\(\int (c+d x^2)^2 (a+b x^8)^p \, dx\) [310]

Optimal result

Integrand size = 19, antiderivative size = 151 \[ \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx=c^2 x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {2}{3} c d x^3 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+\frac {1}{5} d^2 x^5 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},-p,\frac {13}{8},-\frac {b x^8}{a}\right ) \] Output:

c^2*x*(b*x^8+a)^p*hypergeom([1/8, -p],[9/8],-b*x^8/a)/((1+b*x^8/a)^p)+2/3* 
c*d*x^3*(b*x^8+a)^p*hypergeom([3/8, -p],[11/8],-b*x^8/a)/((1+b*x^8/a)^p)+1 
/5*d^2*x^5*(b*x^8+a)^p*hypergeom([5/8, -p],[13/8],-b*x^8/a)/((1+b*x^8/a)^p 
)
 

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.70 \[ \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx=\frac {1}{15} x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (15 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+d x^2 \left (10 c \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+3 d x^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{8},-p,\frac {13}{8},-\frac {b x^8}{a}\right )\right )\right ) \] Input:

Integrate[(c + d*x^2)^2*(a + b*x^8)^p,x]
 

Output:

(x*(a + b*x^8)^p*(15*c^2*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)] + d 
*x^2*(10*c*Hypergeometric2F1[3/8, -p, 11/8, -((b*x^8)/a)] + 3*d*x^2*Hyperg 
eometric2F1[5/8, -p, 13/8, -((b*x^8)/a)])))/(15*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 151, 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 = {2424, 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.

\(\displaystyle \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx\)

\(\Big \downarrow \) 2424

\(\displaystyle \int \left (\left (c^2+d^2 x^4\right ) \left (a+b x^8\right )^p+2 c d x^2 \left (a+b x^8\right )^p\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle c^2 x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {2}{3} c d x^3 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+\frac {1}{5} d^2 x^5 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},-p,\frac {13}{8},-\frac {b x^8}{a}\right )\)

Input:

Int[(c + d*x^2)^2*(a + b*x^8)^p,x]
 

Output:

(c^2*x*(a + b*x^8)^p*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)])/(1 + ( 
b*x^8)/a)^p + (2*c*d*x^3*(a + b*x^8)^p*Hypergeometric2F1[3/8, -p, 11/8, -( 
(b*x^8)/a)])/(3*(1 + (b*x^8)/a)^p) + (d^2*x^5*(a + b*x^8)^p*Hypergeometric 
2F1[5/8, -p, 13/8, -((b*x^8)/a)])/(5*(1 + (b*x^8)/a)^p)
 

Defintions of rubi rules used

rule 2009

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

rule 2424

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, 
 x], j, k}, Int[Sum[x^j*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2 
*((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, p}, 
 x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]
 

Maple [F]

\[\int \left (d \,x^{2}+c \right )^{2} \left (b \,x^{8}+a \right )^{p}d x\]

Input:

int((d*x^2+c)^2*(b*x^8+a)^p,x)
 

Output:

int((d*x^2+c)^2*(b*x^8+a)^p,x)
 

Fricas [F]

\[ \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{2} + c\right )}^{2} {\left (b x^{8} + a\right )}^{p} \,d x } \] Input:

integrate((d*x^2+c)^2*(b*x^8+a)^p,x, algorithm="fricas")
 

Output:

integral((d^2*x^4 + 2*c*d*x^2 + c^2)*(b*x^8 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 178.63 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.79 \[ \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx=\frac {a^{p} c^{2} x \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{8}, - p \\ \frac {9}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {9}{8}\right )} + \frac {a^{p} c d x^{3} \Gamma \left (\frac {3}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{8}, - p \\ \frac {11}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{8}\right )} + \frac {a^{p} d^{2} x^{5} \Gamma \left (\frac {5}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{8}, - p \\ \frac {13}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {13}{8}\right )} \] Input:

integrate((d*x**2+c)**2*(b*x**8+a)**p,x)
 

Output:

a**p*c**2*x*gamma(1/8)*hyper((1/8, -p), (9/8,), b*x**8*exp_polar(I*pi)/a)/ 
(8*gamma(9/8)) + a**p*c*d*x**3*gamma(3/8)*hyper((3/8, -p), (11/8,), b*x**8 
*exp_polar(I*pi)/a)/(4*gamma(11/8)) + a**p*d**2*x**5*gamma(5/8)*hyper((5/8 
, -p), (13/8,), b*x**8*exp_polar(I*pi)/a)/(8*gamma(13/8))
 

Maxima [F]

\[ \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{2} + c\right )}^{2} {\left (b x^{8} + a\right )}^{p} \,d x } \] Input:

integrate((d*x^2+c)^2*(b*x^8+a)^p,x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^2*(b*x^8 + a)^p, x)
 

Giac [F]

\[ \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{2} + c\right )}^{2} {\left (b x^{8} + a\right )}^{p} \,d x } \] Input:

integrate((d*x^2+c)^2*(b*x^8+a)^p,x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)^2*(b*x^8 + a)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx=\int {\left (b\,x^8+a\right )}^p\,{\left (d\,x^2+c\right )}^2 \,d x \] Input:

int((a + b*x^8)^p*(c + d*x^2)^2,x)
 

Output:

int((a + b*x^8)^p*(c + d*x^2)^2, x)
 

Reduce [F]

\[ \int \left (c+d x^2\right )^2 \left (a+b x^8\right )^p \, dx=\text {too large to display} \] Input:

int((d*x^2+c)^2*(b*x^8+a)^p,x)
 

Output:

(64*(a + b*x**8)**p*c**2*p**2*x + 64*(a + b*x**8)**p*c**2*p*x + 15*(a + b* 
x**8)**p*c**2*x + 128*(a + b*x**8)**p*c*d*p**2*x**3 + 96*(a + b*x**8)**p*c 
*d*p*x**3 + 10*(a + b*x**8)**p*c*d*x**3 + 64*(a + b*x**8)**p*d**2*p**2*x** 
5 + 32*(a + b*x**8)**p*d**2*p*x**5 + 3*(a + b*x**8)**p*d**2*x**5 + 262144* 
int((a + b*x**8)**p/(512*a*p**3 + 576*a*p**2 + 184*a*p + 15*a + 512*b*p**3 
*x**8 + 576*b*p**2*x**8 + 184*b*p*x**8 + 15*b*x**8),x)*a*c**2*p**6 + 55705 
6*int((a + b*x**8)**p/(512*a*p**3 + 576*a*p**2 + 184*a*p + 15*a + 512*b*p* 
*3*x**8 + 576*b*p**2*x**8 + 184*b*p*x**8 + 15*b*x**8),x)*a*c**2*p**5 + 450 
560*int((a + b*x**8)**p/(512*a*p**3 + 576*a*p**2 + 184*a*p + 15*a + 512*b* 
p**3*x**8 + 576*b*p**2*x**8 + 184*b*p*x**8 + 15*b*x**8),x)*a*c**2*p**4 + 1 
71008*int((a + b*x**8)**p/(512*a*p**3 + 576*a*p**2 + 184*a*p + 15*a + 512* 
b*p**3*x**8 + 576*b*p**2*x**8 + 184*b*p*x**8 + 15*b*x**8),x)*a*c**2*p**3 + 
 29760*int((a + b*x**8)**p/(512*a*p**3 + 576*a*p**2 + 184*a*p + 15*a + 512 
*b*p**3*x**8 + 576*b*p**2*x**8 + 184*b*p*x**8 + 15*b*x**8),x)*a*c**2*p**2 
+ 1800*int((a + b*x**8)**p/(512*a*p**3 + 576*a*p**2 + 184*a*p + 15*a + 512 
*b*p**3*x**8 + 576*b*p**2*x**8 + 184*b*p*x**8 + 15*b*x**8),x)*a*c**2*p + 2 
62144*int(((a + b*x**8)**p*x**4)/(512*a*p**3 + 576*a*p**2 + 184*a*p + 15*a 
 + 512*b*p**3*x**8 + 576*b*p**2*x**8 + 184*b*p*x**8 + 15*b*x**8),x)*a*d**2 
*p**6 + 425984*int(((a + b*x**8)**p*x**4)/(512*a*p**3 + 576*a*p**2 + 184*a 
*p + 15*a + 512*b*p**3*x**8 + 576*b*p**2*x**8 + 184*b*p*x**8 + 15*b*x**...