Integrand size = 25, antiderivative size = 59 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {a (A b-a B)}{3 b^3 (a+b x)^3}-\frac {A b-2 a B}{2 b^3 (a+b x)^2}-\frac {B}{b^3 (a+b x)} \]
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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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {27, 78} \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {A b-2 a B}{2 b^3 (a+b x)^2}+\frac {a (A b-a B)}{3 b^3 (a+b x)^3}-\frac {B}{b^3 (a+b x)} \]
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Rule 27
Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \frac {x (A+B x)}{(a+b x)^4} \, dx \\ & = \int \left (\frac {a (-A b+a B)}{b^2 (a+b x)^4}+\frac {A b-2 a B}{b^2 (a+b x)^3}+\frac {B}{b^2 (a+b x)^2}\right ) \, dx \\ & = \frac {a (A b-a B)}{3 b^3 (a+b x)^3}-\frac {A b-2 a B}{2 b^3 (a+b x)^2}-\frac {B}{b^3 (a+b x)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.71 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {2 a^2 B+3 b^2 x (A+2 B x)+a b (A+6 B x)}{6 b^3 (a+b x)^3} \]
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Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80
method | result | size |
norman | \(\frac {-\frac {B \,x^{2}}{b}-\frac {\left (A b +2 B a \right ) x}{2 b^{2}}-\frac {a \left (A b +2 B a \right )}{6 b^{3}}}{\left (b x +a \right )^{3}}\) | \(47\) |
default | \(-\frac {A b -2 B a}{2 b^{3} \left (b x +a \right )^{2}}-\frac {B}{b^{3} \left (b x +a \right )}+\frac {a \left (A b -B a \right )}{3 b^{3} \left (b x +a \right )^{3}}\) | \(56\) |
gosper | \(-\frac {6 b^{2} B \,x^{2}+3 A \,b^{2} x +6 B a b x +A b a +2 B \,a^{2}}{6 \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right ) b^{3}}\) | \(64\) |
parallelrisch | \(-\frac {6 b^{2} B \,x^{2}+3 A \,b^{2} x +6 B a b x +A b a +2 B \,a^{2}}{6 \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right ) b^{3}}\) | \(64\) |
risch | \(\frac {-\frac {B \,x^{2}}{b}-\frac {\left (A b +2 B a \right ) x}{2 b^{2}}-\frac {a \left (A b +2 B a \right )}{6 b^{3}}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}\) | \(65\) |
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Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {6 \, B b^{2} x^{2} + 2 \, B a^{2} + A a b + 3 \, {\left (2 \, B a b + A b^{2}\right )} x}{6 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \]
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Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {- A a b - 2 B a^{2} - 6 B b^{2} x^{2} + x \left (- 3 A b^{2} - 6 B a b\right )}{6 a^{3} b^{3} + 18 a^{2} b^{4} x + 18 a b^{5} x^{2} + 6 b^{6} x^{3}} \]
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Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {6 \, B b^{2} x^{2} + 2 \, B a^{2} + A a b + 3 \, {\left (2 \, B a b + A b^{2}\right )} x}{6 \, {\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.76 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {6 \, B b^{2} x^{2} + 6 \, B a b x + 3 \, A b^{2} x + 2 \, B a^{2} + A a b}{6 \, {\left (b x + a\right )}^{3} b^{3}} \]
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Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\frac {B\,x^2}{b}+\frac {a\,\left (A\,b+2\,B\,a\right )}{6\,b^3}+\frac {x\,\left (A\,b+2\,B\,a\right )}{2\,b^2}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \]
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