Integrand size = 68, antiderivative size = 23 \[ \int \frac {20 x+e^{2 x} \left (50+10 x^2\right )+e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{-25-5 x^2+\left (5+x^2\right ) \log \left (25+10 x^2+x^4\right )} \, dx=-e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right ) \]
[Out]
Time = 0.38 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6873, 6857, 2225, 2525, 2437, 2339, 29} \[ \int \frac {20 x+e^{2 x} \left (50+10 x^2\right )+e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{-25-5 x^2+\left (5+x^2\right ) \log \left (25+10 x^2+x^4\right )} \, dx=5 \log \left (5-\log \left (\left (x^2+5\right )^2\right )\right )-e^{2 x} \]
[In]
[Out]
Rule 29
Rule 2225
Rule 2339
Rule 2437
Rule 2525
Rule 6857
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {-20 x-e^{2 x} \left (50+10 x^2\right )-e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{\left (5+x^2\right ) \left (5-\log \left (\left (5+x^2\right )^2\right )\right )} \, dx \\ & = \int \left (-2 e^{2 x}+\frac {20 x}{\left (5+x^2\right ) \left (-5+\log \left (\left (5+x^2\right )^2\right )\right )}\right ) \, dx \\ & = -\left (2 \int e^{2 x} \, dx\right )+20 \int \frac {x}{\left (5+x^2\right ) \left (-5+\log \left (\left (5+x^2\right )^2\right )\right )} \, dx \\ & = -e^{2 x}+10 \text {Subst}\left (\int \frac {1}{(5+x) \left (-5+\log \left ((5+x)^2\right )\right )} \, dx,x,x^2\right ) \\ & = -e^{2 x}+10 \text {Subst}\left (\int \frac {1}{x \left (-5+\log \left (x^2\right )\right )} \, dx,x,5+x^2\right ) \\ & = -e^{2 x}+5 \text {Subst}\left (\int \frac {1}{x} \, dx,x,-5+\log \left (\left (5+x^2\right )^2\right )\right ) \\ & = -e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right ) \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {20 x+e^{2 x} \left (50+10 x^2\right )+e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{-25-5 x^2+\left (5+x^2\right ) \log \left (25+10 x^2+x^4\right )} \, dx=-e^{2 x}+5 \log \left (5-\log \left (\left (5+x^2\right )^2\right )\right ) \]
[In]
[Out]
Time = 1.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
default | \(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{4}+10 x^{2}+25\right )-5\right )\) | \(24\) |
norman | \(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{4}+10 x^{2}+25\right )-5\right )\) | \(24\) |
parallelrisch | \(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{4}+10 x^{2}+25\right )-5\right )\) | \(24\) |
parts | \(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{4}+10 x^{2}+25\right )-5\right )\) | \(24\) |
risch | \(-{\mathrm e}^{2 x}+5 \ln \left (\ln \left (x^{2}+5\right )-\frac {i \left (\pi {\operatorname {csgn}\left (i \left (x^{2}+5\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+5\right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left (x^{2}+5\right )\right ) {\operatorname {csgn}\left (i \left (x^{2}+5\right )^{2}\right )}^{2}+\pi {\operatorname {csgn}\left (i \left (x^{2}+5\right )^{2}\right )}^{3}-10 i\right )}{4}\right )\) | \(88\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {20 x+e^{2 x} \left (50+10 x^2\right )+e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{-25-5 x^2+\left (5+x^2\right ) \log \left (25+10 x^2+x^4\right )} \, dx=-e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{4} + 10 \, x^{2} + 25\right ) - 5\right ) \]
[In]
[Out]
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {20 x+e^{2 x} \left (50+10 x^2\right )+e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{-25-5 x^2+\left (5+x^2\right ) \log \left (25+10 x^2+x^4\right )} \, dx=- e^{2 x} + 5 \log {\left (\log {\left (x^{4} + 10 x^{2} + 25 \right )} - 5 \right )} \]
[In]
[Out]
none
Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {20 x+e^{2 x} \left (50+10 x^2\right )+e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{-25-5 x^2+\left (5+x^2\right ) \log \left (25+10 x^2+x^4\right )} \, dx=-e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{2} + 5\right ) - \frac {5}{2}\right ) \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {20 x+e^{2 x} \left (50+10 x^2\right )+e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{-25-5 x^2+\left (5+x^2\right ) \log \left (25+10 x^2+x^4\right )} \, dx=-e^{\left (2 \, x\right )} + 5 \, \log \left (\log \left (x^{4} + 10 \, x^{2} + 25\right ) - 5\right ) \]
[In]
[Out]
Time = 9.96 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {20 x+e^{2 x} \left (50+10 x^2\right )+e^{2 x} \left (-10-2 x^2\right ) \log \left (25+10 x^2+x^4\right )}{-25-5 x^2+\left (5+x^2\right ) \log \left (25+10 x^2+x^4\right )} \, dx=5\,\ln \left (\ln \left (x^4+10\,x^2+25\right )-5\right )-{\mathrm {e}}^{2\,x} \]
[In]
[Out]