Integrand size = 19, antiderivative size = 77 \[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx=x \operatorname {AppellF1}\left (\frac {1}{4},-p,2,\frac {5}{4},-b x^4,x^4\right )+\frac {2}{3} x^3 \operatorname {AppellF1}\left (\frac {3}{4},-p,2,\frac {7}{4},-b x^4,x^4\right )+\frac {1}{5} x^5 \operatorname {AppellF1}\left (\frac {5}{4},-p,2,\frac {9}{4},-b x^4,x^4\right ) \] Output:
x*AppellF1(1/4,2,-p,5/4,x^4,-b*x^4)+2/3*x^3*AppellF1(3/4,2,-p,7/4,x^4,-b*x ^4)+1/5*x^5*AppellF1(5/4,2,-p,9/4,x^4,-b*x^4)
\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx=\int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx \] Input:
Integrate[(1 + b*x^4)^p/(1 - x^2)^2,x]
Output:
Integrate[(1 + b*x^4)^p/(1 - x^2)^2, x]
Time = 0.45 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1569, 2009}
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.
\(\displaystyle \int \frac {\left (b x^4+1\right )^p}{\left (1-x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1569 |
\(\displaystyle \int \left (\frac {x^4 \left (b x^4+1\right )^p}{\left (x^4-1\right )^2}+\frac {\left (b x^4+1\right )^p}{\left (x^4-1\right )^2}+\frac {2 x^2 \left (b x^4+1\right )^p}{\left (x^4-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x \operatorname {AppellF1}\left (\frac {1}{4},2,-p,\frac {5}{4},x^4,-b x^4\right )+\frac {1}{5} x^5 \operatorname {AppellF1}\left (\frac {5}{4},2,-p,\frac {9}{4},x^4,-b x^4\right )+\frac {2}{3} x^3 \operatorname {AppellF1}\left (\frac {3}{4},2,-p,\frac {7}{4},x^4,-b x^4\right )\) |
Input:
Int[(1 + b*x^4)^p/(1 - x^2)^2,x]
Output:
x*AppellF1[1/4, 2, -p, 5/4, x^4, -(b*x^4)] + (2*x^3*AppellF1[3/4, 2, -p, 7 /4, x^4, -(b*x^4)])/3 + (x^5*AppellF1[5/4, 2, -p, 9/4, x^4, -(b*x^4)])/5
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int [ExpandIntegrand[(a + c*x^4)^p, (d/(d^2 - e^2*x^4) - e*(x^2/(d^2 - e^2*x^4) ))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] && ! IntegerQ[p] && ILtQ[q, 0]
\[\int \frac {\left (b \,x^{4}+1\right )^{p}}{\left (-x^{2}+1\right )^{2}}d x\]
Input:
int((b*x^4+1)^p/(-x^2+1)^2,x)
Output:
int((b*x^4+1)^p/(-x^2+1)^2,x)
\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx=\int { \frac {{\left (b x^{4} + 1\right )}^{p}}{{\left (x^{2} - 1\right )}^{2}} \,d x } \] Input:
integrate((b*x^4+1)^p/(-x^2+1)^2,x, algorithm="fricas")
Output:
integral((b*x^4 + 1)^p/(x^4 - 2*x^2 + 1), x)
Timed out. \[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate((b*x**4+1)**p/(-x**2+1)**2,x)
Output:
Timed out
\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx=\int { \frac {{\left (b x^{4} + 1\right )}^{p}}{{\left (x^{2} - 1\right )}^{2}} \,d x } \] Input:
integrate((b*x^4+1)^p/(-x^2+1)^2,x, algorithm="maxima")
Output:
integrate((b*x^4 + 1)^p/(x^2 - 1)^2, x)
\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx=\int { \frac {{\left (b x^{4} + 1\right )}^{p}}{{\left (x^{2} - 1\right )}^{2}} \,d x } \] Input:
integrate((b*x^4+1)^p/(-x^2+1)^2,x, algorithm="giac")
Output:
integrate((b*x^4 + 1)^p/(x^2 - 1)^2, x)
Timed out. \[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx=\int \frac {{\left (b\,x^4+1\right )}^p}{{\left (x^2-1\right )}^2} \,d x \] Input:
int((b*x^4 + 1)^p/(x^2 - 1)^2,x)
Output:
int((b*x^4 + 1)^p/(x^2 - 1)^2, x)
\[ \int \frac {\left (1+b x^4\right )^p}{\left (1-x^2\right )^2} \, dx=\int \frac {\left (b \,x^{4}+1\right )^{p}}{x^{4}-2 x^{2}+1}d x \] Input:
int((b*x^4+1)^p/(-x^2+1)^2,x)
Output:
int((b*x**4 + 1)**p/(x**4 - 2*x**2 + 1),x)