Integrand size = 22, antiderivative size = 133 \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \sqrt [3]{-1+4 x+4 x^2}}{\sqrt {3}}\right )}{4 \sqrt [3]{2}}-\frac {\log \left (2+2^{2/3} \sqrt [3]{-1+4 x+4 x^2}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (-2+2^{2/3} \sqrt [3]{-1+4 x+4 x^2}-\sqrt [3]{2} \left (-1+4 x+4 x^2\right )^{2/3}\right )}{8 \sqrt [3]{2}} \]
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Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.66, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {708, 272, 58, 631, 210, 31} \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-2^{2/3} \sqrt [3]{(2 x+1)^2-2}}{\sqrt {3}}\right )}{4 \sqrt [3]{2}}+\frac {\log (2 x+1)}{4 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{(2 x+1)^2-2}+\sqrt [3]{2}\right )}{8 \sqrt [3]{2}} \]
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Rule 31
Rule 58
Rule 210
Rule 272
Rule 631
Rule 708
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt [3]{-2+x^2}} \, dx,x,1+2 x\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt [3]{-2+x} x} \, dx,x,(1+2 x)^2\right ) \\ & = \frac {\log (1+2 x)}{4 \sqrt [3]{2}}+\frac {3}{8} \text {Subst}\left (\int \frac {1}{2^{2/3}-\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{-2+(1+2 x)^2}\right )-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt [3]{2}+x} \, dx,x,\sqrt [3]{-2+(1+2 x)^2}\right )}{8 \sqrt [3]{2}} \\ & = \frac {\log (1+2 x)}{4 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{2}+\sqrt [3]{-2+(1+2 x)^2}\right )}{8 \sqrt [3]{2}}+\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2^{2/3} \sqrt [3]{-2+(1+2 x)^2}\right )}{4 \sqrt [3]{2}} \\ & = -\frac {\sqrt {3} \arctan \left (\frac {1-2^{2/3} \sqrt [3]{-2+(1+2 x)^2}}{\sqrt {3}}\right )}{4 \sqrt [3]{2}}+\frac {\log (1+2 x)}{4 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{2}+\sqrt [3]{-2+(1+2 x)^2}\right )}{8 \sqrt [3]{2}} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\frac {-2 \sqrt {3} \arctan \left (\frac {1-2^{2/3} \sqrt [3]{-1+4 x+4 x^2}}{\sqrt {3}}\right )-2 \log \left (2+2^{2/3} \sqrt [3]{-1+4 x+4 x^2}\right )+\log \left (-2+2^{2/3} \sqrt [3]{-1+4 x+4 x^2}-\sqrt [3]{2} \left (-1+4 x+4 x^2\right )^{2/3}\right )}{8 \sqrt [3]{2}} \]
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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 | \(\text {Expression too large to display}\) | \(1229\) |
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Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\frac {1}{8} \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{6}} {\left (2 \, \sqrt {2} \left (-1\right )^{\frac {1}{3}} {\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}} + 2^{\frac {5}{6}}\right )}\right ) - \frac {1}{16} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{8} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}}\right ) \]
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\[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\int \frac {1}{\left (2 x + 1\right ) \sqrt [3]{4 x^{2} + 4 x - 1}}\, dx \]
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\[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\int { \frac {1}{{\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )}} \,d x } \]
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\[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\int { \frac {1}{{\left (4 \, x^{2} + 4 \, x - 1\right )}^{\frac {1}{3}} {\left (2 \, x + 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {1}{(1+2 x) \sqrt [3]{-1+4 x+4 x^2}} \, dx=\int \frac {1}{\left (2\,x+1\right )\,{\left (4\,x^2+4\,x-1\right )}^{1/3}} \,d x \]
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