3.2.83 \(\int (1-x^2)^2 (1+b x^4)^p \, dx\) [183]

3.2.83.1 Optimal result

Integrand size = 19, antiderivative size = 86 \[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=\frac {x \left (1+b x^4\right )^{1+p}}{b (5+4 p)}-\frac {(1-b (5+4 p)) x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )}{b (5+4 p)}-\frac {2}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right ) \]

output

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

3.2.83.2 Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-\frac {2}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )+\frac {1}{5} x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-p,\frac {9}{4},-b x^4\right ) \]

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
 

3.2.83.3 Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09, 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))
 

3.2.83.3.1 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]
 

3.2.83.4 Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.65

method	result	size
meijerg	\(\frac {x^{5} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {5}{4},-p ;\frac {9}{4};-b \,x^{4}\right )}{5}-\frac {2 x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{4},-p ;\frac {7}{4};-b \,x^{4}\right )}{3}+x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},-p ;\frac {5}{4};-b \,x^{4}\right )\)	\(56\)

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)
 

3.2.83.5 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)
 

3.2.83.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 27.98 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09 \[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=\frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} - \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

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))
 

3.2.83.7 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)
 

3.2.83.8 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)
 

3.2.83.9 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)