Integrand size = 16, antiderivative size = 90 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=-\frac {a (2 A b-3 a B) x}{b^4}+\frac {(A b-2 a B) x^2}{2 b^3}+\frac {B x^3}{3 b^2}+\frac {a^3 (A b-a B)}{b^5 (a+b x)}+\frac {a^2 (3 A b-4 a B) \log (a+b x)}{b^5} \]
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Time = 0.05 (sec) , antiderivative size = 90, 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} \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {a^3 (A b-a B)}{b^5 (a+b x)}+\frac {a^2 (3 A b-4 a B) \log (a+b x)}{b^5}-\frac {a x (2 A b-3 a B)}{b^4}+\frac {x^2 (A b-2 a B)}{2 b^3}+\frac {B x^3}{3 b^2} \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a (-2 A b+3 a B)}{b^4}+\frac {(A b-2 a B) x}{b^3}+\frac {B x^2}{b^2}+\frac {a^3 (-A b+a B)}{b^4 (a+b x)^2}-\frac {a^2 (-3 A b+4 a B)}{b^4 (a+b x)}\right ) \, dx \\ & = -\frac {a (2 A b-3 a B) x}{b^4}+\frac {(A b-2 a B) x^2}{2 b^3}+\frac {B x^3}{3 b^2}+\frac {a^3 (A b-a B)}{b^5 (a+b x)}+\frac {a^2 (3 A b-4 a B) \log (a+b x)}{b^5} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {6 a b (-2 A b+3 a B) x+3 b^2 (A b-2 a B) x^2+2 b^3 B x^3+\frac {6 a^3 (A b-a B)}{a+b x}+6 a^2 (3 A b-4 a B) \log (a+b x)}{6 b^5} \]
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Time = 1.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00
method | result | size |
default | \(-\frac {-\frac {1}{3} b^{2} B \,x^{3}-\frac {1}{2} A \,b^{2} x^{2}+B a b \,x^{2}+2 a A b x -3 a^{2} B x}{b^{4}}+\frac {a^{2} \left (3 A b -4 B a \right ) \ln \left (b x +a \right )}{b^{5}}+\frac {a^{3} \left (A b -B a \right )}{b^{5} \left (b x +a \right )}\) | \(90\) |
norman | \(\frac {\frac {a \left (3 a^{2} b A -4 a^{3} B \right )}{b^{5}}+\frac {B \,x^{4}}{3 b}+\frac {\left (3 A b -4 B a \right ) x^{3}}{6 b^{2}}-\frac {a \left (3 A b -4 B a \right ) x^{2}}{2 b^{3}}}{b x +a}+\frac {a^{2} \left (3 A b -4 B a \right ) \ln \left (b x +a \right )}{b^{5}}\) | \(96\) |
risch | \(\frac {B \,x^{3}}{3 b^{2}}+\frac {A \,x^{2}}{2 b^{2}}-\frac {B a \,x^{2}}{b^{3}}-\frac {2 a A x}{b^{3}}+\frac {3 a^{2} B x}{b^{4}}+\frac {a^{3} A}{b^{4} \left (b x +a \right )}-\frac {a^{4} B}{b^{5} \left (b x +a \right )}+\frac {3 a^{2} \ln \left (b x +a \right ) A}{b^{4}}-\frac {4 a^{3} \ln \left (b x +a \right ) B}{b^{5}}\) | \(109\) |
parallelrisch | \(\frac {2 B \,x^{4} b^{4}+3 A \,x^{3} b^{4}-4 B \,x^{3} a \,b^{3}+18 A \ln \left (b x +a \right ) x \,a^{2} b^{2}-9 A \,x^{2} a \,b^{3}-24 B \ln \left (b x +a \right ) x \,a^{3} b +12 B \,x^{2} a^{2} b^{2}+18 A \ln \left (b x +a \right ) a^{3} b -24 B \ln \left (b x +a \right ) a^{4}+18 A \,a^{3} b -24 B \,a^{4}}{6 b^{5} \left (b x +a \right )}\) | \(132\) |
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Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.56 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {2 \, B b^{4} x^{4} - 6 \, B a^{4} + 6 \, A a^{3} b - {\left (4 \, B a b^{3} - 3 \, A b^{4}\right )} x^{3} + 3 \, {\left (4 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{2} + 6 \, {\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} x - 6 \, {\left (4 \, B a^{4} - 3 \, A a^{3} b + {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{6} x + a b^{5}\right )}} \]
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Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {B x^{3}}{3 b^{2}} - \frac {a^{2} \left (- 3 A b + 4 B a\right ) \log {\left (a + b x \right )}}{b^{5}} + x^{2} \left (\frac {A}{2 b^{2}} - \frac {B a}{b^{3}}\right ) + x \left (- \frac {2 A a}{b^{3}} + \frac {3 B a^{2}}{b^{4}}\right ) + \frac {A a^{3} b - B a^{4}}{a b^{5} + b^{6} x} \]
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Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=-\frac {B a^{4} - A a^{3} b}{b^{6} x + a b^{5}} + \frac {2 \, B b^{2} x^{3} - 3 \, {\left (2 \, B a b - A b^{2}\right )} x^{2} + 6 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} x}{6 \, b^{4}} - \frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x + a\right )}{b^{5}} \]
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Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.60 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=\frac {{\left (b x + a\right )}^{3} {\left (2 \, B - \frac {3 \, {\left (4 \, B a b - A b^{2}\right )}}{{\left (b x + a\right )} b} + \frac {18 \, {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )}}{{\left (b x + a\right )}^{2} b^{2}}\right )}}{6 \, b^{5}} + \frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{5}} - \frac {\frac {B a^{4} b^{3}}{b x + a} - \frac {A a^{3} b^{4}}{b x + a}}{b^{8}} \]
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Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.28 \[ \int \frac {x^3 (A+B x)}{(a+b x)^2} \, dx=x^2\,\left (\frac {A}{2\,b^2}-\frac {B\,a}{b^3}\right )-x\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )-\frac {\ln \left (a+b\,x\right )\,\left (4\,B\,a^3-3\,A\,a^2\,b\right )}{b^5}+\frac {B\,x^3}{3\,b^2}-\frac {B\,a^4-A\,a^3\,b}{b\,\left (x\,b^5+a\,b^4\right )} \]
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