Integrand size = 21, antiderivative size = 62 \[ \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx=\frac {x \sqrt {1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {3}{2},2,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c^2 \sqrt {a+b x^3}} \] Output:
x*(1+b*x^3/a)^(1/2)*AppellF1(1/3,3/2,2,4/3,-b*x^3/a,-d*x^3/c)/a/c^2/(b*x^3 +a)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(380\) vs. \(2(62)=124\).
Time = 10.53 (sec) , antiderivative size = 380, normalized size of antiderivative = 6.13 \[ \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx=\frac {x \left (b d (2 b c+a d) x^3 \sqrt {1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {c \left (64 a c \left (3 a^2 d^2+a b d \left (-6 c+d x^3\right )+b^2 c \left (3 c+2 d x^3\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-24 x^3 \left (a^2 d^2+a b d^2 x^3+2 b^2 c \left (c+d x^3\right )\right ) \left (2 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{\left (c+d x^3\right ) \left (8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-3 x^3 \left (2 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )}{24 a c^2 (b c-a d)^2 \sqrt {a+b x^3}} \] Input:
Integrate[1/((a + b*x^3)^(3/2)*(c + d*x^3)^2),x]
Output:
(x*(b*d*(2*b*c + 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)] + (c*(64*a*c*(3*a^2*d^2 + a*b*d*(-6*c + d*x^3) + b^2*c*(3*c + 2*d*x^3))*AppellF1[1/3, 1/2, 1, 4/3, -((b*x^3)/a), -((d*x^3 )/c)] - 24*x^3*(a^2*d^2 + a*b*d^2*x^3 + 2*b^2*c*(c + d*x^3))*(2*a*d*Appell F1[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)*(8*a*c*AppellF1[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)])))))/(24*a*c^2*(b*c - a*d)^2*Sqrt[a + b*x^3])
Time = 0.30 (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.
\(\displaystyle \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\sqrt {\frac {b x^3}{a}+1} \int \frac {1}{\left (\frac {b x^3}{a}+1\right )^{3/2} \left (d x^3+c\right )^2}dx}{a \sqrt {a+b x^3}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \sqrt {\frac {b x^3}{a}+1} \operatorname {AppellF1}\left (\frac {1}{3},\frac {3}{2},2,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c^2 \sqrt {a+b x^3}}\) |
Input:
Int[1/((a + b*x^3)^(3/2)*(c + d*x^3)^2),x]
Output:
(x*Sqrt[1 + (b*x^3)/a]*AppellF1[1/3, 3/2, 2, 4/3, -((b*x^3)/a), -((d*x^3)/ c)])/(a*c^2*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 = 1.30 (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)^(3/2)/(d*x^3+c)^2,x,method=_RETURNVERBOSE)
Output:
1/3/c*d^2/(a*d-b*c)^2*x*(b*x^3+a)^(1/2)/(d*x^3+c)+2/3*b^2*x/a/(a*d-b*c)^2/ ((x^3+a/b)*b)^(1/2)-2/3*I*(1/6*b*d/(a*d-b*c)^2/c+1/3*b^2/(a*d-b*c)^2/a)*3^ (1/2)/b*(-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))-1/18*I /c/b^2*d*2^(1/2)*sum((-4*a*d+13*b*c)/(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)))^(1/2)),_alpha=RootOf(_Z^3*d+c...
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x^3+a)^(3/2)/(d*x^3+c)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{\frac {3}{2}} \left (c + d x^{3}\right )^{2}}\, dx \] Input:
integrate(1/(b*x**3+a)**(3/2)/(d*x**3+c)**2,x)
Output:
Integral(1/((a + b*x**3)**(3/2)*(c + d*x**3)**2), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} {\left (d x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate(1/(b*x^3+a)^(3/2)/(d*x^3+c)^2,x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)^(3/2)*(d*x^3 + c)^2), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} {\left (d x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate(1/(b*x^3+a)^(3/2)/(d*x^3+c)^2,x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^(3/2)*(d*x^3 + c)^2), x)
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{3/2}\,{\left (d\,x^3+c\right )}^2} \,d x \] Input:
int(1/((a + b*x^3)^(3/2)*(c + d*x^3)^2),x)
Output:
int(1/((a + b*x^3)^(3/2)*(c + d*x^3)^2), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{3/2} \left (c+d x^3\right )^2} \, dx=\int \frac {\sqrt {b \,x^{3}+a}}{b^{2} d^{2} x^{12}+2 a b \,d^{2} x^{9}+2 b^{2} c d \,x^{9}+a^{2} d^{2} x^{6}+4 a b c d \,x^{6}+b^{2} c^{2} x^{6}+2 a^{2} c d \,x^{3}+2 a b \,c^{2} x^{3}+a^{2} c^{2}}d x \] Input:
int(1/(b*x^3+a)^(3/2)/(d*x^3+c)^2,x)
Output:
int(sqrt(a + b*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)