Integrand size = 31, antiderivative size = 76 \[ \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx=\frac {x \left (c+d x^{2 n}\right )^p \left (1+\frac {d x^{2 n}}{c}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2 n},1,-p,\frac {1}{2} \left (2+\frac {1}{n}\right ),\frac {b^2 x^{2 n}}{a^2},-\frac {d x^{2 n}}{c}\right )}{a^2} \] Output:
x*(c+d*x^(2*n))^p*AppellF1(1/2/n,1,-p,1+1/2/n,b^2*x^(2*n)/a^2,-d*x^(2*n)/c )/a^2/((1+d*x^(2*n)/c)^p)
Leaf count is larger than twice the leaf count of optimal. \(258\) vs. \(2(76)=152\).
Time = 0.52 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.39 \[ \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx=\frac {a^2 c (1+2 n) x \left (c+d x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1}{2 n},-p,1,1+\frac {1}{2 n},-\frac {d x^{2 n}}{c},\frac {b^2 x^{2 n}}{a^2}\right )}{\left (a^2-b^2 x^{2 n}\right ) \left (2 a^2 d n p x^{2 n} \operatorname {AppellF1}\left (1+\frac {1}{2 n},1-p,1,2+\frac {1}{2 n},-\frac {d x^{2 n}}{c},\frac {b^2 x^{2 n}}{a^2}\right )+2 b^2 c n x^{2 n} \operatorname {AppellF1}\left (1+\frac {1}{2 n},-p,2,2+\frac {1}{2 n},-\frac {d x^{2 n}}{c},\frac {b^2 x^{2 n}}{a^2}\right )+a^2 c (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2 n},-p,1,1+\frac {1}{2 n},-\frac {d x^{2 n}}{c},\frac {b^2 x^{2 n}}{a^2}\right )\right )} \] Input:
Integrate[(c + d*x^(2*n))^p/((a - b*x^n)*(a + b*x^n)),x]
Output:
(a^2*c*(1 + 2*n)*x*(c + d*x^(2*n))^p*AppellF1[1/(2*n), -p, 1, 1 + 1/(2*n), -((d*x^(2*n))/c), (b^2*x^(2*n))/a^2])/((a^2 - b^2*x^(2*n))*(2*a^2*d*n*p*x ^(2*n)*AppellF1[1 + 1/(2*n), 1 - p, 1, 2 + 1/(2*n), -((d*x^(2*n))/c), (b^2 *x^(2*n))/a^2] + 2*b^2*c*n*x^(2*n)*AppellF1[1 + 1/(2*n), -p, 2, 2 + 1/(2*n ), -((d*x^(2*n))/c), (b^2*x^(2*n))/a^2] + a^2*c*(1 + 2*n)*AppellF1[1/(2*n) , -p, 1, 1 + 1/(2*n), -((d*x^(2*n))/c), (b^2*x^(2*n))/a^2]))
Time = 0.45 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2036, 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.
\(\displaystyle \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx\) |
\(\Big \downarrow \) 2036 |
\(\displaystyle \int \frac {\left (c+d x^{2 n}\right )^p}{a^2-b^2 x^{2 n}}dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \left (c+d x^{2 n}\right )^p \left (\frac {d x^{2 n}}{c}+1\right )^{-p} \int \frac {\left (\frac {d x^{2 n}}{c}+1\right )^p}{a^2-b^2 x^{2 n}}dx\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \left (c+d x^{2 n}\right )^p \left (\frac {d x^{2 n}}{c}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2 n},1,-p,\frac {1}{2} \left (2+\frac {1}{n}\right ),\frac {b^2 x^{2 n}}{a^2},-\frac {d x^{2 n}}{c}\right )}{a^2}\) |
Input:
Int[(c + d*x^(2*n))^p/((a - b*x^n)*(a + b*x^n)),x]
Output:
(x*(c + d*x^(2*n))^p*AppellF1[1/(2*n), 1, -p, (2 + n^(-1))/2, (b^2*x^(2*n) )/a^2, -((d*x^(2*n))/c)])/(a^2*(1 + (d*x^(2*n))/c)^p)
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])
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])
Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p _.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.), x_Symbol] :> Int[u*(a1*a2 + b1*b2 *x^n)^p*(c + d*x^n)^q, x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x] && E qQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && Gt Q[a2, 0]))
\[\int \frac {\left (c +d \,x^{2 n}\right )^{p}}{\left (a -b \,x^{n}\right ) \left (a +b \,x^{n}\right )}d x\]
Input:
int((c+d*x^(2*n))^p/(a-b*x^n)/(a+b*x^n),x)
Output:
int((c+d*x^(2*n))^p/(a-b*x^n)/(a+b*x^n),x)
\[ \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx=\int { -\frac {{\left (d x^{2 \, n} + c\right )}^{p}}{{\left (b x^{n} + a\right )} {\left (b x^{n} - a\right )}} \,d x } \] Input:
integrate((c+d*x^(2*n))^p/(a-b*x^n)/(a+b*x^n),x, algorithm="fricas")
Output:
integral(-(d*x^(2*n) + c)^p/(b^2*x^(2*n) - a^2), x)
Exception generated. \[ \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:
integrate((c+d*x**(2*n))**p/(a-b*x**n)/(a+b*x**n),x)
Output:
Exception raised: HeuristicGCDFailed >> no luck
\[ \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx=\int { -\frac {{\left (d x^{2 \, n} + c\right )}^{p}}{{\left (b x^{n} + a\right )} {\left (b x^{n} - a\right )}} \,d x } \] Input:
integrate((c+d*x^(2*n))^p/(a-b*x^n)/(a+b*x^n),x, algorithm="maxima")
Output:
-integrate((d*x^(2*n) + c)^p/((b*x^n + a)*(b*x^n - a)), x)
\[ \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx=\int { -\frac {{\left (d x^{2 \, n} + c\right )}^{p}}{{\left (b x^{n} + a\right )} {\left (b x^{n} - a\right )}} \,d x } \] Input:
integrate((c+d*x^(2*n))^p/(a-b*x^n)/(a+b*x^n),x, algorithm="giac")
Output:
integrate(-(d*x^(2*n) + c)^p/((b*x^n + a)*(b*x^n - a)), x)
Timed out. \[ \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx=-\int -\frac {{\left (c+d\,x^{2\,n}\right )}^p}{a^2-b^2\,x^{2\,n}} \,d x \] Input:
int((c + d*x^(2*n))^p/((a + b*x^n)*(a - b*x^n)),x)
Output:
-int(-(c + d*x^(2*n))^p/(a^2 - b^2*x^(2*n)), x)
\[ \int \frac {\left (c+d x^{2 n}\right )^p}{\left (a-b x^n\right ) \left (a+b x^n\right )} \, dx=-\left (\int \frac {\left (c +d \,x^{2 n}\right )^{p}}{x^{2 n} b^{2}-a^{2}}d x \right ) \] Input:
int((c+d*x^(2*n))^p/(a-b*x^n)/(a+b*x^n),x)
Output:
- int((x**(2*n)*d + c)**p/(x**(2*n)*b**2 - a**2),x)