Integrand size = 22, antiderivative size = 94 \[ \int \frac {x^3}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=-\frac {3}{4} \left (1-x^2\right )^{2/3}-\frac {3 \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}+\frac {3 \log \left (3+x^2\right )}{4\ 2^{2/3}}-\frac {9 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}} \]
-3/4*(-x^2+1)^(2/3)+3/8*ln(x^2+3)*2^(1/3)-9/8*ln(2^(2/3)-(-x^2+1)^(1/3))*2 ^(1/3)-3/4*arctan(1/3*(1+(-2*x^2+2)^(1/3))*3^(1/2))*3^(1/2)*2^(1/3)
Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.12 \[ \int \frac {x^3}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=-\frac {3}{8} \left (2 \left (1-x^2\right )^{2/3}+2 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{2-2 x^2}}{\sqrt {3}}\right )+2 \sqrt [3]{2} \log \left (-2+\sqrt [3]{2-2 x^2}\right )-\sqrt [3]{2} \log \left (4+2 \sqrt [3]{2-2 x^2}+\left (2-2 x^2\right )^{2/3}\right )\right ) \]
(-3*(2*(1 - x^2)^(2/3) + 2*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/ Sqrt[3]] + 2*2^(1/3)*Log[-2 + (2 - 2*x^2)^(1/3)] - 2^(1/3)*Log[4 + 2*(2 - 2*x^2)^(1/3) + (2 - 2*x^2)^(2/3)]))/8
Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {354, 90, 67, 16, 1082, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^3}{\sqrt [3]{1-x^2} \left (x^2+3\right )} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^2}{\sqrt [3]{1-x^2} \left (x^2+3\right )}dx^2\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} \left (-3 \int \frac {1}{\sqrt [3]{1-x^2} \left (x^2+3\right )}dx^2-\frac {3}{2} \left (1-x^2\right )^{2/3}\right )\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {1}{2} \left (-3 \left (-\frac {3 \int \frac {1}{2^{2/3}-\sqrt [3]{1-x^2}}d\sqrt [3]{1-x^2}}{2\ 2^{2/3}}+\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{1-x^2}+2 \sqrt [3]{2}}d\sqrt [3]{1-x^2}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}\right )-\frac {3}{2} \left (1-x^2\right )^{2/3}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (-3 \left (\frac {3}{2} \int \frac {1}{x^4+2^{2/3} \sqrt [3]{1-x^2}+2 \sqrt [3]{2}}d\sqrt [3]{1-x^2}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )-\frac {3}{2} \left (1-x^2\right )^{2/3}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (-3 \left (-\frac {3 \int \frac {1}{-x^4-3}d\left (\sqrt [3]{2} \sqrt [3]{1-x^2}+1\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )-\frac {3}{2} \left (1-x^2\right )^{2/3}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (-3 \left (\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} \sqrt [3]{1-x^2}+1}{\sqrt {3}}\right )}{2^{2/3}}-\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{2\ 2^{2/3}}\right )-\frac {3}{2} \left (1-x^2\right )^{2/3}\right )\) |
((-3*(1 - x^2)^(2/3))/2 - 3*((Sqrt[3]*ArcTan[(1 + 2^(1/3)*(1 - x^2)^(1/3)) /Sqrt[3]])/2^(2/3) - Log[3 + x^2]/(2*2^(2/3)) + (3*Log[2^(2/3) - (1 - x^2) ^(1/3)])/(2*2^(2/3))))/2
3.11.9.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[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]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Time = 8.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03
method | result | size |
pseudoelliptic | \(-\frac {3 \left (-x^{2}+1\right )^{\frac {2}{3}}}{4}-\frac {3 \,2^{\frac {1}{3}} \ln \left (\left (-x^{2}+1\right )^{\frac {1}{3}}-2^{\frac {2}{3}}\right )}{4}+\frac {3 \,2^{\frac {1}{3}} \ln \left (\left (-x^{2}+1\right )^{\frac {2}{3}}+2^{\frac {2}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}+2 \,2^{\frac {1}{3}}\right )}{8}-\frac {3 \sqrt {3}\, 2^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3}\, \left (1+2^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {1}{3}}\right )}{3}\right )}{4}\) | \(97\) |
trager | \(\text {Expression too large to display}\) | \(642\) |
risch | \(\text {Expression too large to display}\) | \(648\) |
-3/4*(-x^2+1)^(2/3)-3/4*2^(1/3)*ln((-x^2+1)^(1/3)-2^(2/3))+3/8*2^(1/3)*ln( (-x^2+1)^(2/3)+2^(2/3)*(-x^2+1)^(1/3)+2*2^(1/3))-3/4*3^(1/2)*2^(1/3)*arcta n(1/3*3^(1/2)*(1+2^(1/3)*(-x^2+1)^(1/3)))
Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.30 \[ \int \frac {x^3}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=-\frac {3}{4} \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (2 \, \left (-1\right )^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 4^{\frac {1}{3}}\right )}\right ) - \frac {3}{16} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} - 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {3}{8} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {3}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} \]
-3/4*4^(1/6)*sqrt(3)*(-1)^(1/3)*arctan(1/6*4^(1/6)*sqrt(3)*(2*(-1)^(1/3)*( -x^2 + 1)^(1/3) - 4^(1/3))) - 3/16*4^(2/3)*(-1)^(1/3)*log(4^(1/3)*(-1)^(2/ 3)*(-x^2 + 1)^(1/3) - 4^(2/3)*(-1)^(1/3) + (-x^2 + 1)^(2/3)) + 3/8*4^(2/3) *(-1)^(1/3)*log(-4^(1/3)*(-1)^(2/3) + (-x^2 + 1)^(1/3)) - 3/4*(-x^2 + 1)^( 2/3)
\[ \int \frac {x^3}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\int \frac {x^{3}}{\sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int \frac {x^3}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=-\frac {3}{8} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) - \frac {3}{8} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {3}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} \]
-3/8*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^( 1/3))) + 3/16*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^ (2/3)) - 3/8*4^(2/3)*log(-4^(1/3) + (-x^2 + 1)^(1/3)) - 3/4*(-x^2 + 1)^(2/ 3)
Time = 0.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int \frac {x^3}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=-\frac {3}{8} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) - \frac {3}{8} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {3}{4} \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}} \]
-3/8*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^( 1/3))) + 3/16*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^ (2/3)) - 3/8*4^(2/3)*log(4^(1/3) - (-x^2 + 1)^(1/3)) - 3/4*(-x^2 + 1)^(2/3 )
Time = 5.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.24 \[ \int \frac {x^3}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=-\frac {3\,2^{1/3}\,\ln \left (\frac {81\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {81\,2^{2/3}}{4}\right )}{4}-\frac {3\,{\left (1-x^2\right )}^{2/3}}{4}-\frac {3\,2^{1/3}\,\ln \left (\frac {81\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {81\,2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8}+\frac {3\,2^{1/3}\,\ln \left (\frac {81\,{\left (1-x^2\right )}^{1/3}}{4}-\frac {81\,2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8} \]