Integrand size = 23, antiderivative size = 18 \[ \int \frac {1}{15} \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx=\frac {1}{3} x \left (1+e^4+x\right ) \left (-\frac {4}{5}+5 x\right ) \]
[Out]
Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.89, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12} \[ \int \frac {1}{15} \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx=\frac {5 x^3}{3}+\frac {7 x^2}{5}-\frac {4 x}{15}+\frac {1}{375} e^4 (2-25 x)^2 \]
[In]
[Out]
Rule 12
Rubi steps \begin{align*} \text {integral}& = \frac {1}{15} \int \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx \\ & = \frac {1}{375} e^4 (2-25 x)^2-\frac {4 x}{15}+\frac {7 x^2}{5}+\frac {5 x^3}{3} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {1}{15} \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx=\frac {1}{15} \left (-4 x-4 e^4 x+21 x^2+25 e^4 x^2+25 x^3\right ) \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28
method | result | size |
gosper | \(\frac {x \left (25 x \,{\mathrm e}^{4}+25 x^{2}-4 \,{\mathrm e}^{4}+21 x -4\right )}{15}\) | \(23\) |
norman | \(\left (-\frac {4 \,{\mathrm e}^{4}}{15}-\frac {4}{15}\right ) x +\left (\frac {5 \,{\mathrm e}^{4}}{3}+\frac {7}{5}\right ) x^{2}+\frac {5 x^{3}}{3}\) | \(25\) |
default | \(\frac {5 x^{2} {\mathrm e}^{4}}{3}+\frac {5 x^{3}}{3}-\frac {4 x \,{\mathrm e}^{4}}{15}+\frac {7 x^{2}}{5}-\frac {4 x}{15}\) | \(27\) |
risch | \(\frac {5 x^{2} {\mathrm e}^{4}}{3}+\frac {5 x^{3}}{3}-\frac {4 x \,{\mathrm e}^{4}}{15}+\frac {7 x^{2}}{5}-\frac {4 x}{15}\) | \(27\) |
parallelrisch | \(\frac {5 x^{2} {\mathrm e}^{4}}{3}+\frac {5 x^{3}}{3}-\frac {4 x \,{\mathrm e}^{4}}{15}+\frac {7 x^{2}}{5}-\frac {4 x}{15}\) | \(27\) |
parts | \(\frac {5 x^{2} {\mathrm e}^{4}}{3}+\frac {5 x^{3}}{3}-\frac {4 x \,{\mathrm e}^{4}}{15}+\frac {7 x^{2}}{5}-\frac {4 x}{15}\) | \(27\) |
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (13) = 26\).
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {1}{15} \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx=\frac {5}{3} \, x^{3} + \frac {7}{5} \, x^{2} + \frac {1}{15} \, {\left (25 \, x^{2} - 4 \, x\right )} e^{4} - \frac {4}{15} \, x \]
[In]
[Out]
Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {1}{15} \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx=\frac {5 x^{3}}{3} + x^{2} \cdot \left (\frac {7}{5} + \frac {5 e^{4}}{3}\right ) + x \left (- \frac {4 e^{4}}{15} - \frac {4}{15}\right ) \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (13) = 26\).
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {1}{15} \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx=\frac {5}{3} \, x^{3} + \frac {7}{5} \, x^{2} + \frac {1}{15} \, {\left (25 \, x^{2} - 4 \, x\right )} e^{4} - \frac {4}{15} \, x \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (13) = 26\).
Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int \frac {1}{15} \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx=\frac {5}{3} \, x^{3} + \frac {7}{5} \, x^{2} + \frac {1}{15} \, {\left (25 \, x^{2} - 4 \, x\right )} e^{4} - \frac {4}{15} \, x \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1}{15} \left (-4+42 x+75 x^2+e^4 (-4+50 x)\right ) \, dx=\frac {x\,\left (25\,x-4\right )\,\left (x+{\mathrm {e}}^4+1\right )}{15} \]
[In]
[Out]