Integrand size = 18, antiderivative size = 140 \[ \int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx=\frac {3}{2} \sqrt [3]{6+2 x+x^2}-\frac {1}{2} \sqrt {3} \sqrt [3]{5} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{6+2 x+x^2}}{\sqrt {3} \sqrt [3]{5}}\right )+\frac {1}{2} \sqrt [3]{5} \log \left (-5+5^{2/3} \sqrt [3]{6+2 x+x^2}\right )-\frac {1}{4} \sqrt [3]{5} \log \left (5+5^{2/3} \sqrt [3]{6+2 x+x^2}+\sqrt [3]{5} \left (6+2 x+x^2\right )^{2/3}\right ) \]
3/2*(x^2+2*x+6)^(1/3)-1/2*3^(1/2)*5^(1/3)*arctan(1/3*3^(1/2)+2/15*(x^2+2*x +6)^(1/3)*3^(1/2)*5^(2/3))+1/2*5^(1/3)*ln(-5+5^(2/3)*(x^2+2*x+6)^(1/3))-1/ 4*5^(1/3)*ln(5+5^(2/3)*(x^2+2*x+6)^(1/3)+5^(1/3)*(x^2+2*x+6)^(2/3))
Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx=\frac {1}{4} \left (6 \sqrt [3]{6+2 x+x^2}-2 \sqrt {3} \sqrt [3]{5} \arctan \left (\frac {5+2\ 5^{2/3} \sqrt [3]{6+2 x+x^2}}{5 \sqrt {3}}\right )+2 \sqrt [3]{5} \log \left (-5+5^{2/3} \sqrt [3]{6+2 x+x^2}\right )-\sqrt [3]{5} \log \left (5+5^{2/3} \sqrt [3]{6+2 x+x^2}+\sqrt [3]{5} \left (6+2 x+x^2\right )^{2/3}\right )\right ) \]
(6*(6 + 2*x + x^2)^(1/3) - 2*Sqrt[3]*5^(1/3)*ArcTan[(5 + 2*5^(2/3)*(6 + 2* x + x^2)^(1/3))/(5*Sqrt[3])] + 2*5^(1/3)*Log[-5 + 5^(2/3)*(6 + 2*x + x^2)^ (1/3)] - 5^(1/3)*Log[5 + 5^(2/3)*(6 + 2*x + x^2)^(1/3) + 5^(1/3)*(6 + 2*x + x^2)^(2/3)])/4
Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.62, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1118, 243, 60, 69, 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.
| \(\displaystyle \int \frac {\sqrt [3]{x^2+2 x+6}}{x+1} \, dx\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle \int \frac {\sqrt [3]{(x+1)^2+5}}{x+1}d(x+1)\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt [3]{x+6}}{(x+1)^2}d(x+1)^2\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (5 \int \frac {1}{(x+1)^2 (x+6)^{2/3}}d(x+1)^2+3 \sqrt [3]{x+6}\right )\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {1}{2} \left (5 \left (-\frac {3 \int \frac {1}{-x+\sqrt [3]{5}-1}d\sqrt [3]{x+6}}{2\ 5^{2/3}}-\frac {3 \int \frac {1}{(x+1)^4+\sqrt [3]{5} \sqrt [3]{x+6}+5^{2/3}}d\sqrt [3]{x+6}}{2 \sqrt [3]{5}}-\frac {\log \left ((x+1)^2\right )}{2\ 5^{2/3}}\right )+3 \sqrt [3]{x+6}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} \left (5 \left (-\frac {3 \int \frac {1}{(x+1)^4+\sqrt [3]{5} \sqrt [3]{x+6}+5^{2/3}}d\sqrt [3]{x+6}}{2 \sqrt [3]{5}}+\frac {3 \log \left (-x+\sqrt [3]{5}-1\right )}{2\ 5^{2/3}}-\frac {\log \left ((x+1)^2\right )}{2\ 5^{2/3}}\right )+3 \sqrt [3]{x+6}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} \left (5 \left (\frac {3 \int \frac {1}{-(x+1)^4-3}d\left (\frac {2 \sqrt [3]{x+6}}{\sqrt [3]{5}}+1\right )}{5^{2/3}}+\frac {3 \log \left (-x+\sqrt [3]{5}-1\right )}{2\ 5^{2/3}}-\frac {\log \left ((x+1)^2\right )}{2\ 5^{2/3}}\right )+3 \sqrt [3]{x+6}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (5 \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{x+6}}{\sqrt [3]{5}}+1}{\sqrt {3}}\right )}{5^{2/3}}+\frac {3 \log \left (-x+\sqrt [3]{5}-1\right )}{2\ 5^{2/3}}-\frac {\log \left ((x+1)^2\right )}{2\ 5^{2/3}}\right )+3 \sqrt [3]{x+6}\right )\) |
(3*(6 + x)^(1/3) + 5*(-((Sqrt[3]*ArcTan[(1 + (2*(6 + x)^(1/3))/5^(1/3))/Sq rt[3]])/5^(2/3)) + (3*Log[-1 + 5^(1/3) - x])/(2*5^(2/3)) - Log[(1 + x)^2]/ (2*5^(2/3))))/2
3.20.71.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Simp[3/(2*b*q) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 /3)], x] - Simp[3/(2*b*q^2) 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_)^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_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(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]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[1/e Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
Time = 13.96 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.74
| method | result | size |
| pseudoelliptic | \(\frac {3 \left (x^{2}+2 x +6\right )^{\frac {1}{3}}}{2}+\frac {5^{\frac {1}{3}} \ln \left (\left (x^{2}+2 x +6\right )^{\frac {1}{3}}-5^{\frac {1}{3}}\right )}{2}-\frac {5^{\frac {1}{3}} \ln \left (\left (x^{2}+2 x +6\right )^{\frac {2}{3}}+5^{\frac {1}{3}} \left (x^{2}+2 x +6\right )^{\frac {1}{3}}+5^{\frac {2}{3}}\right )}{4}-\frac {\sqrt {3}\, 5^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3}}{3}+\frac {2 \left (x^{2}+2 x +6\right )^{\frac {1}{3}} \sqrt {3}\, 5^{\frac {2}{3}}}{15}\right )}{2}\) | \(103\) |
| trager | \(\text {Expression too large to display}\) | \(848\) |
| risch | \(\text {Expression too large to display}\) | \(2698\) |
3/2*(x^2+2*x+6)^(1/3)+1/2*5^(1/3)*ln((x^2+2*x+6)^(1/3)-5^(1/3))-1/4*5^(1/3 )*ln((x^2+2*x+6)^(2/3)+5^(1/3)*(x^2+2*x+6)^(1/3)+5^(2/3))-1/2*3^(1/2)*5^(1 /3)*arctan(1/3*3^(1/2)+2/15*(x^2+2*x+6)^(1/3)*3^(1/2)*5^(2/3))
Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx=-\frac {1}{2} \cdot 5^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {2}{15} \cdot 5^{\frac {2}{3}} \sqrt {3} {\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \cdot 5^{\frac {1}{3}} \log \left (5^{\frac {2}{3}} + 5^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}} + {\left (x^{2} + 2 \, x + 6\right )}^{\frac {2}{3}}\right ) + \frac {1}{2} \cdot 5^{\frac {1}{3}} \log \left (-5^{\frac {1}{3}} + {\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}} \]
-1/2*5^(1/3)*sqrt(3)*arctan(2/15*5^(2/3)*sqrt(3)*(x^2 + 2*x + 6)^(1/3) + 1 /3*sqrt(3)) - 1/4*5^(1/3)*log(5^(2/3) + 5^(1/3)*(x^2 + 2*x + 6)^(1/3) + (x ^2 + 2*x + 6)^(2/3)) + 1/2*5^(1/3)*log(-5^(1/3) + (x^2 + 2*x + 6)^(1/3)) + 3/2*(x^2 + 2*x + 6)^(1/3)
\[ \int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx=\int \frac {\sqrt [3]{x^{2} + 2 x + 6}}{x + 1}\, dx \]
\[ \int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx=\int { \frac {{\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}}}{x + 1} \,d x } \]
\[ \int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx=\int { \frac {{\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}}}{x + 1} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx=\int \frac {{\left (x^2+2\,x+6\right )}^{1/3}}{x+1} \,d x \]