Integrand size = 19, antiderivative size = 123 \[ \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx=\frac {x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{8},-p,1,\frac {9}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c}-\frac {d x^5 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{8},-p,1,\frac {13}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{5 c^2} \] Output:
x*(b*x^8+a)^p*AppellF1(1/8,1,-p,9/8,d^2*x^8/c^2,-b*x^8/a)/c/((1+b*x^8/a)^p )-1/5*d*x^5*(b*x^8+a)^p*AppellF1(5/8,1,-p,13/8,d^2*x^8/c^2,-b*x^8/a)/c^2/( (1+b*x^8/a)^p)
\[ \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx=\int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx \] Input:
Integrate[(a + b*x^8)^p/(c + d*x^4),x]
Output:
Integrate[(a + b*x^8)^p/(c + d*x^4), x]
Time = 0.31 (sec) , antiderivative size = 123, 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 = {1768, 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.
\(\displaystyle \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx\) |
\(\Big \downarrow \) 1768 |
\(\displaystyle \int \left (\frac {c \left (a+b x^8\right )^p}{c^2-d^2 x^8}+\frac {d x^4 \left (a+b x^8\right )^p}{d^2 x^8-c^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{8},-p,1,\frac {9}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c}-\frac {d x^5 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{8},-p,1,\frac {13}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{5 c^2}\) |
Input:
Int[(a + b*x^8)^p/(c + d*x^4),x]
Output:
(x*(a + b*x^8)^p*AppellF1[1/8, -p, 1, 9/8, -((b*x^8)/a), (d^2*x^8)/c^2])/( c*(1 + (b*x^8)/a)^p) - (d*x^5*(a + b*x^8)^p*AppellF1[5/8, -p, 1, 13/8, -(( b*x^8)/a), (d^2*x^8)/c^2])/(5*c^2*(1 + (b*x^8)/a)^p)
Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2*n))^p, (d/(d^2 - e^2*x^(2*n)) - e*(x^n/ (d^2 - e^2*x^(2*n))))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[n 2, 2*n] && NeQ[c*d^2 + a*e^2, 0] && !IntegerQ[p] && ILtQ[q, 0]
\[\int \frac {\left (b \,x^{8}+a \right )^{p}}{x^{4} d +c}d x\]
Input:
int((b*x^8+a)^p/(d*x^4+c),x)
Output:
int((b*x^8+a)^p/(d*x^4+c),x)
\[ \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{d x^{4} + c} \,d x } \] Input:
integrate((b*x^8+a)^p/(d*x^4+c),x, algorithm="fricas")
Output:
integral((b*x^8 + a)^p/(d*x^4 + c), x)
Timed out. \[ \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx=\text {Timed out} \] Input:
integrate((b*x**8+a)**p/(d*x**4+c),x)
Output:
Timed out
\[ \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{d x^{4} + c} \,d x } \] Input:
integrate((b*x^8+a)^p/(d*x^4+c),x, algorithm="maxima")
Output:
integrate((b*x^8 + a)^p/(d*x^4 + c), x)
\[ \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{d x^{4} + c} \,d x } \] Input:
integrate((b*x^8+a)^p/(d*x^4+c),x, algorithm="giac")
Output:
integrate((b*x^8 + a)^p/(d*x^4 + c), x)
Timed out. \[ \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx=\int \frac {{\left (b\,x^8+a\right )}^p}{d\,x^4+c} \,d x \] Input:
int((a + b*x^8)^p/(c + d*x^4),x)
Output:
int((a + b*x^8)^p/(c + d*x^4), x)
\[ \int \frac {\left (a+b x^8\right )^p}{c+d x^4} \, dx=\int \frac {\left (b \,x^{8}+a \right )^{p}}{d \,x^{4}+c}d x \] Input:
int((b*x^8+a)^p/(d*x^4+c),x)
Output:
int((a + b*x**8)**p/(c + d*x**4),x)