\(\int \frac {1}{\sqrt {a+b x^3} (c+d x^3) (e+f x^3)} \, dx\) [3]

Optimal result

Integrand size = 30, antiderivative size = 142 \[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right ) \left (e+f x^3\right )} \, dx=\frac {d x \sqrt {1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c (d e-c f) \sqrt {a+b x^3}}-\frac {f x \sqrt {1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {f x^3}{e}\right )}{e (d e-c f) \sqrt {a+b x^3}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2917\) vs. \(2(142)=284\).

Time = 16.18 (sec) , antiderivative size = 2917, normalized size of antiderivative = 20.54 \[ \int \frac {1}{\sqrt {a+b x^3} \left (c+d x^3\right ) \left (e+f x^3\right )} \, dx=\text {Result too large to show} \] Input:

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

Output:

(2*(((-1)^(1/3)*a^(1/3))/b^(1/3) + ((-1)^(2/3)*a^(1/3))/b^(1/3))*Sqrt[(a^( 
1/3)/b^(1/3) + x)/(a^(1/3)/b^(1/3) + ((-1)^(1/3)*a^(1/3))/b^(1/3))]*Sqrt[( 
(-(((-1)^(2/3)*a^(1/3))/b^(1/3)) - x)*(-(((-1)^(1/3)*a^(1/3))/b^(1/3)) + x 
))/(((-1)^(1/3)*a^(1/3))/b^(1/3) + ((-1)^(2/3)*a^(1/3))/b^(1/3))^2]*Ellipt 
icPi[((a^(1/3) + (-1)^(1/3)*a^(1/3))*d^(1/3))/(-(b^(1/3)*c^(1/3)) + a^(1/3 
)*d^(1/3)), ArcSin[Sqrt[-(((-1)^(2/3)*((-1)^(1/3)*a^(1/3) - b^(1/3)*x))/(( 
1 + (-1)^(1/3))*a^(1/3)))]], (-1)^(1/3)])/((c^(1/3)/d^(1/3) + ((-1)^(1/3)* 
c^(1/3))/d^(1/3))*(((-1)^(1/3)*c^(1/3))/d^(1/3) + ((-1)^(2/3)*c^(1/3))/d^( 
1/3))*((-1)^(1/3)*c^(1/3) - ((-1)^(1/3)*a^(1/3)*d^(1/3))/b^(1/3))*d^(2/3)* 
(((-1)^(1/3)*c^(1/3))/d^(1/3) + e^(1/3)/f^(1/3))*(((-1)^(1/3)*c^(1/3))/d^( 
1/3) - ((-1)^(1/3)*e^(1/3))/f^(1/3))*(((-1)^(1/3)*c^(1/3))/d^(1/3) + ((-1) 
^(2/3)*e^(1/3))/f^(1/3))*f*Sqrt[a + b*x^3]) + (2*(((-1)^(1/3)*a^(1/3))/b^( 
1/3) + ((-1)^(2/3)*a^(1/3))/b^(1/3))*Sqrt[(a^(1/3)/b^(1/3) + x)/(a^(1/3)/b 
^(1/3) + ((-1)^(1/3)*a^(1/3))/b^(1/3))]*Sqrt[((-(((-1)^(2/3)*a^(1/3))/b^(1 
/3)) - x)*(-(((-1)^(1/3)*a^(1/3))/b^(1/3)) + x))/(((-1)^(1/3)*a^(1/3))/b^( 
1/3) + ((-1)^(2/3)*a^(1/3))/b^(1/3))^2]*EllipticPi[((a^(1/3) + (-1)^(1/3)* 
a^(1/3))*d^(1/3))/((-1)^(1/3)*b^(1/3)*c^(1/3) + a^(1/3)*d^(1/3)), ArcSin[S 
qrt[-(((-1)^(2/3)*((-1)^(1/3)*a^(1/3) - b^(1/3)*x))/((1 + (-1)^(1/3))*a^(1 
/3)))]], (-1)^(1/3)])/((c^(1/3)/d^(1/3) - ((-1)^(2/3)*c^(1/3))/d^(1/3))*(- 
(((-1)^(1/3)*c^(1/3))/d^(1/3)) - ((-1)^(2/3)*c^(1/3))/d^(1/3))*(-((-1)^...
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1029, 937, 936}

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.

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

\(\Big \downarrow \) 1029

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 936

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 937

Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q 
}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 1029

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_ 
.)*(x_)^(n_))^(r_), x_Symbol] :> Simp[b/(b*c - a*d)   Int[(a + b*x^n)^p*(c 
+ d*x^n)^(q + 1)*(e + f*x^n)^r, x], x] - Simp[d/(b*c - a*d)   Int[(a + b*x^ 
n)^(p + 1)*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c, d, e, f, 
n, q}, x] && ILtQ[p, 0] && LeQ[q, -1]
 

Maple [C] (warning: unable to verify)

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

Time = 0.77 (sec) , antiderivative size = 880, normalized size of antiderivative = 6.20

method	result	size
default	\(\text {Expression too large to display}\)	\(880\)
elliptic	\(\text {Expression too large to display}\)	\(3236\)

Input:

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

Output:

1/3*I*f/(c*f-d*e)/b^2*2^(1/2)*sum(1/_alpha^2/(a*f-b*e)*(-a*b^2)^(1/3)*(1/2 
*I*b*(2*x+1/b*(-I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^ 
(1/2)*(b*(x-1/b*(-a*b^2)^(1/3))/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3 
)))^(1/2)*(-1/2*I*b*(2*x+1/b*(I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3)))/(- 
a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*(I*(-a*b^2)^(1/3)*_alpha*3^(1/2)*b-I*3 
^(1/2)*(-a*b^2)^(2/3)+2*_alpha^2*b^2-(-a*b^2)^(1/3)*_alpha*b-(-a*b^2)^(2/3 
))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b 
^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),-1/2/b*f*(2*I*3^(1/2)*(-a*b^2)^ 
(1/3)*_alpha^2*b-I*3^(1/2)*(-a*b^2)^(2/3)*_alpha+I*3^(1/2)*a*b-3*(-a*b^2)^ 
(2/3)*_alpha-3*a*b)/(a*f-b*e),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2) 
^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*f+e))-1/ 
3*I*d/(c*f-d*e)/b^2*2^(1/2)*sum(1/_alpha^2/(a*d-b*c)*(-a*b^2)^(1/3)*(1/2*I 
*b*(2*x+1/b*(-I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1 
/2)*(b*(x-1/b*(-a*b^2)^(1/3))/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)) 
)^(1/2)*(-1/2*I*b*(2*x+1/b*(I*3^(1/2)*(-a*b^2)^(1/3)+(-a*b^2)^(1/3)))/(-a* 
b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*(I*(-a*b^2)^(1/3)*_alpha*3^(1/2)*b-I*3^( 
1/2)*(-a*b^2)^(2/3)+2*_alpha^2*b^2-(-a*b^2)^(1/3)*_alpha*b-(-a*b^2)^(2/3)) 
*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2 
)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),-1/2/b*d*(2*I*3^(1/2)*(-a*b^2)^(1 
/3)*_alpha^2*b-I*3^(1/2)*(-a*b^2)^(2/3)*_alpha+I*3^(1/2)*a*b-3*(-a*b^2)...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt(a + b*x**3)*(c + d*x**3)*(e + f*x**3)), x)
 

Maxima [F]

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

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

Output:

integrate(1/(sqrt(b*x^3 + a)*(d*x^3 + c)*(f*x^3 + e)), x)
 

Giac [F]

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

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

Output:

integrate(1/(sqrt(b*x^3 + a)*(d*x^3 + c)*(f*x^3 + e)), x)
 

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

int(sqrt(a + b*x**3)/(a*c*e + a*c*f*x**3 + a*d*e*x**3 + a*d*f*x**6 + b*c*e 
*x**3 + b*c*f*x**6 + b*d*e*x**6 + b*d*f*x**9),x)