\(\int \frac {1}{(a+b x^n) (c+d x^n)^3} \, dx\) [87]

Optimal result

Integrand size = 19, antiderivative size = 210 \[ \int \frac {1}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=-\frac {d x}{2 c (b c-a d) n \left (c+d x^n\right )^2}-\frac {d (a d (1-2 n)-b (c-4 c n)) x}{2 c^2 (b c-a d)^2 n^2 \left (c+d x^n\right )}+\frac {b^3 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a (b c-a d)^3}-\frac {d \left (a^2 d^2 \left (1-3 n+2 n^2\right )-2 a b c d \left (1-4 n+3 n^2\right )+b^2 c^2 \left (1-5 n+6 n^2\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{2 c^3 (b c-a d)^3 n^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\frac {x \left (-a c^2 d (b c-a d)^2 n+a c d (b c-a d) (a d (-1+2 n)+b (c-4 c n)) \left (c+d x^n\right )+2 b^3 c^3 n^2 \left (c+d x^n\right )^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )-a d \left (a^2 d^2 \left (1-3 n+2 n^2\right )-2 a b c d \left (1-4 n+3 n^2\right )+b^2 c^2 \left (1-5 n+6 n^2\right )\right ) \left (c+d x^n\right )^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )\right )}{2 a c^3 (b c-a d)^3 n^2 \left (c+d x^n\right )^2} \] Input:

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

Output:

(x*(-(a*c^2*d*(b*c - a*d)^2*n) + a*c*d*(b*c - a*d)*(a*d*(-1 + 2*n) + b*(c 
- 4*c*n))*(c + d*x^n) + 2*b^3*c^3*n^2*(c + d*x^n)^2*Hypergeometric2F1[1, n 
^(-1), 1 + n^(-1), -((b*x^n)/a)] - a*d*(a^2*d^2*(1 - 3*n + 2*n^2) - 2*a*b* 
c*d*(1 - 4*n + 3*n^2) + b^2*c^2*(1 - 5*n + 6*n^2))*(c + d*x^n)^2*Hypergeom 
etric2F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)]))/(2*a*c^3*(b*c - a*d)^3*n^2 
*(c + d*x^n)^2)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {931, 1024, 1020, 778}

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.

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

\(\Big \downarrow \) 931

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

\(\Big \downarrow \) 1024

\(\displaystyle \frac {\frac {\int \frac {-b d (b c (1-4 n)-a d (1-2 n)) (1-n) x^n+2 b^2 c^2 n^2+a^2 d^2 \left (2 n^2-3 n+1\right )-a b c d \left (4 n^2-5 n+1\right )}{\left (b x^n+a\right ) \left (d x^n+c\right )}dx}{c n (b c-a d)}+\frac {d x (b c (1-4 n)-a d (1-2 n))}{c n (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}-\frac {d x}{2 c n (b c-a d) \left (c+d x^n\right )^2}\)

\(\Big \downarrow \) 1020

\(\displaystyle \frac {\frac {\frac {2 b^3 c^2 n^2 \int \frac {1}{b x^n+a}dx}{b c-a d}-\frac {d \left (a^2 d^2 \left (2 n^2-3 n+1\right )-2 a b c d \left (3 n^2-4 n+1\right )+b^2 c^2 \left (6 n^2-5 n+1\right )\right ) \int \frac {1}{d x^n+c}dx}{b c-a d}}{c n (b c-a d)}+\frac {d x (b c (1-4 n)-a d (1-2 n))}{c n (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}-\frac {d x}{2 c n (b c-a d) \left (c+d x^n\right )^2}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {\frac {\frac {2 b^3 c^2 n^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a (b c-a d)}-\frac {d x \left (a^2 d^2 \left (2 n^2-3 n+1\right )-2 a b c d \left (3 n^2-4 n+1\right )+b^2 c^2 \left (6 n^2-5 n+1\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{c (b c-a d)}}{c n (b c-a d)}+\frac {d x (b c (1-4 n)-a d (1-2 n))}{c n (b c-a d) \left (c+d x^n\right )}}{2 c n (b c-a d)}-\frac {d x}{2 c n (b c-a d) \left (c+d x^n\right )^2}\)

Input:

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

Output:

-1/2*(d*x)/(c*(b*c - a*d)*n*(c + d*x^n)^2) + ((d*(b*c*(1 - 4*n) - a*d*(1 - 
 2*n))*x)/(c*(b*c - a*d)*n*(c + d*x^n)) + ((2*b^3*c^2*n^2*x*Hypergeometric 
2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*(b*c - a*d)) - (d*(a^2*d^2*(1 
 - 3*n + 2*n^2) - 2*a*b*c*d*(1 - 4*n + 3*n^2) + b^2*c^2*(1 - 5*n + 6*n^2)) 
*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)])/(c*(b*c - a*d)) 
)/(c*(b*c - a*d)*n))/(2*c*(b*c - a*d)*n)
 

Defintions of rubi rules used

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 931

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(-b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - 
 a*d))), x] + Simp[1/(a*n*(p + 1)*(b*c - a*d))   Int[(a + b*x^n)^(p + 1)*(c 
 + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, 
 x], x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, 
-1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, 
 c, d, n, p, q, x]
 

rule 1020

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^( 
n_))), x_Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^n), x], x 
] - Simp[(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b 
, c, d, e, f, n}, x]
 

rule 1024

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f 
_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c 
+ d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*( 
p + 1))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b 
*c - a*d)*(p + 1) + d*(b*e - a*f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

-b^3*integrate(-1/(a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3 + 
 (b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*x^n), x) + ((6*n^ 
2 - 5*n + 1)*b^2*c^2*d - 2*(3*n^2 - 4*n + 1)*a*b*c*d^2 + (2*n^2 - 3*n + 1) 
*a^2*d^3)*integrate(-1/2/(b^3*c^6*n^2 - 3*a*b^2*c^5*d*n^2 + 3*a^2*b*c^4*d^ 
2*n^2 - a^3*c^3*d^3*n^2 + (b^3*c^5*d*n^2 - 3*a*b^2*c^4*d^2*n^2 + 3*a^2*b*c 
^3*d^3*n^2 - a^3*c^2*d^4*n^2)*x^n), x) - 1/2*((b*c*d^2*(4*n - 1) - a*d^3*( 
2*n - 1))*x*x^n + (b*c^2*d*(5*n - 1) - a*c*d^2*(3*n - 1))*x)/(b^2*c^6*n^2 
- 2*a*b*c^5*d*n^2 + a^2*c^4*d^2*n^2 + (b^2*c^4*d^2*n^2 - 2*a*b*c^3*d^3*n^2 
 + a^2*c^2*d^4*n^2)*x^(2*n) + 2*(b^2*c^5*d*n^2 - 2*a*b*c^4*d^2*n^2 + a^2*c 
^3*d^3*n^2)*x^n)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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