\(\int (1-x^2)^3 (1+b x^4)^p \, dx\) [487]

Optimal result

Integrand size = 19, antiderivative size = 118 \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=\frac {3 x \left (1+b x^4\right )^{1+p}}{b (5+4 p)}-\frac {x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+\left (1-\frac {3}{5 b+4 b p}\right ) x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-\left (1-\frac {1}{7 b+4 b p}\right ) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.73 \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )+\frac {3}{5} x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-b x^4\right )-\frac {1}{7} x^7 \operatorname {Hypergeometric2F1}\left (\frac {7}{4},-p,\frac {11}{4},-b x^4\right ) \] Input:

Integrate[(1 - x^2)^3*(1 + b*x^4)^p,x]
 

Output:

x*Hypergeometric2F1[1/4, -p, 5/4, -(b*x^4)] - x^3*Hypergeometric2F1[3/4, - 
p, 7/4, -(b*x^4)] + (3*x^5*Hypergeometric2F1[5/4, -p, 9/4, -(b*x^4)])/5 - 
(x^7*Hypergeometric2F1[7/4, -p, 11/4, -(b*x^4)])/7
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1519, 2432, 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.

\(\displaystyle \int \left (1-x^2\right )^3 \left (b x^4+1\right )^p \, dx\)

\(\Big \downarrow \) 1519

\(\displaystyle \frac {\int \left (b x^4+1\right )^p \left (3 b (4 p+7) x^4+3 (1-b (4 p+7)) x^2+b (4 p+7)\right )dx}{b (4 p+7)}-\frac {x^3 \left (b x^4+1\right )^{p+1}}{b (4 p+7)}\)

\(\Big \downarrow \) 2432

\(\displaystyle \frac {\int \left (3 b (4 p+7) x^4 \left (b x^4+1\right )^p+3 (1-b (4 p+7)) x^2 \left (b x^4+1\right )^p+b (4 p+7) \left (b x^4+1\right )^p\right )dx}{b (4 p+7)}-\frac {x^3 \left (b x^4+1\right )^{p+1}}{b (4 p+7)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b (4 p+7) x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )+\frac {3}{5} b (4 p+7) x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-b x^4\right )+x^3 (1-b (4 p+7)) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )}{b (4 p+7)}-\frac {x^3 \left (b x^4+1\right )^{p+1}}{b (4 p+7)}\)

Input:

Int[(1 - x^2)^3*(1 + b*x^4)^p,x]
 

Output:

-((x^3*(1 + b*x^4)^(1 + p))/(b*(7 + 4*p))) + (b*(7 + 4*p)*x*Hypergeometric 
2F1[1/4, -p, 5/4, -(b*x^4)] + (1 - b*(7 + 4*p))*x^3*Hypergeometric2F1[3/4, 
 -p, 7/4, -(b*x^4)] + (3*b*(7 + 4*p)*x^5*Hypergeometric2F1[5/4, -p, 9/4, - 
(b*x^4)])/5)/(b*(7 + 4*p))
 

Defintions of rubi rules used

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]
 

rule 2432

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ 
Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly 
Q[Pq, x^n])
 

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64

method	result	size
meijerg	\(-\frac {x^{7} \operatorname {hypergeom}\left (\left [\frac {7}{4}, -p \right ], \left [\frac {11}{4}\right ], -b \,x^{4}\right )}{7}+\frac {3 x^{5} \operatorname {hypergeom}\left (\left [\frac {5}{4}, -p \right ], \left [\frac {9}{4}\right ], -b \,x^{4}\right )}{5}-x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, -p \right ], \left [\frac {7}{4}\right ], -b \,x^{4}\right )+x \operatorname {hypergeom}\left (\left [\frac {1}{4}, -p \right ], \left [\frac {5}{4}\right ], -b \,x^{4}\right )\)	\(75\)

Input:

int((-x^2+1)^3*(b*x^4+1)^p,x,method=_RETURNVERBOSE)
 

Output:

-1/7*x^7*hypergeom([7/4,-p],[11/4],-b*x^4)+3/5*x^5*hypergeom([5/4,-p],[9/4 
],-b*x^4)-x^3*hypergeom([3/4,-p],[7/4],-b*x^4)+x*hypergeom([1/4,-p],[5/4], 
-b*x^4)
 

Fricas [F]

\[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )}^{3} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:

integrate((-x^2+1)^3*(b*x^4+1)^p,x, algorithm="fricas")
 

Output:

integral(-(x^6 - 3*x^4 + 3*x^2 - 1)*(b*x^4 + 1)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 51.95 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.09 \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=- \frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, - p \\ \frac {11}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} + \frac {3 x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} - \frac {3 x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((-x**2+1)**3*(b*x**4+1)**p,x)
 

Output:

-x**7*gamma(7/4)*hyper((7/4, -p), (11/4,), b*x**4*exp_polar(I*pi))/(4*gamm 
a(11/4)) + 3*x**5*gamma(5/4)*hyper((5/4, -p), (9/4,), b*x**4*exp_polar(I*p 
i))/(4*gamma(9/4)) - 3*x**3*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**4*exp 
_polar(I*pi))/(4*gamma(7/4)) + x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x** 
4*exp_polar(I*pi))/(4*gamma(5/4))
 

Maxima [F]

\[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )}^{3} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:

integrate((-x^2+1)^3*(b*x^4+1)^p,x, algorithm="maxima")
 

Output:

-integrate((x^2 - 1)^3*(b*x^4 + 1)^p, x)
 

Giac [F]

\[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=\int { -{\left (x^{2} - 1\right )}^{3} {\left (b x^{4} + 1\right )}^{p} \,d x } \] Input:

integrate((-x^2+1)^3*(b*x^4+1)^p,x, algorithm="giac")
 

Output:

integrate(-(x^2 - 1)^3*(b*x^4 + 1)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=-\int {\left (x^2-1\right )}^3\,{\left (b\,x^4+1\right )}^p \,d x \] Input:

int(-(x^2 - 1)^3*(b*x^4 + 1)^p,x)
 

Output:

-int((x^2 - 1)^3*(b*x^4 + 1)^p, x)
 

Reduce [F]

\[ \int \left (1-x^2\right )^3 \left (1+b x^4\right )^p \, dx=\text {too large to display} \] Input:

int((-x^2+1)^3*(b*x^4+1)^p,x)
 

Output:

( - 64*(b*x**4 + 1)**p*b*p**3*x**7 + 192*(b*x**4 + 1)**p*b*p**3*x**5 - 192 
*(b*x**4 + 1)**p*b*p**3*x**3 + 64*(b*x**4 + 1)**p*b*p**3*x - 144*(b*x**4 + 
 1)**p*b*p**2*x**7 + 528*(b*x**4 + 1)**p*b*p**2*x**5 - 624*(b*x**4 + 1)**p 
*b*p**2*x**3 + 240*(b*x**4 + 1)**p*b*p**2*x - 92*(b*x**4 + 1)**p*b*p*x**7 
+ 372*(b*x**4 + 1)**p*b*p*x**5 - 564*(b*x**4 + 1)**p*b*p*x**3 + 284*(b*x** 
4 + 1)**p*b*p*x - 15*(b*x**4 + 1)**p*b*x**7 + 63*(b*x**4 + 1)**p*b*x**5 - 
105*(b*x**4 + 1)**p*b*x**3 + 105*(b*x**4 + 1)**p*b*x - 64*(b*x**4 + 1)**p* 
p**3*x**3 + 192*(b*x**4 + 1)**p*p**3*x - 96*(b*x**4 + 1)**p*p**2*x**3 + 48 
0*(b*x**4 + 1)**p*p**2*x - 20*(b*x**4 + 1)**p*p*x**3 + 252*(b*x**4 + 1)**p 
*p*x + 65536*int((b*x**4 + 1)**p/(256*b*p**4*x**4 + 1024*b*p**3*x**4 + 137 
6*b*p**2*x**4 + 704*b*p*x**4 + 105*b*x**4 + 256*p**4 + 1024*p**3 + 1376*p* 
*2 + 704*p + 105),x)*b*p**8 + 507904*int((b*x**4 + 1)**p/(256*b*p**4*x**4 
+ 1024*b*p**3*x**4 + 1376*b*p**2*x**4 + 704*b*p*x**4 + 105*b*x**4 + 256*p* 
*4 + 1024*p**3 + 1376*p**2 + 704*p + 105),x)*b*p**7 + 1626112*int((b*x**4 
+ 1)**p/(256*b*p**4*x**4 + 1024*b*p**3*x**4 + 1376*b*p**2*x**4 + 704*b*p*x 
**4 + 105*b*x**4 + 256*p**4 + 1024*p**3 + 1376*p**2 + 704*p + 105),x)*b*p* 
*6 + 2771968*int((b*x**4 + 1)**p/(256*b*p**4*x**4 + 1024*b*p**3*x**4 + 137 
6*b*p**2*x**4 + 704*b*p*x**4 + 105*b*x**4 + 256*p**4 + 1024*p**3 + 1376*p* 
*2 + 704*p + 105),x)*b*p**5 + 2695936*int((b*x**4 + 1)**p/(256*b*p**4*x**4 
 + 1024*b*p**3*x**4 + 1376*b*p**2*x**4 + 704*b*p*x**4 + 105*b*x**4 + 25...