Integrand size = 24, antiderivative size = 65 \[ \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {c x^4 \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {4}{3},2,-\frac {3}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a^2 \sqrt {1+\frac {d x^3}{c}}} \] Output:
1/4*c*x^4*(d*x^3+c)^(1/2)*AppellF1(4/3,2,-3/2,7/3,-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. \(338\) vs. \(2(65)=130\).
Time = 10.32 (sec) , antiderivative size = 338, normalized size of antiderivative = 5.20 \[ \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {x^4 \left (\frac {d (43 b c-55 a d) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{a}+\frac {8 \left (-8 a c d \left (11 a d+b \left (c+6 d x^3\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-3 \left (c+d x^3\right ) \left (5 b c-11 a d-6 b d x^3\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}\right )}{120 b^2 \sqrt {c+d x^3}} \] Input:
Integrate[(x^3*(c + d*x^3)^(3/2))/(a + b*x^3)^2,x]
Output:
(x^4*((d*(43*b*c - 55*a*d)*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)])/a + (8*(-8*a*c*d*(11*a*d + b*(c + 6*d*x^3))*A ppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] - 3*(c + d*x^3)*(5*b *c - 11*a*d - 6*b*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)])))))/(120*b^ 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.
| \(\displaystyle \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 1013 |
| \(\displaystyle \frac {c \sqrt {c+d x^3} \int \frac {x^3 \left (\frac {d x^3}{c}+1\right )^{3/2}}{\left (b x^3+a\right )^2}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle \frac {c x^4 \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {4}{3},2,-\frac {3}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a^2 \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[(x^3*(c + d*x^3)^(3/2))/(a + b*x^3)^2,x]
Output:
(c*x^4*Sqrt[c + d*x^3]*AppellF1[4/3, 2, -3/2, 7/3, -((b*x^3)/a), -((d*x^3) /c)])/(4*a^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 = 4.67 (sec) , antiderivative size = 808, normalized size of antiderivative = 12.43
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(808\) |
| risch | \(\text {Expression too large to display}\) | \(1563\) |
| default | \(\text {Expression too large to display}\) | \(1587\) |
Input:
int(x^3*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x,method=_RETURNVERBOSE)
Output:
1/3*(a*d-b*c)*x/b^2*(d*x^3+c)^(1/2)/(b*x^3+a)+2/5*x/b^2*d*(d*x^3+c)^(1/2)- 2/3*I*(-11/6*(a*d-b*c)*d/b^3-2/5*c*d/b^2)*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/b^3/d^2*2^(1/2)*sum((-11*a^2* d^2+13*a*b*c*d-2*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))*Ellipti cPi(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)*_alph a-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. \[ \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x^3*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int \frac {x^{3} \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (a + b x^{3}\right )^{2}}\, dx \] Input:
integrate(x**3*(d*x**3+c)**(3/2)/(b*x**3+a)**2,x)
Output:
Integral(x**3*(c + d*x**3)**(3/2)/(a + b*x**3)**2, x)
\[ \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{3}}{{\left (b x^{3} + a\right )}^{2}} \,d x } \] Input:
integrate(x^3*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)*x^3/(b*x^3 + a)^2, x)
\[ \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{3}}{{\left (b x^{3} + a\right )}^{2}} \,d x } \] Input:
integrate(x^3*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)*x^3/(b*x^3 + a)^2, x)
Timed out. \[ \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int \frac {x^3\,{\left (d\,x^3+c\right )}^{3/2}}{{\left (b\,x^3+a\right )}^2} \,d x \] Input:
int((x^3*(c + d*x^3)^(3/2))/(a + b*x^3)^2,x)
Output:
int((x^3*(c + d*x^3)^(3/2))/(a + b*x^3)^2, x)
\[ \int \frac {x^3 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\text {too large to display} \] Input:
int(x^3*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x)
Output:
( - 16*sqrt(c + d*x**3)*a*c*d*x + 10*sqrt(c + d*x**3)*a*d**2*x**4 + 10*sqr t(c + d*x**3)*b*c**2*x - 8*sqrt(c + d*x**3)*b*c*d*x**4 + 80*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**4*c**2*d**2 - 114*in t(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*b*c**3 *d + 80*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*b*c**2*d**2*x**3 + 40*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**2*c**4 - 114*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**2*c**3*d*x**3 + 40*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...