\(\int \frac {1}{x^2 (a+b x^3) (c+d x^3)^{3/2}} \, dx\) [569]

Optimal result

Integrand size = 24, antiderivative size = 65 \[ \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (-\frac {1}{3},1,\frac {3}{2},\frac {2}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c x \sqrt {c+d x^3}} \] Output:

-(1+d*x^3/c)^(1/2)*AppellF1(-1/3,1,3/2,2/3,-b*x^3/a,-d*x^3/c)/a/c/x/(d*x^3 
+c)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(193\) vs. \(2(65)=130\).

Time = 10.17 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.97 \[ \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {20 a \left (-3 b c \left (c+d x^3\right )+a d \left (3 c+5 d x^3\right )\right )-5 \left (6 b^2 c^2-3 a b c d+5 a^2 d^2\right ) x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+2 b d (3 b c-5 a d) x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{60 a^2 c^2 (b c-a d) x \sqrt {c+d x^3}} \] Input:

Integrate[1/(x^2*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
 

Output:

(20*a*(-3*b*c*(c + d*x^3) + a*d*(3*c + 5*d*x^3)) - 5*(6*b^2*c^2 - 3*a*b*c* 
d + 5*a^2*d^2)*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3 
)/c), -((b*x^3)/a)] + 2*b*d*(3*b*c - 5*a*d)*x^6*Sqrt[1 + (d*x^3)/c]*Appell 
F1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)])/(60*a^2*c^2*(b*c - a*d)* 
x*Sqrt[c + d*x^3])
 

Rubi [A] (verified)

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.

\(\displaystyle \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{x^2 \left (b x^3+a\right ) \left (\frac {d x^3}{c}+1\right )^{3/2}}dx}{c \sqrt {c+d x^3}}\)

\(\Big \downarrow \) 1012

\(\displaystyle -\frac {\sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (-\frac {1}{3},1,\frac {3}{2},\frac {2}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c x \sqrt {c+d x^3}}\)

Input:

Int[1/(x^2*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
 

Output:

-((Sqrt[1 + (d*x^3)/c]*AppellF1[-1/3, 1, 3/2, 2/3, -((b*x^3)/a), -((d*x^3) 
/c)])/(a*c*x*Sqrt[c + d*x^3]))
 

Defintions of rubi rules used

rule 1012

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])
 

rule 1013

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])
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 6.

Time = 3.10 (sec) , antiderivative size = 952, normalized size of antiderivative = 14.65

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(952\)
risch	\(\text {Expression too large to display}\)	\(1382\)
default	\(\text {Expression too large to display}\)	\(1392\)

Input:

int(1/x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/c^2/a*(d*x^3+c)^(1/2)/x-2/3*d^2*x^2/c^2/(a*d-b*c)/((x^3+c/d)*d)^(1/2)-2 
/3*I*(1/2/a/c^2*d+1/3*d^2/c^2/(a*d-b*c))*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)*((-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)))-1/3*I*b^2/a/d^2*2^(1/2)*sum(1/(a*d-b*c)^2/_alpha*(-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)*_alph 
a*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*...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{x^{2} \left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/x**2/(b*x**3+a)/(d*x**3+c)**(3/2),x)
                                                                                    
                                                                                    
 

Output:

Integral(1/(x**2*(a + b*x**3)*(c + d*x**3)**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^2), x)
 

Giac [F]

\[ \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="giac")
 

Output:

integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{x^2\,\left (b\,x^3+a\right )\,{\left (d\,x^3+c\right )}^{3/2}} \,d x \] Input:

int(1/(x^2*(a + b*x^3)*(c + d*x^3)^(3/2)),x)
 

Output:

int(1/(x^2*(a + b*x^3)*(c + d*x^3)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^2 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}-5 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b c d x -5 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b \,d^{2} x^{4}-5 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) a c d x -5 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) a \,d^{2} x^{4}-2 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b \,c^{2} x -2 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{b \,d^{2} x^{9}+a \,d^{2} x^{6}+2 b c d \,x^{6}+2 a c d \,x^{3}+b \,c^{2} x^{3}+a \,c^{2}}d x \right ) b c d \,x^{4}}{2 a c x \left (d \,x^{3}+c \right )} \] Input:

int(1/x^2/(b*x^3+a)/(d*x^3+c)^(3/2),x)
 

Output:

( - 2*sqrt(c + d*x**3) - 5*int((sqrt(c + d*x**3)*x**4)/(a*c**2 + 2*a*c*d*x 
**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*b*c*d*x - 
 5*int((sqrt(c + d*x**3)*x**4)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c* 
*2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*b*d**2*x**4 - 5*int((sqrt(c + d*x 
**3)*x)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 
+ b*d**2*x**9),x)*a*c*d*x - 5*int((sqrt(c + d*x**3)*x)/(a*c**2 + 2*a*c*d*x 
**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*a*d**2*x* 
*4 - 2*int((sqrt(c + d*x**3)*x)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c 
**2*x**3 + 2*b*c*d*x**6 + b*d**2*x**9),x)*b*c**2*x - 2*int((sqrt(c + d*x** 
3)*x)/(a*c**2 + 2*a*c*d*x**3 + a*d**2*x**6 + b*c**2*x**3 + 2*b*c*d*x**6 + 
b*d**2*x**9),x)*b*c*d*x**4)/(2*a*c*x*(c + d*x**3))