Integrand size = 24, antiderivative size = 65 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx=-\frac {c \sqrt {c+d x^3} \operatorname {AppellF1}\left (-\frac {2}{3},2,-\frac {3}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 x^2 \sqrt {1+\frac {d x^3}{c}}} \] Output:
-1/2*c*(d*x^3+c)^(1/2)*AppellF1(-2/3,2,-3/2,1/3,-b*x^3/a,-d*x^3/c)/a^2/x^2 /(1+d*x^3/c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(370\) vs. \(2(65)=130\).
Time = 10.29 (sec) , antiderivative size = 370, normalized size of antiderivative = 5.69 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx=\frac {-d (5 b c-2 a d) x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {8 a \left (4 a c \left (10 b c x^3 \left (3 c+d x^3\right )+a \left (6 c^2-15 c d x^3-4 d^2 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-3 x^3 \left (c+d x^3\right ) \left (3 a c+5 b c x^3-2 a d x^3\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}}{48 a^3 x^2 \sqrt {c+d x^3}} \] Input:
Integrate[(c + d*x^3)^(3/2)/(x^3*(a + b*x^3)^2),x]
Output:
(-(d*(5*b*c - 2*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*(10*b*c*x^3*(3*c + d*x^3) + a*(6* c^2 - 15*c*d*x^3 - 4*d^2*x^6))*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -( (b*x^3)/a)] - 3*x^3*(c + d*x^3)*(3*a*c + 5*b*c*x^3 - 2*a*d*x^3)*(2*b*c*App ellF1[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)]))))/(48*a^3*x^2*Sqrt[c + d*x^3])
Time = 0.35 (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.
| \(\displaystyle \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 1013 |
| \(\displaystyle \frac {c \sqrt {c+d x^3} \int \frac {\left (\frac {d x^3}{c}+1\right )^{3/2}}{x^3 \left (b x^3+a\right )^2}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle -\frac {c \sqrt {c+d x^3} \operatorname {AppellF1}\left (-\frac {2}{3},2,-\frac {3}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 x^2 \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[(c + d*x^3)^(3/2)/(x^3*(a + b*x^3)^2),x]
Output:
-1/2*(c*Sqrt[c + d*x^3]*AppellF1[-2/3, 2, -3/2, 1/3, -((b*x^3)/a), -((d*x^ 3)/c)])/(a^2*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 = 5.29 (sec) , antiderivative size = 815, normalized size of antiderivative = 12.54
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(815\) |
| risch | \(\text {Expression too large to display}\) | \(1548\) |
| default | \(\text {Expression too large to display}\) | \(1902\) |
Input:
int((d*x^3+c)^(3/2)/x^3/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/2*c/a^2*(d*x^3+c)^(1/2)/x^2+1/3*(a*d-b*c)/a^2*x*(d*x^3+c)^(1/2)/(b*x^3+ a)-2/3*I*(-1/4*d*c/a^2+1/6*d*(a*d-b*c)/a^2/b)*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^2/b/d^2*2^(1/2)*sum((a^ 2*d^2-11*a*b*c*d+10*b^2*c^2)/_alpha^2/(a*d-b*c)*(-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))*Elli pticPi(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)*_a lpha-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. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate((d*x^3+c)^(3/2)/x^3/(b*x^3+a)^2,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate((d*x**3+c)**(3/2)/x**3/(b*x**3+a)**2,x)
Output:
Timed out
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )}^{2} x^{3}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/x^3/(b*x^3+a)^2,x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)/((b*x^3 + a)^2*x^3), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )}^{2} x^{3}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/x^3/(b*x^3+a)^2,x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)/((b*x^3 + a)^2*x^3), x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{3/2}}{x^3\,{\left (b\,x^3+a\right )}^2} \,d x \] Input:
int((c + d*x^3)^(3/2)/(x^3*(a + b*x^3)^2),x)
Output:
int((c + d*x^3)^(3/2)/(x^3*(a + b*x^3)^2), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^3 \left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int((d*x^3+c)^(3/2)/x^3/(b*x^3+a)^2,x)
Output:
( - 2*sqrt(c + d*x**3)*c + 7*int(sqrt(c + d*x**3)/(a**3*c*d + a**3*d**2*x* *3 + 10*a**2*b*c**2 + 12*a**2*b*c*d*x**3 + 2*a**2*b*d**2*x**6 + 20*a*b**2* c**2*x**3 + 21*a*b**2*c*d*x**6 + a*b**2*d**2*x**9 + 10*b**3*c**2*x**6 + 10 *b**3*c*d*x**9),x)*a**3*c*d**2*x**2 + 60*int(sqrt(c + d*x**3)/(a**3*c*d + a**3*d**2*x**3 + 10*a**2*b*c**2 + 12*a**2*b*c*d*x**3 + 2*a**2*b*d**2*x**6 + 20*a*b**2*c**2*x**3 + 21*a*b**2*c*d*x**6 + a*b**2*d**2*x**9 + 10*b**3*c* *2*x**6 + 10*b**3*c*d*x**9),x)*a**2*b*c**2*d*x**2 + 7*int(sqrt(c + d*x**3) /(a**3*c*d + a**3*d**2*x**3 + 10*a**2*b*c**2 + 12*a**2*b*c*d*x**3 + 2*a**2 *b*d**2*x**6 + 20*a*b**2*c**2*x**3 + 21*a*b**2*c*d*x**6 + a*b**2*d**2*x**9 + 10*b**3*c**2*x**6 + 10*b**3*c*d*x**9),x)*a**2*b*c*d**2*x**5 - 100*int(s qrt(c + d*x**3)/(a**3*c*d + a**3*d**2*x**3 + 10*a**2*b*c**2 + 12*a**2*b*c* d*x**3 + 2*a**2*b*d**2*x**6 + 20*a*b**2*c**2*x**3 + 21*a*b**2*c*d*x**6 + a *b**2*d**2*x**9 + 10*b**3*c**2*x**6 + 10*b**3*c*d*x**9),x)*a*b**2*c**3*x** 2 + 60*int(sqrt(c + d*x**3)/(a**3*c*d + a**3*d**2*x**3 + 10*a**2*b*c**2 + 12*a**2*b*c*d*x**3 + 2*a**2*b*d**2*x**6 + 20*a*b**2*c**2*x**3 + 21*a*b**2* c*d*x**6 + a*b**2*d**2*x**9 + 10*b**3*c**2*x**6 + 10*b**3*c*d*x**9),x)*a*b **2*c**2*d*x**5 - 100*int(sqrt(c + d*x**3)/(a**3*c*d + a**3*d**2*x**3 + 10 *a**2*b*c**2 + 12*a**2*b*c*d*x**3 + 2*a**2*b*d**2*x**6 + 20*a*b**2*c**2*x* *3 + 21*a*b**2*c*d*x**6 + a*b**2*d**2*x**9 + 10*b**3*c**2*x**6 + 10*b**3*c *d*x**9),x)*b**3*c**3*x**5 + 4*int((sqrt(c + d*x**3)*x**3)/(a**3*c*d + ...