\(\int \frac {(a+b x^4)^p}{(c+d x)^2} \, dx\) [222]

Optimal result

Integrand size = 17, antiderivative size = 452 \[ \int \frac {\left (a+b x^4\right )^p}{(c+d x)^2} \, dx=\frac {x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,2,\frac {5}{4},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{c^2}-\frac {d x^2 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{c^3}+\frac {d^2 x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},-p,2,\frac {7}{4},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{c^4}+\frac {3 d^4 x^5 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{4},-p,2,\frac {9}{4},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{5 c^6}-\frac {d^5 x^6 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,2,\frac {5}{2},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{3 c^7}+\frac {d^6 x^7 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {7}{4},-p,2,\frac {11}{4},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{7 c^8}-\frac {b c^3 d^3 \left (a+b x^4\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,\frac {d^4 \left (a+b x^4\right )}{b c^4+a d^4}\right )}{\left (b c^4+a d^4\right )^2 (1+p)} \] Output:

x*(b*x^4+a)^p*AppellF1(1/4,2,-p,5/4,d^4*x^4/c^4,-b*x^4/a)/c^2/((1+b*x^4/a) 
^p)-d*x^2*(b*x^4+a)^p*AppellF1(1/2,2,-p,3/2,d^4*x^4/c^4,-b*x^4/a)/c^3/((1+ 
b*x^4/a)^p)+d^2*x^3*(b*x^4+a)^p*AppellF1(3/4,2,-p,7/4,d^4*x^4/c^4,-b*x^4/a 
)/c^4/((1+b*x^4/a)^p)+3/5*d^4*x^5*(b*x^4+a)^p*AppellF1(5/4,2,-p,9/4,d^4*x^ 
4/c^4,-b*x^4/a)/c^6/((1+b*x^4/a)^p)-1/3*d^5*x^6*(b*x^4+a)^p*AppellF1(3/2,2 
,-p,5/2,d^4*x^4/c^4,-b*x^4/a)/c^7/((1+b*x^4/a)^p)+1/7*d^6*x^7*(b*x^4+a)^p* 
AppellF1(7/4,2,-p,11/4,d^4*x^4/c^4,-b*x^4/a)/c^8/((1+b*x^4/a)^p)-b*c^3*d^3 
*(b*x^4+a)^(p+1)*hypergeom([2, p+1],[2+p],d^4*(b*x^4+a)/(a*d^4+b*c^4))/(a* 
d^4+b*c^4)^2/(p+1)
 

Mathematica [F]

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

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

Output:

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

Rubi [F]

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.

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

\(\Big \downarrow \) 2584

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

\(\Big \downarrow \) 2255

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1577

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

\(\Big \downarrow \) 505

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

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {\left (c^2+d^2 x^2\right ) \left (b x^4+a\right )^p}{\left (c^2-d^2 x^2\right )^2}dx-c d \left (\frac {x^2 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{c^4}+\frac {d^4 x^6 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,2,\frac {5}{2},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{3 c^8}+\frac {b c^2 d^2 \left (a+b x^4\right )^{p+1} \operatorname {Hypergeometric2F1}\left (2,p+1,p+2,\frac {d^4 \left (b x^4+a\right )}{b c^4+a d^4}\right )}{(p+1) \left (a d^4+b c^4\right )^2}\right )\)

\(\Big \downarrow \) 2261

\(\displaystyle \int \frac {\left (c^2+d^2 x^2\right ) \left (b x^4+a\right )^p}{\left (c^2-d^2 x^2\right )^2}dx-c d \left (\frac {x^2 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{c^4}+\frac {d^4 x^6 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,2,\frac {5}{2},-\frac {b x^4}{a},\frac {d^4 x^4}{c^4}\right )}{3 c^8}+\frac {b c^2 d^2 \left (a+b x^4\right )^{p+1} \operatorname {Hypergeometric2F1}\left (2,p+1,p+2,\frac {d^4 \left (b x^4+a\right )}{b c^4+a d^4}\right )}{(p+1) \left (a d^4+b c^4\right )^2}\right )\)

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 505

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^( 
-n), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && PosQ[a/b]
 

rule 1577

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, c, d, e, p, q}, x]
 

rule 2009

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

rule 2255

Int[(Pr_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] 
 :> Module[{r = Expon[Pr, x], k}, Int[Sum[Coeff[Pr, x, 2*k]*x^(2*k), {k, 0, 
 r/2}]*(d + e*x^2)^q*(a + c*x^4)^p, x] + Int[x*Sum[Coeff[Pr, x, 2*k + 1]*x^ 
(2*k), {k, 0, (r - 1)/2}]*(d + e*x^2)^q*(a + c*x^4)^p, x]] /; FreeQ[{a, c, 
d, e, p, q}, x] && PolyQ[Pr, x] &&  !PolyQ[Pr, x^2]
 

rule 2261

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Unintegrable[Px*(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, 
p, q}, x] && PolyQ[Px, x]
 

rule 2584

Int[((c_) + (d_.)*(x_)^(n_.))^(q_)*((a_) + (b_.)*(x_)^(nn_.))^(p_), x_Symbo 
l] :> Int[ExpandToSum[(c - d*x^n)^(-q), x]*((a + b*x^nn)^p/(c^2 - d^2*x^(2* 
n))^(-q)), x] /; FreeQ[{a, b, c, d, n, nn, p}, x] &&  !IntegerQ[p] && ILtQ[ 
q, 0] && IGtQ[Log[2, nn/n], 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

((a + b*x**4)**p + 16*int((a + b*x**4)**p/(4*a*c**2*p - a*c**2 + 8*a*c*d*p 
*x - 2*a*c*d*x + 4*a*d**2*p*x**2 - a*d**2*x**2 + 4*b*c**2*p*x**4 - b*c**2* 
x**4 + 8*b*c*d*p*x**5 - 2*b*c*d*x**5 + 4*b*d**2*p*x**6 - b*d**2*x**6),x)*a 
*c*d*p**2 - 4*int((a + b*x**4)**p/(4*a*c**2*p - a*c**2 + 8*a*c*d*p*x - 2*a 
*c*d*x + 4*a*d**2*p*x**2 - a*d**2*x**2 + 4*b*c**2*p*x**4 - b*c**2*x**4 + 8 
*b*c*d*p*x**5 - 2*b*c*d*x**5 + 4*b*d**2*p*x**6 - b*d**2*x**6),x)*a*c*d*p + 
 16*int((a + b*x**4)**p/(4*a*c**2*p - a*c**2 + 8*a*c*d*p*x - 2*a*c*d*x + 4 
*a*d**2*p*x**2 - a*d**2*x**2 + 4*b*c**2*p*x**4 - b*c**2*x**4 + 8*b*c*d*p*x 
**5 - 2*b*c*d*x**5 + 4*b*d**2*p*x**6 - b*d**2*x**6),x)*a*d**2*p**2*x - 4*i 
nt((a + b*x**4)**p/(4*a*c**2*p - a*c**2 + 8*a*c*d*p*x - 2*a*c*d*x + 4*a*d* 
*2*p*x**2 - a*d**2*x**2 + 4*b*c**2*p*x**4 - b*c**2*x**4 + 8*b*c*d*p*x**5 - 
 2*b*c*d*x**5 + 4*b*d**2*p*x**6 - b*d**2*x**6),x)*a*d**2*p*x - 16*int(((a 
+ b*x**4)**p*x**3)/(4*a*c**2*p - a*c**2 + 8*a*c*d*p*x - 2*a*c*d*x + 4*a*d* 
*2*p*x**2 - a*d**2*x**2 + 4*b*c**2*p*x**4 - b*c**2*x**4 + 8*b*c*d*p*x**5 - 
 2*b*c*d*x**5 + 4*b*d**2*p*x**6 - b*d**2*x**6),x)*b*c**2*p**2 + 4*int(((a 
+ b*x**4)**p*x**3)/(4*a*c**2*p - a*c**2 + 8*a*c*d*p*x - 2*a*c*d*x + 4*a*d* 
*2*p*x**2 - a*d**2*x**2 + 4*b*c**2*p*x**4 - b*c**2*x**4 + 8*b*c*d*p*x**5 - 
 2*b*c*d*x**5 + 4*b*d**2*p*x**6 - b*d**2*x**6),x)*b*c**2*p - 16*int(((a + 
b*x**4)**p*x**3)/(4*a*c**2*p - a*c**2 + 8*a*c*d*p*x - 2*a*c*d*x + 4*a*d**2 
*p*x**2 - a*d**2*x**2 + 4*b*c**2*p*x**4 - b*c**2*x**4 + 8*b*c*d*p*x**5 ...