Integrand size = 15, antiderivative size = 72 \[ \int \sqrt {e^{a+b x}} x^3 \, dx=-\frac {96 \sqrt {e^{a+b x}}}{b^4}+\frac {48 \sqrt {e^{a+b x}} x}{b^3}-\frac {12 \sqrt {e^{a+b x}} x^2}{b^2}+\frac {2 \sqrt {e^{a+b x}} x^3}{b} \]
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Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2207, 2225} \[ \int \sqrt {e^{a+b x}} x^3 \, dx=-\frac {96 \sqrt {e^{a+b x}}}{b^4}+\frac {48 x \sqrt {e^{a+b x}}}{b^3}-\frac {12 x^2 \sqrt {e^{a+b x}}}{b^2}+\frac {2 x^3 \sqrt {e^{a+b x}}}{b} \]
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Rule 2207
Rule 2225
Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {e^{a+b x}} x^3}{b}-\frac {6 \int \sqrt {e^{a+b x}} x^2 \, dx}{b} \\ & = -\frac {12 \sqrt {e^{a+b x}} x^2}{b^2}+\frac {2 \sqrt {e^{a+b x}} x^3}{b}+\frac {24 \int \sqrt {e^{a+b x}} x \, dx}{b^2} \\ & = \frac {48 \sqrt {e^{a+b x}} x}{b^3}-\frac {12 \sqrt {e^{a+b x}} x^2}{b^2}+\frac {2 \sqrt {e^{a+b x}} x^3}{b}-\frac {48 \int \sqrt {e^{a+b x}} \, dx}{b^3} \\ & = -\frac {96 \sqrt {e^{a+b x}}}{b^4}+\frac {48 \sqrt {e^{a+b x}} x}{b^3}-\frac {12 \sqrt {e^{a+b x}} x^2}{b^2}+\frac {2 \sqrt {e^{a+b x}} x^3}{b} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.51 \[ \int \sqrt {e^{a+b x}} x^3 \, dx=\frac {2 \sqrt {e^{a+b x}} \left (-48+24 b x-6 b^2 x^2+b^3 x^3\right )}{b^4} \]
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Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.49
method | result | size |
gosper | \(\frac {2 \left (b^{3} x^{3}-6 b^{2} x^{2}+24 b x -48\right ) \sqrt {{\mathrm e}^{b x +a}}}{b^{4}}\) | \(35\) |
risch | \(\frac {2 \left (b^{3} x^{3}-6 b^{2} x^{2}+24 b x -48\right ) \sqrt {{\mathrm e}^{b x +a}}}{b^{4}}\) | \(35\) |
parallelrisch | \(\frac {2 \sqrt {{\mathrm e}^{b x +a}}\, x^{3} b^{3}-12 \sqrt {{\mathrm e}^{b x +a}}\, x^{2} b^{2}+48 b \sqrt {{\mathrm e}^{b x +a}}\, x -96 \sqrt {{\mathrm e}^{b x +a}}}{b^{4}}\) | \(60\) |
meijerg | \(\frac {16 \,{\mathrm e}^{-2 a -\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}} \sqrt {{\mathrm e}^{b x +a}}\, \left (6-\frac {\left (-\frac {b^{3} x^{3} {\mathrm e}^{\frac {3 a}{2}}}{2}+3 b^{2} x^{2} {\mathrm e}^{a}-12 b x \,{\mathrm e}^{\frac {a}{2}}+24\right ) {\mathrm e}^{\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}}}{4}\right )}{b^{4}}\) | \(72\) |
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.49 \[ \int \sqrt {e^{a+b x}} x^3 \, dx=\frac {2 \, {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 24 \, b x - 48\right )} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{b^{4}} \]
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Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.58 \[ \int \sqrt {e^{a+b x}} x^3 \, dx=\begin {cases} \frac {\left (2 b^{3} x^{3} - 12 b^{2} x^{2} + 48 b x - 96\right ) \sqrt {e^{a + b x}}}{b^{4}} & \text {for}\: b^{4} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases} \]
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Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67 \[ \int \sqrt {e^{a+b x}} x^3 \, dx=\frac {2 \, {\left (b^{3} x^{3} e^{\left (\frac {1}{2} \, a\right )} - 6 \, b^{2} x^{2} e^{\left (\frac {1}{2} \, a\right )} + 24 \, b x e^{\left (\frac {1}{2} \, a\right )} - 48 \, e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (\frac {1}{2} \, b x\right )}}{b^{4}} \]
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Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.49 \[ \int \sqrt {e^{a+b x}} x^3 \, dx=\frac {2 \, {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 24 \, b x - 48\right )} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{b^{4}} \]
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Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.51 \[ \int \sqrt {e^{a+b x}} x^3 \, dx=\sqrt {{\mathrm {e}}^{a+b\,x}}\,\left (\frac {48\,x}{b^3}-\frac {96}{b^4}+\frac {2\,x^3}{b}-\frac {12\,x^2}{b^2}\right ) \]
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