*c*d-91*b^2*c^2)/(a*d-b*c )^3/_alpha^2*(-a*b^2)^(1/3)*(1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)-I*3^(1/2)*(- a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)*(b*(x-1/b*(-a*b^2)^(1/3))/(-3*(-a*b^2 )^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)))^(1/2)*(-1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3 )+I*3^(1/2)*(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*(I*(-a* b^2)^(1/3)*_alpha*3^(1/2)*b-I*3^(1/2)*(-a*b^2)^(2/3)+2*_alpha^2*b^2-(-a*b^ 2)^(1/3)*_alpha*b-(-a*b^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b ^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2), -1/2/b*d*(2*I*(-a*b^2)^(1/3)*3^(1/2)*_alpha^2*b-I*(-a*b^2)^(2/3)*3^(1/2)*_ alpha+I*3^(1/2)*a*b-3*(-a*b^2)^(2/3)*_alpha-3*a*b)/(a*d-b*c),(I*3^(1/2)/b* (-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^...
Timed out. \[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x^3+a)^(1/2)/(d*x^3+c)^3,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x**3+a)**(1/2)/(d*x**3+c)**3,x)
Output:
Timed out
\[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a} {\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:
integrate(1/(b*x^3+a)^(1/2)/(d*x^3+c)^3,x, algorithm="maxima")
Output:
integrate(1/(sqrt(b*x^3 + a)*(d*x^3 + c)^3), x)
\[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a} {\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:
integrate(1/(b*x^3+a)^(1/2)/(d*x^3+c)^3,x, algorithm="giac")
Output:
integrate(1/(sqrt(b*x^3 + a)*(d*x^3 + c)^3), x)
Timed out. \[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx=\int \frac {1}{\sqrt {b\,x^3+a}\,{\left (d\,x^3+c\right )}^3} \,d x \] Input:
int(1/((a + b*x^3)^(1/2)*(c + d*x^3)^3),x)
Output:
int(1/((a + b*x^3)^(1/2)*(c + d*x^3)^3), x)
\[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right )^3} \, dx=\int \frac {\sqrt {b \,x^{3}+a}}{b \,d^{3} x^{12}+a \,d^{3} x^{9}+3 b c \,d^{2} x^{9}+3 a c \,d^{2} x^{6}+3 b \,c^{2} d \,x^{6}+3 a \,c^{2} d \,x^{3}+b \,c^{3} x^{3}+a \,c^{3}}d x \] Input:
int(1/(b*x^3+a)^(1/2)/(d*x^3+c)^3,x)
Output:
int(sqrt(a + b*x**3)/(a*c**3 + 3*a*c**2*d*x**3 + 3*a*c*d**2*x**6 + a*d**3* x**9 + b*c**3*x**3 + 3*b*c**2*d*x**6 + 3*b*c*d**2*x**9 + b*d**3*x**12),x)