\(\int \frac {(c+d x^n)^{1-\frac {1}{n}}}{(a+b x^n)^3} \, dx\) [141]

Optimal result

Integrand size = 25, antiderivative size = 130 \[ \int \frac {\left (c+d x^n\right )^{1-\frac {1}{n}}}{\left (a+b x^n\right )^3} \, dx=\frac {b x \left (c+d x^n\right )^{2-\frac {1}{n}}}{2 a (b c-a d) n \left (a+b x^n\right )^2}-\frac {c \left (2 a d-b c \left (2-\frac {1}{n}\right )\right ) x \left (c+d x^n\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {(b c-a d) x^n}{a \left (c+d x^n\right )}\right )}{2 a^3 (b c-a d)} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 8.83 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.56 \[ \int \frac {\left (c+d x^n\right )^{1-\frac {1}{n}}}{\left (a+b x^n\right )^3} \, dx=-\frac {x \left (c+d x^n\right )^{2-\frac {1}{n}} \left (\left (2 a n+b (-1+2 n) x^n\right ) \Phi \left (\frac {(-b c+a d) x^n}{a \left (c+d x^n\right )},1,-1+\frac {1}{n}\right )-b x^n \Phi \left (\frac {(-b c+a d) x^n}{a \left (c+d x^n\right )},1,1+\frac {1}{n}\right )-2 \left (a n+b (-1+n) x^n\right ) \Phi \left (\frac {(-b c+a d) x^n}{a \left (c+d x^n\right )},1,\frac {1}{n}\right )\right )}{2 n \left (a+b x^n\right )^2 \left (-a \left (c+d x^n\right ) \left (a n+b (-1+n) x^n\right ) \Phi \left (\frac {(-b c+a d) x^n}{a \left (c+d x^n\right )},1,-1+\frac {1}{n}\right )+x^n \left (-b (b c-a d) x^n \Phi \left (\frac {(-b c+a d) x^n}{a \left (c+d x^n\right )},1,1+\frac {1}{n}\right )+\left (a^2 d n-b^2 c (-1+n) x^n+a b \left (-c (1+n)+d (-2+n) x^n\right )\right ) \Phi \left (\frac {(-b c+a d) x^n}{a \left (c+d x^n\right )},1,\frac {1}{n}\right )\right )\right )} \] Input:

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

Output:

-1/2*(x*(c + d*x^n)^(2 - n^(-1))*((2*a*n + b*(-1 + 2*n)*x^n)*HurwitzLerchP 
hi[((-(b*c) + a*d)*x^n)/(a*(c + d*x^n)), 1, -1 + n^(-1)] - b*x^n*HurwitzLe 
rchPhi[((-(b*c) + a*d)*x^n)/(a*(c + d*x^n)), 1, 1 + n^(-1)] - 2*(a*n + b*( 
-1 + n)*x^n)*HurwitzLerchPhi[((-(b*c) + a*d)*x^n)/(a*(c + d*x^n)), 1, n^(- 
1)]))/(n*(a + b*x^n)^2*(-(a*(c + d*x^n)*(a*n + b*(-1 + n)*x^n)*HurwitzLerc 
hPhi[((-(b*c) + a*d)*x^n)/(a*(c + d*x^n)), 1, -1 + n^(-1)]) + x^n*(-(b*(b* 
c - a*d)*x^n*HurwitzLerchPhi[((-(b*c) + a*d)*x^n)/(a*(c + d*x^n)), 1, 1 + 
n^(-1)]) + (a^2*d*n - b^2*c*(-1 + n)*x^n + a*b*(-(c*(1 + n)) + d*(-2 + n)* 
x^n))*HurwitzLerchPhi[((-(b*c) + a*d)*x^n)/(a*(c + d*x^n)), 1, n^(-1)])))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {907, 904}

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.

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

\(\Big \downarrow \) 907

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

\(\Big \downarrow \) 904

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 904

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*(x/(c^(p + 1)*(c + d*x^n)^(1/n)))*Hypergeometric2F1[1/n, -p, 1 
+ 1/n, (-(b*c - a*d))*(x^n/(a*(c + d*x^n)))], x] /; FreeQ[{a, b, c, d, n, q 
}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && ILtQ[p, 0]
 

rule 907

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[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d)) 
  Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q} 
, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  ! 
LtQ[q, -1]) && NeQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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