Integrand size = 16, antiderivative size = 62 \[ \int \frac {c+d x}{x^3 (a+b x)} \, dx=-\frac {c}{2 a x^2}+\frac {b c-a d}{a^2 x}+\frac {b (b c-a d) \log (x)}{a^3}-\frac {b (b c-a d) \log (a+b x)}{a^3} \]
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Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {c+d x}{x^3 (a+b x)} \, dx=\frac {b \log (x) (b c-a d)}{a^3}-\frac {b (b c-a d) \log (a+b x)}{a^3}+\frac {b c-a d}{a^2 x}-\frac {c}{2 a x^2} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c}{a x^3}+\frac {-b c+a d}{a^2 x^2}-\frac {b (-b c+a d)}{a^3 x}+\frac {b^2 (-b c+a d)}{a^3 (a+b x)}\right ) \, dx \\ & = -\frac {c}{2 a x^2}+\frac {b c-a d}{a^2 x}+\frac {b (b c-a d) \log (x)}{a^3}-\frac {b (b c-a d) \log (a+b x)}{a^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.94 \[ \int \frac {c+d x}{x^3 (a+b x)} \, dx=\frac {-\frac {a (a c-2 b c x+2 a d x)}{x^2}+2 b (b c-a d) \log (x)+2 b (-b c+a d) \log (a+b x)}{2 a^3} \]
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Time = 0.43 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00
method | result | size |
default | \(-\frac {c}{2 a \,x^{2}}-\frac {a d -b c}{a^{2} x}-\frac {\left (a d -b c \right ) b \ln \left (x \right )}{a^{3}}+\frac {\left (a d -b c \right ) b \ln \left (b x +a \right )}{a^{3}}\) | \(62\) |
norman | \(\frac {-\frac {c}{2 a}-\frac {\left (a d -b c \right ) x}{a^{2}}}{x^{2}}+\frac {\left (a d -b c \right ) b \ln \left (b x +a \right )}{a^{3}}-\frac {\left (a d -b c \right ) b \ln \left (x \right )}{a^{3}}\) | \(62\) |
parallelrisch | \(-\frac {2 \ln \left (x \right ) x^{2} a b d -2 \ln \left (x \right ) x^{2} b^{2} c -2 \ln \left (b x +a \right ) x^{2} a b d +2 \ln \left (b x +a \right ) x^{2} b^{2} c +2 a^{2} d x -2 a b c x +a^{2} c}{2 a^{3} x^{2}}\) | \(78\) |
risch | \(\frac {-\frac {c}{2 a}-\frac {\left (a d -b c \right ) x}{a^{2}}}{x^{2}}-\frac {b \ln \left (x \right ) d}{a^{2}}+\frac {b^{2} \ln \left (x \right ) c}{a^{3}}+\frac {b \ln \left (-b x -a \right ) d}{a^{2}}-\frac {b^{2} \ln \left (-b x -a \right ) c}{a^{3}}\) | \(79\) |
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Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10 \[ \int \frac {c+d x}{x^3 (a+b x)} \, dx=-\frac {2 \, {\left (b^{2} c - a b d\right )} x^{2} \log \left (b x + a\right ) - 2 \, {\left (b^{2} c - a b d\right )} x^{2} \log \left (x\right ) + a^{2} c - 2 \, {\left (a b c - a^{2} d\right )} x}{2 \, a^{3} x^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (53) = 106\).
Time = 0.23 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.11 \[ \int \frac {c+d x}{x^3 (a+b x)} \, dx=\frac {- a c + x \left (- 2 a d + 2 b c\right )}{2 a^{2} x^{2}} - \frac {b \left (a d - b c\right ) \log {\left (x + \frac {a^{2} b d - a b^{2} c - a b \left (a d - b c\right )}{2 a b^{2} d - 2 b^{3} c} \right )}}{a^{3}} + \frac {b \left (a d - b c\right ) \log {\left (x + \frac {a^{2} b d - a b^{2} c + a b \left (a d - b c\right )}{2 a b^{2} d - 2 b^{3} c} \right )}}{a^{3}} \]
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Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x}{x^3 (a+b x)} \, dx=-\frac {{\left (b^{2} c - a b d\right )} \log \left (b x + a\right )}{a^{3}} + \frac {{\left (b^{2} c - a b d\right )} \log \left (x\right )}{a^{3}} - \frac {a c - 2 \, {\left (b c - a d\right )} x}{2 \, a^{2} x^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.21 \[ \int \frac {c+d x}{x^3 (a+b x)} \, dx=\frac {{\left (b^{2} c - a b d\right )} \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {{\left (b^{3} c - a b^{2} d\right )} \log \left ({\left | b x + a \right |}\right )}{a^{3} b} - \frac {a^{2} c - 2 \, {\left (a b c - a^{2} d\right )} x}{2 \, a^{3} x^{2}} \]
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Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18 \[ \int \frac {c+d x}{x^3 (a+b x)} \, dx=-\frac {\frac {c}{2\,a}+\frac {x\,\left (a\,d-b\,c\right )}{a^2}}{x^2}-\frac {2\,b\,\mathrm {atanh}\left (\frac {b\,\left (a\,d-b\,c\right )\,\left (a+2\,b\,x\right )}{a\,\left (b^2\,c-a\,b\,d\right )}\right )\,\left (a\,d-b\,c\right )}{a^3} \]
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