Integrand size = 29, antiderivative size = 51 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {d^2 x}{b^2}-\frac {(b c-a d)^2}{b^3 (a+b x)}+\frac {2 d (b c-a d) \log (a+b x)}{b^3} \]
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Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {640, 45} \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=-\frac {(b c-a d)^2}{b^3 (a+b x)}+\frac {2 d (b c-a d) \log (a+b x)}{b^3}+\frac {d^2 x}{b^2} \]
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Rule 45
Rule 640
Rubi steps \begin{align*} \text {integral}& = \int \frac {(c+d x)^2}{(a+b x)^2} \, dx \\ & = \int \left (\frac {d^2}{b^2}+\frac {(b c-a d)^2}{b^2 (a+b x)^2}+\frac {2 d (b c-a d)}{b^2 (a+b x)}\right ) \, dx \\ & = \frac {d^2 x}{b^2}-\frac {(b c-a d)^2}{b^3 (a+b x)}+\frac {2 d (b c-a d) \log (a+b x)}{b^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {b d^2 x-\frac {(b c-a d)^2}{a+b x}+2 d (b c-a d) \log (a+b x)}{b^3} \]
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Time = 2.70 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24
method | result | size |
default | \(\frac {d^{2} x}{b^{2}}-\frac {2 d \left (a d -b c \right ) \ln \left (b x +a \right )}{b^{3}}-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{b^{3} \left (b x +a \right )}\) | \(63\) |
risch | \(\frac {d^{2} x}{b^{2}}-\frac {2 d^{2} \ln \left (b x +a \right ) a}{b^{3}}+\frac {2 d \ln \left (b x +a \right ) c}{b^{2}}-\frac {a^{2} d^{2}}{b^{3} \left (b x +a \right )}+\frac {2 a c d}{b^{2} \left (b x +a \right )}-\frac {c^{2}}{b \left (b x +a \right )}\) | \(86\) |
parallelrisch | \(-\frac {2 \ln \left (b x +a \right ) x a b \,d^{2}-2 \ln \left (b x +a \right ) x \,b^{2} c d -d^{2} x^{2} b^{2}+2 \ln \left (b x +a \right ) a^{2} d^{2}-2 \ln \left (b x +a \right ) a b c d +2 a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{b^{3} \left (b x +a \right )}\) | \(100\) |
norman | \(\frac {b \,d^{2} x^{4}-\frac {a^{2} \left (4 a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{b^{3}}-\frac {\left (7 a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{b}-\frac {a \left (10 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) x}{b^{2}}}{\left (b x +a \right )^{3}}-\frac {2 d \left (a d -b c \right ) \ln \left (b x +a \right )}{b^{3}}\) | \(129\) |
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Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {b^{2} d^{2} x^{2} + a b d^{2} x - b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} + 2 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \]
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Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {- a^{2} d^{2} + 2 a b c d - b^{2} c^{2}}{a b^{3} + b^{4} x} + \frac {d^{2} x}{b^{2}} - \frac {2 d \left (a d - b c\right ) \log {\left (a + b x \right )}}{b^{3}} \]
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Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.31 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {d^{2} x}{b^{2}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{b^{4} x + a b^{3}} + \frac {2 \, {\left (b c d - a d^{2}\right )} \log \left (b x + a\right )}{b^{3}} \]
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Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {d^{2} x}{b^{2}} + \frac {2 \, {\left (b c d - a d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{{\left (b x + a\right )} b^{3}} \]
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Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx=\frac {d^2\,x}{b^2}-\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{b\,\left (x\,b^3+a\,b^2\right )}-\frac {\ln \left (a+b\,x\right )\,\left (2\,a\,d^2-2\,b\,c\,d\right )}{b^3} \]
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