\(\int (1-x^2)^2 (1+b x^4)^p \, dx\) [488]

Optimal result

Integrand size = 19, antiderivative size = 79 \[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=\frac {x \left (1+b x^4\right )^{1+p}}{b (5+4 p)}+\left (1-\frac {1}{5 b+4 b p}\right ) x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-\frac {2}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.82 \[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-\frac {2}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )+\frac {1}{5} x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-b x^4\right ) \] Input:

Integrate[(1 - x^2)^2*(1 + b*x^4)^p,x]
 

Output:

x*Hypergeometric2F1[1/4, -p, 5/4, -(b*x^4)] - (2*x^3*Hypergeometric2F1[3/4 
, -p, 7/4, -(b*x^4)])/3 + (x^5*Hypergeometric2F1[5/4, -p, 9/4, -(b*x^4)])/ 
5
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {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.

\(\displaystyle \int \left (1-x^2\right )^2 \left (b x^4+1\right )^p \, dx\)

\(\Big \downarrow \) 1519

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1516

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

\(\Big \downarrow \) 2009

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

Input:

Int[(1 - x^2)^2*(1 + b*x^4)^p,x]
 

Output:

(x*(1 + b*x^4)^(1 + p))/(b*(5 + 4*p)) - ((1 - b*(5 + 4*p))*x*Hypergeometri 
c2F1[1/4, -p, 5/4, -(b*x^4)] + (2*b*(5 + 4*p)*x^3*Hypergeometric2F1[3/4, - 
p, 7/4, -(b*x^4)])/3)/(b*(5 + 4*p))
 

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 2009

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

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.71

method	result	size
meijerg	\(\frac {x^{5} \operatorname {hypergeom}\left (\left [\frac {5}{4}, -p \right ], \left [\frac {9}{4}\right ], -b \,x^{4}\right )}{5}-\frac {2 x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, -p \right ], \left [\frac {7}{4}\right ], -b \,x^{4}\right )}{3}+x \operatorname {hypergeom}\left (\left [\frac {1}{4}, -p \right ], \left [\frac {5}{4}\right ], -b \,x^{4}\right )\)	\(56\)

Input:

int((-x^2+1)^2*(b*x^4+1)^p,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

\[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=\int { {\left (x^{2} - 1\right )}^{2} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:

integrate((-x^2+1)^2*(b*x^4+1)^p,x, algorithm="fricas")
 

Output:

integral((x^4 - 2*x^2 + 1)*(b*x^4 + 1)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 28.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.19 \[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=\frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} - \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((-x**2+1)**2*(b*x**4+1)**p,x)
 

Output:

x**5*gamma(5/4)*hyper((5/4, -p), (9/4,), b*x**4*exp_polar(I*pi))/(4*gamma( 
9/4)) - x**3*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**4*exp_polar(I*pi))/( 
2*gamma(7/4)) + x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**4*exp_polar(I*p 
i))/(4*gamma(5/4))
 

Maxima [F]

\[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=\int { {\left (x^{2} - 1\right )}^{2} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:

integrate((-x^2+1)^2*(b*x^4+1)^p,x, algorithm="maxima")
 

Output:

integrate((x^2 - 1)^2*(b*x^4 + 1)^p, x)
 

Giac [F]

\[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=\int { {\left (x^{2} - 1\right )}^{2} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:

integrate((-x^2+1)^2*(b*x^4+1)^p,x, algorithm="giac")
 

Output:

integrate((x^2 - 1)^2*(b*x^4 + 1)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=\int {\left (x^2-1\right )}^2\,{\left (b\,x^4+1\right )}^p \,d x \] Input:

int((x^2 - 1)^2*(b*x^4 + 1)^p,x)
 

Output:

int((x^2 - 1)^2*(b*x^4 + 1)^p, x)
 

Reduce [F]

\[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx =\text {Too large to display} \] Input:

int((-x^2+1)^2*(b*x^4+1)^p,x)
 

Output:

(16*(b*x**4 + 1)**p*b*p**2*x**5 - 32*(b*x**4 + 1)**p*b*p**2*x**3 + 16*(b*x 
**4 + 1)**p*b*p**2*x + 16*(b*x**4 + 1)**p*b*p*x**5 - 48*(b*x**4 + 1)**p*b* 
p*x**3 + 32*(b*x**4 + 1)**p*b*p*x + 3*(b*x**4 + 1)**p*b*x**5 - 10*(b*x**4 
+ 1)**p*b*x**3 + 15*(b*x**4 + 1)**p*b*x + 16*(b*x**4 + 1)**p*p**2*x + 12*( 
b*x**4 + 1)**p*p*x + 4096*int((b*x**4 + 1)**p/(64*b*p**3*x**4 + 144*b*p**2 
*x**4 + 92*b*p*x**4 + 15*b*x**4 + 64*p**3 + 144*p**2 + 92*p + 15),x)*b*p** 
6 + 17408*int((b*x**4 + 1)**p/(64*b*p**3*x**4 + 144*b*p**2*x**4 + 92*b*p*x 
**4 + 15*b*x**4 + 64*p**3 + 144*p**2 + 92*p + 15),x)*b*p**5 + 28160*int((b 
*x**4 + 1)**p/(64*b*p**3*x**4 + 144*b*p**2*x**4 + 92*b*p*x**4 + 15*b*x**4 
+ 64*p**3 + 144*p**2 + 92*p + 15),x)*b*p**4 + 21376*int((b*x**4 + 1)**p/(6 
4*b*p**3*x**4 + 144*b*p**2*x**4 + 92*b*p*x**4 + 15*b*x**4 + 64*p**3 + 144* 
p**2 + 92*p + 15),x)*b*p**3 + 7440*int((b*x**4 + 1)**p/(64*b*p**3*x**4 + 1 
44*b*p**2*x**4 + 92*b*p*x**4 + 15*b*x**4 + 64*p**3 + 144*p**2 + 92*p + 15) 
,x)*b*p**2 + 900*int((b*x**4 + 1)**p/(64*b*p**3*x**4 + 144*b*p**2*x**4 + 9 
2*b*p*x**4 + 15*b*x**4 + 64*p**3 + 144*p**2 + 92*p + 15),x)*b*p - 1024*int 
((b*x**4 + 1)**p/(64*b*p**3*x**4 + 144*b*p**2*x**4 + 92*b*p*x**4 + 15*b*x* 
*4 + 64*p**3 + 144*p**2 + 92*p + 15),x)*p**5 - 3072*int((b*x**4 + 1)**p/(6 
4*b*p**3*x**4 + 144*b*p**2*x**4 + 92*b*p*x**4 + 15*b*x**4 + 64*p**3 + 144* 
p**2 + 92*p + 15),x)*p**4 - 3200*int((b*x**4 + 1)**p/(64*b*p**3*x**4 + 144 
*b*p**2*x**4 + 92*b*p*x**4 + 15*b*x**4 + 64*p**3 + 144*p**2 + 92*p + 15...