Integrand size = 24, antiderivative size = 63 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx=-\frac {c \sqrt {c+d x^3} \operatorname {AppellF1}\left (-\frac {1}{3},1,-\frac {3}{2},\frac {2}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a x \sqrt {1+\frac {d x^3}{c}}} \] Output:
-c*(d*x^3+c)^(1/2)*AppellF1(-1/3,1,-3/2,2/3,-b*x^3/a,-d*x^3/c)/a/x/(1+d*x^ 3/c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(148\) vs. \(2(63)=126\).
Time = 10.14 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.35 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx=\frac {-20 a c \left (c+d x^3\right )+5 c (-2 b c+5 a d) x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+2 d (b c+2 a d) x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{20 a^2 x \sqrt {c+d x^3}} \] Input:
Integrate[(c + d*x^3)^(3/2)/(x^2*(a + b*x^3)),x]
Output:
(-20*a*c*(c + d*x^3) + 5*c*(-2*b*c + 5*a*d)*x^3*Sqrt[1 + (d*x^3)/c]*Appell F1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), -((b*x^3)/a)] + 2*d*(b*c + 2*a*d)*x^6*S qrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)]) /(20*a^2*x*Sqrt[c + d*x^3])
Time = 0.36 (sec) , antiderivative size = 63, 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.
| \(\displaystyle \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx\) |
| \(\Big \downarrow \) 1013 |
| \(\displaystyle \frac {c \sqrt {c+d x^3} \int \frac {\left (\frac {d x^3}{c}+1\right )^{3/2}}{x^2 \left (b x^3+a\right )}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle -\frac {c \sqrt {c+d x^3} \operatorname {AppellF1}\left (-\frac {1}{3},1,-\frac {3}{2},\frac {2}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a x \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[(c + d*x^3)^(3/2)/(x^2*(a + b*x^3)),x]
Output:
-((c*Sqrt[c + d*x^3]*AppellF1[-1/3, 1, -3/2, 2/3, -((b*x^3)/a), -((d*x^3)/ c)])/(a*x*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 = 2.90 (sec) , antiderivative size = 920, normalized size of antiderivative = 14.60
| method | result | size |
| risch | \(\text {Expression too large to display}\) | \(920\) |
| elliptic | \(\text {Expression too large to display}\) | \(924\) |
| default | \(\text {Expression too large to display}\) | \(1404\) |
Input:
int((d*x^3+c)^(3/2)/x^2/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
-c/a*(d*x^3+c)^(1/2)/x+1/2/a*(-2/3*I*(2*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)*((-3/2/d*(- c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*EllipticE(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/d*(-c*d^2)^(1/3)*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)))+2/3*I*(a^2*d^2-2*a*b*c*d+b^2*c^2)/b/d^2*2^(1/ 2)*sum(1/_alpha/(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*_alph a^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)...
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx=\text {Timed out} \] Input:
integrate((d*x^3+c)^(3/2)/x^2/(b*x^3+a),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx=\int \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{x^{2} \left (a + b x^{3}\right )}\, dx \] Input:
integrate((d*x**3+c)**(3/2)/x**2/(b*x**3+a),x)
Output:
Integral((c + d*x**3)**(3/2)/(x**2*(a + b*x**3)), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )} x^{2}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/x^2/(b*x^3+a),x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)/((b*x^3 + a)*x^2), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (b x^{3} + a\right )} x^{2}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/x^2/(b*x^3+a),x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)/((b*x^3 + a)*x^2), x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{3/2}}{x^2\,\left (b\,x^3+a\right )} \,d x \] Input:
int((c + d*x^3)^(3/2)/(x^2*(a + b*x^3)),x)
Output:
int((c + d*x^3)^(3/2)/(x^2*(a + b*x^3)), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^2 \left (a+b x^3\right )} \, dx=\frac {4 \sqrt {d \,x^{3}+c}\, c d +5 \left (\int \frac {\sqrt {d \,x^{3}+c}}{a b \,d^{2} x^{8}-2 b^{2} c d \,x^{8}+a^{2} d^{2} x^{5}-a b c d \,x^{5}-2 b^{2} c^{2} x^{5}+a^{2} c d \,x^{2}-2 a b \,c^{2} x^{2}}d x \right ) a^{2} c^{2} d^{2} x -12 \left (\int \frac {\sqrt {d \,x^{3}+c}}{a b \,d^{2} x^{8}-2 b^{2} c d \,x^{8}+a^{2} d^{2} x^{5}-a b c d \,x^{5}-2 b^{2} c^{2} x^{5}+a^{2} c d \,x^{2}-2 a b \,c^{2} x^{2}}d x \right ) a b \,c^{3} d x +4 \left (\int \frac {\sqrt {d \,x^{3}+c}}{a b \,d^{2} x^{8}-2 b^{2} c d \,x^{8}+a^{2} d^{2} x^{5}-a b c d \,x^{5}-2 b^{2} c^{2} x^{5}+a^{2} c d \,x^{2}-2 a b \,c^{2} x^{2}}d x \right ) b^{2} c^{4} x +\left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{a b \,d^{2} x^{6}-2 b^{2} c d \,x^{6}+a^{2} d^{2} x^{3}-a b c d \,x^{3}-2 b^{2} c^{2} x^{3}+a^{2} c d -2 a b \,c^{2}}d x \right ) a^{2} d^{4} x -6 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{a b \,d^{2} x^{6}-2 b^{2} c d \,x^{6}+a^{2} d^{2} x^{3}-a b c d \,x^{3}-2 b^{2} c^{2} x^{3}+a^{2} c d -2 a b \,c^{2}}d x \right ) a b c \,d^{3} x +8 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{a b \,d^{2} x^{6}-2 b^{2} c d \,x^{6}+a^{2} d^{2} x^{3}-a b c d \,x^{3}-2 b^{2} c^{2} x^{3}+a^{2} c d -2 a b \,c^{2}}d x \right ) b^{2} c^{2} d^{2} x}{x \left (a d -2 b c \right )} \] Input:
int((d*x^3+c)^(3/2)/x^2/(b*x^3+a),x)
Output:
(4*sqrt(c + d*x**3)*c*d + 5*int(sqrt(c + d*x**3)/(a**2*c*d*x**2 + a**2*d** 2*x**5 - 2*a*b*c**2*x**2 - a*b*c*d*x**5 + a*b*d**2*x**8 - 2*b**2*c**2*x**5 - 2*b**2*c*d*x**8),x)*a**2*c**2*d**2*x - 12*int(sqrt(c + d*x**3)/(a**2*c* d*x**2 + a**2*d**2*x**5 - 2*a*b*c**2*x**2 - a*b*c*d*x**5 + a*b*d**2*x**8 - 2*b**2*c**2*x**5 - 2*b**2*c*d*x**8),x)*a*b*c**3*d*x + 4*int(sqrt(c + d*x* *3)/(a**2*c*d*x**2 + a**2*d**2*x**5 - 2*a*b*c**2*x**2 - a*b*c*d*x**5 + a*b *d**2*x**8 - 2*b**2*c**2*x**5 - 2*b**2*c*d*x**8),x)*b**2*c**4*x + int((sqr t(c + d*x**3)*x**4)/(a**2*c*d + a**2*d**2*x**3 - 2*a*b*c**2 - a*b*c*d*x**3 + a*b*d**2*x**6 - 2*b**2*c**2*x**3 - 2*b**2*c*d*x**6),x)*a**2*d**4*x - 6* int((sqrt(c + d*x**3)*x**4)/(a**2*c*d + a**2*d**2*x**3 - 2*a*b*c**2 - a*b* c*d*x**3 + a*b*d**2*x**6 - 2*b**2*c**2*x**3 - 2*b**2*c*d*x**6),x)*a*b*c*d* *3*x + 8*int((sqrt(c + d*x**3)*x**4)/(a**2*c*d + a**2*d**2*x**3 - 2*a*b*c* *2 - a*b*c*d*x**3 + a*b*d**2*x**6 - 2*b**2*c**2*x**3 - 2*b**2*c*d*x**6),x) *b**2*c**2*d**2*x)/(x*(a*d - 2*b*c))