Integrand size = 19, antiderivative size = 58 \[ \int x^{2 n-3 (1+n)} (a+b x)^n \, dx=-\frac {x^{-2-n} (a+b x)^{1+n}}{a (2+n)}+\frac {b x^{-1-n} (a+b x)^{1+n}}{a^2 (1+n) (2+n)} \]
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Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int x^{2 n-3 (1+n)} (a+b x)^n \, dx=\frac {b x^{-n-1} (a+b x)^{n+1}}{a^2 (n+1) (n+2)}-\frac {x^{-n-2} (a+b x)^{n+1}}{a (n+2)} \]
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Rule 37
Rule 47
Rubi steps \begin{align*} \text {integral}& = -\frac {x^{-2-n} (a+b x)^{1+n}}{a (2+n)}-\frac {b \int x^{-2-n} (a+b x)^n \, dx}{a (2+n)} \\ & = -\frac {x^{-2-n} (a+b x)^{1+n}}{a (2+n)}+\frac {b x^{-1-n} (a+b x)^{1+n}}{a^2 (1+n) (2+n)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.69 \[ \int x^{2 n-3 (1+n)} (a+b x)^n \, dx=-\frac {x^{-2-n} (a+a n-b x) (a+b x)^{1+n}}{a^2 (1+n) (2+n)} \]
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Time = 0.00 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71
method | result | size |
gosper | \(-\frac {x^{-2-n} \left (b x +a \right )^{1+n} \left (a n -b x +a \right )}{a^{2} \left (1+n \right ) \left (2+n \right )}\) | \(41\) |
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Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.10 \[ \int x^{2 n-3 (1+n)} (a+b x)^n \, dx=-\frac {{\left (a b n x^{2} - b^{2} x^{3} + {\left (a^{2} n + a^{2}\right )} x\right )} {\left (b x + a\right )}^{n} x^{-n - 3}}{a^{2} n^{2} + 3 \, a^{2} n + 2 \, a^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (48) = 96\).
Time = 2.03 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.31 \[ \int x^{2 n-3 (1+n)} (a+b x)^n \, dx=- \frac {a a^{n} n x^{- n - 2} \left (1 + \frac {b x}{a}\right )^{n + 2} \Gamma \left (- n - 2\right )}{a \Gamma \left (- n\right ) + b x \Gamma \left (- n\right )} - \frac {a a^{n} x^{- n - 2} \left (1 + \frac {b x}{a}\right )^{n + 2} \Gamma \left (- n - 2\right )}{a \Gamma \left (- n\right ) + b x \Gamma \left (- n\right )} + \frac {a^{n} b x x^{- n - 2} \left (1 + \frac {b x}{a}\right )^{n + 2} \Gamma \left (- n - 2\right )}{a \Gamma \left (- n\right ) + b x \Gamma \left (- n\right )} \]
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\[ \int x^{2 n-3 (1+n)} (a+b x)^n \, dx=\int { {\left (b x + a\right )}^{n} x^{-n - 3} \,d x } \]
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\[ \int x^{2 n-3 (1+n)} (a+b x)^n \, dx=\int { {\left (b x + a\right )}^{n} x^{-n - 3} \,d x } \]
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Time = 0.00 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.48 \[ \int x^{2 n-3 (1+n)} (a+b x)^n \, dx=-{\left (a+b\,x\right )}^n\,\left (\frac {x\,\left (n+1\right )}{x^{n+3}\,\left (n^2+3\,n+2\right )}-\frac {b^2\,x^3}{a^2\,x^{n+3}\,\left (n^2+3\,n+2\right )}+\frac {b\,n\,x^2}{a\,x^{n+3}\,\left (n^2+3\,n+2\right )}\right ) \]
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