Integrand size = 11, antiderivative size = 58 \[ \int \frac {x^4}{(a+b x)^2} \, dx=\frac {3 a^2 x}{b^4}-\frac {a x^2}{b^3}+\frac {x^3}{3 b^2}-\frac {a^4}{b^5 (a+b x)}-\frac {4 a^3 \log (a+b x)}{b^5} \]
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Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {x^4}{(a+b x)^2} \, dx=-\frac {a^4}{b^5 (a+b x)}-\frac {4 a^3 \log (a+b x)}{b^5}+\frac {3 a^2 x}{b^4}-\frac {a x^2}{b^3}+\frac {x^3}{3 b^2} \]
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Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 a^2}{b^4}-\frac {2 a x}{b^3}+\frac {x^2}{b^2}+\frac {a^4}{b^4 (a+b x)^2}-\frac {4 a^3}{b^4 (a+b x)}\right ) \, dx \\ & = \frac {3 a^2 x}{b^4}-\frac {a x^2}{b^3}+\frac {x^3}{3 b^2}-\frac {a^4}{b^5 (a+b x)}-\frac {4 a^3 \log (a+b x)}{b^5} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{(a+b x)^2} \, dx=\frac {9 a^2 b x-3 a b^2 x^2+b^3 x^3-\frac {3 a^4}{a+b x}-12 a^3 \log (a+b x)}{3 b^5} \]
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Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {\frac {1}{3} b^{2} x^{3}-a b \,x^{2}+3 a^{2} x}{b^{4}}-\frac {4 a^{3} \ln \left (b x +a \right )}{b^{5}}-\frac {a^{4}}{b^{5} \left (b x +a \right )}\) | \(57\) |
risch | \(\frac {3 a^{2} x}{b^{4}}-\frac {a \,x^{2}}{b^{3}}+\frac {x^{3}}{3 b^{2}}-\frac {a^{4}}{b^{5} \left (b x +a \right )}-\frac {4 a^{3} \ln \left (b x +a \right )}{b^{5}}\) | \(57\) |
norman | \(\frac {\frac {x^{4}}{3 b}-\frac {2 a \,x^{3}}{3 b^{2}}-\frac {4 a^{4}}{b^{5}}+\frac {2 a^{2} x^{2}}{b^{3}}}{b x +a}-\frac {4 a^{3} \ln \left (b x +a \right )}{b^{5}}\) | \(61\) |
parallelrisch | \(-\frac {-b^{4} x^{4}+2 a \,b^{3} x^{3}+12 \ln \left (b x +a \right ) x \,a^{3} b -6 a^{2} b^{2} x^{2}+12 a^{4} \ln \left (b x +a \right )+12 a^{4}}{3 b^{5} \left (b x +a \right )}\) | \(71\) |
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none
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.26 \[ \int \frac {x^4}{(a+b x)^2} \, dx=\frac {b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4} - 12 \, {\left (a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{3 \, {\left (b^{6} x + a b^{5}\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{(a+b x)^2} \, dx=- \frac {a^{4}}{a b^{5} + b^{6} x} - \frac {4 a^{3} \log {\left (a + b x \right )}}{b^{5}} + \frac {3 a^{2} x}{b^{4}} - \frac {a x^{2}}{b^{3}} + \frac {x^{3}}{3 b^{2}} \]
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none
Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02 \[ \int \frac {x^4}{(a+b x)^2} \, dx=-\frac {a^{4}}{b^{6} x + a b^{5}} - \frac {4 \, a^{3} \log \left (b x + a\right )}{b^{5}} + \frac {b^{2} x^{3} - 3 \, a b x^{2} + 9 \, a^{2} x}{3 \, b^{4}} \]
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none
Time = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.36 \[ \int \frac {x^4}{(a+b x)^2} \, dx=-\frac {{\left (b x + a\right )}^{3} {\left (\frac {6 \, a}{b x + a} - \frac {18 \, a^{2}}{{\left (b x + a\right )}^{2}} - 1\right )}}{3 \, b^{5}} + \frac {4 \, a^{3} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b^{5}} - \frac {a^{4}}{{\left (b x + a\right )} b^{5}} \]
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Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.07 \[ \int \frac {x^4}{(a+b x)^2} \, dx=\frac {x^3}{3\,b^2}-\frac {4\,a^3\,\ln \left (a+b\,x\right )}{b^5}-\frac {a\,x^2}{b^3}+\frac {3\,a^2\,x}{b^4}-\frac {a^4}{b\,\left (x\,b^5+a\,b^4\right )} \]
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