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

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

[Out]

-3/10*x^4*(-x^2+1)^(2/3)/(x^2+3)+9/40*(-x^2+1)^(2/3)*(14*x^2+69)/(x^2+3)-99/32*ln(x^2+3)*2^(1/3)+297/32*ln(2^(
2/3)-(-x^2+1)^(1/3))*2^(1/3)+99/16*arctan(1/3*(1+(-2*x^2+2)^(1/3))*3^(1/2))*3^(1/2)*2^(1/3)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 133, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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 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]

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 = 0.19, size = 120, normalized size = 0.90 \[ \frac {3}{80} \left (\frac {6 \left (1-x^2\right )^{2/3} \left (14 x^2+69\right )}{x^2+3}+\frac {165 \left (-\log \left (x^2+3\right )+3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{2-2 x^2}+1}{\sqrt {3}}\right )\right )}{2^{2/3}}-\frac {8 \left (1-x^2\right )^{2/3} x^4}{x^2+3}\right ) \]

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

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 133, normalized size = 1.00 \[ \frac {3 \, {\left (660 \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (x^{2} + 3\right )} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 165 \cdot 4^{\frac {2}{3}} {\left (x^{2} + 3\right )} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + 330 \cdot 4^{\frac {2}{3}} {\left (x^{2} + 3\right )} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - 8 \, {\left (4 \, x^{4} - 42 \, x^{2} - 207\right )} {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right )}}{320 \, {\left (x^{2} + 3\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.48, size = 126, normalized size = 0.95 \[ \frac {99}{32} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{10} \, {\left (-x^{2} + 1\right )}^{\frac {5}{3}} - \frac {99}{64} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {99}{32} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + \frac {15}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} + \frac {27 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]

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]

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)

________________________________________________________________________________________

maple [C]  time = 5.98, size = 770, normalized size = 5.79 \[ \text {result too large to display} \]

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]

3/40*(4*x^6-46*x^4-165*x^2+207)/(x^2+3)/(-x^2+1)^(1/3)+99/16*RootOf(_Z^3-2)*ln((-96*RootOf(RootOf(_Z^3-2)^2+4*
_Z*RootOf(_Z^3-2)+16*_Z^2)^2*RootOf(_Z^3-2)^2*x^2-8*x^2*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_
Z^3-2)+16*_Z^2)-168*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2)-42*Root
Of(_Z^3-2)^2*(-x^2+1)^(1/3)-60*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2-5*RootOf(_Z^3-2)*x^2+4
2*(-x^2+1)^(2/3)+252*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)+21*RootOf(_Z^3-2))/(x^2+3))-99/16*ln
((64*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)^2*RootOf(_Z^3-2)^2*x^2-8*x^2*RootOf(_Z^3-2)^3*RootOf
(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)-168*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+
16*_Z^2)*RootOf(_Z^3-2)-42*RootOf(_Z^3-2)^2*(-x^2+1)^(1/3)-8*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z
^2)*x^2+RootOf(_Z^3-2)*x^2+42*(-x^2+1)^(2/3)+168*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)-21*RootO
f(_Z^3-2))/(x^2+3))*RootOf(_Z^3-2)-99/4*ln((64*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)^2*RootOf(_
Z^3-2)^2*x^2-8*x^2*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)-168*(-x^2+1)^(1/3)*Ro
otOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2)-42*RootOf(_Z^3-2)^2*(-x^2+1)^(1/3)-8*RootOf(
RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2+RootOf(_Z^3-2)*x^2+42*(-x^2+1)^(2/3)+168*RootOf(RootOf(_Z^3-
2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)-21*RootOf(_Z^3-2))/(x^2+3))*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_
Z^2)

________________________________________________________________________________________

maxima [A]  time = 1.98, size = 126, normalized size = 0.95 \[ \frac {99}{32} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{10} \, {\left (-x^{2} + 1\right )}^{\frac {5}{3}} - \frac {99}{64} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {99}{32} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) + \frac {15}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} + \frac {27 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} + 3\right )}} \]

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)

________________________________________________________________________________________

mupad [B]  time = 0.92, size = 148, normalized size = 1.11 \[ \frac {99\,2^{1/3}\,\ln \left (\frac {88209\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {88209\,2^{2/3}}{64}\right )}{16}+\frac {27\,{\left (1-x^2\right )}^{2/3}}{8\,\left (x^2+3\right )}+\frac {15\,{\left (1-x^2\right )}^{2/3}}{4}+\frac {3\,{\left (1-x^2\right )}^{5/3}}{10}+\frac {99\,2^{1/3}\,\ln \left (\frac {88209\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {88209\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{32}-\frac {99\,2^{1/3}\,\ln \left (\frac {88209\,{\left (1-x^2\right )}^{1/3}}{64}-\frac {88209\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{256}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(99*2^(1/3)*log((88209*(1 - x^2)^(1/3))/64 - (88209*2^(2/3))/64))/16 + (27*(1 - x^2)^(2/3))/(8*(x^2 + 3)) + (1
5*(1 - x^2)^(2/3))/4 + (3*(1 - x^2)^(5/3))/10 + (99*2^(1/3)*log((88209*(1 - x^2)^(1/3))/64 - (88209*2^(2/3)*(3
^(1/2)*1i - 1)^2)/256)*(3^(1/2)*1i - 1))/32 - (99*2^(1/3)*log((88209*(1 - x^2)^(1/3))/64 - (88209*2^(2/3)*(3^(
1/2)*1i + 1)^2)/256)*(3^(1/2)*1i + 1))/32

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{7}}{\sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )^{2}}\, dx \]

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]

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

________________________________________________________________________________________