Integrand size = 30, antiderivative size = 204 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\frac {2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x) (d+e x)^{7/2}}-\frac {6 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac {2 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) (d+e x)^{3/2}}-\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}} \]
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Time = 0.04 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {660, 45} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) (d+e x)^{3/2}}-\frac {6 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{7 e^4 (a+b x) (d+e x)^{7/2}}-\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}} \]
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Rule 45
Rule 660
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^3}{(d+e x)^{9/2}} \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b^3 (b d-a e)^3}{e^3 (d+e x)^{9/2}}+\frac {3 b^4 (b d-a e)^2}{e^3 (d+e x)^{7/2}}-\frac {3 b^5 (b d-a e)}{e^3 (d+e x)^{5/2}}+\frac {b^6}{e^3 (d+e x)^{3/2}}\right ) \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x) (d+e x)^{7/2}}-\frac {6 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac {2 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) (d+e x)^{3/2}}-\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (5 a^3 e^3+3 a^2 b e^2 (2 d+7 e x)+a b^2 e \left (8 d^2+28 d e x+35 e^2 x^2\right )+b^3 \left (16 d^3+56 d^2 e x+70 d e^2 x^2+35 e^3 x^3\right )\right )}{35 e^4 (a+b x) (d+e x)^{7/2}} \]
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Time = 2.62 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(-\frac {2 \left (35 e^{3} x^{3} b^{3}+35 x^{2} a \,b^{2} e^{3}+70 x^{2} b^{3} d \,e^{2}+21 a^{2} b \,e^{3} x +28 x a \,b^{2} d \,e^{2}+56 b^{3} d^{2} e x +5 a^{3} e^{3}+6 a^{2} b d \,e^{2}+8 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \left (e x +d \right )^{\frac {7}{2}} e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
default | \(-\frac {2 \left (35 e^{3} x^{3} b^{3}+35 x^{2} a \,b^{2} e^{3}+70 x^{2} b^{3} d \,e^{2}+21 a^{2} b \,e^{3} x +28 x a \,b^{2} d \,e^{2}+56 b^{3} d^{2} e x +5 a^{3} e^{3}+6 a^{2} b d \,e^{2}+8 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \left (e x +d \right )^{\frac {7}{2}} e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
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Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (35 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} + 8 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} + 5 \, a^{3} e^{3} + 35 \, {\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (8 \, b^{3} d^{2} e + 4 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{35 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \]
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\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {9}{2}}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (35 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} + 8 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} + 5 \, a^{3} e^{3} + 35 \, {\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (8 \, b^{3} d^{2} e + 4 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )}}{35 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )} \sqrt {e x + d}} \]
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Time = 0.33 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{3} b^{3} \mathrm {sgn}\left (b x + a\right ) - 35 \, {\left (e x + d\right )}^{2} b^{3} d \mathrm {sgn}\left (b x + a\right ) + 21 \, {\left (e x + d\right )} b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 5 \, b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, {\left (e x + d\right )}^{2} a b^{2} e \mathrm {sgn}\left (b x + a\right ) - 42 \, {\left (e x + d\right )} a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + 15 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 21 \, {\left (e x + d\right )} a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) - 15 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{35 \, {\left (e x + d\right )}^{\frac {7}{2}} e^{4}} \]
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Time = 10.33 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {10\,a^3\,e^3+12\,a^2\,b\,d\,e^2+16\,a\,b^2\,d^2\,e+32\,b^3\,d^3}{35\,b\,e^7}+\frac {2\,x\,\left (3\,a^2\,e^2+4\,a\,b\,d\,e+8\,b^2\,d^2\right )}{5\,e^6}+\frac {2\,b^2\,x^3}{e^4}+\frac {2\,b\,x^2\,\left (a\,e+2\,b\,d\right )}{e^5}\right )}{x^4\,\sqrt {d+e\,x}+\frac {a\,d^3\,\sqrt {d+e\,x}}{b\,e^3}+\frac {x^3\,\left (35\,a\,e^7+105\,b\,d\,e^6\right )\,\sqrt {d+e\,x}}{35\,b\,e^7}+\frac {3\,d\,x^2\,\left (a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}+\frac {d^2\,x\,\left (3\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^3}} \]
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