Integrand size = 35, antiderivative size = 54 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {\left (c d^2-a e^2\right ) (a e+c d x)^4}{4 c^2 d^2}+\frac {e (a e+c d x)^5}{5 c^2 d^2} \]
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Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {640, 45} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {\left (c d^2-a e^2\right ) (a e+c d x)^4}{4 c^2 d^2}+\frac {e (a e+c d x)^5}{5 c^2 d^2} \]
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Rule 45
Rule 640
Rubi steps \begin{align*} \text {integral}& = \int (a e+c d x)^3 (d+e x) \, dx \\ & = \int \left (\frac {\left (c d^2-a e^2\right ) (a e+c d x)^3}{c d}+\frac {e (a e+c d x)^4}{c d}\right ) \, dx \\ & = \frac {\left (c d^2-a e^2\right ) (a e+c d x)^4}{4 c^2 d^2}+\frac {e (a e+c d x)^5}{5 c^2 d^2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {1}{20} x \left (10 a^3 e^3 (2 d+e x)+10 a^2 c d e^2 x (3 d+2 e x)+5 a c^2 d^2 e x^2 (4 d+3 e x)+c^3 d^3 x^3 (5 d+4 e x)\right ) \]
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Time = 2.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.83
method | result | size |
risch | \(\frac {1}{5} d^{3} e \,c^{3} x^{5}+\frac {3}{4} x^{4} d^{2} e^{2} c^{2} a +\frac {1}{4} c^{3} d^{4} x^{4}+a^{2} c d \,e^{3} x^{3}+a \,c^{2} d^{3} e \,x^{3}+\frac {1}{2} x^{2} e^{4} a^{3}+\frac {3}{2} x^{2} d^{2} e^{2} a^{2} c +a^{3} d \,e^{3} x\) | \(99\) |
parallelrisch | \(\frac {1}{5} d^{3} e \,c^{3} x^{5}+\frac {3}{4} x^{4} d^{2} e^{2} c^{2} a +\frac {1}{4} c^{3} d^{4} x^{4}+a^{2} c d \,e^{3} x^{3}+a \,c^{2} d^{3} e \,x^{3}+\frac {1}{2} x^{2} e^{4} a^{3}+\frac {3}{2} x^{2} d^{2} e^{2} a^{2} c +a^{3} d \,e^{3} x\) | \(99\) |
gosper | \(\frac {x \left (4 d^{3} e \,c^{3} x^{4}+15 d^{2} e^{2} c^{2} a \,x^{3}+5 d^{4} c^{3} x^{3}+20 a^{2} c d \,e^{3} x^{2}+20 d^{3} e \,c^{2} a \,x^{2}+10 e^{4} a^{3} x +30 d^{2} e^{2} a^{2} c x +20 a^{3} d \,e^{3}\right )}{20}\) | \(100\) |
default | \(\frac {d^{3} e \,c^{3} x^{5}}{5}+\frac {\left (2 d^{2} e^{2} c^{2} a +c^{2} d^{2} \left (e^{2} a +c \,d^{2}\right )\right ) x^{4}}{4}+\frac {\left (a^{2} c d \,e^{3}+2 a c d e \left (e^{2} a +c \,d^{2}\right )+d^{3} e \,c^{2} a \right ) x^{3}}{3}+\frac {\left (a^{2} e^{2} \left (e^{2} a +c \,d^{2}\right )+2 d^{2} e^{2} a^{2} c \right ) x^{2}}{2}+a^{3} d \,e^{3} x\) | \(136\) |
norman | \(\frac {\left (\frac {3}{2} d \,e^{4} a^{3}+\frac {3}{2} d^{3} e^{2} a^{2} c \right ) x^{2}+\left (\frac {3}{4} e^{3} a \,c^{2} d^{2}+\frac {9}{20} d^{4} e \,c^{3}\right ) x^{5}+\left (\frac {1}{2} a^{3} e^{5}+\frac {5}{2} d^{2} e^{3} a^{2} c +d^{4} a \,c^{2} e \right ) x^{3}+\left (d \,e^{4} a^{2} c +\frac {7}{4} d^{3} e^{2} c^{2} a +\frac {1}{4} d^{5} c^{3}\right ) x^{4}+d^{2} a^{3} e^{3} x +\frac {e^{2} c^{3} d^{3} x^{6}}{5}}{e x +d}\) | \(155\) |
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Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {1}{5} \, c^{3} d^{3} e x^{5} + a^{3} d e^{3} x + \frac {1}{4} \, {\left (c^{3} d^{4} + 3 \, a c^{2} d^{2} e^{2}\right )} x^{4} + {\left (a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} x^{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (49) = 98\).
Time = 0.06 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=a^{3} d e^{3} x + \frac {c^{3} d^{3} e x^{5}}{5} + x^{4} \cdot \left (\frac {3 a c^{2} d^{2} e^{2}}{4} + \frac {c^{3} d^{4}}{4}\right ) + x^{3} \left (a^{2} c d e^{3} + a c^{2} d^{3} e\right ) + x^{2} \left (\frac {a^{3} e^{4}}{2} + \frac {3 a^{2} c d^{2} e^{2}}{2}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.76 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {1}{5} \, c^{3} d^{3} e x^{5} + a^{3} d e^{3} x + \frac {1}{4} \, {\left (c^{3} d^{4} + 3 \, a c^{2} d^{2} e^{2}\right )} x^{4} + {\left (a c^{2} d^{3} e + a^{2} c d e^{3}\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}\right )} x^{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (50) = 100\).
Time = 0.26 (sec) , antiderivative size = 174, normalized size of antiderivative = 3.22 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=\frac {{\left (4 \, c^{3} d^{3} - \frac {15 \, c^{3} d^{4}}{e x + d} + \frac {20 \, c^{3} d^{5}}{{\left (e x + d\right )}^{2}} - \frac {10 \, c^{3} d^{6}}{{\left (e x + d\right )}^{3}} + \frac {15 \, a c^{2} d^{2} e^{2}}{e x + d} - \frac {40 \, a c^{2} d^{3} e^{2}}{{\left (e x + d\right )}^{2}} + \frac {30 \, a c^{2} d^{4} e^{2}}{{\left (e x + d\right )}^{3}} + \frac {20 \, a^{2} c d e^{4}}{{\left (e x + d\right )}^{2}} - \frac {30 \, a^{2} c d^{2} e^{4}}{{\left (e x + d\right )}^{3}} + \frac {10 \, a^{3} e^{6}}{{\left (e x + d\right )}^{3}}\right )} {\left (e x + d\right )}^{5}}{20 \, e^{4}} \]
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Time = 9.95 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^2} \, dx=x^2\,\left (\frac {a^3\,e^4}{2}+\frac {3\,c\,a^2\,d^2\,e^2}{2}\right )+x^4\,\left (\frac {c^3\,d^4}{4}+\frac {3\,a\,c^2\,d^2\,e^2}{4}\right )+\frac {c^3\,d^3\,e\,x^5}{5}+a^3\,d\,e^3\,x+a\,c\,d\,e\,x^3\,\left (c\,d^2+a\,e^2\right ) \]
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