Integrand size = 27, antiderivative size = 86 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=-\frac {a^4 A}{4 x^4}-\frac {a^3 (4 A b+a B)}{3 x^3}-\frac {a^2 b (3 A b+2 a B)}{x^2}-\frac {2 a b^2 (2 A b+3 a B)}{x}+b^4 B x+b^3 (A b+4 a B) \log (x) \]
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Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {27, 77} \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=-\frac {a^4 A}{4 x^4}-\frac {a^3 (a B+4 A b)}{3 x^3}-\frac {a^2 b (2 a B+3 A b)}{x^2}+b^3 \log (x) (4 a B+A b)-\frac {2 a b^2 (3 a B+2 A b)}{x}+b^4 B x \]
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Rule 27
Rule 77
Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b x)^4 (A+B x)}{x^5} \, dx \\ & = \int \left (b^4 B+\frac {a^4 A}{x^5}+\frac {a^3 (4 A b+a B)}{x^4}+\frac {2 a^2 b (3 A b+2 a B)}{x^3}+\frac {2 a b^2 (2 A b+3 a B)}{x^2}+\frac {b^3 (A b+4 a B)}{x}\right ) \, dx \\ & = -\frac {a^4 A}{4 x^4}-\frac {a^3 (4 A b+a B)}{3 x^3}-\frac {a^2 b (3 A b+2 a B)}{x^2}-\frac {2 a b^2 (2 A b+3 a B)}{x}+b^4 B x+b^3 (A b+4 a B) \log (x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=-\frac {4 a A b^3}{x}+b^4 B x-\frac {3 a^2 b^2 (A+2 B x)}{x^2}-\frac {2 a^3 b (2 A+3 B x)}{3 x^3}-\frac {a^4 (3 A+4 B x)}{12 x^4}+b^3 (A b+4 a B) \log (x) \]
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Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97
method | result | size |
default | \(-\frac {a^{4} A}{4 x^{4}}-\frac {a^{3} \left (4 A b +B a \right )}{3 x^{3}}-\frac {a^{2} b \left (3 A b +2 B a \right )}{x^{2}}-\frac {2 a \,b^{2} \left (2 A b +3 B a \right )}{x}+b^{4} B x +b^{3} \left (A b +4 B a \right ) \ln \left (x \right )\) | \(83\) |
risch | \(b^{4} B x +\frac {\left (-4 A \,b^{3} a -6 B \,a^{2} b^{2}\right ) x^{3}+\left (-3 A \,a^{2} b^{2}-2 B \,a^{3} b \right ) x^{2}+\left (-\frac {4}{3} A \,a^{3} b -\frac {1}{3} B \,a^{4}\right ) x -\frac {A \,a^{4}}{4}}{x^{4}}+A \ln \left (x \right ) b^{4}+4 B \ln \left (x \right ) a \,b^{3}\) | \(93\) |
norman | \(\frac {\left (-\frac {4}{3} A \,a^{3} b -\frac {1}{3} B \,a^{4}\right ) x +\left (-4 A \,b^{3} a -6 B \,a^{2} b^{2}\right ) x^{3}+\left (-3 A \,a^{2} b^{2}-2 B \,a^{3} b \right ) x^{2}+b^{4} B \,x^{5}-\frac {A \,a^{4}}{4}}{x^{4}}+\left (A \,b^{4}+4 B \,b^{3} a \right ) \ln \left (x \right )\) | \(95\) |
parallelrisch | \(\frac {12 A \ln \left (x \right ) x^{4} b^{4}+48 B \ln \left (x \right ) x^{4} a \,b^{3}+12 b^{4} B \,x^{5}-48 a A \,b^{3} x^{3}-72 x^{3} B \,a^{2} b^{2}-36 a^{2} A \,b^{2} x^{2}-24 x^{2} B \,a^{3} b -16 a^{3} A b x -4 a^{4} B x -3 A \,a^{4}}{12 x^{4}}\) | \(104\) |
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Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=\frac {12 \, B b^{4} x^{5} + 12 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} \log \left (x\right ) - 3 \, A a^{4} - 24 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} - 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} - 4 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \]
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Time = 0.65 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=B b^{4} x + b^{3} \left (A b + 4 B a\right ) \log {\left (x \right )} + \frac {- 3 A a^{4} + x^{3} \left (- 48 A a b^{3} - 72 B a^{2} b^{2}\right ) + x^{2} \left (- 36 A a^{2} b^{2} - 24 B a^{3} b\right ) + x \left (- 16 A a^{3} b - 4 B a^{4}\right )}{12 x^{4}} \]
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Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=B b^{4} x + {\left (4 \, B a b^{3} + A b^{4}\right )} \log \left (x\right ) - \frac {3 \, A a^{4} + 24 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 4 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=B b^{4} x + {\left (4 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | x \right |}\right ) - \frac {3 \, A a^{4} + 24 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + 12 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 4 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x}{12 \, x^{4}} \]
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Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x^5} \, dx=\ln \left (x\right )\,\left (A\,b^4+4\,B\,a\,b^3\right )-\frac {x\,\left (\frac {B\,a^4}{3}+\frac {4\,A\,b\,a^3}{3}\right )+\frac {A\,a^4}{4}+x^2\,\left (2\,B\,a^3\,b+3\,A\,a^2\,b^2\right )+x^3\,\left (6\,B\,a^2\,b^2+4\,A\,a\,b^3\right )}{x^4}+B\,b^4\,x \]
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