3.183 \(\int (1-x^2)^2 (1+b x^4)^p \, dx\)

Optimal. Leaf size=86 \[ -\frac{x (1-b (4 p+5)) \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-b x^4\right )}{b (4 p+5)}-\frac{2}{3} x^3 \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-b x^4\right )+\frac{x \left (b x^4+1\right )^{p+1}}{b (4 p+5)} \]

[Out]

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

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Rubi [A]  time = 0.0683756, antiderivative size = 79, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1207, 1204, 245, 364} \[ x \left (1-\frac{1}{4 b p+5 b}\right ) \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-b x^4\right )-\frac{2}{3} x^3 \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-b x^4\right )+\frac{x \left (b x^4+1\right )^{p+1}}{b (4 p+5)} \]

Antiderivative was successfully verified.

[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 1207

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]

Rule 1204

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 245

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 364

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

Rubi steps

\begin{align*} \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{\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]  time = 0.0107821, size = 65, normalized size = 0.76 \[ \frac{1}{5} x^5 \, _2F_1\left (\frac{5}{4},-p;\frac{9}{4};-b x^4\right )-\frac{2}{3} x^3 \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-b x^4\right )+x \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-b x^4\right ) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.045, size = 56, normalized size = 0.7 \begin{align*}{\frac{{x}^{5}}{5}{\mbox{$_2$F$_1$}({\frac{5}{4}},-p;\,{\frac{9}{4}};\,-b{x}^{4})}}-{\frac{2\,{x}^{3}}{3}{\mbox{$_2$F$_1$}({\frac{3}{4}},-p;\,{\frac{7}{4}};\,-b{x}^{4})}}+x{\mbox{$_2$F$_1$}({\frac{1}{4}},-p;\,{\frac{5}{4}};\,-b{x}^{4})} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (x^{2} - 1\right )}^{2}{\left (b x^{4} + 1\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (x^{4} - 2 \, x^{2} + 1\right )}{\left (b x^{4} + 1\right )}^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [C]  time = 99.1881, size = 94, normalized size = 1.09 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (x^{2} - 1\right )}^{2}{\left (b x^{4} + 1\right )}^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)