Integrand size = 29, antiderivative size = 273 \[ \int \frac {x^{-1-n (-3+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx=\frac {x^{-n p} \left (a+b x^n\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )}{d^3 n p}-\frac {a x^{n (1-p)} \left (a+b x^n\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )}{c d^2 n (1-p)}-\frac {a^2 x^{n (2-p)} \left (a+b x^n\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (3,2-p,3-p,\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )}{c^2 d n (2-p)}-\frac {x^{-n p} \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {b x^n}{a}\right )}{d^3 n p} \] Output:
(a+b*x^n)^p*hypergeom([1, -p],[1-p],(-a*d+b*c)*x^n/c/(a+b*x^n))/d^3/n/p/(x ^(n*p))-a*x^(n*(1-p))*(a+b*x^n)^(-1+p)*hypergeom([2, 1-p],[-p+2],(-a*d+b*c )*x^n/c/(a+b*x^n))/c/d^2/n/(1-p)-a^2*x^(n*(-p+2))*(a+b*x^n)^(-2+p)*hyperge om([3, -p+2],[3-p],(-a*d+b*c)*x^n/c/(a+b*x^n))/c^2/d/n/(-p+2)-(a+b*x^n)^p* hypergeom([-p, -p],[1-p],-b*x^n/a)/d^3/n/p/(x^(n*p))/((1+b*x^n/a)^p)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.29 \[ \int \frac {x^{-1-n (-3+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx=-\frac {x^{-n (-3+p)} \left (a+b x^n\right )^p \left (\frac {a+b x^n}{a}\right )^{-p} \operatorname {AppellF1}\left (3-p,-p,3,4-p,-\frac {b x^n}{a},-\frac {d x^n}{c}\right )}{c^3 n (-3+p)} \] Input:
Integrate[(x^(-1 - n*(-3 + p))*(a + b*x^n)^p)/(c + d*x^n)^3,x]
Output:
-(((a + b*x^n)^p*AppellF1[3 - p, -p, 3, 4 - p, -((b*x^n)/a), -((d*x^n)/c)] )/(c^3*n*(-3 + p)*x^(n*(-3 + p))*((a + b*x^n)/a)^p))
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1013, 1012}
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.
\(\displaystyle \int \frac {x^{-n (p-3)-1} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \int \frac {x^{n (3-p)-1} \left (\frac {b x^n}{a}+1\right )^p}{\left (d x^n+c\right )^3}dx\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^{n (3-p)} \left (a+b x^n\right )^p \left (\frac {b x^n}{a}+1\right )^{-p} \operatorname {AppellF1}\left (3-p,-p,3,4-p,-\frac {b x^n}{a},-\frac {d x^n}{c}\right )}{c^3 n (3-p)}\) |
Input:
Int[(x^(-1 - n*(-3 + p))*(a + b*x^n)^p)/(c + d*x^n)^3,x]
Output:
(x^(n*(3 - p))*(a + b*x^n)^p*AppellF1[3 - p, -p, 3, 4 - p, -((b*x^n)/a), - ((d*x^n)/c)])/(c^3*n*(3 - p)*(1 + (b*x^n)/a)^p)
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((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[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {x^{-1-n \left (-3+p \right )} \left (a +b \,x^{n}\right )^{p}}{\left (c +d \,x^{n}\right )^{3}}d x\]
Input:
int(x^(-1-n*(-3+p))*(a+b*x^n)^p/(c+d*x^n)^3,x)
Output:
int(x^(-1-n*(-3+p))*(a+b*x^n)^p/(c+d*x^n)^3,x)
\[ \int \frac {x^{-1-n (-3+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{p} x^{-n {\left (p - 3\right )} - 1}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \] Input:
integrate(x^(-1-n*(-3+p))*(a+b*x^n)^p/(c+d*x^n)^3,x, algorithm="fricas")
Output:
integral((b*x^n + a)^p*x^(-n*p + 3*n - 1)/(d^3*x^(3*n) + 3*c*d^2*x^(2*n) + 3*c^2*d*x^n + c^3), x)
Timed out. \[ \int \frac {x^{-1-n (-3+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx=\text {Timed out} \] Input:
integrate(x**(-1-n*(-3+p))*(a+b*x**n)**p/(c+d*x**n)**3,x)
Output:
Timed out
\[ \int \frac {x^{-1-n (-3+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{p} x^{-n {\left (p - 3\right )} - 1}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \] Input:
integrate(x^(-1-n*(-3+p))*(a+b*x^n)^p/(c+d*x^n)^3,x, algorithm="maxima")
Output:
integrate((b*x^n + a)^p*x^(-n*(p - 3) - 1)/(d*x^n + c)^3, x)
\[ \int \frac {x^{-1-n (-3+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{p} x^{-n {\left (p - 3\right )} - 1}}{{\left (d x^{n} + c\right )}^{3}} \,d x } \] Input:
integrate(x^(-1-n*(-3+p))*(a+b*x^n)^p/(c+d*x^n)^3,x, algorithm="giac")
Output:
integrate((b*x^n + a)^p*x^(-n*(p - 3) - 1)/(d*x^n + c)^3, x)
Timed out. \[ \int \frac {x^{-1-n (-3+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx=\int \frac {{\left (a+b\,x^n\right )}^p}{x^{n\,\left (p-3\right )+1}\,{\left (c+d\,x^n\right )}^3} \,d x \] Input:
int((a + b*x^n)^p/(x^(n*(p - 3) + 1)*(c + d*x^n)^3),x)
Output:
int((a + b*x^n)^p/(x^(n*(p - 3) + 1)*(c + d*x^n)^3), x)
\[ \int \frac {x^{-1-n (-3+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^3} \, dx=\int \frac {x^{3 n} \left (x^{n} b +a \right )^{p}}{x^{n p +3 n} d^{3} x +3 x^{n p +2 n} c \,d^{2} x +3 x^{n p +n} c^{2} d x +x^{n p} c^{3} x}d x \] Input:
int(x^(-1-n*(-3+p))*(a+b*x^n)^p/(c+d*x^n)^3,x)
Output:
int((x**(3*n)*(x**n*b + a)**p)/(x**(n*p + 3*n)*d**3*x + 3*x**(n*p + 2*n)*c *d**2*x + 3*x**(n*p + n)*c**2*d*x + x**(n*p)*c**3*x),x)