Integrand size = 19, antiderivative size = 63 \[ \int \frac {1}{(a+b x) (a c-b c x)^3} \, dx=\frac {1}{4 a b c^3 (a-b x)^2}+\frac {1}{4 a^2 b c^3 (a-b x)}+\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{4 a^3 b c^3} \]
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Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {46, 214} \[ \int \frac {1}{(a+b x) (a c-b c x)^3} \, dx=\frac {\text {arctanh}\left (\frac {b x}{a}\right )}{4 a^3 b c^3}+\frac {1}{4 a^2 b c^3 (a-b x)}+\frac {1}{4 a b c^3 (a-b x)^2} \]
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Rule 46
Rule 214
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2 a c^3 (a-b x)^3}+\frac {1}{4 a^2 c^3 (a-b x)^2}+\frac {1}{4 a^2 c^3 \left (a^2-b^2 x^2\right )}\right ) \, dx \\ & = \frac {1}{4 a b c^3 (a-b x)^2}+\frac {1}{4 a^2 b c^3 (a-b x)}+\frac {\int \frac {1}{a^2-b^2 x^2} \, dx}{4 a^2 c^3} \\ & = \frac {1}{4 a b c^3 (a-b x)^2}+\frac {1}{4 a^2 b c^3 (a-b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{4 a^3 b c^3} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(a+b x) (a c-b c x)^3} \, dx=\frac {2 a (2 a-b x)-(a-b x)^2 \log (a-b x)+(a-b x)^2 \log (a+b x)}{8 a^3 b c^3 (a-b x)^2} \]
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Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02
method | result | size |
risch | \(\frac {-\frac {x}{4 a^{2}}+\frac {1}{2 b a}}{c^{3} \left (-b x +a \right )^{2}}-\frac {\ln \left (-b x +a \right )}{8 a^{3} c^{3} b}+\frac {\ln \left (b x +a \right )}{8 a^{3} c^{3} b}\) | \(64\) |
default | \(\frac {\frac {\ln \left (b x +a \right )}{8 a^{3} b}-\frac {\ln \left (-b x +a \right )}{8 a^{3} b}+\frac {1}{4 a^{2} b \left (-b x +a \right )}+\frac {1}{4 b a \left (-b x +a \right )^{2}}}{c^{3}}\) | \(67\) |
norman | \(\frac {\frac {3 x}{4 a^{2} c}-\frac {b \,x^{2}}{2 a^{3} c}}{c^{2} \left (-b x +a \right )^{2}}-\frac {\ln \left (-b x +a \right )}{8 a^{3} c^{3} b}+\frac {\ln \left (b x +a \right )}{8 a^{3} c^{3} b}\) | \(71\) |
parallelrisch | \(\frac {-\ln \left (b x -a \right ) x^{2} b^{2}+b^{2} \ln \left (b x +a \right ) x^{2}+2 \ln \left (b x -a \right ) x a b -2 \ln \left (b x +a \right ) x a b -4 b^{2} x^{2}-a^{2} \ln \left (b x -a \right )+a^{2} \ln \left (b x +a \right )+6 a b x}{8 a^{3} c^{3} \left (b x -a \right )^{2} b}\) | \(111\) |
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Time = 0.22 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.56 \[ \int \frac {1}{(a+b x) (a c-b c x)^3} \, dx=-\frac {2 \, a b x - 4 \, a^{2} - {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \log \left (b x + a\right ) + {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \log \left (b x - a\right )}{8 \, {\left (a^{3} b^{3} c^{3} x^{2} - 2 \, a^{4} b^{2} c^{3} x + a^{5} b c^{3}\right )}} \]
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Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(a+b x) (a c-b c x)^3} \, dx=- \frac {- 2 a + b x}{4 a^{4} b c^{3} - 8 a^{3} b^{2} c^{3} x + 4 a^{2} b^{3} c^{3} x^{2}} - \frac {\frac {\log {\left (- \frac {a}{b} + x \right )}}{8} - \frac {\log {\left (\frac {a}{b} + x \right )}}{8}}{a^{3} b c^{3}} \]
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Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(a+b x) (a c-b c x)^3} \, dx=-\frac {b x - 2 \, a}{4 \, {\left (a^{2} b^{3} c^{3} x^{2} - 2 \, a^{3} b^{2} c^{3} x + a^{4} b c^{3}\right )}} + \frac {\log \left (b x + a\right )}{8 \, a^{3} b c^{3}} - \frac {\log \left (b x - a\right )}{8 \, a^{3} b c^{3}} \]
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Time = 0.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(a+b x) (a c-b c x)^3} \, dx=\frac {\log \left ({\left | b x + a \right |}\right )}{8 \, a^{3} b c^{3}} - \frac {\log \left ({\left | b x - a \right |}\right )}{8 \, a^{3} b c^{3}} - \frac {a b x - 2 \, a^{2}}{4 \, {\left (b x - a\right )}^{2} a^{3} b c^{3}} \]
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Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a+b x) (a c-b c x)^3} \, dx=\frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{4\,a^3\,b\,c^3}-\frac {\frac {x}{4\,a^2}-\frac {1}{2\,a\,b}}{a^2\,c^3-2\,a\,b\,c^3\,x+b^2\,c^3\,x^2} \]
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