Integrand size = 26, antiderivative size = 71 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 (b d-a e)^2 (d+e x)^{3/2}}{3 e^3}-\frac {4 b (b d-a e) (d+e x)^{5/2}}{5 e^3}+\frac {2 b^2 (d+e x)^{7/2}}{7 e^3} \]
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Time = 0.02 (sec) , antiderivative size = 71, 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} \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=-\frac {4 b (d+e x)^{5/2} (b d-a e)}{5 e^3}+\frac {2 (d+e x)^{3/2} (b d-a e)^2}{3 e^3}+\frac {2 b^2 (d+e x)^{7/2}}{7 e^3} \]
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Rule 27
Rule 45
Rubi steps \begin{align*} \text {integral}& = \int (a+b x)^2 \sqrt {d+e x} \, dx \\ & = \int \left (\frac {(-b d+a e)^2 \sqrt {d+e x}}{e^2}-\frac {2 b (b d-a e) (d+e x)^{3/2}}{e^2}+\frac {b^2 (d+e x)^{5/2}}{e^2}\right ) \, dx \\ & = \frac {2 (b d-a e)^2 (d+e x)^{3/2}}{3 e^3}-\frac {4 b (b d-a e) (d+e x)^{5/2}}{5 e^3}+\frac {2 b^2 (d+e x)^{7/2}}{7 e^3} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 (d+e x)^{3/2} \left (35 a^2 e^2+14 a b e (-2 d+3 e x)+b^2 \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3} \]
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Time = 2.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(\frac {2 \left (\left (\frac {3}{7} b^{2} x^{2}+\frac {6}{5} a b x +a^{2}\right ) e^{2}-\frac {4 b \left (\frac {3 b x}{7}+a \right ) d e}{5}+\frac {8 b^{2} d^{2}}{35}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3 e^{3}}\) | \(54\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (15 x^{2} b^{2} e^{2}+42 x a b \,e^{2}-12 b^{2} d e x +35 a^{2} e^{2}-28 a b d e +8 b^{2} d^{2}\right )}{105 e^{3}}\) | \(63\) |
derivativedivides | \(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 a e b -2 b^{2} d \right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{3}}\) | \(70\) |
default | \(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 a e b -2 b^{2} d \right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{3}}\) | \(70\) |
trager | \(\frac {2 \left (15 b^{2} e^{3} x^{3}+42 a b \,e^{3} x^{2}+3 b^{2} d \,e^{2} x^{2}+35 a^{2} e^{3} x +14 a b d \,e^{2} x -4 b^{2} d^{2} e x +35 a^{2} d \,e^{2}-28 a b \,d^{2} e +8 b^{2} d^{3}\right ) \sqrt {e x +d}}{105 e^{3}}\) | \(100\) |
risch | \(\frac {2 \left (15 b^{2} e^{3} x^{3}+42 a b \,e^{3} x^{2}+3 b^{2} d \,e^{2} x^{2}+35 a^{2} e^{3} x +14 a b d \,e^{2} x -4 b^{2} d^{2} e x +35 a^{2} d \,e^{2}-28 a b \,d^{2} e +8 b^{2} d^{3}\right ) \sqrt {e x +d}}{105 e^{3}}\) | \(100\) |
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Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (15 \, b^{2} e^{3} x^{3} + 8 \, b^{2} d^{3} - 28 \, a b d^{2} e + 35 \, a^{2} d e^{2} + 3 \, {\left (b^{2} d e^{2} + 14 \, a b e^{3}\right )} x^{2} - {\left (4 \, b^{2} d^{2} e - 14 \, a b d e^{2} - 35 \, a^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, e^{3}} \]
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Time = 0.79 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.55 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (2 a b e - 2 b^{2} d\right )}{5 e^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (a^{2} e^{2} - 2 a b d e + b^{2} d^{2}\right )}{3 e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\\sqrt {d} \left (a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{2} - 42 \, {\left (b^{2} d - a b e\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{105 \, e^{3}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (59) = 118\).
Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.82 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (105 \, \sqrt {e x + d} a^{2} d + 35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} + \frac {70 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b d}{e} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2} d}{e^{2}} + \frac {14 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a b}{e} + \frac {3 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b^{2}}{e^{2}}\right )}}{105 \, e} \]
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Time = 9.56 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2\,{\left (d+e\,x\right )}^{3/2}\,\left (15\,b^2\,{\left (d+e\,x\right )}^2+35\,a^2\,e^2+35\,b^2\,d^2-42\,b^2\,d\,\left (d+e\,x\right )+42\,a\,b\,e\,\left (d+e\,x\right )-70\,a\,b\,d\,e\right )}{105\,e^3} \]
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