Integrand size = 21, antiderivative size = 60 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {c x \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {1}{3},2,-\frac {3}{2},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 \sqrt {1+\frac {d x^3}{c}}} \] Output:
c*x*(d*x^3+c)^(1/2)*AppellF1(1/3,2,-3/2,4/3,-b*x^3/a,-d*x^3/c)/a^2/(1+d*x^ 3/c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(339\) vs. \(2(60)=120\).
Time = 10.26 (sec) , antiderivative size = 339, normalized size of antiderivative = 5.65 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {x \left (d (b c+5 a d) x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {a \left (-64 a c \left (-a d^2 x^3+b c \left (3 c+d x^3\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+24 (b c-a d) x^3 \left (c+d x^3\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}\right )}{24 a^2 b \sqrt {c+d x^3}} \] Input:
Integrate[(c + d*x^3)^(3/2)/(a + b*x^3)^2,x]
Output:
(x*(d*(b*c + 5*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*(-(a*d^2*x^3) + b*c*(3*c + d*x^3))* AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] + 24*(b*c - a*d)*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*Appe llF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)])))))/(24*a^2*b*Sqrt[c + d*x^3])
Time = 0.31 (sec) , antiderivative size = 60, 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.
| \(\displaystyle \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 937 |
| \(\displaystyle \frac {c \sqrt {c+d x^3} \int \frac {\left (\frac {d x^3}{c}+1\right )^{3/2}}{\left (b x^3+a\right )^2}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
| \(\Big \downarrow \) 936 |
| \(\displaystyle \frac {c x \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {1}{3},2,-\frac {3}{2},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[(c + d*x^3)^(3/2)/(a + b*x^3)^2,x]
Output:
(c*x*Sqrt[c + d*x^3]*AppellF1[1/3, 2, -3/2, 4/3, -((b*x^3)/a), -((d*x^3)/c )])/(a^2*Sqrt[1 + (d*x^3)/c])
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 = 1.14 (sec) , antiderivative size = 801, normalized size of antiderivative = 13.35
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(801\) |
| elliptic | \(\text {Expression too large to display}\) | \(801\) |
Input:
int((d*x^3+c)^(3/2)/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
-1/3*(a*d-b*c)/b/a*x*(d*x^3+c)^(1/2)/(b*x^3+a)-2/3*I*(d^2/b^2-1/6/b^2*d*(a *d-b*c)/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/b^2/d^2*2^(1/2)*sum((5*a^2*d^2-a*b*c*d-4*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))*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. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate((d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (a + b x^{3}\right )^{2}}\, dx \] Input:
integrate((d*x**3+c)**(3/2)/(b*x**3+a)**2,x)
Output:
Integral((c + d*x**3)**(3/2)/(a + b*x**3)**2, x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )}^{2}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)/(b*x^3 + a)^2, x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )}^{2}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)/(b*x^3 + a)^2, x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{3/2}}{{\left (b\,x^3+a\right )}^2} \,d x \] Input:
int((c + d*x^3)^(3/2)/(a + b*x^3)^2,x)
Output:
int((c + d*x^3)^(3/2)/(a + b*x^3)^2, x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:
int((d*x^3+c)^(3/2)/(b*x^3+a)^2,x)
Output:
(4*sqrt(c + d*x**3)*c*d*x + 5*int(sqrt(c + d*x**3)/(5*a**3*c*d + 5*a**3*d* *2*x**3 - 4*a**2*b*c**2 + 6*a**2*b*c*d*x**3 + 10*a**2*b*d**2*x**6 - 8*a*b* *2*c**2*x**3 - 3*a*b**2*c*d*x**6 + 5*a*b**2*d**2*x**9 - 4*b**3*c**2*x**6 - 4*b**3*c*d*x**9),x)*a**3*c**2*d**2 - 24*int(sqrt(c + d*x**3)/(5*a**3*c*d + 5*a**3*d**2*x**3 - 4*a**2*b*c**2 + 6*a**2*b*c*d*x**3 + 10*a**2*b*d**2*x* *6 - 8*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + 5*a*b**2*d**2*x**9 - 4*b**3* c**2*x**6 - 4*b**3*c*d*x**9),x)*a**2*b*c**3*d + 5*int(sqrt(c + d*x**3)/(5* a**3*c*d + 5*a**3*d**2*x**3 - 4*a**2*b*c**2 + 6*a**2*b*c*d*x**3 + 10*a**2* b*d**2*x**6 - 8*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + 5*a*b**2*d**2*x**9 - 4*b**3*c**2*x**6 - 4*b**3*c*d*x**9),x)*a**2*b*c**2*d**2*x**3 + 16*int(sq rt(c + d*x**3)/(5*a**3*c*d + 5*a**3*d**2*x**3 - 4*a**2*b*c**2 + 6*a**2*b*c *d*x**3 + 10*a**2*b*d**2*x**6 - 8*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + 5 *a*b**2*d**2*x**9 - 4*b**3*c**2*x**6 - 4*b**3*c*d*x**9),x)*a*b**2*c**4 - 2 4*int(sqrt(c + d*x**3)/(5*a**3*c*d + 5*a**3*d**2*x**3 - 4*a**2*b*c**2 + 6* a**2*b*c*d*x**3 + 10*a**2*b*d**2*x**6 - 8*a*b**2*c**2*x**3 - 3*a*b**2*c*d* x**6 + 5*a*b**2*d**2*x**9 - 4*b**3*c**2*x**6 - 4*b**3*c*d*x**9),x)*a*b**2* c**3*d*x**3 + 16*int(sqrt(c + d*x**3)/(5*a**3*c*d + 5*a**3*d**2*x**3 - 4*a **2*b*c**2 + 6*a**2*b*c*d*x**3 + 10*a**2*b*d**2*x**6 - 8*a*b**2*c**2*x**3 - 3*a*b**2*c*d*x**6 + 5*a*b**2*d**2*x**9 - 4*b**3*c**2*x**6 - 4*b**3*c*d*x **9),x)*b**3*c**4*x**3 + 25*int((sqrt(c + d*x**3)*x**6)/(5*a**3*c*d + 5...