Integrand size = 24, antiderivative size = 59 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx=-\frac {(b d-a e)^2}{2 e^3 (d+e x)^2}+\frac {2 b (b d-a e)}{e^3 (d+e x)}+\frac {b^2 \log (d+e x)}{e^3} \]
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Time = 0.03 (sec) , antiderivative size = 59, 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} \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx=\frac {2 b (b d-a e)}{e^3 (d+e x)}-\frac {(b d-a e)^2}{2 e^3 (d+e x)^2}+\frac {b^2 \log (d+e x)}{e^3} \]
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Rule 27
Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b x)^2}{(d+e x)^3} \, dx \\ & = \int \left (\frac {(-b d+a e)^2}{e^2 (d+e x)^3}-\frac {2 b (b d-a e)}{e^2 (d+e x)^2}+\frac {b^2}{e^2 (d+e x)}\right ) \, dx \\ & = -\frac {(b d-a e)^2}{2 e^3 (d+e x)^2}+\frac {2 b (b d-a e)}{e^3 (d+e x)}+\frac {b^2 \log (d+e x)}{e^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx=\frac {\frac {(b d-a e) (3 b d+a e+4 b e x)}{(d+e x)^2}+2 b^2 \log (d+e x)}{2 e^3} \]
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Time = 2.40 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {-\frac {2 b \left (a e -b d \right ) x}{e^{2}}-\frac {a^{2} e^{2}+2 a b d e -3 b^{2} d^{2}}{2 e^{3}}}{\left (e x +d \right )^{2}}+\frac {b^{2} \ln \left (e x +d \right )}{e^{3}}\) | \(66\) |
norman | \(\frac {-\frac {a^{2} e^{2}+2 a b d e -3 b^{2} d^{2}}{2 e^{3}}-\frac {2 \left (a e b -b^{2} d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {b^{2} \ln \left (e x +d \right )}{e^{3}}\) | \(68\) |
default | \(-\frac {2 b \left (a e -b d \right )}{e^{3} \left (e x +d \right )}-\frac {a^{2} e^{2}-2 a b d e +b^{2} d^{2}}{2 e^{3} \left (e x +d \right )^{2}}+\frac {b^{2} \ln \left (e x +d \right )}{e^{3}}\) | \(69\) |
parallelrisch | \(\frac {2 \ln \left (e x +d \right ) x^{2} b^{2} e^{2}+4 \ln \left (e x +d \right ) x \,b^{2} d e +2 \ln \left (e x +d \right ) b^{2} d^{2}-4 x a b \,e^{2}+4 b^{2} d e x -a^{2} e^{2}-2 a b d e +3 b^{2} d^{2}}{2 e^{3} \left (e x +d \right )^{2}}\) | \(97\) |
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Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.69 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx=\frac {3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]
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Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx=\frac {b^{2} \log {\left (d + e x \right )}}{e^{3}} + \frac {- a^{2} e^{2} - 2 a b d e + 3 b^{2} d^{2} + x \left (- 4 a b e^{2} + 4 b^{2} d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \]
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Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx=\frac {3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac {b^{2} \log \left (e x + d\right )}{e^{3}} \]
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Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx=\frac {b^{2} \log \left ({\left | e x + d \right |}\right )}{e^{3}} + \frac {4 \, {\left (b^{2} d - a b e\right )} x + \frac {3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2}}{e}}{2 \, {\left (e x + d\right )}^{2} e^{2}} \]
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Time = 10.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.31 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^3} \, dx=\frac {b^2\,\ln \left (d+e\,x\right )}{e^3}-\frac {\frac {a^2\,e^2+2\,a\,b\,d\,e-3\,b^2\,d^2}{2\,e^3}+\frac {2\,b\,x\,\left (a\,e-b\,d\right )}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2} \]
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