Integrand size = 24, antiderivative size = 65 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx=-\frac {c \sqrt {c+d x^3} \operatorname {AppellF1}\left (-\frac {2}{3},1,-\frac {3}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a x^2 \sqrt {1+\frac {d x^3}{c}}} \] Output:
-1/2*c*(d*x^3+c)^(1/2)*AppellF1(-2/3,1,-3/2,1/3,-b*x^3/a,-d*x^3/c)/a/x^2/( 1+d*x^3/c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(343\) vs. \(2(65)=130\).
Time = 10.39 (sec) , antiderivative size = 343, normalized size of antiderivative = 5.28 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx=-\frac {d (b c-4 a d) x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {8 a c \left (-4 a c \left (2 a c+6 b c x^3-5 a d x^3+2 b d x^6\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}}{16 a^2 x^2 \sqrt {c+d x^3}} \] Input:
Integrate[(c + d*x^3)^(3/2)/(x^3*(a + b*x^3)),x]
Output:
-1/16*(d*(b*c - 4*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*c*(-4*a*c*(2*a*c + 6*b*c*x^3 - 5*a*d*x^ 3 + 2*b*d*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)*(c + 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)]))))/(a^2* x^2*Sqrt[c + d*x^3])
Time = 0.34 (sec) , antiderivative size = 65, 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.
| \(\displaystyle \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx\) |
| \(\Big \downarrow \) 1013 |
| \(\displaystyle \frac {c \sqrt {c+d x^3} \int \frac {\left (\frac {d x^3}{c}+1\right )^{3/2}}{x^3 \left (b x^3+a\right )}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle -\frac {c \sqrt {c+d x^3} \operatorname {AppellF1}\left (-\frac {2}{3},1,-\frac {3}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a x^2 \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[(c + d*x^3)^(3/2)/(x^3*(a + b*x^3)),x]
Output:
-1/2*(c*Sqrt[c + d*x^3]*AppellF1[-2/3, 1, -3/2, 1/3, -((b*x^3)/a), -((d*x^ 3)/c)])/(a*x^2*Sqrt[1 + (d*x^3)/c])
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 = 2.48 (sec) , antiderivative size = 769, normalized size of antiderivative = 11.83
| method | result | size |
| risch | \(\text {Expression too large to display}\) | \(769\) |
| elliptic | \(\text {Expression too large to display}\) | \(772\) |
| default | \(\text {Expression too large to display}\) | \(1096\) |
Input:
int((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
-1/2*c/a*(d*x^3+c)^(1/2)/x^2+1/4/a*(-2/3*I*(4*a*d-b*c)/b*3^(1/2)*(-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)*Ellip ticF(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))+4/3*I*(a^2*d^2-2*a*b*c* d+b^2*c^2)/b/d^2*2^(1/2)*sum(1/(a*d-b*c)/_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))*El lipticPi(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. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx=\text {Timed out} \] Input:
integrate((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx=\int \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{x^{3} \left (a + b x^{3}\right )}\, dx \] Input:
integrate((d*x**3+c)**(3/2)/x**3/(b*x**3+a),x)
Output:
Integral((c + d*x**3)**(3/2)/(x**3*(a + b*x**3)), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )} x^{3}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)/((b*x^3 + a)*x^3), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )} x^{3}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)/((b*x^3 + a)*x^3), x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{3/2}}{x^3\,\left (b\,x^3+a\right )} \,d x \] Input:
int((c + d*x^3)^(3/2)/(x^3*(a + b*x^3)),x)
Output:
int((c + d*x^3)^(3/2)/(x^3*(a + b*x^3)), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )} \, dx=\frac {-4 \sqrt {d \,x^{3}+c}\, c d -7 \left (\int \frac {\sqrt {d \,x^{3}+c}}{a b \,d^{2} x^{9}+4 b^{2} c d \,x^{9}+a^{2} d^{2} x^{6}+5 a b c d \,x^{6}+4 b^{2} c^{2} x^{6}+a^{2} c d \,x^{3}+4 a b \,c^{2} x^{3}}d x \right ) a^{2} c^{2} d^{2} x^{2}-24 \left (\int \frac {\sqrt {d \,x^{3}+c}}{a b \,d^{2} x^{9}+4 b^{2} c d \,x^{9}+a^{2} d^{2} x^{6}+5 a b c d \,x^{6}+4 b^{2} c^{2} x^{6}+a^{2} c d \,x^{3}+4 a b \,c^{2} x^{3}}d x \right ) a b \,c^{3} d \,x^{2}+16 \left (\int \frac {\sqrt {d \,x^{3}+c}}{a b \,d^{2} x^{9}+4 b^{2} c d \,x^{9}+a^{2} d^{2} x^{6}+5 a b c d \,x^{6}+4 b^{2} c^{2} x^{6}+a^{2} c d \,x^{3}+4 a b \,c^{2} x^{3}}d x \right ) b^{2} c^{4} x^{2}+\left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{a b \,d^{2} x^{6}+4 b^{2} c d \,x^{6}+a^{2} d^{2} x^{3}+5 a b c d \,x^{3}+4 b^{2} c^{2} x^{3}+a^{2} c d +4 a b \,c^{2}}d x \right ) a^{2} d^{4} x^{2}+6 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{a b \,d^{2} x^{6}+4 b^{2} c d \,x^{6}+a^{2} d^{2} x^{3}+5 a b c d \,x^{3}+4 b^{2} c^{2} x^{3}+a^{2} c d +4 a b \,c^{2}}d x \right ) a b c \,d^{3} x^{2}+8 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{a b \,d^{2} x^{6}+4 b^{2} c d \,x^{6}+a^{2} d^{2} x^{3}+5 a b c d \,x^{3}+4 b^{2} c^{2} x^{3}+a^{2} c d +4 a b \,c^{2}}d x \right ) b^{2} c^{2} d^{2} x^{2}}{x^{2} \left (a d +4 b c \right )} \] Input:
int((d*x^3+c)^(3/2)/x^3/(b*x^3+a),x)
Output:
( - 4*sqrt(c + d*x**3)*c*d - 7*int(sqrt(c + d*x**3)/(a**2*c*d*x**3 + a**2* d**2*x**6 + 4*a*b*c**2*x**3 + 5*a*b*c*d*x**6 + a*b*d**2*x**9 + 4*b**2*c**2 *x**6 + 4*b**2*c*d*x**9),x)*a**2*c**2*d**2*x**2 - 24*int(sqrt(c + d*x**3)/ (a**2*c*d*x**3 + a**2*d**2*x**6 + 4*a*b*c**2*x**3 + 5*a*b*c*d*x**6 + a*b*d **2*x**9 + 4*b**2*c**2*x**6 + 4*b**2*c*d*x**9),x)*a*b*c**3*d*x**2 + 16*int (sqrt(c + d*x**3)/(a**2*c*d*x**3 + a**2*d**2*x**6 + 4*a*b*c**2*x**3 + 5*a* b*c*d*x**6 + a*b*d**2*x**9 + 4*b**2*c**2*x**6 + 4*b**2*c*d*x**9),x)*b**2*c **4*x**2 + int((sqrt(c + d*x**3)*x**3)/(a**2*c*d + a**2*d**2*x**3 + 4*a*b* c**2 + 5*a*b*c*d*x**3 + a*b*d**2*x**6 + 4*b**2*c**2*x**3 + 4*b**2*c*d*x**6 ),x)*a**2*d**4*x**2 + 6*int((sqrt(c + d*x**3)*x**3)/(a**2*c*d + a**2*d**2* x**3 + 4*a*b*c**2 + 5*a*b*c*d*x**3 + a*b*d**2*x**6 + 4*b**2*c**2*x**3 + 4* b**2*c*d*x**6),x)*a*b*c*d**3*x**2 + 8*int((sqrt(c + d*x**3)*x**3)/(a**2*c* d + a**2*d**2*x**3 + 4*a*b*c**2 + 5*a*b*c*d*x**3 + a*b*d**2*x**6 + 4*b**2* c**2*x**3 + 4*b**2*c*d*x**6),x)*b**2*c**2*d**2*x**2)/(x**2*(a*d + 4*b*c))