Integrand size = 26, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{7/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2}{80 c^3 d (b d+2 c d x)^{5/2}}+\frac {b^2-4 a c}{8 c^3 d^3 \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{48 c^3 d^5} \]
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Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {697} \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {b^2-4 a c}{8 c^3 d^3 \sqrt {b d+2 c d x}}-\frac {\left (b^2-4 a c\right )^2}{80 c^3 d (b d+2 c d x)^{5/2}}+\frac {(b d+2 c d x)^{3/2}}{48 c^3 d^5} \]
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Rule 697
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (-b^2+4 a c\right )^2}{16 c^2 (b d+2 c d x)^{7/2}}+\frac {-b^2+4 a c}{8 c^2 d^2 (b d+2 c d x)^{3/2}}+\frac {\sqrt {b d+2 c d x}}{16 c^2 d^4}\right ) \, dx \\ & = -\frac {\left (b^2-4 a c\right )^2}{80 c^3 d (b d+2 c d x)^{5/2}}+\frac {b^2-4 a c}{8 c^3 d^3 \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{48 c^3 d^5} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {-3 b^4+24 a b^2 c-48 a^2 c^2+30 b^2 (b+2 c x)^2-120 a c (b+2 c x)^2+5 (b+2 c x)^4}{240 c^3 d (d (b+2 c x))^{5/2}} \]
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Time = 2.58 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {2 d^{2} \left (4 a c -b^{2}\right )}{\sqrt {2 c d x +b d}}}{16 c^{3} d^{5}}\) | \(84\) |
default | \(\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {2 d^{2} \left (4 a c -b^{2}\right )}{\sqrt {2 c d x +b d}}}{16 c^{3} d^{5}}\) | \(84\) |
pseudoelliptic | \(-\frac {-\frac {5 c^{4} x^{4}}{3}+10 x^{2} \left (-\frac {b x}{3}+a \right ) c^{3}+\left (-5 b^{2} x^{2}+10 a b x +a^{2}\right ) c^{2}+2 \left (-\frac {5 b x}{3}+a \right ) b^{2} c -\frac {2 b^{4}}{3}}{5 \sqrt {d \left (2 c x +b \right )}\, d^{3} \left (2 c x +b \right )^{2} c^{3}}\) | \(88\) |
gosper | \(-\frac {\left (2 c x +b \right ) \left (-5 c^{4} x^{4}-10 b \,c^{3} x^{3}+30 x^{2} c^{3} a -15 b^{2} c^{2} x^{2}+30 a b \,c^{2} x -10 b^{3} c x +3 a^{2} c^{2}+6 a \,b^{2} c -2 b^{4}\right )}{15 c^{3} \left (2 c d x +b d \right )^{\frac {7}{2}}}\) | \(96\) |
trager | \(-\frac {\left (-5 c^{4} x^{4}-10 b \,c^{3} x^{3}+30 x^{2} c^{3} a -15 b^{2} c^{2} x^{2}+30 a b \,c^{2} x -10 b^{3} c x +3 a^{2} c^{2}+6 a \,b^{2} c -2 b^{4}\right ) \sqrt {2 c d x +b d}}{15 d^{4} c^{3} \left (2 c x +b \right )^{3}}\) | \(101\) |
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Time = 0.37 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {{\left (5 \, c^{4} x^{4} + 10 \, b c^{3} x^{3} + 2 \, b^{4} - 6 \, a b^{2} c - 3 \, a^{2} c^{2} + 15 \, {\left (b^{2} c^{2} - 2 \, a c^{3}\right )} x^{2} + 10 \, {\left (b^{3} c - 3 \, a b c^{2}\right )} x\right )} \sqrt {2 \, c d x + b d}}{15 \, {\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 688 vs. \(2 (82) = 164\).
Time = 0.51 (sec) , antiderivative size = 688, normalized size of antiderivative = 7.82 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{7/2}} \, dx=\begin {cases} - \frac {3 a^{2} c^{2} \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} - \frac {6 a b^{2} c \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} - \frac {30 a b c^{2} x \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} - \frac {30 a c^{3} x^{2} \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} + \frac {2 b^{4} \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} + \frac {10 b^{3} c x \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} + \frac {15 b^{2} c^{2} x^{2} \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} + \frac {10 b c^{3} x^{3} \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} + \frac {5 c^{4} x^{4} \sqrt {b d + 2 c d x}}{15 b^{3} c^{3} d^{4} + 90 b^{2} c^{4} d^{4} x + 180 b c^{5} d^{4} x^{2} + 120 c^{6} d^{4} x^{3}} & \text {for}\: c \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}}{\left (b d\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {\frac {5 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{c^{2} d^{4}} + \frac {3 \, {\left (10 \, {\left (2 \, c d x + b d\right )}^{2} {\left (b^{2} - 4 \, a c\right )} - {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{2}\right )}}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{2} d^{2}}}{240 \, c d} \]
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Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{48 \, c^{3} d^{5}} - \frac {b^{4} d^{2} - 8 \, a b^{2} c d^{2} + 16 \, a^{2} c^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{2} b^{2} + 40 \, {\left (2 \, c d x + b d\right )}^{2} a c}{80 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{3} d^{3}} \]
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Time = 9.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {-3\,a^2\,c^2-6\,a\,b^2\,c-30\,a\,b\,c^2\,x-30\,a\,c^3\,x^2+2\,b^4+10\,b^3\,c\,x+15\,b^2\,c^2\,x^2+10\,b\,c^3\,x^3+5\,c^4\,x^4}{15\,c^3\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}} \]
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