3.1019 \(\int \frac{x^7}{\sqrt [3]{1-x^2} (3+x^2)^2} \, dx\)

Optimal. Leaf size=133 \[ -\frac{3 \left (1-x^2\right )^{2/3} x^4}{10 \left (x^2+3\right )}+\frac{9 \left (1-x^2\right )^{2/3} \left (14 x^2+69\right )}{40 \left (x^2+3\right )}-\frac{99 \log \left (x^2+3\right )}{16\ 2^{2/3}}+\frac{297 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac{99 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{8\ 2^{2/3}} \]

[Out]

(-3*x^4*(1 - x^2)^(2/3))/(10*(3 + x^2)) + (9*(1 - x^2)^(2/3)*(69 + 14*x^2))/(40*(3 + x^2)) + (99*Sqrt[3]*ArcTa
n[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(8*2^(2/3)) - (99*Log[3 + x^2])/(16*2^(2/3)) + (297*Log[2^(2/3) - (1 - x^2
)^(1/3)])/(16*2^(2/3))

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Rubi [A]  time = 0.0857265, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {446, 100, 146, 55, 617, 204, 31} \[ -\frac{3 \left (1-x^2\right )^{2/3} x^4}{10 \left (x^2+3\right )}+\frac{9 \left (1-x^2\right )^{2/3} \left (14 x^2+69\right )}{40 \left (x^2+3\right )}-\frac{99 \log \left (x^2+3\right )}{16\ 2^{2/3}}+\frac{297 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac{99 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{8\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x^7/((1 - x^2)^(1/3)*(3 + x^2)^2),x]

[Out]

(-3*x^4*(1 - x^2)^(2/3))/(10*(3 + x^2)) + (9*(1 - x^2)^(2/3)*(69 + 14*x^2))/(40*(3 + x^2)) + (99*Sqrt[3]*ArcTa
n[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(8*2^(2/3)) - (99*Log[3 + x^2])/(16*2^(2/3)) + (297*Log[2^(2/3) - (1 - x^2
)^(1/3)])/(16*2^(2/3))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{x^7}{\sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3}{\sqrt [3]{1-x} (3+x)^2} \, dx,x,x^2\right )\\ &=-\frac{3 x^4 \left (1-x^2\right )^{2/3}}{10 \left (3+x^2\right )}-\frac{3}{10} \operatorname{Subst}\left (\int \frac{x (-6+7 x)}{\sqrt [3]{1-x} (3+x)^2} \, dx,x,x^2\right )\\ &=-\frac{3 x^4 \left (1-x^2\right )^{2/3}}{10 \left (3+x^2\right )}+\frac{9 \left (1-x^2\right )^{2/3} \left (69+14 x^2\right )}{40 \left (3+x^2\right )}+\frac{99}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac{3 x^4 \left (1-x^2\right )^{2/3}}{10 \left (3+x^2\right )}+\frac{9 \left (1-x^2\right )^{2/3} \left (69+14 x^2\right )}{40 \left (3+x^2\right )}-\frac{99 \log \left (3+x^2\right )}{16\ 2^{2/3}}+\frac{297}{16} \operatorname{Subst}\left (\int \frac{1}{2 \sqrt [3]{2}+2^{2/3} x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac{297 \operatorname{Subst}\left (\int \frac{1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}\\ &=-\frac{3 x^4 \left (1-x^2\right )^{2/3}}{10 \left (3+x^2\right )}+\frac{9 \left (1-x^2\right )^{2/3} \left (69+14 x^2\right )}{40 \left (3+x^2\right )}-\frac{99 \log \left (3+x^2\right )}{16\ 2^{2/3}}+\frac{297 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}-\frac{297 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{8\ 2^{2/3}}\\ &=-\frac{3 x^4 \left (1-x^2\right )^{2/3}}{10 \left (3+x^2\right )}+\frac{9 \left (1-x^2\right )^{2/3} \left (69+14 x^2\right )}{40 \left (3+x^2\right )}+\frac{99 \sqrt{3} \tan ^{-1}\left (\frac{1+\sqrt [3]{2-2 x^2}}{\sqrt{3}}\right )}{8\ 2^{2/3}}-\frac{99 \log \left (3+x^2\right )}{16\ 2^{2/3}}+\frac{297 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.168505, size = 120, normalized size = 0.9 \[ \frac{3}{80} \left (-\frac{8 \left (1-x^2\right )^{2/3} x^4}{x^2+3}+\frac{6 \left (1-x^2\right )^{2/3} \left (14 x^2+69\right )}{x^2+3}+\frac{165 \left (-\log \left (x^2+3\right )+3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )\right )}{2^{2/3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/((1 - x^2)^(1/3)*(3 + x^2)^2),x]

[Out]

(3*((-8*x^4*(1 - x^2)^(2/3))/(3 + x^2) + (6*(1 - x^2)^(2/3)*(69 + 14*x^2))/(3 + x^2) + (165*(2*Sqrt[3]*ArcTan[
(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]] - Log[3 + x^2] + 3*Log[2^(2/3) - (1 - x^2)^(1/3)]))/2^(2/3)))/80

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Maple [F]  time = 0.048, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{7}}{ \left ({x}^{2}+3 \right ) ^{2}}{\frac{1}{\sqrt [3]{-{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(-x^2+1)^(1/3)/(x^2+3)^2,x)

[Out]

int(x^7/(-x^2+1)^(1/3)/(x^2+3)^2,x)

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Maxima [A]  time = 1.49337, size = 170, normalized size = 1.28 \begin{align*} \frac{99}{32} \cdot 4^{\frac{2}{3}} \sqrt{3} \arctan \left (\frac{1}{12} \cdot 4^{\frac{2}{3}} \sqrt{3}{\left (4^{\frac{1}{3}} + 2 \,{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right )}\right ) + \frac{3}{10} \,{\left (-x^{2} + 1\right )}^{\frac{5}{3}} - \frac{99}{64} \cdot 4^{\frac{2}{3}} \log \left (4^{\frac{2}{3}} + 4^{\frac{1}{3}}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{2}{3}}\right ) + \frac{99}{32} \cdot 4^{\frac{2}{3}} \log \left (-4^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) + \frac{15}{4} \,{\left (-x^{2} + 1\right )}^{\frac{2}{3}} + \frac{27 \,{\left (-x^{2} + 1\right )}^{\frac{2}{3}}}{8 \,{\left (x^{2} + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(-x^2+1)^(1/3)/(x^2+3)^2,x, algorithm="maxima")

[Out]

99/32*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) + 3/10*(-x^2 + 1)^(5/3) - 99
/64*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 99/32*4^(2/3)*log(-4^(1/3) + (-x^2 +
1)^(1/3)) + 15/4*(-x^2 + 1)^(2/3) + 27/8*(-x^2 + 1)^(2/3)/(x^2 + 3)

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Fricas [A]  time = 1.56803, size = 394, normalized size = 2.96 \begin{align*} \frac{3 \,{\left (660 \cdot 4^{\frac{1}{6}} \sqrt{3}{\left (x^{2} + 3\right )} \arctan \left (\frac{1}{6} \cdot 4^{\frac{1}{6}} \sqrt{3}{\left (4^{\frac{1}{3}} + 2 \,{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right )}\right ) - 165 \cdot 4^{\frac{2}{3}}{\left (x^{2} + 3\right )} \log \left (4^{\frac{2}{3}} + 4^{\frac{1}{3}}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{2}{3}}\right ) + 330 \cdot 4^{\frac{2}{3}}{\left (x^{2} + 3\right )} \log \left (-4^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) - 8 \,{\left (4 \, x^{4} - 42 \, x^{2} - 207\right )}{\left (-x^{2} + 1\right )}^{\frac{2}{3}}\right )}}{320 \,{\left (x^{2} + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(-x^2+1)^(1/3)/(x^2+3)^2,x, algorithm="fricas")

[Out]

3/320*(660*4^(1/6)*sqrt(3)*(x^2 + 3)*arctan(1/6*4^(1/6)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) - 165*4^(2/3)*
(x^2 + 3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 330*4^(2/3)*(x^2 + 3)*log(-4^(1/3) + (-
x^2 + 1)^(1/3)) - 8*(4*x^4 - 42*x^2 - 207)*(-x^2 + 1)^(2/3))/(x^2 + 3)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(-x**2+1)**(1/3)/(x**2+3)**2,x)

[Out]

Exception raised: ValueError

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(-x^2+1)^(1/3)/(x^2+3)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError