\(\int \frac {1}{(a+b x^4)^{4/3} (c+d x^4)^2} \, dx\) [74]

Optimal result

Integrand size = 21, antiderivative size = 62 \[ \int \frac {1}{\left (a+b x^4\right )^{4/3} \left (c+d x^4\right )^2} \, dx=\frac {x \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {4}{3},2,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a c^2 \sqrt [3]{a+b x^4}} \] Output:

x*(1+b*x^4/a)^(1/3)*AppellF1(1/4,4/3,2,5/4,-b*x^4/a,-d*x^4/c)/a/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. \(380\) vs. \(2(62)=124\).

Time = 10.58 (sec) , antiderivative size = 380, normalized size of antiderivative = 6.13 \[ \int \frac {1}{\left (a+b x^4\right )^{4/3} \left (c+d x^4\right )^2} \, dx=\frac {x \left (b d (3 b c+a d) x^4 \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{3},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+\frac {c \left (225 a c \left (4 a^2 d^2+a b d \left (-8 c+d x^4\right )+b^2 c \left (4 c+3 d x^4\right )\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{3},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-60 x^4 \left (a^2 d^2+a b d^2 x^4+3 b^2 c \left (c+d x^4\right )\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{3},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {4}{3},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}{\left (c+d x^4\right ) \left (15 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{3},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-4 x^4 \left (3 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{3},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {4}{3},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}\right )}{60 a c^2 (b c-a d)^2 \sqrt [3]{a+b x^4}} \] Input:

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

Output:

(x*(b*d*(3*b*c + a*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)] + (c*(225*a*c*(4*a^2*d^2 + a*b*d*(-8*c + d*x^ 
4) + b^2*c*(4*c + 3*d*x^4))*AppellF1[1/4, 1/3, 1, 5/4, -((b*x^4)/a), -((d* 
x^4)/c)] - 60*x^4*(a^2*d^2 + a*b*d^2*x^4 + 3*b^2*c*(c + d*x^4))*(3*a*d*App 
ellF1[5/4, 1/3, 2, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + b*c*AppellF1[5/4, 4/ 
3, 1, 9/4, -((b*x^4)/a), -((d*x^4)/c)])))/((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*AppellF1[5/4, 4/3, 1, 9/4, - 
((b*x^4)/a), -((d*x^4)/c)])))))/(60*a*c^2*(b*c - a*d)^2*(a + b*x^4)^(1/3))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 62, 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.

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 936

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b x^4\right )^{4/3} \left (c+d x^4\right )^2} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{3}} a \,c^{2}+2 \left (b \,x^{4}+a \right )^{\frac {1}{3}} a c d \,x^{4}+\left (b \,x^{4}+a \right )^{\frac {1}{3}} a \,d^{2} x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,c^{2} x^{4}+2 \left (b \,x^{4}+a \right )^{\frac {1}{3}} b c d \,x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{3}} b \,d^{2} x^{12}}d x \] Input:

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

Output:

int(1/((a + b*x**4)**(1/3)*a*c**2 + 2*(a + b*x**4)**(1/3)*a*c*d*x**4 + (a 
+ b*x**4)**(1/3)*a*d**2*x**8 + (a + b*x**4)**(1/3)*b*c**2*x**4 + 2*(a + b* 
x**4)**(1/3)*b*c*d*x**8 + (a + b*x**4)**(1/3)*b*d**2*x**12),x)