Integrand size = 34, antiderivative size = 28 \[ \int \frac {-1+3 x-3 x^2+x^3}{1+4 x+6 x^2+4 x^3+x^4} \, dx=\frac {8}{3 (1+x)^3}-\frac {6}{(1+x)^2}+\frac {6}{1+x}+\log (1+x) \]
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Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1694, 45} \[ \int \frac {-1+3 x-3 x^2+x^3}{1+4 x+6 x^2+4 x^3+x^4} \, dx=\frac {6}{x+1}-\frac {6}{(x+1)^2}+\frac {8}{3 (x+1)^3}+\log (x+1) \]
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Rule 45
Rule 1694
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {(-2+x)^3}{x^4} \, dx,x,1+x\right ) \\ & = \text {Subst}\left (\int \left (-\frac {8}{x^4}+\frac {12}{x^3}-\frac {6}{x^2}+\frac {1}{x}\right ) \, dx,x,1+x\right ) \\ & = \frac {8}{3 (1+x)^3}-\frac {6}{(1+x)^2}+\frac {6}{1+x}+\log (1+x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-1+3 x-3 x^2+x^3}{1+4 x+6 x^2+4 x^3+x^4} \, dx=\frac {2 \left (4+9 x+9 x^2\right )}{3 (1+x)^3}+\log (1+x) \]
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Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79
method | result | size |
norman | \(\frac {6 x +6 x^{2}+\frac {8}{3}}{\left (x +1\right )^{3}}+\ln \left (x +1\right )\) | \(22\) |
default | \(\frac {8}{3 \left (x +1\right )^{3}}-\frac {6}{\left (x +1\right )^{2}}+\frac {6}{x +1}+\ln \left (x +1\right )\) | \(27\) |
risch | \(\frac {6 x +6 x^{2}+\frac {8}{3}}{x^{3}+3 x^{2}+3 x +1}+\ln \left (x +1\right )\) | \(32\) |
parallelrisch | \(\frac {3 \ln \left (x +1\right ) x^{3}+8+9 \ln \left (x +1\right ) x^{2}+9 \ln \left (x +1\right ) x +18 x^{2}+3 \ln \left (x +1\right )+18 x}{3 x^{3}+9 x^{2}+9 x +3}\) | \(59\) |
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Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-1+3 x-3 x^2+x^3}{1+4 x+6 x^2+4 x^3+x^4} \, dx=\frac {18 \, x^{2} + 3 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} \log \left (x + 1\right ) + 18 \, x + 8}{3 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )}} \]
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Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-1+3 x-3 x^2+x^3}{1+4 x+6 x^2+4 x^3+x^4} \, dx=\frac {18 x^{2} + 18 x + 8}{3 x^{3} + 9 x^{2} + 9 x + 3} + \log {\left (x + 1 \right )} \]
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Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-1+3 x-3 x^2+x^3}{1+4 x+6 x^2+4 x^3+x^4} \, dx=\frac {2 \, {\left (9 \, x^{2} + 9 \, x + 4\right )}}{3 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )}} + \log \left (x + 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-1+3 x-3 x^2+x^3}{1+4 x+6 x^2+4 x^3+x^4} \, dx=\frac {2 \, {\left (9 \, x^{2} + 9 \, x + 4\right )}}{3 \, {\left (x + 1\right )}^{3}} + \log \left ({\left | x + 1 \right |}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-1+3 x-3 x^2+x^3}{1+4 x+6 x^2+4 x^3+x^4} \, dx=\ln \left (x+1\right )+\frac {6\,x^2+6\,x+\frac {8}{3}}{{\left (x+1\right )}^3} \]
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