\(\int \frac {1}{\sqrt [3]{a+b x^4} (c+d x^4)^2} \, dx\) [70]

Optimal result

Integrand size = 21, antiderivative size = 59 \[ \int \frac {1}{\sqrt [3]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\frac {x \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{3},2,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2 \sqrt [3]{a+b x^4}} \] Output:

x*(1+b*x^4/a)^(1/3)*AppellF1(1/4,1/3,2,5/4,-b*x^4/a,-d*x^4/c)/c^2/(b*x^4+a 
)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(335\) vs. \(2(59)=118\).

Time = 10.43 (sec) , antiderivative size = 335, normalized size of antiderivative = 5.68 \[ \int \frac {1}{\sqrt [3]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\frac {x \left (\frac {b d x^4 \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{3},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{-b c+a d}+\frac {c \left (225 a c \left (-4 b c+4 a d+b d x^4\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{3},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-60 d x^4 \left (a+b x^4\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{3},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {4}{3},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}{(b c-a d) \left (c+d x^4\right ) \left (-15 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{3},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+4 x^4 \left (3 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{3},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {4}{3},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}\right )}{60 c^2 \sqrt [3]{a+b x^4}} \] Input:

Integrate[1/((a + b*x^4)^(1/3)*(c + d*x^4)^2),x]
 

Output:

(x*((b*d*x^4*(1 + (b*x^4)/a)^(1/3)*AppellF1[5/4, 1/3, 1, 9/4, -((b*x^4)/a) 
, -((d*x^4)/c)])/(-(b*c) + a*d) + (c*(225*a*c*(-4*b*c + 4*a*d + b*d*x^4)*A 
ppellF1[1/4, 1/3, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] - 60*d*x^4*(a + b*x^ 
4)*(3*a*d*AppellF1[5/4, 1/3, 2, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + b*c*App 
ellF1[5/4, 4/3, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))/((b*c - a*d)*(c + d 
*x^4)*(-15*a*c*AppellF1[1/4, 1/3, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + 4* 
x^4*(3*a*d*AppellF1[5/4, 1/3, 2, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + b*c*Ap 
pellF1[5/4, 4/3, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))))/(60*c^2*(a + b*x 
^4)^(1/3))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {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.

\(\displaystyle \int \frac {1}{\sqrt [3]{a+b x^4} \left (c+d x^4\right )^2} \, dx\)

\(\Big \downarrow \) 937

\(\displaystyle \frac {\sqrt [3]{\frac {b x^4}{a}+1} \int \frac {1}{\sqrt [3]{\frac {b x^4}{a}+1} \left (d x^4+c\right )^2}dx}{\sqrt [3]{a+b x^4}}\)

\(\Big \downarrow \) 936

\(\displaystyle \frac {x \sqrt [3]{\frac {b x^4}{a}+1} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{3},2,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2 \sqrt [3]{a+b x^4}}\)

Input:

Int[1/((a + b*x^4)^(1/3)*(c + d*x^4)^2),x]
 

Output:

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

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])
 

Maple [F]

\[\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} \left (d \,x^{4}+c \right )^{2}}d x\]

Input:

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

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x^4+a)^(1/3)/(d*x^4+c)^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate(1/(b*x**4+a)**(1/3)/(d*x**4+c)**2,x)
 

Output:

Integral(1/((a + b*x**4)**(1/3)*(c + d*x**4)**2), x)
 

Maxima [F]

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

integrate(1/(b*x^4+a)^(1/3)/(d*x^4+c)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(1/(b*x^4+a)^(1/3)/(d*x^4+c)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt [3]{a+b x^4} \left (c+d x^4\right )^2} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} c^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{3}} c d \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{3}} d^{2} x^{8}}d x \] Input:

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

Output:

int(1/((a + b*x**4)**(1/3)*c**2 + 2*(a + b*x**4)**(1/3)*c*d*x**4 + (a + b* 
x**4)**(1/3)*d**2*x**8),x)