\(\int \frac {x^3 \sqrt {c+d x^3}}{(a+b x^3)^2} \, dx\) [638]

Optimal result

Integrand size = 24, antiderivative size = 64 \[ \int \frac {x^3 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=\frac {x^4 \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {4}{3},2,-\frac {1}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a^2 \sqrt {1+\frac {d x^3}{c}}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(235\) vs. \(2(64)=128\).

Time = 10.25 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.67 \[ \int \frac {x^3 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=\frac {x \left (\frac {5 d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{a}+\frac {8 \left (-c-d x^3+\frac {8 a c^2 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}\right )}{a+b x^3}\right )}{24 b \sqrt {c+d x^3}} \] Input:

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

Output:

(x*((5*d*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), 
-((b*x^3)/a)])/a + (8*(-c - d*x^3 + (8*a*c^2*AppellF1[1/3, 1/2, 1, 4/3, -( 
(d*x^3)/c), -((b*x^3)/a)])/(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)]))) 
)/(a + b*x^3)))/(24*b*Sqrt[c + d*x^3])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 64, 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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

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

Input:

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

Output:

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

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 = 2.08 (sec) , antiderivative size = 748, normalized size of antiderivative = 11.69

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(748\)
default	\(\text {Expression too large to display}\)	\(1468\)

Input:

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

Output:

-1/3*x/b*(d*x^3+c)^(1/2)/(b*x^3+a)-5/9*I/b^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/b^2/d^2*2^(1/2)*sum((5*a*d-2 
*b*c)/_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))*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. \[ \int \frac {x^3 \sqrt {c+d x^3}}{\left (a+b x^3\right )^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*sqrt(c + d*x**3)*c*x - 10*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*c**2*d + 8*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*c**3 - 10*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*c**2*d*x**3 + 8*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*b**2*c**3*x**3 + 25*int 
((sqrt(c + d*x**3)*x**6)/(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* 
d**3 - 35*int((sqrt(c + d*x**3)*x**6)/(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*c*d**2 + 25*int((sqrt(c + d*x**3)*x**6)/(5*a**3*c*d + 5*...