Integrand size = 18, antiderivative size = 12 \[ \int e^x \left (-30 x-3 x^2+4 x^3\right ) \, dx=e^x x^2 (-15+4 x) \]
[Out]
Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.42, number of steps used = 12, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1608, 2227, 2207, 2225} \[ \int e^x \left (-30 x-3 x^2+4 x^3\right ) \, dx=4 e^x x^3-15 e^x x^2 \]
[In]
[Out]
Rule 1608
Rule 2207
Rule 2225
Rule 2227
Rubi steps \begin{align*} \text {integral}& = \int e^x x \left (-30-3 x+4 x^2\right ) \, dx \\ & = \int \left (-30 e^x x-3 e^x x^2+4 e^x x^3\right ) \, dx \\ & = -\left (3 \int e^x x^2 \, dx\right )+4 \int e^x x^3 \, dx-30 \int e^x x \, dx \\ & = -30 e^x x-3 e^x x^2+4 e^x x^3+6 \int e^x x \, dx-12 \int e^x x^2 \, dx+30 \int e^x \, dx \\ & = 30 e^x-24 e^x x-15 e^x x^2+4 e^x x^3-6 \int e^x \, dx+24 \int e^x x \, dx \\ & = 24 e^x-15 e^x x^2+4 e^x x^3-24 \int e^x \, dx \\ & = -15 e^x x^2+4 e^x x^3 \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int e^x \left (-30 x-3 x^2+4 x^3\right ) \, dx=e^x x^2 (-15+4 x) \]
[In]
[Out]
Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00
method | result | size |
gosper | \(x^{2} \left (4 x -15\right ) {\mathrm e}^{x}\) | \(12\) |
risch | \(\left (4 x^{3}-15 x^{2}\right ) {\mathrm e}^{x}\) | \(15\) |
default | \(-15 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\) | \(16\) |
norman | \(-15 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\) | \(16\) |
parallelrisch | \(-15 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\) | \(16\) |
parts | \(-15 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x^{3}\) | \(16\) |
meijerg | \(-\left (-4 x^{3}+12 x^{2}-24 x +24\right ) {\mathrm e}^{x}-\left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}+15 \left (2-2 x \right ) {\mathrm e}^{x}\) | \(44\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int e^x \left (-30 x-3 x^2+4 x^3\right ) \, dx={\left (4 \, x^{3} - 15 \, x^{2}\right )} e^{x} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int e^x \left (-30 x-3 x^2+4 x^3\right ) \, dx=\left (4 x^{3} - 15 x^{2}\right ) e^{x} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (11) = 22\).
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 3.08 \[ \int e^x \left (-30 x-3 x^2+4 x^3\right ) \, dx=4 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} - 3 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 30 \, {\left (x - 1\right )} e^{x} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int e^x \left (-30 x-3 x^2+4 x^3\right ) \, dx={\left (4 \, x^{3} - 15 \, x^{2}\right )} e^{x} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int e^x \left (-30 x-3 x^2+4 x^3\right ) \, dx=x^2\,{\mathrm {e}}^x\,\left (4\,x-15\right ) \]
[In]
[Out]