Integrand size = 25, antiderivative size = 171 \[ \int \frac {1+x}{\left (1+4 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\frac {\arctan \left (\frac {\sqrt {3} \sqrt [3]{1-x^3}}{2^{2/3}-2^{2/3} x+\sqrt [3]{1-x^3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (-2^{2/3}+2^{2/3} x+2 \sqrt [3]{1-x^3}\right )}{3\ 2^{2/3}}-\frac {\log \left (-\sqrt [3]{2}+2 \sqrt [3]{2} x-\sqrt [3]{2} x^2+\left (-2^{2/3}+2^{2/3} x\right ) \sqrt [3]{1-x^3}-2 \left (1-x^3\right )^{2/3}\right )}{6\ 2^{2/3}} \]
1/6*arctan(3^(1/2)*(-x^3+1)^(1/3)/(2^(2/3)-2^(2/3)*x+(-x^3+1)^(1/3)))*2^(1 /3)*3^(1/2)+1/6*ln(-2^(2/3)+2^(2/3)*x+2*(-x^3+1)^(1/3))*2^(1/3)-1/12*ln(-2 ^(1/3)+2*2^(1/3)*x-2^(1/3)*x^2+(-2^(2/3)+2^(2/3)*x)*(-x^3+1)^(1/3)-2*(-x^3 +1)^(2/3))*2^(1/3)
Time = 1.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.91 \[ \int \frac {1+x}{\left (1+4 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1-x^3}}{2^{2/3}-2^{2/3} x+\sqrt [3]{1-x^3}}\right )+2 \log \left (-2^{2/3}+2^{2/3} x+2 \sqrt [3]{1-x^3}\right )-\log \left (-\sqrt [3]{2}+2 \sqrt [3]{2} x-\sqrt [3]{2} x^2+2^{2/3} (-1+x) \sqrt [3]{1-x^3}-2 \left (1-x^3\right )^{2/3}\right )}{6\ 2^{2/3}} \]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*(1 - x^3)^(1/3))/(2^(2/3) - 2^(2/3)*x + (1 - x^ 3)^(1/3))] + 2*Log[-2^(2/3) + 2^(2/3)*x + 2*(1 - x^3)^(1/3)] - Log[-2^(1/3 ) + 2*2^(1/3)*x - 2^(1/3)*x^2 + 2^(2/3)*(-1 + x)*(1 - x^3)^(1/3) - 2*(1 - x^3)^(2/3)])/(6*2^(2/3))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.39 (sec) , antiderivative size = 559, normalized size of antiderivative = 3.27, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {7279, 2009}
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.
| \(\displaystyle \int \frac {x+1}{\left (x^2+4 x+1\right ) \sqrt [3]{1-x^3}} \, dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {1-\frac {1}{\sqrt {3}}}{\left (2 x-2 \sqrt {3}+4\right ) \sqrt [3]{1-x^3}}+\frac {1+\frac {1}{\sqrt {3}}}{\left (2 x+2 \sqrt {3}+4\right ) \sqrt [3]{1-x^3}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{12} \left (9+5 \sqrt {3}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-\frac {x^3}{\left (2-\sqrt {3}\right )^3}\right )-\frac {1}{12} \left (9-5 \sqrt {3}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-\frac {x^3}{\left (2+\sqrt {3}\right )^3}\right )-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{3 \left (9-5 \sqrt {3}\right )} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt {3}}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{3 \left (9+5 \sqrt {3}\right )} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{1-x^3}}{\sqrt [3]{3 \left (9-5 \sqrt {3}\right )}}+1}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{1-x^3}}{\sqrt [3]{3 \left (9+5 \sqrt {3}\right )}}+1}{\sqrt {3}}\right )}{3\ 2^{2/3} \sqrt {3}}-\frac {\log \left (x^3-15 \sqrt {3}+26\right )}{18\ 2^{2/3}}-\frac {\log \left (x^3+15 \sqrt {3}+26\right )}{18\ 2^{2/3}}-\frac {\log \left (8 x^3+8 \left (2-\sqrt {3}\right )^3\right )}{18\ 2^{2/3}}-\frac {\log \left (8 x^3+8 \left (2+\sqrt {3}\right )^3\right )}{18\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{3 \left (9-5 \sqrt {3}\right )}-\sqrt [3]{1-x^3}\right )}{6\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{3 \left (9+5 \sqrt {3}\right )}-\sqrt [3]{1-x^3}\right )}{6\ 2^{2/3}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{3 \left (9-5 \sqrt {3}\right )} x\right )}{6\ 2^{2/3}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{3 \left (9+5 \sqrt {3}\right )} x\right )}{6\ 2^{2/3}}\) |
-1/12*((9 + 5*Sqrt[3])*x^2*AppellF1[2/3, 1/3, 1, 5/3, x^3, -(x^3/(2 - Sqrt [3])^3)]) - ((9 - 5*Sqrt[3])*x^2*AppellF1[2/3, 1/3, 1, 5/3, x^3, -(x^3/(2 + Sqrt[3])^3)])/12 - ArcTan[(1 - (2*(3*(9 - 5*Sqrt[3]))^(1/3)*x)/(1 - x^3) ^(1/3))/Sqrt[3]]/(3*2^(2/3)*Sqrt[3]) - ArcTan[(1 - (2*(3*(9 + 5*Sqrt[3]))^ (1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]]/(3*2^(2/3)*Sqrt[3]) + ArcTan[(1 + (2*(1 - x^3)^(1/3))/(3*(9 - 5*Sqrt[3]))^(1/3))/Sqrt[3]]/(3*2^(2/3)*Sqrt[3]) + A rcTan[(1 + (2*(1 - x^3)^(1/3))/(3*(9 + 5*Sqrt[3]))^(1/3))/Sqrt[3]]/(3*2^(2 /3)*Sqrt[3]) - Log[26 - 15*Sqrt[3] + x^3]/(18*2^(2/3)) - Log[26 + 15*Sqrt[ 3] + x^3]/(18*2^(2/3)) - Log[8*(2 - Sqrt[3])^3 + 8*x^3]/(18*2^(2/3)) - Log [8*(2 + Sqrt[3])^3 + 8*x^3]/(18*2^(2/3)) + Log[(3*(9 - 5*Sqrt[3]))^(1/3) - (1 - x^3)^(1/3)]/(6*2^(2/3)) + Log[(3*(9 + 5*Sqrt[3]))^(1/3) - (1 - x^3)^ (1/3)]/(6*2^(2/3)) + Log[-((3*(9 - 5*Sqrt[3]))^(1/3)*x) - (1 - x^3)^(1/3)] /(6*2^(2/3)) + Log[-((3*(9 + 5*Sqrt[3]))^(1/3)*x) - (1 - x^3)^(1/3)]/(6*2^ (2/3))
3.23.61.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 7.40 (sec) , antiderivative size = 1176, normalized size of antiderivative = 6.88
RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*ln((108*(-x^3+1)^(2/3 )*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^2+36 *RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^3*x-5 40*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^2 *x+15*(-x^3+1)^(1/3)*RootOf(_Z^3-2)^2*x-108*(-x^3+1)^(1/3)*RootOf(RootOf(_ Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)*x-15*(-x^3+1)^(1/3)*R ootOf(_Z^3-2)^2+108*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^ 3-2)+36*_Z^2)*RootOf(_Z^3-2)-14*RootOf(_Z^3-2)*x^2+210*RootOf(RootOf(_Z^3- 2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^2-30*(-x^3+1)^(2/3)+4*RootOf(_Z^3-2)*x -60*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x-14*RootOf(_Z^3- 2)+210*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2))/(x^2+4*x+1))- 1/6*ln(-(108*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36 *_Z^2)*RootOf(_Z^3-2)^2+126*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36 *_Z^2)*RootOf(_Z^3-2)^3*x+540*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+ 36*_Z^2)^2*RootOf(_Z^3-2)^2*x-33*(-x^3+1)^(1/3)*RootOf(_Z^3-2)^2*x-108*(-x ^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z ^3-2)*x+33*(-x^3+1)^(1/3)*RootOf(_Z^3-2)^2+108*(-x^3+1)^(1/3)*RootOf(RootO f(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)+49*RootOf(_Z^3-2)* x^2+210*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^2+66*(-x^3+ 1)^(2/3)+28*RootOf(_Z^3-2)*x+120*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z...
Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (134) = 268\).
Time = 5.54 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.85 \[ \int \frac {1+x}{\left (1+4 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\frac {1}{18} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {4^{\frac {1}{6}} {\left (12 \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (2 \, x^{4} + 7 \, x^{3} + 7 \, x + 2\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 4^{\frac {1}{3}} \sqrt {3} {\left (91 \, x^{6} - 42 \, x^{5} + 105 \, x^{4} - 92 \, x^{3} + 105 \, x^{2} - 42 \, x + 91\right )} - 12 \, \sqrt {3} {\left (19 \, x^{5} - 29 \, x^{4} + 28 \, x^{3} - 28 \, x^{2} + 29 \, x - 19\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{6 \, {\left (53 \, x^{6} - 174 \, x^{5} + 111 \, x^{4} - 196 \, x^{3} + 111 \, x^{2} - 174 \, x + 53\right )}}\right ) - \frac {1}{72} \cdot 4^{\frac {2}{3}} \log \left (\frac {6 \cdot 4^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} {\left (2 \, x^{2} - x + 2\right )} + 4^{\frac {1}{3}} {\left (19 \, x^{4} - 10 \, x^{3} + 18 \, x^{2} - 10 \, x + 19\right )} - 6 \, {\left (5 \, x^{3} - 3 \, x^{2} + 3 \, x - 5\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{4} + 8 \, x^{3} + 18 \, x^{2} + 8 \, x + 1}\right ) + \frac {1}{36} \cdot 4^{\frac {2}{3}} \log \left (\frac {4^{\frac {2}{3}} {\left (x^{2} + 4 \, x + 1\right )} - 6 \cdot 4^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 12 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2} + 4 \, x + 1}\right ) \]
1/18*4^(1/6)*sqrt(3)*arctan(1/6*4^(1/6)*(12*4^(2/3)*sqrt(3)*(2*x^4 + 7*x^3 + 7*x + 2)*(-x^3 + 1)^(2/3) + 4^(1/3)*sqrt(3)*(91*x^6 - 42*x^5 + 105*x^4 - 92*x^3 + 105*x^2 - 42*x + 91) - 12*sqrt(3)*(19*x^5 - 29*x^4 + 28*x^3 - 2 8*x^2 + 29*x - 19)*(-x^3 + 1)^(1/3))/(53*x^6 - 174*x^5 + 111*x^4 - 196*x^3 + 111*x^2 - 174*x + 53)) - 1/72*4^(2/3)*log((6*4^(2/3)*(-x^3 + 1)^(2/3)*( 2*x^2 - x + 2) + 4^(1/3)*(19*x^4 - 10*x^3 + 18*x^2 - 10*x + 19) - 6*(5*x^3 - 3*x^2 + 3*x - 5)*(-x^3 + 1)^(1/3))/(x^4 + 8*x^3 + 18*x^2 + 8*x + 1)) + 1/36*4^(2/3)*log((4^(2/3)*(x^2 + 4*x + 1) - 6*4^(1/3)*(-x^3 + 1)^(1/3)*(x - 1) - 12*(-x^3 + 1)^(2/3))/(x^2 + 4*x + 1))
\[ \int \frac {1+x}{\left (1+4 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int \frac {x + 1}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{2} + 4 x + 1\right )}\, dx \]
\[ \int \frac {1+x}{\left (1+4 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int { \frac {x + 1}{{\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} + 4 \, x + 1\right )}} \,d x } \]
\[ \int \frac {1+x}{\left (1+4 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int { \frac {x + 1}{{\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} + 4 \, x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1+x}{\left (1+4 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int \frac {x+1}{{\left (1-x^3\right )}^{1/3}\,\left (x^2+4\,x+1\right )} \,d x \]