Integrand size = 24, antiderivative size = 55 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {b^2-4 a c}{12 c^2 d (b d+2 c d x)^{3/2}}+\frac {\sqrt {b d+2 c d x}}{4 c^2 d^3} \]
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Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {697} \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {b^2-4 a c}{12 c^2 d (b d+2 c d x)^{3/2}}+\frac {\sqrt {b d+2 c d x}}{4 c^2 d^3} \]
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Rule 697
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-b^2+4 a c}{4 c (b d+2 c d x)^{5/2}}+\frac {1}{4 c d^2 \sqrt {b d+2 c d x}}\right ) \, dx \\ & = \frac {b^2-4 a c}{12 c^2 d (b d+2 c d x)^{3/2}}+\frac {\sqrt {b d+2 c d x}}{4 c^2 d^3} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {b^2-4 a c+3 (b+2 c x)^2}{12 c^2 d (d (b+2 c x))^{3/2}} \]
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Time = 2.35 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82
method | result | size |
gosper | \(-\frac {\left (2 c x +b \right ) \left (-3 c^{2} x^{2}-3 b c x +a c -b^{2}\right )}{3 c^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}\) | \(45\) |
derivativedivides | \(\frac {\sqrt {2 c d x +b d}-\frac {d^{2} \left (4 a c -b^{2}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{4 c^{2} d^{3}}\) | \(47\) |
default | \(\frac {\sqrt {2 c d x +b d}-\frac {d^{2} \left (4 a c -b^{2}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{4 c^{2} d^{3}}\) | \(47\) |
pseudoelliptic | \(-\frac {-3 c^{2} x^{2}+\left (-3 b x +a \right ) c -b^{2}}{3 \sqrt {d \left (2 c x +b \right )}\, d^{2} \left (2 c x +b \right ) c^{2}}\) | \(49\) |
trager | \(-\frac {\left (-3 c^{2} x^{2}-3 b c x +a c -b^{2}\right ) \sqrt {2 c d x +b d}}{3 d^{3} c^{2} \left (2 c x +b \right )^{2}}\) | \(50\) |
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Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {{\left (3 \, c^{2} x^{2} + 3 \, b c x + b^{2} - a c\right )} \sqrt {2 \, c d x + b d}}{3 \, {\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (49) = 98\).
Time = 0.32 (sec) , antiderivative size = 235, normalized size of antiderivative = 4.27 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx=\begin {cases} - \frac {a c \sqrt {b d + 2 c d x}}{3 b^{2} c^{2} d^{3} + 12 b c^{3} d^{3} x + 12 c^{4} d^{3} x^{2}} + \frac {b^{2} \sqrt {b d + 2 c d x}}{3 b^{2} c^{2} d^{3} + 12 b c^{3} d^{3} x + 12 c^{4} d^{3} x^{2}} + \frac {3 b c x \sqrt {b d + 2 c d x}}{3 b^{2} c^{2} d^{3} + 12 b c^{3} d^{3} x + 12 c^{4} d^{3} x^{2}} + \frac {3 c^{2} x^{2} \sqrt {b d + 2 c d x}}{3 b^{2} c^{2} d^{3} + 12 b c^{3} d^{3} x + 12 c^{4} d^{3} x^{2}} & \text {for}\: c \neq 0 \\\frac {a x + \frac {b x^{2}}{2}}{\left (b d\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {\frac {b^{2} - 4 \, a c}{{\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c} + \frac {3 \, \sqrt {2 \, c d x + b d}}{c d^{2}}}{12 \, c d} \]
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Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {b^{2} - 4 \, a c}{12 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c^{2} d} + \frac {\sqrt {2 \, c d x + b d}}{4 \, c^{2} d^{3}} \]
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Time = 9.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.67 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {3\,{\left (b+2\,c\,x\right )}^2-4\,a\,c+b^2}{12\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}} \]
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