\(\int \frac {1}{x^2 (a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\) [671]

Optimal result

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

-(1+d*x^3/c)^(1/2)*AppellF1(-1/3,2,3/2,2/3,-b*x^3/a,-d*x^3/c)/a^2/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. \(308\) vs. \(2(65)=130\).

Time = 10.31 (sec) , antiderivative size = 308, normalized size of antiderivative = 4.74 \[ \int \frac {1}{x^2 \left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {-20 a \left (4 b^3 c^2 x^3 \left (c+d x^3\right )+a^3 d^2 \left (3 c+5 d x^3\right )+3 a b^2 c \left (c^2-c d x^3-2 d^2 x^6\right )+a^2 b d \left (-6 c^2-3 c d x^3+5 d^2 x^6\right )\right )+5 \left (-8 b^3 c^3+21 a b^2 c^2 d-6 a^2 b c d^2+5 a^3 d^3\right ) x^3 \left (a+b x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+2 b d \left (4 b^2 c^2-6 a b c d+5 a^2 d^2\right ) x^6 \left (a+b x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{60 a^3 c^2 (b c-a d)^2 x \left (a+b x^3\right ) \sqrt {c+d x^3}} \] Input:

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

Output:

(-20*a*(4*b^3*c^2*x^3*(c + d*x^3) + a^3*d^2*(3*c + 5*d*x^3) + 3*a*b^2*c*(c 
^2 - c*d*x^3 - 2*d^2*x^6) + a^2*b*d*(-6*c^2 - 3*c*d*x^3 + 5*d^2*x^6)) + 5* 
(-8*b^3*c^3 + 21*a*b^2*c^2*d - 6*a^2*b*c*d^2 + 5*a^3*d^3)*x^3*(a + b*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*(4*b^2*c^2 - 6*a*b*c*d + 5*a^2*d^2)*x^6*(a + b*x^3)*Sqrt[1 + (d*x 
^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)])/(60*a^3*c^2 
*(b*c - a*d)^2*x*(a + b*x^3)*Sqrt[c + d*x^3])
 

Rubi [A] (verified)

Time = 0.36 (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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

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

Input:

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

Output:

-((Sqrt[1 + (d*x^3)/c]*AppellF1[-1/3, 2, 3/2, 2/3, -((b*x^3)/a), -((d*x^3) 
/c)])/(a^2*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 = 6.02 (sec) , antiderivative size = 1019, normalized size of antiderivative = 15.68

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(1019\)
risch	\(\text {Expression too large to display}\)	\(2334\)
default	\(\text {Expression too large to display}\)	\(2383\)

Input:

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

Output:

-2/3*d^3*x^2/c^2/(a*d-b*c)^2/((x^3+c/d)*d)^(1/2)-1/3/(a*d-b*c)^2/a^2*b^3*x 
^2*(d*x^3+c)^(1/2)/(b*x^3+a)-1/c^2/a^2*(d*x^3+c)^(1/2)/x-2/3*I*(1/3/c^2*d^ 
3/(a*d-b*c)^2+1/6*b^2*d/a^2/(a*d-b*c)^2+1/2*d/c^2/a^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)*((-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)))-1/18*I*b^2/a^2/d^2*2^(1/2)*sum((17*a*d- 
8*b*c)/(a*d-b*c)^3/_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)*_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)*...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*sqrt(c + d*x**3) - 11*int((sqrt(c + d*x**3)*x**4)/(a**2*c**2 + 2*a** 
2*c*d*x**3 + a**2*d**2*x**6 + 2*a*b*c**2*x**3 + 4*a*b*c*d*x**6 + 2*a*b*d** 
2*x**9 + b**2*c**2*x**6 + 2*b**2*c*d*x**9 + b**2*d**2*x**12),x)*a*b*c*d*x 
- 11*int((sqrt(c + d*x**3)*x**4)/(a**2*c**2 + 2*a**2*c*d*x**3 + a**2*d**2* 
x**6 + 2*a*b*c**2*x**3 + 4*a*b*c*d*x**6 + 2*a*b*d**2*x**9 + b**2*c**2*x**6 
 + 2*b**2*c*d*x**9 + b**2*d**2*x**12),x)*a*b*d**2*x**4 - 11*int((sqrt(c + 
d*x**3)*x**4)/(a**2*c**2 + 2*a**2*c*d*x**3 + a**2*d**2*x**6 + 2*a*b*c**2*x 
**3 + 4*a*b*c*d*x**6 + 2*a*b*d**2*x**9 + b**2*c**2*x**6 + 2*b**2*c*d*x**9 
+ b**2*d**2*x**12),x)*b**2*c*d*x**4 - 11*int((sqrt(c + d*x**3)*x**4)/(a**2 
*c**2 + 2*a**2*c*d*x**3 + a**2*d**2*x**6 + 2*a*b*c**2*x**3 + 4*a*b*c*d*x** 
6 + 2*a*b*d**2*x**9 + b**2*c**2*x**6 + 2*b**2*c*d*x**9 + b**2*d**2*x**12), 
x)*b**2*d**2*x**7 - 5*int((sqrt(c + d*x**3)*x)/(a**2*c**2 + 2*a**2*c*d*x** 
3 + a**2*d**2*x**6 + 2*a*b*c**2*x**3 + 4*a*b*c*d*x**6 + 2*a*b*d**2*x**9 + 
b**2*c**2*x**6 + 2*b**2*c*d*x**9 + b**2*d**2*x**12),x)*a**2*c*d*x - 5*int( 
(sqrt(c + d*x**3)*x)/(a**2*c**2 + 2*a**2*c*d*x**3 + a**2*d**2*x**6 + 2*a*b 
*c**2*x**3 + 4*a*b*c*d*x**6 + 2*a*b*d**2*x**9 + b**2*c**2*x**6 + 2*b**2*c* 
d*x**9 + b**2*d**2*x**12),x)*a**2*d**2*x**4 - 8*int((sqrt(c + d*x**3)*x)/( 
a**2*c**2 + 2*a**2*c*d*x**3 + a**2*d**2*x**6 + 2*a*b*c**2*x**3 + 4*a*b*c*d 
*x**6 + 2*a*b*d**2*x**9 + b**2*c**2*x**6 + 2*b**2*c*d*x**9 + b**2*d**2*x** 
12),x)*a*b*c**2*x - 13*int((sqrt(c + d*x**3)*x)/(a**2*c**2 + 2*a**2*c*d...