\(\int \frac {x^3}{(a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\) [668]

Optimal result

Integrand size = 24, antiderivative size = 67 \[ \int \frac {x^3}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {x^4 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},2,\frac {3}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a^2 c \sqrt {c+d x^3}} \] Output:

1/4*x^4*(1+d*x^3/c)^(1/2)*AppellF1(4/3,2,3/2,7/3,-b*x^3/a,-d*x^3/c)/a^2/c/ 
(d*x^3+c)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(381\) vs. \(2(67)=134\).

Time = 10.24 (sec) , antiderivative size = 381, normalized size of antiderivative = 5.69 \[ \int \frac {x^3}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {x^4 \left (-8 a b c d \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right ) \left (8 a+\left (a+b x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )+\left (8 a \left (2 a d+b \left (c+3 d x^3\right )\right )+3 b d x^3 \left (a+b x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{8 a (b c-a d)^2 \left (a+b x^3\right ) \sqrt {c+d x^3} \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )} \] Input:

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

Output:

-1/8*(x^4*(-8*a*b*c*d*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a 
)]*(8*a + (a + b*x^3)*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d* 
x^3)/c), -((b*x^3)/a)]) + (8*a*(2*a*d + b*(c + 3*d*x^3)) + 3*b*d*x^3*(a + 
b*x^3)*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -((b*x 
^3)/a)])*(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*c - a*d 
)^2*(a + b*x^3)*Sqrt[c + d*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)])))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 67, 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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {x^4 \sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (\frac {4}{3},2,\frac {3}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a^2 c \sqrt {c+d x^3}}\)

Input:

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

Output:

(x^4*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 2, 3/2, 7/3, -((b*x^3)/a), -((d*x^3 
)/c)])/(4*a^2*c*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 = 2.13 (sec) , antiderivative size = 787, normalized size of antiderivative = 11.75

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(787\)
default	\(\text {Expression too large to display}\)	\(1593\)

Input:

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

Output:

-2/3*d*x/(a*d-b*c)^2/((x^3+c/d)*d)^(1/2)-1/3*b/(a*d-b*c)^2*x*(d*x^3+c)^(1/ 
2)/(b*x^3+a)+1/3*I/(a*d-b*c)^2*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)*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/d^2*2^(1/2)*sum((-7*a*d-2*b*c)/(a*d-b*c)^3 
/_alpha^2*(-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))
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b^{2} d^{2} x^{12}+2 a b \,d^{2} x^{9}+2 b^{2} c d \,x^{9}+a^{2} d^{2} x^{6}+4 a b c d \,x^{6}+b^{2} c^{2} x^{6}+2 a^{2} c d \,x^{3}+2 a b \,c^{2} x^{3}+a^{2} c^{2}}d x \] Input:

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

Output:

int((sqrt(c + d*x**3)*x**3)/(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)