Integrand size = 30, antiderivative size = 39 \[ \int (d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{2+p}}{2 c^2 e (2+p)} \]
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Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {657, 643} \[ \int (d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{p+2}}{2 c^2 e (p+2)} \]
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Rule 643
Rule 657
Rubi steps \begin{align*} \text {integral}& = \frac {\int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{1+p} \, dx}{c} \\ & = \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{2+p}}{2 c^2 e (2+p)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int (d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {(d+e x)^4 \left (c (d+e x)^2\right )^p}{e (4+2 p)} \]
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Time = 2.53 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03
method | result | size |
gosper | \(\frac {\left (e x +d \right )^{4} \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{p}}{2 e \left (2+p \right )}\) | \(40\) |
risch | \(\frac {\left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right ) \left (c \left (e x +d \right )^{2}\right )^{p}}{2 e \left (2+p \right )}\) | \(60\) |
parallelrisch | \(\frac {x^{4} {\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d \,e^{4}+4 x^{3} {\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d^{2} e^{3}+6 x^{2} {\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d^{3} e^{2}+4 x {\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d^{4} e +{\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d^{5}}{2 e \left (2+p \right ) d}\) | \(156\) |
norman | \(\frac {2 d^{3} x \,{\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{2+p}+\frac {d^{4} {\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{2 e \left (2+p \right )}+\frac {e^{3} x^{4} {\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{4+2 p}+\frac {2 d \,e^{2} x^{3} {\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{2+p}+\frac {3 d^{2} e \,x^{2} {\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{2+p}\) | \(187\) |
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Time = 0.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.82 \[ \int (d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )} {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, {\left (e p + 2 \, e\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (34) = 68\).
Time = 0.39 (sec) , antiderivative size = 233, normalized size of antiderivative = 5.97 \[ \int (d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\begin {cases} \frac {x}{c^{2} d} & \text {for}\: e = 0 \wedge p = -2 \\d^{3} x \left (c d^{2}\right )^{p} & \text {for}\: e = 0 \\\frac {\log {\left (\frac {d}{e} + x \right )}}{c^{2} e} & \text {for}\: p = -2 \\\frac {d^{4} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} + \frac {4 d^{3} e x \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} + \frac {6 d^{2} e^{2} x^{2} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} + \frac {4 d e^{3} x^{3} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} + \frac {e^{4} x^{4} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 4 e} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (37) = 74\).
Time = 0.21 (sec) , antiderivative size = 313, normalized size of antiderivative = 8.03 \[ \int (d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (c^{p} e x + c^{p} d\right )} {\left (e x + d\right )}^{2 \, p} d^{3}}{e {\left (2 \, p + 1\right )}} + \frac {3 \, {\left (c^{p} e^{2} {\left (2 \, p + 1\right )} x^{2} + 2 \, c^{p} d e p x - c^{p} d^{2}\right )} {\left (e x + d\right )}^{2 \, p} d^{2}}{2 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} e} + \frac {3 \, {\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} c^{p} e^{3} x^{3} + {\left (2 \, p^{2} + p\right )} c^{p} d e^{2} x^{2} - 2 \, c^{p} d^{2} e p x + c^{p} d^{3}\right )} {\left (e x + d\right )}^{2 \, p} d}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} e} + \frac {{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} c^{p} e^{4} x^{4} + 2 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} c^{p} d e^{3} x^{3} - 3 \, {\left (2 \, p^{2} + p\right )} c^{p} d^{2} e^{2} x^{2} + 6 \, c^{p} d^{3} e p x - 3 \, c^{p} d^{4}\right )} {\left (e x + d\right )}^{2 \, p}}{2 \, {\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} e} \]
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Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.46 \[ \int (d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{2} {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, c^{2} e {\left (p + 2\right )}} \]
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Time = 9.90 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.31 \[ \int (d+e x)^3 \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx={\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p\,\left (\frac {d^4}{2\,e\,\left (p+2\right )}+\frac {e^3\,x^4}{2\,\left (p+2\right )}+\frac {2\,d^3\,x}{p+2}+\frac {3\,d^2\,e\,x^2}{p+2}+\frac {2\,d\,e^2\,x^3}{p+2}\right ) \]
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