Integrand size = 24, antiderivative size = 65 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^6} \, dx=-\frac {(b d-a e)^2}{5 e^3 (d+e x)^5}+\frac {b (b d-a e)}{2 e^3 (d+e x)^4}-\frac {b^2}{3 e^3 (d+e x)^3} \]
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Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 45} \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^6} \, dx=\frac {b (b d-a e)}{2 e^3 (d+e x)^4}-\frac {(b d-a e)^2}{5 e^3 (d+e x)^5}-\frac {b^2}{3 e^3 (d+e x)^3} \]
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Rule 27
Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b x)^2}{(d+e x)^6} \, dx \\ & = \int \left (\frac {(-b d+a e)^2}{e^2 (d+e x)^6}-\frac {2 b (b d-a e)}{e^2 (d+e x)^5}+\frac {b^2}{e^2 (d+e x)^4}\right ) \, dx \\ & = -\frac {(b d-a e)^2}{5 e^3 (d+e x)^5}+\frac {b (b d-a e)}{2 e^3 (d+e x)^4}-\frac {b^2}{3 e^3 (d+e x)^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^6} \, dx=-\frac {6 a^2 e^2+3 a b e (d+5 e x)+b^2 \left (d^2+5 d e x+10 e^2 x^2\right )}{30 e^3 (d+e x)^5} \]
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Time = 2.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95
method | result | size |
gosper | \(-\frac {10 x^{2} b^{2} e^{2}+15 x a b \,e^{2}+5 b^{2} d e x +6 a^{2} e^{2}+3 a b d e +b^{2} d^{2}}{30 e^{3} \left (e x +d \right )^{5}}\) | \(62\) |
risch | \(\frac {-\frac {b^{2} x^{2}}{3 e}-\frac {b \left (3 a e +b d \right ) x}{6 e^{2}}-\frac {6 a^{2} e^{2}+3 a b d e +b^{2} d^{2}}{30 e^{3}}}{\left (e x +d \right )^{5}}\) | \(63\) |
parallelrisch | \(\frac {-10 b^{2} x^{2} e^{4}-15 a b \,e^{4} x -5 b^{2} d \,e^{3} x -6 a^{2} e^{4}-3 a b d \,e^{3}-b^{2} d^{2} e^{2}}{30 e^{5} \left (e x +d \right )^{5}}\) | \(70\) |
default | \(-\frac {a^{2} e^{2}-2 a b d e +b^{2} d^{2}}{5 e^{3} \left (e x +d \right )^{5}}-\frac {b^{2}}{3 e^{3} \left (e x +d \right )^{3}}-\frac {b \left (a e -b d \right )}{2 e^{3} \left (e x +d \right )^{4}}\) | \(71\) |
norman | \(\frac {-\frac {b^{2} x^{2}}{3 e}-\frac {\left (3 a b \,e^{3}+b^{2} d \,e^{2}\right ) x}{6 e^{4}}-\frac {6 a^{2} e^{4}+3 a b d \,e^{3}+b^{2} d^{2} e^{2}}{30 e^{5}}}{\left (e x +d \right )^{5}}\) | \(75\) |
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Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.68 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^6} \, dx=-\frac {10 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 3 \, a b d e + 6 \, a^{2} e^{2} + 5 \, {\left (b^{2} d e + 3 \, a b e^{2}\right )} x}{30 \, {\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (54) = 108\).
Time = 0.52 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.78 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^6} \, dx=\frac {- 6 a^{2} e^{2} - 3 a b d e - b^{2} d^{2} - 10 b^{2} e^{2} x^{2} + x \left (- 15 a b e^{2} - 5 b^{2} d e\right )}{30 d^{5} e^{3} + 150 d^{4} e^{4} x + 300 d^{3} e^{5} x^{2} + 300 d^{2} e^{6} x^{3} + 150 d e^{7} x^{4} + 30 e^{8} x^{5}} \]
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Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.68 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^6} \, dx=-\frac {10 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 3 \, a b d e + 6 \, a^{2} e^{2} + 5 \, {\left (b^{2} d e + 3 \, a b e^{2}\right )} x}{30 \, {\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^6} \, dx=-\frac {10 \, b^{2} e^{2} x^{2} + 5 \, b^{2} d e x + 15 \, a b e^{2} x + b^{2} d^{2} + 3 \, a b d e + 6 \, a^{2} e^{2}}{30 \, {\left (e x + d\right )}^{5} e^{3}} \]
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Time = 9.96 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.65 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^6} \, dx=-\frac {\frac {6\,a^2\,e^2+3\,a\,b\,d\,e+b^2\,d^2}{30\,e^3}+\frac {b^2\,x^2}{3\,e}+\frac {b\,x\,\left (3\,a\,e+b\,d\right )}{6\,e^2}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \]
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