Integrand size = 13, antiderivative size = 49 \[ \int \frac {x^2}{(a+b x)^{5/2}} \, dx=-\frac {2 a^2}{3 b^3 (a+b x)^{3/2}}+\frac {4 a}{b^3 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^3} \]
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Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int \frac {x^2}{(a+b x)^{5/2}} \, dx=-\frac {2 a^2}{3 b^3 (a+b x)^{3/2}}+\frac {4 a}{b^3 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^3} \]
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Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^2}{b^2 (a+b x)^{5/2}}-\frac {2 a}{b^2 (a+b x)^{3/2}}+\frac {1}{b^2 \sqrt {a+b x}}\right ) \, dx \\ & = -\frac {2 a^2}{3 b^3 (a+b x)^{3/2}}+\frac {4 a}{b^3 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x}}{b^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{(a+b x)^{5/2}} \, dx=\frac {2 \left (8 a^2+12 a b x+3 b^2 x^2\right )}{3 b^3 (a+b x)^{3/2}} \]
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Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(\frac {2 b^{2} x^{2}+8 a b x +\frac {16}{3} a^{2}}{b^{3} \left (b x +a \right )^{\frac {3}{2}}}\) | \(32\) |
trager | \(\frac {2 b^{2} x^{2}+8 a b x +\frac {16}{3} a^{2}}{b^{3} \left (b x +a \right )^{\frac {3}{2}}}\) | \(32\) |
pseudoelliptic | \(\frac {2 b^{2} x^{2}+8 a b x +\frac {16}{3} a^{2}}{b^{3} \left (b x +a \right )^{\frac {3}{2}}}\) | \(32\) |
risch | \(\frac {2 \sqrt {b x +a}}{b^{3}}+\frac {2 a \left (6 b x +5 a \right )}{3 b^{3} \left (b x +a \right )^{\frac {3}{2}}}\) | \(35\) |
derivativedivides | \(\frac {2 \sqrt {b x +a}+\frac {4 a}{\sqrt {b x +a}}-\frac {2 a^{2}}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{3}}\) | \(36\) |
default | \(\frac {2 \sqrt {b x +a}+\frac {4 a}{\sqrt {b x +a}}-\frac {2 a^{2}}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{3}}\) | \(36\) |
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none
Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {x^2}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{2} x^{2} + 12 \, a b x + 8 \, a^{2}\right )} \sqrt {b x + a}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (46) = 92\).
Time = 0.64 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.47 \[ \int \frac {x^2}{(a+b x)^{5/2}} \, dx=\begin {cases} \frac {16 a^{2}}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} + \frac {24 a b x}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} + \frac {6 b^{2} x^{2}}{3 a b^{3} \sqrt {a + b x} + 3 b^{4} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
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none
Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {x^2}{(a+b x)^{5/2}} \, dx=\frac {2 \, \sqrt {b x + a}}{b^{3}} + \frac {4 \, a}{\sqrt {b x + a} b^{3}} - \frac {2 \, a^{2}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3}} \]
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none
Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int \frac {x^2}{(a+b x)^{5/2}} \, dx=\frac {2 \, \sqrt {b x + a}}{b^{3}} + \frac {2 \, {\left (6 \, {\left (b x + a\right )} a - a^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3}} \]
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Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{(a+b x)^{5/2}} \, dx=\frac {6\,{\left (a+b\,x\right )}^2+12\,a\,\left (a+b\,x\right )-2\,a^2}{3\,b^3\,{\left (a+b\,x\right )}^{3/2}} \]
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