\(\int \frac {1}{(a+b x^n)^{4/3} (c+d x^n)^2} \, dx\) [118]

Optimal result

Integrand size = 21, antiderivative size = 64 \[ \int \frac {1}{\left (a+b x^n\right )^{4/3} \left (c+d x^n\right )^2} \, dx=\frac {x \sqrt [3]{1+\frac {b x^n}{a}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {4}{3},2,1+\frac {1}{n},-\frac {b x^n}{a},-\frac {d x^n}{c}\right )}{a c^2 \sqrt [3]{a+b x^n}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1294\) vs. \(2(64)=128\).

Time = 1.10 (sec) , antiderivative size = 1294, normalized size of antiderivative = 20.22 \[ \int \frac {1}{\left (a+b x^n\right )^{4/3} \left (c+d x^n\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(x*(27*a*b^2*c^4*(1 + n)^2*AppellF1[n^(-1), 1/3, 1, 1 + n^(-1), -((b*x^n)/ 
a), -((d*x^n)/c)] + 9*a^3*c^2*d^2*(1 + n)^2*AppellF1[n^(-1), 1/3, 1, 1 + n 
^(-1), -((b*x^n)/a), -((d*x^n)/c)] - 9*a*b^2*c^4*n*(1 + n)^2*AppellF1[n^(- 
1), 1/3, 1, 1 + n^(-1), -((b*x^n)/a), -((d*x^n)/c)] + 18*a^2*b*c^3*d*n*(1 
+ n)^2*AppellF1[n^(-1), 1/3, 1, 1 + n^(-1), -((b*x^n)/a), -((d*x^n)/c)] - 
9*a^3*c^2*d^2*n*(1 + n)^2*AppellF1[n^(-1), 1/3, 1, 1 + n^(-1), -((b*x^n)/a 
), -((d*x^n)/c)] + 3*c*(1 + n)*(a*d^2*(a + b*x^n) + 3*b^2*c*(c + d*x^n))*( 
3*a*d*n*x^n*AppellF1[1 + n^(-1), 1/3, 2, 2 + n^(-1), -((b*x^n)/a), -((d*x^ 
n)/c)] + b*c*n*x^n*AppellF1[1 + n^(-1), 4/3, 1, 2 + n^(-1), -((b*x^n)/a), 
-((d*x^n)/c)] - 3*a*c*(1 + n)*AppellF1[n^(-1), 1/3, 1, 1 + n^(-1), -((b*x^ 
n)/a), -((d*x^n)/c)]) - 9*b^2*c*d*x^n*(1 + (b*x^n)/a)^(1/3)*(c + d*x^n)*Ap 
pellF1[1 + n^(-1), 1/3, 1, 2 + n^(-1), -((b*x^n)/a), -((d*x^n)/c)]*(3*a*d* 
n*x^n*AppellF1[1 + n^(-1), 1/3, 2, 2 + n^(-1), -((b*x^n)/a), -((d*x^n)/c)] 
 + b*c*n*x^n*AppellF1[1 + n^(-1), 4/3, 1, 2 + n^(-1), -((b*x^n)/a), -((d*x 
^n)/c)] - 3*a*c*(1 + n)*AppellF1[n^(-1), 1/3, 1, 1 + n^(-1), -((b*x^n)/a), 
 -((d*x^n)/c)]) - 3*a*b*d^2*x^n*(1 + (b*x^n)/a)^(1/3)*(c + d*x^n)*AppellF1 
[1 + n^(-1), 1/3, 1, 2 + n^(-1), -((b*x^n)/a), -((d*x^n)/c)]*(3*a*d*n*x^n* 
AppellF1[1 + n^(-1), 1/3, 2, 2 + n^(-1), -((b*x^n)/a), -((d*x^n)/c)] + b*c 
*n*x^n*AppellF1[1 + n^(-1), 4/3, 1, 2 + n^(-1), -((b*x^n)/a), -((d*x^n)/c) 
] - 3*a*c*(1 + n)*AppellF1[n^(-1), 1/3, 1, 1 + n^(-1), -((b*x^n)/a), -(...
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 936

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x^n + a)^(2/3)/(b^2*d^2*x^(4*n) + a^2*c^2 + 2*(b^2*c*d + a*b*d 
^2)*x^(3*n) + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^(2*n) + 2*(a*b*c^2 + a^2*c 
*d)*x^n), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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