Integrand size = 17, antiderivative size = 64 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\frac {a^2 x^n}{b^3 n}-\frac {a x^{2 n}}{2 b^2 n}+\frac {x^{3 n}}{3 b n}-\frac {a^3 \log \left (a+b x^n\right )}{b^4 n} \]
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Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {272, 45} \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=-\frac {a^3 \log \left (a+b x^n\right )}{b^4 n}+\frac {a^2 x^n}{b^3 n}-\frac {a x^{2 n}}{2 b^2 n}+\frac {x^{3 n}}{3 b n} \]
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Rule 45
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^3}{a+b x} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a^2}{b^3}-\frac {a x}{b^2}+\frac {x^2}{b}-\frac {a^3}{b^3 (a+b x)}\right ) \, dx,x,x^n\right )}{n} \\ & = \frac {a^2 x^n}{b^3 n}-\frac {a x^{2 n}}{2 b^2 n}+\frac {x^{3 n}}{3 b n}-\frac {a^3 \log \left (a+b x^n\right )}{b^4 n} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\frac {b x^n \left (6 a^2-3 a b x^n+2 b^2 x^{2 n}\right )-6 a^3 \log \left (a+b x^n\right )}{6 b^4 n} \]
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Time = 4.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98
method | result | size |
risch | \(\frac {x^{3 n}}{3 b n}-\frac {a \,x^{2 n}}{2 b^{2} n}+\frac {a^{2} x^{n}}{b^{3} n}-\frac {a^{3} \ln \left (x^{n}+\frac {a}{b}\right )}{b^{4} n}\) | \(63\) |
norman | \(\frac {a^{2} {\mathrm e}^{n \ln \left (x \right )}}{b^{3} n}+\frac {{\mathrm e}^{3 n \ln \left (x \right )}}{3 b n}-\frac {a \,{\mathrm e}^{2 n \ln \left (x \right )}}{2 b^{2} n}-\frac {a^{3} \ln \left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{b^{4} n}\) | \(69\) |
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Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\frac {2 \, b^{3} x^{3 \, n} - 3 \, a b^{2} x^{2 \, n} + 6 \, a^{2} b x^{n} - 6 \, a^{3} \log \left (b x^{n} + a\right )}{6 \, b^{4} n} \]
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Time = 1.52 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\begin {cases} \frac {\log {\left (x \right )}}{a} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x x^{4 n - 1}}{4 a n} & \text {for}\: b = 0 \\\frac {\log {\left (x \right )}}{a + b} & \text {for}\: n = 0 \\- \frac {a^{3} \log {\left (\frac {a}{b} + x^{n} \right )}}{b^{4} n} + \frac {a^{2} x^{n}}{b^{3} n} - \frac {a x^{2 n}}{2 b^{2} n} + \frac {x^{3 n}}{3 b n} & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=-\frac {a^{3} \log \left (\frac {b x^{n} + a}{b}\right )}{b^{4} n} + \frac {2 \, b^{2} x^{3 \, n} - 3 \, a b x^{2 \, n} + 6 \, a^{2} x^{n}}{6 \, b^{3} n} \]
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\[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\int { \frac {x^{4 \, n - 1}}{b x^{n} + a} \,d x } \]
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Timed out. \[ \int \frac {x^{-1+4 n}}{a+b x^n} \, dx=\int \frac {x^{4\,n-1}}{a+b\,x^n} \,d x \]
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