Integrand size = 24, antiderivative size = 67 \[ \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx=-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{5 a b (a+b x)^4}-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{15 a^2 b (a+b x)^3} \]
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Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {673, 665} \[ \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx=-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{15 a^2 b (a+b x)^3}-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{5 a b (a+b x)^4} \]
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Rule 665
Rule 673
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2-b^2 x^2\right )^{3/2}}{5 a b (a+b x)^4}+\frac {\int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^3} \, dx}{5 a} \\ & = -\frac {\left (a^2-b^2 x^2\right )^{3/2}}{5 a b (a+b x)^4}-\frac {\left (a^2-b^2 x^2\right )^{3/2}}{15 a^2 b (a+b x)^3} \\ \end{align*}
Time = 0.43 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx=\frac {\sqrt {a^2-b^2 x^2} \left (-4 a^2+3 a b x+b^2 x^2\right )}{15 a^2 b (a+b x)^3} \]
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Time = 2.37 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(-\frac {\left (-b x +a \right ) \left (b x +4 a \right ) \sqrt {-b^{2} x^{2}+a^{2}}}{15 \left (b x +a \right )^{3} a^{2} b}\) | \(43\) |
trager | \(-\frac {\left (-b^{2} x^{2}-3 a b x +4 a^{2}\right ) \sqrt {-b^{2} x^{2}+a^{2}}}{15 a^{2} \left (b x +a \right )^{3} b}\) | \(49\) |
default | \(\frac {-\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {3}{2}}}{5 a b \left (x +\frac {a}{b}\right )^{4}}-\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {3}{2}}}{15 a^{2} \left (x +\frac {a}{b}\right )^{3}}}{b^{4}}\) | \(93\) |
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none
Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.55 \[ \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx=-\frac {4 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 12 \, a^{2} b x + 4 \, a^{3} - {\left (b^{2} x^{2} + 3 \, a b x - 4 \, a^{2}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{15 \, {\left (a^{2} b^{4} x^{3} + 3 \, a^{3} b^{3} x^{2} + 3 \, a^{4} b^{2} x + a^{5} b\right )}} \]
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\[ \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx=\int \frac {\sqrt {- \left (- a + b x\right ) \left (a + b x\right )}}{\left (a + b x\right )^{4}}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (59) = 118\).
Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx=-\frac {2 \, \sqrt {-b^{2} x^{2} + a^{2}}}{5 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{15 \, {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{15 \, {\left (a^{2} b^{2} x + a^{3} b\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (59) = 118\).
Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.46 \[ \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}}{b^{2} x} + \frac {25 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}^{2}}{b^{4} x^{2}} + \frac {15 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}^{3}}{b^{6} x^{3}} + \frac {15 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}^{4}}{b^{8} x^{4}} + 4\right )}}{15 \, a^{2} {\left (\frac {a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}}{b^{2} x} + 1\right )}^{5} {\left | b \right |}} \]
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Time = 9.82 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^4} \, dx=\frac {\sqrt {a^2-b^2\,x^2}\,\left (-4\,a^2+3\,a\,b\,x+b^2\,x^2\right )}{15\,a^2\,b\,{\left (a+b\,x\right )}^3} \]
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