Integrand size = 30, antiderivative size = 24 \[ \int \frac {(d+e x)^m}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {(d+e x)^{-5+m}}{c^3 e (5-m)} \]
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Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 12, 32} \[ \int \frac {(d+e x)^m}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=-\frac {(d+e x)^{m-5}}{c^3 e (5-m)} \]
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Rule 12
Rule 27
Rule 32
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^{-6+m}}{c^3} \, dx \\ & = \frac {\int (d+e x)^{-6+m} \, dx}{c^3} \\ & = -\frac {(d+e x)^{-5+m}}{c^3 e (5-m)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^m}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {(d+e x)^{-5+m}}{c^3 e (-5+m)} \]
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Time = 2.76 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {\left (e x +d \right )^{m}}{c^{3} e \left (-5+m \right ) \left (e x +d \right )^{5}}\) | \(27\) |
parallelrisch | \(\frac {\left (e x +d \right )^{m}}{c^{3} e \left (-5+m \right ) \left (e x +d \right )^{5}}\) | \(27\) |
norman | \(\frac {{\mathrm e}^{m \ln \left (e x +d \right )}}{c^{3} e \left (-5+m \right ) \left (e x +d \right )^{5}}\) | \(29\) |
gosper | \(\frac {\left (e x +d \right )^{-1+m}}{c^{3} e \left (-5+m \right ) \left (x^{2} e^{2}+2 d e x +d^{2}\right )^{2}}\) | \(40\) |
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Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (21) = 42\).
Time = 0.38 (sec) , antiderivative size = 156, normalized size of antiderivative = 6.50 \[ \int \frac {(d+e x)^m}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {{\left (e x + d\right )}^{m}}{c^{3} d^{5} e m - 5 \, c^{3} d^{5} e + {\left (c^{3} e^{6} m - 5 \, c^{3} e^{6}\right )} x^{5} + 5 \, {\left (c^{3} d e^{5} m - 5 \, c^{3} d e^{5}\right )} x^{4} + 10 \, {\left (c^{3} d^{2} e^{4} m - 5 \, c^{3} d^{2} e^{4}\right )} x^{3} + 10 \, {\left (c^{3} d^{3} e^{3} m - 5 \, c^{3} d^{3} e^{3}\right )} x^{2} + 5 \, {\left (c^{3} d^{4} e^{2} m - 5 \, c^{3} d^{4} e^{2}\right )} x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (17) = 34\).
Time = 1.78 (sec) , antiderivative size = 201, normalized size of antiderivative = 8.38 \[ \int \frac {(d+e x)^m}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\begin {cases} \frac {x}{c^{3} d} & \text {for}\: e = 0 \wedge m = 5 \\\frac {d^{m} x}{c^{3} d^{6}} & \text {for}\: e = 0 \\\frac {\log {\left (\frac {d}{e} + x \right )}}{c^{3} e} & \text {for}\: m = 5 \\\frac {\left (d + e x\right )^{m}}{c^{3} d^{5} e m - 5 c^{3} d^{5} e + 5 c^{3} d^{4} e^{2} m x - 25 c^{3} d^{4} e^{2} x + 10 c^{3} d^{3} e^{3} m x^{2} - 50 c^{3} d^{3} e^{3} x^{2} + 10 c^{3} d^{2} e^{4} m x^{3} - 50 c^{3} d^{2} e^{4} x^{3} + 5 c^{3} d e^{5} m x^{4} - 25 c^{3} d e^{5} x^{4} + c^{3} e^{6} m x^{5} - 5 c^{3} e^{6} x^{5}} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (21) = 42\).
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 4.12 \[ \int \frac {(d+e x)^m}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {{\left (e x + d\right )}^{m}}{c^{3} e^{6} {\left (m - 5\right )} x^{5} + 5 \, c^{3} d e^{5} {\left (m - 5\right )} x^{4} + 10 \, c^{3} d^{2} e^{4} {\left (m - 5\right )} x^{3} + 10 \, c^{3} d^{3} e^{3} {\left (m - 5\right )} x^{2} + 5 \, c^{3} d^{4} e^{2} {\left (m - 5\right )} x + c^{3} d^{5} e {\left (m - 5\right )}} \]
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\[ \int \frac {(d+e x)^m}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{3}} \,d x } \]
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Time = 9.81 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.00 \[ \int \frac {(d+e x)^m}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {{\left (d+e\,x\right )}^m}{c^3\,e^6\,\left (m-5\right )\,\left (x^5+\frac {d^5}{e^5}+\frac {5\,d\,x^4}{e}+\frac {5\,d^4\,x}{e^4}+\frac {10\,d^2\,x^3}{e^2}+\frac {10\,d^3\,x^2}{e^3}\right )} \]
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