Optimal. Leaf size=25 \[ \left (2+\frac {5}{x}\right )^2+\frac {\log \left (\log \left (x+\frac {1+x}{4}\right )\right )}{x} \]
[Out]
________________________________________________________________________________________
Rubi [F]
time = 0.31, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{\left (x^3+5 x^4\right ) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x^2+\left (-50-270 x-100 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right )+\left (-x-5 x^2\right ) \log \left (\frac {1}{4} (1+5 x)\right ) \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x^3 (1+5 x) \log \left (\frac {1}{4} (1+5 x)\right )} \, dx\\ &=\int \left (\frac {5}{x (1+5 x) \log \left (\frac {1}{4}+\frac {5 x}{4}\right )}-\frac {50+20 x+x \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x^3}\right ) \, dx\\ &=5 \int \frac {1}{x (1+5 x) \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-\int \frac {50+20 x+x \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x^3} \, dx\\ &=5 \int \left (\frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )}-\frac {5}{(1+5 x) \log \left (\frac {1}{4}+\frac {5 x}{4}\right )}\right ) \, dx-\int \left (\frac {10 (5+2 x)}{x^3}+\frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2}\right ) \, dx\\ &=5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-10 \int \frac {5+2 x}{x^3} \, dx-25 \int \frac {1}{(1+5 x) \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx\\ &=\frac {(5+2 x)^2}{x^2}+5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-20 \text {Subst}\left (\int \frac {1}{4 x \log (x)} \, dx,x,\frac {1}{4}+\frac {5 x}{4}\right )-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx\\ &=\frac {(5+2 x)^2}{x^2}+5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-5 \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,\frac {1}{4}+\frac {5 x}{4}\right )-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx\\ &=\frac {(5+2 x)^2}{x^2}+5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-5 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {1}{4} (1+5 x)\right )\right )-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx\\ &=\frac {(5+2 x)^2}{x^2}-5 \log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )+5 \int \frac {1}{x \log \left (\frac {1}{4}+\frac {5 x}{4}\right )} \, dx-\int \frac {\log \left (\log \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A]
time = 0.14, size = 26, normalized size = 1.04 \begin {gather*} \frac {25}{x^2}+\frac {20}{x}+\frac {\log \left (\log \left (\frac {1}{4} (1+5 x)\right )\right )}{x} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A]
time = 1.98, size = 23, normalized size = 0.92
method | result | size |
risch | \(\frac {\ln \left (\ln \left (\frac {1}{4}+\frac {5 x}{4}\right )\right )}{x}+\frac {20 x +25}{x^{2}}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [A]
time = 0.54, size = 23, normalized size = 0.92 \begin {gather*} \frac {x \log \left (-2 \, \log \left (2\right ) + \log \left (5 \, x + 1\right )\right ) + 20 \, x + 25}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A]
time = 0.39, size = 18, normalized size = 0.72 \begin {gather*} \frac {x \log \left (\log \left (\frac {5}{4} \, x + \frac {1}{4}\right )\right ) + 20 \, x + 25}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [A]
time = 0.11, size = 22, normalized size = 0.88 \begin {gather*} \frac {\log {\left (\log {\left (\frac {5 x}{4} + \frac {1}{4} \right )} \right )}}{x} - \frac {- 20 x - 25}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A]
time = 0.44, size = 22, normalized size = 0.88 \begin {gather*} \frac {\log \left (\log \left (\frac {5}{4} \, x + \frac {1}{4}\right )\right )}{x} + \frac {5 \, {\left (4 \, x + 5\right )}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Mupad [B]
time = 5.80, size = 17, normalized size = 0.68 \begin {gather*} \frac {x\,\left (\ln \left (\ln \left (\frac {5\,x}{4}+\frac {1}{4}\right )\right )+20\right )+25}{x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________