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

Optimal result

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

[Out]

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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1221, 1218, 251, 371} \[ \int \left (1-x^2\right )^2 \left (1+b x^4\right )^p \, dx=x \left (1-\frac {1}{4 b p+5 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right )-\frac {2}{3} x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-b x^4\right )+\frac {x \left (b x^4+1\right )^{p+1}}{b (4 p+5)} \]

[In]

Int[(1 - x^2)^2*(1 + b*x^4)^p,x]

[Out]

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

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1218

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(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 1221

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e^q*x^(2*q - 3)*((a + c*x^4)^(p +
1)/(c*(4*p + 2*q + 1))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {x \left (1+b x^4\right )^{1+p}}{b (5+4 p)}+\frac {\int \left (-1+b (5+4 p)-2 b (5+4 p) x^2\right ) \left (1+b x^4\right )^p \, dx}{b (5+4 p)} \\ & = \frac {x \left (1+b x^4\right )^{1+p}}{b (5+4 p)}+\frac {\int \left ((-1+b (5+4 p)) \left (1+b x^4\right )^p-2 b (5+4 p) x^2 \left (1+b x^4\right )^p\right ) \, dx}{b (5+4 p)} \\ & = \frac {x \left (1+b x^4\right )^{1+p}}{b (5+4 p)}-2 \int x^2 \left (1+b x^4\right )^p \, dx+\left (1-\frac {1}{5 b+4 b p}\right ) \int \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 \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-b x^4\right )-\frac {2}{3} x^3 \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-b x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

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*Hyper
geometric2F1[5/4, -p, 9/4, -(b*x^4)])/5

Maple [A] (verified)

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

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

[In]

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

[Out]

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
*pi))/(4*gamma(5/4))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((x^2 - 1)^2*(b*x^4 + 1)^p,x)

[Out]

int((x^2 - 1)^2*(b*x^4 + 1)^p, x)