Integrand size = 24, antiderivative size = 67 \[ \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {3}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a c x^2 \sqrt {c+d x^3}} \] Output:
-1/2*(1+d*x^3/c)^(1/2)*AppellF1(-2/3,1,3/2,1/3,-b*x^3/a,-d*x^3/c)/a/c/x^2/ (d*x^3+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(408\) vs. \(2(67)=134\).
Time = 10.42 (sec) , antiderivative size = 408, normalized size of antiderivative = 6.09 \[ \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {b d (3 b c-7 a d) x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {8 a \left (-4 a c \left (-6 b^2 c x^3 \left (3 c+d x^3\right )+3 a^2 d \left (2 c+7 d x^3\right )+a b \left (-6 c^2-3 c d x^3+14 d^2 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (a+b x^3\right ) \left (-3 b c \left (c+d x^3\right )+a d \left (3 c+7 d x^3\right )\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}}{48 a^2 c^2 (-b c+a d) x^2 \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^3*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
Output:
(b*d*(3*b*c - 7*a*d)*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -( (d*x^3)/c), -((b*x^3)/a)] + (8*a*(-4*a*c*(-6*b^2*c*x^3*(3*c + d*x^3) + 3*a ^2*d*(2*c + 7*d*x^3) + a*b*(-6*c^2 - 3*c*d*x^3 + 14*d^2*x^6))*AppellF1[1/3 , 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] + 3*x^3*(a + b*x^3)*(-3*b*c*(c + d*x^3) + a*d*(3*c + 7*d*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)])))/((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)]))))/(4 8*a^2*c^2*(-(b*c) + a*d)*x^2*Sqrt[c + d*x^3])
Time = 0.35 (sec) , antiderivative size = 67, 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 = {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.
| \(\displaystyle \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1013 |
| \(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{x^3 \left (b x^3+a\right ) \left (\frac {d x^3}{c}+1\right )^{3/2}}dx}{c \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle -\frac {\sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {3}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a c x^2 \sqrt {c+d x^3}}\) |
Input:
Int[1/(x^3*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
Output:
-1/2*(Sqrt[1 + (d*x^3)/c]*AppellF1[-2/3, 1, 3/2, 1/3, -((b*x^3)/a), -((d*x ^3)/c)])/(a*c*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 = 3.21 (sec) , antiderivative size = 798, normalized size of antiderivative = 11.91
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(798\) |
| risch | \(\text {Expression too large to display}\) | \(1076\) |
| default | \(\text {Expression too large to display}\) | \(1084\) |
Input:
int(1/x^3/(b*x^3+a)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2/c^2/a*(d*x^3+c)^(1/2)/x^2-2/3*d^2*x/c^2/(a*d-b*c)/((x^3+c/d)*d)^(1/2) -2/3*I*(-1/4/a/c^2*d-1/3*d^2/c^2/(a*d-b*c))*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))-1/3*I*b^2/a/d^2*2^(1/2)*sum(1/(a*d -b*c)^2/_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)*_alph a*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. \[ \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/x^3/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x**3/(b*x**3+a)/(d*x**3+c)**(3/2),x)
Output:
Integral(1/(x**3*(a + b*x**3)*(c + d*x**3)**(3/2)), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^3), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{x^3\,\left (b\,x^3+a\right )\,{\left (d\,x^3+c\right )}^{3/2}} \,d x \] Input:
int(1/(x^3*(a + b*x^3)*(c + d*x^3)^(3/2)),x)
Output:
int(1/(x^3*(a + b*x^3)*(c + d*x^3)^(3/2)), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}-7 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) a c d \,x^{2}-7 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) a \,d^{2} x^{5}-4 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b \,c^{2} x^{2}-4 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b c d \,x^{5}-7 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b c d \,x^{2}-7 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b \,d^{2} x^{5}}{4 a c \,x^{2} \left (d \,x^{3}+c \right )} \] Input:
int(1/x^3/(b*x^3+a)/(d*x^3+c)^(3/2),x)
Output:
( - 2*sqrt(c + d*x**3) - 7*int(sqrt(c + d*x**3)/(a*c**2 + 2*a*c*d*x**3 + a *d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*a*c*d*x**2 - 7*i nt(sqrt(c + d*x**3)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2 *b*c*d*x**6 + b*d**2*x**9),x)*a*d**2*x**5 - 4*int(sqrt(c + d*x**3)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9), x)*b*c**2*x**2 - 4*int(sqrt(c + d*x**3)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x* *6 + b*c**2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*b*c*d*x**5 - 7*int((sqrt (c + d*x**3)*x**3)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2* b*c*d*x**6 + b*d**2*x**9),x)*b*c*d*x**2 - 7*int((sqrt(c + d*x**3)*x**3)/(a *c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 + b*d**2*x **9),x)*b*d**2*x**5)/(4*a*c*x**2*(c + d*x**3))