Integrand size = 22, antiderivative size = 133 \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \sqrt [3]{-1+4 x+4 x^2}}{\sqrt {3}}\right )}{4 \sqrt [3]{2}}-\frac {\log \left (2+2^{2/3} \sqrt [3]{-1+4 x+4 x^2}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (-2+2^{2/3} \sqrt [3]{-1+4 x+4 x^2}-\sqrt [3]{2} \left (-1+4 x+4 x^2\right )^{2/3}\right )}{8 \sqrt [3]{2}} \]
1/8*3^(1/2)*arctan(-1/3*3^(1/2)+1/3*2^(2/3)*(4*x^2+4*x-1)^(1/3)*3^(1/2))*2 ^(2/3)-1/8*ln(2+2^(2/3)*(4*x^2+4*x-1)^(1/3))*2^(2/3)+1/16*ln(-2+2^(2/3)*(4 *x^2+4*x-1)^(1/3)-2^(1/3)*(4*x^2+4*x-1)^(2/3))*2^(2/3)
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\frac {-2 \sqrt {3} \arctan \left (\frac {1-2^{2/3} \sqrt [3]{-1+4 x+4 x^2}}{\sqrt {3}}\right )-2 \log \left (2+2^{2/3} \sqrt [3]{-1+4 x+4 x^2}\right )+\log \left (-2+2^{2/3} \sqrt [3]{-1+4 x+4 x^2}-\sqrt [3]{2} \left (-1+4 x+4 x^2\right )^{2/3}\right )}{8 \sqrt [3]{2}} \]
(-2*Sqrt[3]*ArcTan[(1 - 2^(2/3)*(-1 + 4*x + 4*x^2)^(1/3))/Sqrt[3]] - 2*Log [2 + 2^(2/3)*(-1 + 4*x + 4*x^2)^(1/3)] + Log[-2 + 2^(2/3)*(-1 + 4*x + 4*x^ 2)^(1/3) - 2^(1/3)*(-1 + 4*x + 4*x^2)^(2/3)])/(8*2^(1/3))
Time = 0.22 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.63, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1118, 243, 68, 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.
\(\displaystyle \int \frac {1}{(2 x+1) \sqrt [3]{4 x^2+4 x-1}} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \frac {1}{2} \int \frac {1}{(2 x+1) \sqrt [3]{(2 x+1)^2-2}}d(2 x+1)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{4} \int \frac {1}{\sqrt [3]{2 x-1} (2 x+1)^2}d(2 x+1)^2\) |
\(\Big \downarrow \) 68 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 \int \frac {1}{\sqrt [3]{2 x-1}+\sqrt [3]{2}}d\sqrt [3]{2 x-1}}{2 \sqrt [3]{2}}+\frac {3}{2} \int \frac {1}{(2 x+1)^4-\sqrt [3]{2} \sqrt [3]{2 x-1}+2^{2/3}}d\sqrt [3]{2 x-1}+\frac {\log \left ((2 x+1)^2\right )}{2 \sqrt [3]{2}}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{4} \left (\frac {3}{2} \int \frac {1}{(2 x+1)^4-\sqrt [3]{2} \sqrt [3]{2 x-1}+2^{2/3}}d\sqrt [3]{2 x-1}+\frac {\log \left ((2 x+1)^2\right )}{2 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{2 x-1}+\sqrt [3]{2}\right )}{2 \sqrt [3]{2}}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{4} \left (\frac {3 \int \frac {1}{-(2 x+1)^4-3}d\left (1-2^{2/3} \sqrt [3]{2 x-1}\right )}{\sqrt [3]{2}}+\frac {\log \left ((2 x+1)^2\right )}{2 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{2 x-1}+\sqrt [3]{2}\right )}{2 \sqrt [3]{2}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{4} \left (-\frac {\sqrt {3} \arctan \left (\frac {1-2^{2/3} \sqrt [3]{2 x-1}}{\sqrt {3}}\right )}{\sqrt [3]{2}}+\frac {\log \left ((2 x+1)^2\right )}{2 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{2 x-1}+\sqrt [3]{2}\right )}{2 \sqrt [3]{2}}\right )\) |
(-((Sqrt[3]*ArcTan[(1 - 2^(2/3)*(-1 + 2*x)^(1/3))/Sqrt[3]])/2^(1/3)) + Log [(1 + 2*x)^2]/(2*2^(1/3)) - (3*Log[2^(1/3) + (-1 + 2*x)^(1/3)])/(2*2^(1/3) ))/4
3.20.6.3.1 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] && NegQ[(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 = 6.58 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {2}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (4 x^{2}+4 x -1\right )^{\frac {1}{3}}-1\right )}{3}\right )-2 \ln \left (\left (4 x^{2}+4 x -1\right )^{\frac {1}{3}}+2^{\frac {1}{3}}\right )+\ln \left (\left (4 x^{2}+4 x -1\right )^{\frac {2}{3}}-2^{\frac {1}{3}} \left (4 x^{2}+4 x -1\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )\right )}{16}\) | \(89\) |
trager | \(\text {Expression too large to display}\) | \(1229\) |
1/16*2^(2/3)*(2*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(4*x^2+4*x-1)^(1/3)-1) )-2*ln((4*x^2+4*x-1)^(1/3)+2^(1/3))+ln((4*x^2+4*x-1)^(2/3)-2^(1/3)*(4*x^2+ 4*x-1)^(1/3)+2^(2/3)))
Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\frac {1}{8} \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{6}} {\left (2 \, \sqrt {2} \left (-1\right )^{\frac {1}{3}} {\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}} + 2^{\frac {5}{6}}\right )}\right ) - \frac {1}{16} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}}\right ) \]
1/8*sqrt(3)*2^(2/3)*(-1)^(1/3)*arctan(1/6*sqrt(3)*2^(1/6)*(2*sqrt(2)*(-1)^ (1/3)*(4*x^2 + 4*x - 1)^(1/3) + 2^(5/6))) - 1/16*2^(2/3)*(-1)^(1/3)*log(-2 ^(1/3)*(-1)^(2/3)*(4*x^2 + 4*x - 1)^(1/3) - 2^(2/3)*(-1)^(1/3) + (4*x^2 + 4*x - 1)^(2/3)) + 1/8*2^(2/3)*(-1)^(1/3)*log(2^(1/3)*(-1)^(2/3) + (4*x^2 + 4*x - 1)^(1/3))
\[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\int \frac {1}{\left (2 x + 1\right ) \sqrt [3]{4 x^{2} + 4 x - 1}}\, dx \]
\[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\int { \frac {1}{{\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )}} \,d x } \]
\[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\int { \frac {1}{{\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\int \frac {1}{\left (2\,x+1\right )\,{\left (4\,x^2+4\,x-1\right )}^{1/3}} \,d x \]