\(\int x (c+d x^4)^2 (a+b x^8)^p \, dx\) [14]

Optimal result

Integrand size = 20, antiderivative size = 159 \[ \int x \left (c+d x^4\right )^2 \left (a+b x^8\right )^p \, dx=\frac {d^2 x^2 \left (a+b x^8\right )^{1+p}}{2 b (5+4 p)}-\frac {\left (a d^2-b c^2 (5+4 p)\right ) x^2 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )}{2 b (5+4 p)}+\frac {1}{3} c d x^6 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^8}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.68 \[ \int x \left (c+d x^4\right )^2 \left (a+b x^8\right )^p \, dx=\frac {1}{30} x^2 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (15 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )+d x^4 \left (10 c \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^8}{a}\right )+3 d x^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-\frac {b x^8}{a}\right )\right )\right ) \] Input:

Integrate[x*(c + d*x^4)^2*(a + b*x^8)^p,x]
 

Output:

(x^2*(a + b*x^8)^p*(15*c^2*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^8)/a)] + 
 d*x^4*(10*c*Hypergeometric2F1[3/4, -p, 7/4, -((b*x^8)/a)] + 3*d*x^4*Hyper 
geometric2F1[5/4, -p, 9/4, -((b*x^8)/a)])))/(30*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1815, 1519, 25, 1516, 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.

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

\(\Big \downarrow \) 1815

\(\displaystyle \frac {1}{2} \int \left (d x^4+c\right )^2 \left (b x^8+a\right )^pdx^2\)

\(\Big \downarrow \) 1519

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

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {d^2 x^2 \left (a+b x^8\right )^{p+1}}{b (4 p+5)}-\frac {\int \left (-2 b c d (4 p+5) x^4+a d^2-b c^2 (4 p+5)\right ) \left (b x^8+a\right )^pdx^2}{b (4 p+5)}\right )\)

\(\Big \downarrow \) 1516

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {d^2 x^2 \left (a+b x^8\right )^{p+1}}{b (4 p+5)}-\frac {x^2 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \left (a d^2-b c^2 (4 p+5)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )-\frac {2}{3} b c d (4 p+5) x^6 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^8}{a}\right )}{b (4 p+5)}\right )\)

Input:

Int[x*(c + d*x^4)^2*(a + b*x^8)^p,x]
 

Output:

((d^2*x^2*(a + b*x^8)^(1 + p))/(b*(5 + 4*p)) - (((a*d^2 - b*c^2*(5 + 4*p)) 
*x^2*(a + b*x^8)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^8)/a)])/(1 + (b* 
x^8)/a)^p - (2*b*c*d*(5 + 4*p)*x^6*(a + b*x^8)^p*Hypergeometric2F1[3/4, -p 
, 7/4, -((b*x^8)/a)])/(3*(1 + (b*x^8)/a)^p))/(b*(5 + 4*p)))/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1516

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[Expa 
ndIntegrand[(d + e*x^2)*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e}, x] && 
NeQ[c*d^2 + a*e^2, 0]
 

rule 1519

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Sim 
p[e^q*x^(2*q - 3)*((a + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Simp[1/(c 
*(4*p + 2*q + 1))   Int[(a + c*x^4)^p*ExpandToSum[c*(4*p + 2*q + 1)*(d + e* 
x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x 
], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[q, 1]
 

rule 1815

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_ 
.), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/ 
k - 1)*(d + e*x^(n/k))^q*(a + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] 
/; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IntegerQ[m 
]
 

rule 2009

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

Maple [F]

\[\int x \left (x^{4} d +c \right )^{2} \left (b \,x^{8}+a \right )^{p}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int x \left (c+d x^4\right )^2 \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{4} + c\right )}^{2} {\left (b x^{8} + a\right )}^{p} x \,d x } \] Input:

integrate(x*(d*x^4+c)^2*(b*x^8+a)^p,x, algorithm="fricas")
 

Output:

integral((d^2*x^9 + 2*c*d*x^5 + c^2*x)*(b*x^8 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 178.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.76 \[ \int x \left (c+d x^4\right )^2 \left (a+b x^8\right )^p \, dx=\frac {a^{p} c^{2} x^{2} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} c d x^{6} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {a^{p} d^{2} x^{10} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {9}{4}\right )} \] Input:

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

Output:

a**p*c**2*x**2*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**8*exp_polar(I*pi)/ 
a)/(8*gamma(5/4)) + a**p*c*d*x**6*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x* 
*8*exp_polar(I*pi)/a)/(4*gamma(7/4)) + a**p*d**2*x**10*gamma(5/4)*hyper((5 
/4, -p), (9/4,), b*x**8*exp_polar(I*pi)/a)/(8*gamma(9/4))
 

Maxima [F]

\[ \int x \left (c+d x^4\right )^2 \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{4} + c\right )}^{2} {\left (b x^{8} + a\right )}^{p} x \,d x } \] Input:

integrate(x*(d*x^4+c)^2*(b*x^8+a)^p,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int x \left (c+d x^4\right )^2 \left (a+b x^8\right )^p \, dx=\int { {\left (d x^{4} + c\right )}^{2} {\left (b x^{8} + a\right )}^{p} x \,d x } \] Input:

integrate(x*(d*x^4+c)^2*(b*x^8+a)^p,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x \left (c+d x^4\right )^2 \left (a+b x^8\right )^p \, dx=\int x\,{\left (b\,x^8+a\right )}^p\,{\left (d\,x^4+c\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x \left (c+d x^4\right )^2 \left (a+b x^8\right )^p \, dx =\text {Too large to display} \] Input:

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

Output:

(16*(a + b*x**8)**p*a*d**2*p**2*x**2 + 12*(a + b*x**8)**p*a*d**2*p*x**2 + 
16*(a + b*x**8)**p*b*c**2*p**2*x**2 + 32*(a + b*x**8)**p*b*c**2*p*x**2 + 1 
5*(a + b*x**8)**p*b*c**2*x**2 + 32*(a + b*x**8)**p*b*c*d*p**2*x**6 + 48*(a 
 + b*x**8)**p*b*c*d*p*x**6 + 10*(a + b*x**8)**p*b*c*d*x**6 + 16*(a + b*x** 
8)**p*b*d**2*p**2*x**10 + 16*(a + b*x**8)**p*b*d**2*p*x**10 + 3*(a + b*x** 
8)**p*b*d**2*x**10 + 16384*int(((a + b*x**8)**p*x**5)/(64*a*p**3 + 144*a*p 
**2 + 92*a*p + 15*a + 64*b*p**3*x**8 + 144*b*p**2*x**8 + 92*b*p*x**8 + 15* 
b*x**8),x)*a*b*c*d*p**6 + 61440*int(((a + b*x**8)**p*x**5)/(64*a*p**3 + 14 
4*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**8 + 144*b*p**2*x**8 + 92*b*p*x**8 
+ 15*b*x**8),x)*a*b*c*d*p**5 + 83968*int(((a + b*x**8)**p*x**5)/(64*a*p**3 
 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**8 + 144*b*p**2*x**8 + 92*b*p* 
x**8 + 15*b*x**8),x)*a*b*c*d*p**4 + 50688*int(((a + b*x**8)**p*x**5)/(64*a 
*p**3 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**8 + 144*b*p**2*x**8 + 92 
*b*p*x**8 + 15*b*x**8),x)*a*b*c*d*p**3 + 13120*int(((a + b*x**8)**p*x**5)/ 
(64*a*p**3 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**8 + 144*b*p**2*x**8 
 + 92*b*p*x**8 + 15*b*x**8),x)*a*b*c*d*p**2 + 1200*int(((a + b*x**8)**p*x* 
*5)/(64*a*p**3 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**8 + 144*b*p**2* 
x**8 + 92*b*p*x**8 + 15*b*x**8),x)*a*b*c*d*p - 2048*int(((a + b*x**8)**p*x 
)/(64*a*p**3 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**8 + 144*b*p**2*x* 
*8 + 92*b*p*x**8 + 15*b*x**8),x)*a**2*d**2*p**5 - 6144*int(((a + b*x**8...