Integrand size = 19, antiderivative size = 110 \[ \int \frac {1}{(1+3 x) \sqrt [3]{-x+x^3}} \, dx=\frac {1}{4} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-x+x^3}}{1-x+\sqrt [3]{-x+x^3}}\right )+\frac {1}{4} \log \left (-1+x+2 \sqrt [3]{-x+x^3}\right )-\frac {1}{8} \log \left (1-2 x+x^2+(2-2 x) \sqrt [3]{-x+x^3}+4 \left (-x+x^3\right )^{2/3}\right ) \]
1/4*3^(1/2)*arctan(3^(1/2)*(x^3-x)^(1/3)/(1-x+(x^3-x)^(1/3)))+1/4*ln(-1+x+ 2*(x^3-x)^(1/3))-1/8*ln(1-2*x+x^2+(2-2*x)*(x^3-x)^(1/3)+4*(x^3-x)^(2/3))
Time = 15.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(1+3 x) \sqrt [3]{-x+x^3}} \, dx=\frac {1}{8} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x \left (-1+x^2\right )}}{1-x+\sqrt [3]{x \left (-1+x^2\right )}}\right )+2 \log \left (-1+x+2 \sqrt [3]{x \left (-1+x^2\right )}\right )-\log \left (1-2 x+x^2-2 (-1+x) \sqrt [3]{x \left (-1+x^2\right )}+4 \left (x \left (-1+x^2\right )\right )^{2/3}\right )\right ) \]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*(x*(-1 + x^2))^(1/3))/(1 - x + (x*(-1 + x^2))^( 1/3))] + 2*Log[-1 + x + 2*(x*(-1 + x^2))^(1/3)] - Log[1 - 2*x + x^2 - 2*(- 1 + x)*(x*(-1 + x^2))^(1/3) + 4*(x*(-1 + x^2))^(2/3)])/8
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.
| \(\displaystyle \int \frac {1}{(3 x+1) \sqrt [3]{x^3-x}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{x^2-1} \int \frac {1}{\sqrt [3]{x} (3 x+1) \sqrt [3]{x^2-1}}dx}{\sqrt [3]{x^3-x}}\) |
| \(\Big \downarrow \) 616 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2-1} \int \frac {\sqrt [3]{x}}{(3 x+1) \sqrt [3]{x^2-1}}d\sqrt [3]{x}}{\sqrt [3]{x^3-x}}\) |
| \(\Big \downarrow \) 1888 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2-1} \int \frac {\sqrt [3]{x}}{(3 x+1) \sqrt [3]{x^2-1}}d\sqrt [3]{x}}{\sqrt [3]{x^3-x}}\) |
3.17.7.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/e))^n*(a + b*(x^(2*k)/e^2))^p, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && ILtQ[n, 0] && FractionQ[m]
Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ (n_))^(q_.), x_Symbol] :> Unintegrable[(f*x)^m*(d + e*x^n)^q*(a + c*x^(2*n) )^p, x] /; FreeQ[{a, c, d, e, f, m, n, p, q}, x] && EqQ[n2, 2*n]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.30 (sec) , antiderivative size = 656, normalized size of antiderivative = 5.96
1/4*ln(-(7297600*RootOf(4*_Z^2+2*_Z+1)^2*x^2+7258272*RootOf(4*_Z^2+2*_Z+1) *(x^3-x)^(2/3)-19359420*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)*x-18973760*Roo tOf(4*_Z^2+2*_Z+1)^2*x+32081060*RootOf(4*_Z^2+2*_Z+1)*x^2-19359420*(x^3-x) ^(2/3)+19359420*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)-11494278*x*(x^3-x)^(1/ 3)+8757120*RootOf(4*_Z^2+2*_Z+1)^2+40119368*RootOf(4*_Z^2+2*_Z+1)*x-108805 97*x^2+11494278*(x^3-x)^(1/3)-2679436*RootOf(4*_Z^2+2*_Z+1)-10794014*x-288 61)/(1+3*x)^2)-1/4*ln((11915360*RootOf(4*_Z^2+2*_Z+1)^2*x^2+7258272*RootOf (4*_Z^2+2*_Z+1)*(x^3-x)^(2/3)+22988556*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3) *x-30979936*RootOf(4*_Z^2+2*_Z+1)^2*x-47884574*RootOf(4*_Z^2+2*_Z+1)*x^2+2 2988556*(x^3-x)^(2/3)-22988556*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)+9679710 *x*(x^3-x)^(1/3)+14298432*RootOf(4*_Z^2+2*_Z+1)^2-77197364*RootOf(4*_Z^2+2 *_Z+1)*x-9726157*x^2-9679710*(x^3-x)^(1/3)+9770930*RootOf(4*_Z^2+2*_Z+1)-1 3795558*x+1356467)/(1+3*x)^2)-1/2*ln((11915360*RootOf(4*_Z^2+2*_Z+1)^2*x^2 +7258272*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(2/3)+22988556*RootOf(4*_Z^2+2*_Z+1 )*(x^3-x)^(1/3)*x-30979936*RootOf(4*_Z^2+2*_Z+1)^2*x-47884574*RootOf(4*_Z^ 2+2*_Z+1)*x^2+22988556*(x^3-x)^(2/3)-22988556*RootOf(4*_Z^2+2*_Z+1)*(x^3-x )^(1/3)+9679710*x*(x^3-x)^(1/3)+14298432*RootOf(4*_Z^2+2*_Z+1)^2-77197364* RootOf(4*_Z^2+2*_Z+1)*x-9726157*x^2-9679710*(x^3-x)^(1/3)+9770930*RootOf(4 *_Z^2+2*_Z+1)-13795558*x+1356467)/(1+3*x)^2)*RootOf(4*_Z^2+2*_Z+1)
Time = 0.48 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(1+3 x) \sqrt [3]{-x+x^3}} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {286273 \, \sqrt {3} {\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (x - 1\right )} + \sqrt {3} {\left (635653 \, x^{2} + 434719 \, x + 66978\right )} + 539695 \, \sqrt {3} {\left (x^{3} - x\right )}^{\frac {2}{3}}}{1293894 \, x^{2} + 1974837 \, x - 226981}\right ) + \frac {1}{8} \, \log \left (\frac {9 \, x^{2} + 6 \, {\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (x - 1\right )} + 6 \, x + 12 \, {\left (x^{3} - x\right )}^{\frac {2}{3}} + 1}{9 \, x^{2} + 6 \, x + 1}\right ) \]
1/4*sqrt(3)*arctan((286273*sqrt(3)*(x^3 - x)^(1/3)*(x - 1) + sqrt(3)*(6356 53*x^2 + 434719*x + 66978) + 539695*sqrt(3)*(x^3 - x)^(2/3))/(1293894*x^2 + 1974837*x - 226981)) + 1/8*log((9*x^2 + 6*(x^3 - x)^(1/3)*(x - 1) + 6*x + 12*(x^3 - x)^(2/3) + 1)/(9*x^2 + 6*x + 1))
\[ \int \frac {1}{(1+3 x) \sqrt [3]{-x+x^3}} \, dx=\int \frac {1}{\sqrt [3]{x \left (x - 1\right ) \left (x + 1\right )} \left (3 x + 1\right )}\, dx \]
\[ \int \frac {1}{(1+3 x) \sqrt [3]{-x+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (3 \, x + 1\right )}} \,d x } \]
\[ \int \frac {1}{(1+3 x) \sqrt [3]{-x+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (3 \, x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(1+3 x) \sqrt [3]{-x+x^3}} \, dx=\int \frac {1}{{\left (x^3-x\right )}^{1/3}\,\left (3\,x+1\right )} \,d x \]