Integrand size = 26, antiderivative size = 69 \[ \int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx=\frac {2 (b d-a e)^2 \sqrt {d+e x}}{e^3}-\frac {4 b (b d-a e) (d+e x)^{3/2}}{3 e^3}+\frac {2 b^2 (d+e x)^{5/2}}{5 e^3} \]
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Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 45} \[ \int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx=-\frac {4 b (d+e x)^{3/2} (b d-a e)}{3 e^3}+\frac {2 \sqrt {d+e x} (b d-a e)^2}{e^3}+\frac {2 b^2 (d+e x)^{5/2}}{5 e^3} \]
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Rule 27
Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b x)^2}{\sqrt {d+e x}} \, dx \\ & = \int \left (\frac {(-b d+a e)^2}{e^2 \sqrt {d+e x}}-\frac {2 b (b d-a e) \sqrt {d+e x}}{e^2}+\frac {b^2 (d+e x)^{3/2}}{e^2}\right ) \, dx \\ & = \frac {2 (b d-a e)^2 \sqrt {d+e x}}{e^3}-\frac {4 b (b d-a e) (d+e x)^{3/2}}{3 e^3}+\frac {2 b^2 (d+e x)^{5/2}}{5 e^3} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87 \[ \int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (15 a^2 e^2+10 a b e (-2 d+e x)+b^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3} \]
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Time = 2.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {e x +d}\, \left (\left (\frac {1}{5} b^{2} x^{2}+\frac {2}{3} a b x +a^{2}\right ) e^{2}-\frac {4 \left (\frac {b x}{5}+a \right ) b d e}{3}+\frac {8 b^{2} d^{2}}{15}\right )}{e^{3}}\) | \(54\) |
derivativedivides | \(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 \left (a e -b d \right ) b \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -b d \right )^{2} \sqrt {e x +d}}{e^{3}}\) | \(55\) |
default | \(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 \left (a e -b d \right ) b \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -b d \right )^{2} \sqrt {e x +d}}{e^{3}}\) | \(55\) |
gosper | \(\frac {2 \left (3 x^{2} b^{2} e^{2}+10 x a b \,e^{2}-4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {e x +d}}{15 e^{3}}\) | \(63\) |
trager | \(\frac {2 \left (3 x^{2} b^{2} e^{2}+10 x a b \,e^{2}-4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {e x +d}}{15 e^{3}}\) | \(63\) |
risch | \(\frac {2 \left (3 x^{2} b^{2} e^{2}+10 x a b \,e^{2}-4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {e x +d}}{15 e^{3}}\) | \(63\) |
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Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} - 20 \, a b d e + 15 \, a^{2} e^{2} - 2 \, {\left (2 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{3}} \]
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Time = 0.62 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59 \[ \int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 a^{2} \sqrt {d + e x} + \frac {4 a b \left (- d \sqrt {d + e x} + \frac {\left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{e} + \frac {2 b^{2} \left (d^{2} \sqrt {d + e x} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3} + \frac {\left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{e^{2}}}{e} & \text {for}\: e \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}}{\sqrt {d}} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} a^{2} + \frac {10 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2}}{e^{2}}\right )}}{15 \, e} \]
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Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} a^{2} + \frac {10 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2}}{e^{2}}\right )}}{15 \, e} \]
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Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx=\frac {2\,\sqrt {d+e\,x}\,\left (3\,b^2\,{\left (d+e\,x\right )}^2+15\,a^2\,e^2+15\,b^2\,d^2-10\,b^2\,d\,\left (d+e\,x\right )+10\,a\,b\,e\,\left (d+e\,x\right )-30\,a\,b\,d\,e\right )}{15\,e^3} \]
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