Integrand size = 21, antiderivative size = 62 \[ \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {x \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {1}{3},2,\frac {3}{2},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 c \sqrt {c+d x^3}} \] Output:
x*(1+d*x^3/c)^(1/2)*AppellF1(1/3,2,3/2,4/3,-b*x^3/a,-d*x^3/c)/a^2/c/(d*x^3 +c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(381\) vs. \(2(62)=124\).
Time = 10.41 (sec) , antiderivative size = 381, normalized size of antiderivative = 6.15 \[ \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {x \left (b d (b c+2 a d) x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {a \left (64 a c \left (3 a^2 d^2+2 a b d \left (-3 c+d x^3\right )+b^2 c \left (3 c+d x^3\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-24 x^3 \left (2 a^2 d^2+2 a b d^2 x^3+b^2 c \left (c+d x^3\right )\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}\right )}{24 a^2 c (b c-a d)^2 \sqrt {c+d x^3}} \] Input:
Integrate[1/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
(x*(b*d*(b*c + 2*a*d)*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, - ((d*x^3)/c), -((b*x^3)/a)] + (a*(64*a*c*(3*a^2*d^2 + 2*a*b*d*(-3*c + d*x^3 ) + b^2*c*(3*c + d*x^3))*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3 )/a)] - 24*x^3*(2*a^2*d^2 + 2*a*b*d^2*x^3 + b^2*c*(c + d*x^3))*(2*b*c*Appe llF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + a*d*AppellF1[4/3, 3/2 , 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)])))/((a + b*x^3)*(8*a*c*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] - 3*x^3*(2*b*c*AppellF1[4/3, 1/2 , 2, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + a*d*AppellF1[4/3, 3/2, 1, 7/3, -(( d*x^3)/c), -((b*x^3)/a)])))))/(24*a^2*c*(b*c - a*d)^2*Sqrt[c + d*x^3])
Time = 0.31 (sec) , antiderivative size = 62, 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.
\(\displaystyle \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{\left (b x^3+a\right )^2 \left (\frac {d x^3}{c}+1\right )^{3/2}}dx}{c \sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (\frac {1}{3},2,\frac {3}{2},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 c \sqrt {c+d x^3}}\) |
Input:
Int[1/((a + b*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, -((b*x^3)/a), -((d*x^3)/ c)])/(a^2*c*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 = 1.11 (sec) , antiderivative size = 830, normalized size of antiderivative = 13.39
method | result | size |
default | \(\text {Expression too large to display}\) | \(830\) |
elliptic | \(\text {Expression too large to display}\) | \(830\) |
Input:
int(1/(b*x^3+a)^2/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/3*d^2*x/c/(a*d-b*c)^2/((x^3+c/d)*d)^(1/2)+1/3*b^2/(a*d-b*c)^2/a*x*(d*x^3 +c)^(1/2)/(b*x^3+a)-2/3*I*(1/3*d^2/(a*d-b*c)^2/c+1/6*d*b/(a*d-b*c)^2/a)*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))+1/18*I /a/d^2*b*2^(1/2)*sum((13*a*d-4*b*c)/(a*d-b*c)^3/_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),1/2*b/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)/(a*d-b*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*b+a)...
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(b*x**3+a)**2/(d*x**3+c)**(3/2),x)
Output:
Integral(1/((a + b*x**3)**2*(c + d*x**3)**(3/2)), x)
\[ \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)), x)
\[ \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)), x)
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^2\,{\left (d\,x^3+c\right )}^{3/2}} \,d x \] Input:
int(1/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x)
Output:
int(1/((a + b*x^3)^2*(c + d*x^3)^(3/2)), x)
\[ \int \frac {1}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{b^{2} d^{2} x^{12}+2 a b \,d^{2} x^{9}+2 b^{2} c d \,x^{9}+a^{2} d^{2} x^{6}+4 a b c d \,x^{6}+b^{2} c^{2} x^{6}+2 a^{2} c d \,x^{3}+2 a b \,c^{2} x^{3}+a^{2} c^{2}}d x \] Input:
int(1/(b*x^3+a)^2/(d*x^3+c)^(3/2),x)
Output:
int(sqrt(c + d*x**3)/(a**2*c**2 + 2*a**2*c*d*x**3 + a**2*d**2*x**6 + 2*a*b *c**2*x**3 + 4*a*b*c*d*x**6 + 2*a*b*d**2*x**9 + b**2*c**2*x**6 + 2*b**2*c* d*x**9 + b**2*d**2*x**12),x)