\(\int \frac {(a+b x^4)^p}{(c+d x^2)^2} \, dx\) [486]

Optimal result

Integrand size = 19, antiderivative size = 189 \[ \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx=\frac {x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,2,\frac {5}{4},-\frac {b x^4}{a},\frac {d^2 x^4}{c^2}\right )}{c^2}-\frac {2 d x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},-p,2,\frac {7}{4},-\frac {b x^4}{a},\frac {d^2 x^4}{c^2}\right )}{3 c^3}+\frac {d^2 x^5 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},-p,2,\frac {9}{4},-\frac {b x^4}{a},\frac {d^2 x^4}{c^2}\right )}{5 c^4} \] Output:

x*(b*x^4+a)^p*AppellF1(1/4,2,-p,5/4,d^2*x^4/c^2,-b*x^4/a)/c^2/((1+b*x^4/a) 
^p)-2/3*d*x^3*(b*x^4+a)^p*AppellF1(3/4,2,-p,7/4,d^2*x^4/c^2,-b*x^4/a)/c^3/ 
((1+b*x^4/a)^p)+1/5*d^2*x^5*(b*x^4+a)^p*AppellF1(5/4,2,-p,9/4,d^2*x^4/c^2, 
-b*x^4/a)/c^4/((1+b*x^4/a)^p)
 

Mathematica [F]

\[ \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx=\int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx \] Input:

Integrate[(a + b*x^4)^p/(c + d*x^2)^2,x]
 

Output:

Integrate[(a + b*x^4)^p/(c + d*x^2)^2, x]
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 189, 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.

\(\displaystyle \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1569

\(\displaystyle \int \left (\frac {c^2 \left (a+b x^4\right )^p}{\left (c^2-d^2 x^4\right )^2}+\frac {d^2 x^4 \left (a+b x^4\right )^p}{\left (d^2 x^4-c^2\right )^2}-\frac {2 c d x^2 \left (a+b x^4\right )^p}{\left (c^2-d^2 x^4\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,2,\frac {5}{4},-\frac {b x^4}{a},\frac {d^2 x^4}{c^2}\right )}{c^2}+\frac {d^2 x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},-p,2,\frac {9}{4},-\frac {b x^4}{a},\frac {d^2 x^4}{c^2}\right )}{5 c^4}-\frac {2 d x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},-p,2,\frac {7}{4},-\frac {b x^4}{a},\frac {d^2 x^4}{c^2}\right )}{3 c^3}\)

Input:

Int[(a + b*x^4)^p/(c + d*x^2)^2,x]
 

Output:

(x*(a + b*x^4)^p*AppellF1[1/4, -p, 2, 5/4, -((b*x^4)/a), (d^2*x^4)/c^2])/( 
c^2*(1 + (b*x^4)/a)^p) - (2*d*x^3*(a + b*x^4)^p*AppellF1[3/4, -p, 2, 7/4, 
-((b*x^4)/a), (d^2*x^4)/c^2])/(3*c^3*(1 + (b*x^4)/a)^p) + (d^2*x^5*(a + b* 
x^4)^p*AppellF1[5/4, -p, 2, 9/4, -((b*x^4)/a), (d^2*x^4)/c^2])/(5*c^4*(1 + 
 (b*x^4)/a)^p)
 

Defintions of rubi rules used

rule 1569

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {\left (b \,x^{4}+a \right )^{p}}{\left (d \,x^{2}+c \right )^{2}}d x\]

Input:

int((b*x^4+a)^p/(d*x^2+c)^2,x)
 

Output:

int((b*x^4+a)^p/(d*x^2+c)^2,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{p}}{{\left (d x^{2} + c\right )}^{2}} \,d x } \] Input:

integrate((b*x^4+a)^p/(d*x^2+c)^2,x, algorithm="fricas")
 

Output:

integral((b*x^4 + a)^p/(d^2*x^4 + 2*c*d*x^2 + c^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((b*x**4+a)**p/(d*x**2+c)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{p}}{{\left (d x^{2} + c\right )}^{2}} \,d x } \] Input:

integrate((b*x^4+a)^p/(d*x^2+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^4 + a)^p/(d*x^2 + c)^2, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{p}}{{\left (d x^{2} + c\right )}^{2}} \,d x } \] Input:

integrate((b*x^4+a)^p/(d*x^2+c)^2,x, algorithm="giac")
 

Output:

integrate((b*x^4 + a)^p/(d*x^2 + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx=\int \frac {{\left (b\,x^4+a\right )}^p}{{\left (d\,x^2+c\right )}^2} \,d x \] Input:

int((a + b*x^4)^p/(c + d*x^2)^2,x)
 

Output:

int((a + b*x^4)^p/(c + d*x^2)^2, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^4\right )^p}{\left (c+d x^2\right )^2} \, dx=\int \frac {\left (b \,x^{4}+a \right )^{p}}{d^{2} x^{4}+2 c d \,x^{2}+c^{2}}d x \] Input:

int((b*x^4+a)^p/(d*x^2+c)^2,x)
 

Output:

int((a + b*x**4)**p/(c**2 + 2*c*d*x**2 + d**2*x**4),x)