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

Optimal. Leaf size=108 \[ x \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-b x^4\right )+\frac{x^3 (1-b (4 p+7)) \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-b x^4\right )}{b (4 p+7)}+\frac{3}{5} x^5 \, _2F_1\left (\frac{5}{4},-p;\frac{9}{4};-b x^4\right )-\frac{x^3 \left (b x^4+1\right )^{p+1}}{b (4 p+7)} \]

[Out]

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

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Rubi [A]  time = 0.11461, antiderivative size = 103, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1207, 1893, 245, 364} \[ x \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-b x^4\right )-x^3 \left (1-\frac{1}{4 b p+7 b}\right ) \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-b x^4\right )+\frac{3}{5} x^5 \, _2F_1\left (\frac{5}{4},-p;\frac{9}{4};-b x^4\right )-\frac{x^3 \left (b x^4+1\right )^{p+1}}{b (4 p+7)} \]

Antiderivative was successfully verified.

[In]

Int[(1 - x^2)^3*(1 + b*x^4)^p,x]

[Out]

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

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 1893

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] || PolyQ[Pq, x^n])

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 )^3 \left (1+b x^4\right )^p \, dx &=-\frac{x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+\frac{\int \left (1+b x^4\right )^p \left (b (7+4 p)+3 (1-b (7+4 p)) x^2+3 b (7+4 p) x^4\right ) \, dx}{b (7+4 p)}\\ &=-\frac{x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+\frac{\int \left (b (7+4 p) \left (1+b x^4\right )^p+3 (1-b (7+4 p)) x^2 \left (1+b x^4\right )^p+3 b (7+4 p) x^4 \left (1+b x^4\right )^p\right ) \, dx}{b (7+4 p)}\\ &=-\frac{x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+3 \int x^4 \left (1+b x^4\right )^p \, dx-\left (3 \left (1-\frac{1}{7 b+4 b p}\right )\right ) \int x^2 \left (1+b x^4\right )^p \, dx+\int \left (1+b x^4\right )^p \, dx\\ &=-\frac{x^3 \left (1+b x^4\right )^{1+p}}{b (7+4 p)}+x \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-b x^4\right )-\left (1-\frac{1}{7 b+4 b p}\right ) x^3 \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-b x^4\right )+\frac{3}{5} x^5 \, _2F_1\left (\frac{5}{4},-p;\frac{9}{4};-b x^4\right )\\ \end{align*}

Mathematica [A]  time = 0.0172261, size = 86, normalized size = 0.8 \[ -\frac{1}{7} x^7 \, _2F_1\left (\frac{7}{4},-p;\frac{11}{4};-b x^4\right )+\frac{3}{5} x^5 \, _2F_1\left (\frac{5}{4},-p;\frac{9}{4};-b x^4\right )-x^3 \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-b x^4\right )+x \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-b x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*Hypergeometric2F1[1/4, -p, 5/4, -(b*x^4)] - x^3*Hypergeometric2F1[3/4, -p, 7/4, -(b*x^4)] + (3*x^5*Hypergeom
etric2F1[5/4, -p, 9/4, -(b*x^4)])/5 - (x^7*Hypergeometric2F1[7/4, -p, 11/4, -(b*x^4)])/7

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Maple [A]  time = 0.105, size = 75, normalized size = 0.7 \begin{align*} -{\frac{{x}^{7}}{7}{\mbox{$_2$F$_1$}({\frac{7}{4}},-p;\,{\frac{11}{4}};\,-b{x}^{4})}}+{\frac{3\,{x}^{5}}{5}{\mbox{$_2$F$_1$}({\frac{5}{4}},-p;\,{\frac{9}{4}};\,-b{x}^{4})}}-{x}^{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)^3*(b*x^4+1)^p,x)

[Out]

-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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int{\left (x^{2} - 1\right )}^{3}{\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)^3*(b*x^4+1)^p,x, algorithm="maxima")

[Out]

-integrate((x^2 - 1)^3*(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^{6} - 3 \, x^{4} + 3 \, 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)^3*(b*x^4+1)^p,x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 176.712, size = 129, normalized size = 1.19 \begin{align*} - \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**7*gamma(7/4)*hyper((7/4, -p), (11/4,), b*x**4*exp_polar(I*pi))/(4*gamma(11/4)) + 3*x**5*gamma(5/4)*hyper((
5/4, -p), (9/4,), b*x**4*exp_polar(I*pi))/(4*gamma(9/4)) - 3*x**3*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**4*e
xp_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))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -{\left (x^{2} - 1\right )}^{3}{\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)^3*(b*x^4+1)^p,x, algorithm="giac")

[Out]

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