Integrand size = 18, antiderivative size = 70 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx=-\frac {(b d-a e) (B d-A e)}{2 e^3 (d+e x)^2}+\frac {2 b B d-A b e-a B e}{e^3 (d+e x)}+\frac {b B \log (d+e x)}{e^3} \]
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Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx=-\frac {(b d-a e) (B d-A e)}{2 e^3 (d+e x)^2}+\frac {-a B e-A b e+2 b B d}{e^3 (d+e x)}+\frac {b B \log (d+e x)}{e^3} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-b d+a e) (-B d+A e)}{e^2 (d+e x)^3}+\frac {-2 b B d+A b e+a B e}{e^2 (d+e x)^2}+\frac {b B}{e^2 (d+e x)}\right ) \, dx \\ & = -\frac {(b d-a e) (B d-A e)}{2 e^3 (d+e x)^2}+\frac {2 b B d-A b e-a B e}{e^3 (d+e x)}+\frac {b B \log (d+e x)}{e^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx=\frac {-a e (A e+B (d+2 e x))+b (-A e (d+2 e x)+B d (3 d+4 e x))+2 b B (d+e x)^2 \log (d+e x)}{2 e^3 (d+e x)^2} \]
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Time = 2.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03
method | result | size |
norman | \(\frac {-\frac {A a \,e^{2}+A b d e +B a d e -3 b B \,d^{2}}{2 e^{3}}-\frac {\left (A b e +B a e -2 B b d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {b B \ln \left (e x +d \right )}{e^{3}}\) | \(72\) |
risch | \(\frac {-\frac {A a \,e^{2}+A b d e +B a d e -3 b B \,d^{2}}{2 e^{3}}-\frac {\left (A b e +B a e -2 B b d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {b B \ln \left (e x +d \right )}{e^{3}}\) | \(72\) |
default | \(-\frac {A b e +B a e -2 B b d}{e^{3} \left (e x +d \right )}-\frac {A a \,e^{2}-A b d e -B a d e +b B \,d^{2}}{2 e^{3} \left (e x +d \right )^{2}}+\frac {b B \ln \left (e x +d \right )}{e^{3}}\) | \(77\) |
parallelrisch | \(-\frac {-2 B \ln \left (e x +d \right ) x^{2} b \,e^{2}-4 B \ln \left (e x +d \right ) x b d e +2 A x b \,e^{2}-2 B \ln \left (e x +d \right ) b \,d^{2}+2 B x a \,e^{2}-4 B x b d e +A a \,e^{2}+A b d e +B a d e -3 b B \,d^{2}}{2 e^{3} \left (e x +d \right )^{2}}\) | \(102\) |
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Time = 0.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx=\frac {3 \, B b d^{2} - A a e^{2} - {\left (B a + A b\right )} d e + 2 \, {\left (2 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x + 2 \, {\left (B b e^{2} x^{2} + 2 \, B b d e x + B b d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]
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Time = 0.41 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx=\frac {B b \log {\left (d + e x \right )}}{e^{3}} + \frac {- A a e^{2} - A b d e - B a d e + 3 B b d^{2} + x \left (- 2 A b e^{2} - 2 B a e^{2} + 4 B b d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \]
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Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx=\frac {3 \, B b d^{2} - A a e^{2} - {\left (B a + A b\right )} d e + 2 \, {\left (2 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac {B b \log \left (e x + d\right )}{e^{3}} \]
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Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx=\frac {B b \log \left ({\left | e x + d \right |}\right )}{e^{3}} + \frac {2 \, {\left (2 \, B b d - B a e - A b e\right )} x + \frac {3 \, B b d^{2} - B a d e - A b d e - A a e^{2}}{e}}{2 \, {\left (e x + d\right )}^{2} e^{2}} \]
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Time = 1.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx=\frac {B\,b\,\ln \left (d+e\,x\right )}{e^3}-\frac {\frac {A\,a\,e^2-3\,B\,b\,d^2+A\,b\,d\,e+B\,a\,d\,e}{2\,e^3}+\frac {x\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2} \]
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