Integrand size = 30, antiderivative size = 202 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x) (d+e x)^{5/2}}-\frac {2 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) (d+e x)^{3/2}}+\frac {6 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}}+\frac {2 b^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)} \]
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Time = 0.05 (sec) , antiderivative size = 202, 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} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) \sqrt {d+e x}}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) (d+e x)^{3/2}}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x}}{e^4 (a+b x)} \]
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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)^{7/2}} \, 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)^{7/2}}+\frac {3 b^4 (b d-a e)^2}{e^3 (d+e x)^{5/2}}-\frac {3 b^5 (b d-a e)}{e^3 (d+e x)^{3/2}}+\frac {b^6}{e^3 \sqrt {d+e x}}\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}}{5 e^4 (a+b x) (d+e x)^{5/2}}-\frac {2 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) (d+e x)^{3/2}}+\frac {6 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}}+\frac {2 b^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (a^3 e^3+a^2 b e^2 (2 d+5 e x)+a b^2 e \left (8 d^2+20 d e x+15 e^2 x^2\right )-b^3 \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )\right )}{5 e^4 (a+b x) (d+e x)^{5/2}} \]
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Time = 2.51 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(-\frac {2 \left (-5 e^{3} x^{3} b^{3}+15 x^{2} a \,b^{2} e^{3}-30 x^{2} b^{3} d \,e^{2}+5 a^{2} b \,e^{3} x +20 x a \,b^{2} d \,e^{2}-40 b^{3} d^{2} e x +a^{3} e^{3}+2 a^{2} b d \,e^{2}+8 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{5 \left (e x +d \right )^{\frac {5}{2}} e^{4} \left (b x +a \right )^{3}}\) | \(131\) |
default | \(-\frac {2 \left (-5 e^{3} x^{3} b^{3}+15 x^{2} a \,b^{2} e^{3}-30 x^{2} b^{3} d \,e^{2}+5 a^{2} b \,e^{3} x +20 x a \,b^{2} d \,e^{2}-40 b^{3} d^{2} e x +a^{3} e^{3}+2 a^{2} b d \,e^{2}+8 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{5 \left (e x +d \right )^{\frac {5}{2}} e^{4} \left (b x +a \right )^{3}}\) | \(131\) |
risch | \(\frac {2 b^{3} \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{e^{4} \left (b x +a \right )}-\frac {2 \left (15 x^{2} b^{2} e^{2}+5 x a b \,e^{2}+25 b^{2} d e x +a^{2} e^{2}+3 a b d e +11 b^{2} d^{2}\right ) \left (a e -b d \right ) \sqrt {\left (b x +a \right )^{2}}}{5 e^{4} \sqrt {e x +d}\, \left (x^{2} e^{2}+2 d e x +d^{2}\right ) \left (b x +a \right )}\) | \(136\) |
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Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{5 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]
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\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {7}{2}}}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.68 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )}}{5 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )} \sqrt {e x + d}} \]
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Time = 0.34 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, \sqrt {e x + d} b^{3} \mathrm {sgn}\left (b x + a\right )}{e^{4}} + \frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} b^{3} d \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (e x + d\right )} b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 15 \, {\left (e x + d\right )}^{2} a b^{2} e \mathrm {sgn}\left (b x + a\right ) + 10 \, {\left (e x + d\right )} a b^{2} d e \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (e x + d\right )} a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{5 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{4}} \]
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Time = 10.23 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{7/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,a^3\,e^3+4\,a^2\,b\,d\,e^2+16\,a\,b^2\,d^2\,e-32\,b^3\,d^3}{5\,b\,e^6}+\frac {2\,x\,\left (a^2\,e^2+4\,a\,b\,d\,e-8\,b^2\,d^2\right )}{e^5}-\frac {2\,b^2\,x^3}{e^3}+\frac {6\,b\,x^2\,\left (a\,e-2\,b\,d\right )}{e^4}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (5\,a\,e^6+10\,b\,d\,e^5\right )\,\sqrt {d+e\,x}}{5\,b\,e^6}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \]
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