Integrand size = 26, antiderivative size = 26 \[ \int \left (10-18 x-45 x^2-12 x^3+15 x^4+3 \log \left (\frac {5}{4}\right )\right ) \, dx=x+3 (3-x) \left (-3-\left (x+x^2\right )^2-\log \left (\frac {5}{4}\right )\right ) \]
[Out]
Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 1, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \left (10-18 x-45 x^2-12 x^3+15 x^4+3 \log \left (\frac {5}{4}\right )\right ) \, dx=3 x^5-3 x^4-15 x^3-9 x^2+x \left (10+3 \log \left (\frac {5}{4}\right )\right ) \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = -9 x^2-15 x^3-3 x^4+3 x^5+x \left (10+3 \log \left (\frac {5}{4}\right )\right ) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \left (10-18 x-45 x^2-12 x^3+15 x^4+3 \log \left (\frac {5}{4}\right )\right ) \, dx=10 x-9 x^2-15 x^3-3 x^4+3 x^5+3 x \log \left (\frac {5}{4}\right ) \]
[In]
[Out]
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
gosper | \(-x \left (-3 x^{4}+3 x^{3}+15 x^{2}+3 \ln \left (\frac {4}{5}\right )+9 x -10\right )\) | \(28\) |
default | \(-3 x \ln \left (\frac {4}{5}\right )+3 x^{5}-3 x^{4}-15 x^{3}-9 x^{2}+10 x\) | \(30\) |
parallelrisch | \(3 x^{5}-3 x^{4}-15 x^{3}-9 x^{2}+\left (-3 \ln \left (\frac {4}{5}\right )+10\right ) x\) | \(30\) |
parts | \(-3 x \ln \left (\frac {4}{5}\right )+3 x^{5}-3 x^{4}-15 x^{3}-9 x^{2}+10 x\) | \(30\) |
norman | \(\left (3 \ln \left (5\right )-6 \ln \left (2\right )+10\right ) x -9 x^{2}-15 x^{3}-3 x^{4}+3 x^{5}\) | \(34\) |
risch | \(3 x^{5}-3 x^{4}-15 x^{3}-6 x \ln \left (2\right )+3 x \ln \left (5\right )-9 x^{2}+10 x\) | \(35\) |
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \left (10-18 x-45 x^2-12 x^3+15 x^4+3 \log \left (\frac {5}{4}\right )\right ) \, dx=3 \, x^{5} - 3 \, x^{4} - 15 \, x^{3} - 9 \, x^{2} - 3 \, x \log \left (\frac {4}{5}\right ) + 10 \, x \]
[In]
[Out]
Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \left (10-18 x-45 x^2-12 x^3+15 x^4+3 \log \left (\frac {5}{4}\right )\right ) \, dx=3 x^{5} - 3 x^{4} - 15 x^{3} - 9 x^{2} + x \left (- 6 \log {\left (2 \right )} + 3 \log {\left (5 \right )} + 10\right ) \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \left (10-18 x-45 x^2-12 x^3+15 x^4+3 \log \left (\frac {5}{4}\right )\right ) \, dx=3 \, x^{5} - 3 \, x^{4} - 15 \, x^{3} - 9 \, x^{2} - 3 \, x \log \left (\frac {4}{5}\right ) + 10 \, x \]
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \left (10-18 x-45 x^2-12 x^3+15 x^4+3 \log \left (\frac {5}{4}\right )\right ) \, dx=3 \, x^{5} - 3 \, x^{4} - 15 \, x^{3} - 9 \, x^{2} - 3 \, x \log \left (\frac {4}{5}\right ) + 10 \, x \]
[In]
[Out]
Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \left (10-18 x-45 x^2-12 x^3+15 x^4+3 \log \left (\frac {5}{4}\right )\right ) \, dx=3\,x^5-3\,x^4-15\,x^3-9\,x^2+\left (10-3\,\ln \left (\frac {4}{5}\right )\right )\,x \]
[In]
[Out]