Integrand size = 13, antiderivative size = 54 \[ \int \frac {1}{x \left (a+c x^4\right )^3} \, dx=\frac {1}{8 a \left (a+c x^4\right )^2}+\frac {1}{4 a^2 \left (a+c x^4\right )}+\frac {\log (x)}{a^3}-\frac {\log \left (a+c x^4\right )}{4 a^3} \]
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Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 46} \[ \int \frac {1}{x \left (a+c x^4\right )^3} \, dx=-\frac {\log \left (a+c x^4\right )}{4 a^3}+\frac {\log (x)}{a^3}+\frac {1}{4 a^2 \left (a+c x^4\right )}+\frac {1}{8 a \left (a+c x^4\right )^2} \]
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Rule 46
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {1}{x (a+c x)^3} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{a^3 x}-\frac {c}{a (a+c x)^3}-\frac {c}{a^2 (a+c x)^2}-\frac {c}{a^3 (a+c x)}\right ) \, dx,x,x^4\right ) \\ & = \frac {1}{8 a \left (a+c x^4\right )^2}+\frac {1}{4 a^2 \left (a+c x^4\right )}+\frac {\log (x)}{a^3}-\frac {\log \left (a+c x^4\right )}{4 a^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \left (a+c x^4\right )^3} \, dx=\frac {\frac {a \left (3 a+2 c x^4\right )}{\left (a+c x^4\right )^2}+8 \log (x)-2 \log \left (a+c x^4\right )}{8 a^3} \]
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Time = 4.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85
method | result | size |
risch | \(\frac {\frac {c \,x^{4}}{4 a^{2}}+\frac {3}{8 a}}{\left (x^{4} c +a \right )^{2}}+\frac {\ln \left (x \right )}{a^{3}}-\frac {\ln \left (x^{4} c +a \right )}{4 a^{3}}\) | \(46\) |
norman | \(\frac {-\frac {c \,x^{4}}{2 a^{2}}-\frac {3 c^{2} x^{8}}{8 a^{3}}}{\left (x^{4} c +a \right )^{2}}+\frac {\ln \left (x \right )}{a^{3}}-\frac {\ln \left (x^{4} c +a \right )}{4 a^{3}}\) | \(52\) |
default | \(\frac {\ln \left (x \right )}{a^{3}}-\frac {c \left (-\frac {a}{2 c \left (x^{4} c +a \right )}+\frac {\ln \left (x^{4} c +a \right )}{2 c}-\frac {a^{2}}{4 c \left (x^{4} c +a \right )^{2}}\right )}{2 a^{3}}\) | \(60\) |
parallelrisch | \(\frac {8 \ln \left (x \right ) x^{8} c^{2}-2 \ln \left (x^{4} c +a \right ) x^{8} c^{2}-3 c^{2} x^{8}+16 \ln \left (x \right ) x^{4} a c -4 \ln \left (x^{4} c +a \right ) x^{4} a c -4 a \,x^{4} c +8 a^{2} \ln \left (x \right )-2 \ln \left (x^{4} c +a \right ) a^{2}}{8 a^{3} \left (x^{4} c +a \right )^{2}}\) | \(101\) |
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Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.67 \[ \int \frac {1}{x \left (a+c x^4\right )^3} \, dx=\frac {2 \, a c x^{4} + 3 \, a^{2} - 2 \, {\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \log \left (c x^{4} + a\right ) + 8 \, {\left (c^{2} x^{8} + 2 \, a c x^{4} + a^{2}\right )} \log \left (x\right )}{8 \, {\left (a^{3} c^{2} x^{8} + 2 \, a^{4} c x^{4} + a^{5}\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x \left (a+c x^4\right )^3} \, dx=\frac {3 a + 2 c x^{4}}{8 a^{4} + 16 a^{3} c x^{4} + 8 a^{2} c^{2} x^{8}} + \frac {\log {\left (x \right )}}{a^{3}} - \frac {\log {\left (\frac {a}{c} + x^{4} \right )}}{4 a^{3}} \]
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Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x \left (a+c x^4\right )^3} \, dx=\frac {2 \, c x^{4} + 3 \, a}{8 \, {\left (a^{2} c^{2} x^{8} + 2 \, a^{3} c x^{4} + a^{4}\right )}} - \frac {\log \left (c x^{4} + a\right )}{4 \, a^{3}} + \frac {\log \left (x^{4}\right )}{4 \, a^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x \left (a+c x^4\right )^3} \, dx=\frac {\log \left (x^{4}\right )}{4 \, a^{3}} - \frac {\log \left ({\left | c x^{4} + a \right |}\right )}{4 \, a^{3}} + \frac {3 \, c^{2} x^{8} + 8 \, a c x^{4} + 6 \, a^{2}}{8 \, {\left (c x^{4} + a\right )}^{2} a^{3}} \]
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Time = 5.43 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x \left (a+c x^4\right )^3} \, dx=\frac {\ln \left (x\right )}{a^3}+\frac {\frac {3}{8\,a}+\frac {c\,x^4}{4\,a^2}}{a^2+2\,a\,c\,x^4+c^2\,x^8}-\frac {\ln \left (c\,x^4+a\right )}{4\,a^3} \]
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