Integrand size = 81, antiderivative size = 23 \[ \int \frac {\left (7 x+14 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )+\left (-7 e^{2 x}-7 x+\left (-7 e^{2 x} x-7 x^2\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (e^{2 x}+x\right )}{3 e^{3 x} x+3 e^x x^2} \, dx=\frac {7}{3} e^{-x} \log \left (-\frac {10}{x}\right ) \log \left (e^{2 x}+x\right ) \] Output:
7/3*ln(exp(x)^2+x)*ln(-10/x)/exp(x)
Time = 0.49 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (7 x+14 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )+\left (-7 e^{2 x}-7 x+\left (-7 e^{2 x} x-7 x^2\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (e^{2 x}+x\right )}{3 e^{3 x} x+3 e^x x^2} \, dx=\frac {7}{3} e^{-x} \log \left (-\frac {10}{x}\right ) \log \left (e^{2 x}+x\right ) \] Input:
Integrate[((7*x + 14*E^(2*x)*x)*Log[-10/x] + (-7*E^(2*x) - 7*x + (-7*E^(2* x)*x - 7*x^2)*Log[-10/x])*Log[E^(2*x) + x])/(3*E^(3*x)*x + 3*E^x*x^2),x]
Output:
(7*Log[-10/x]*Log[E^(2*x) + x])/(3*E^x)
Time = 3.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {7292, 27, 27, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (\left (-7 x^2-7 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )-7 x-7 e^{2 x}\right ) \log \left (x+e^{2 x}\right )+\left (14 e^{2 x} x+7 x\right ) \log \left (-\frac {10}{x}\right )}{3 e^x x^2+3 e^{3 x} x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x} \left (\left (\left (-7 x^2-7 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )-7 x-7 e^{2 x}\right ) \log \left (x+e^{2 x}\right )+\left (14 e^{2 x} x+7 x\right ) \log \left (-\frac {10}{x}\right )\right )}{3 x \left (x+e^{2 x}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {7 e^{-x} \left (\left (2 e^{2 x} x+x\right ) \log \left (-\frac {10}{x}\right )-\left (x+e^{2 x}+\left (x^2+e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (x+e^{2 x}\right )\right )}{x \left (x+e^{2 x}\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7}{3} \int \frac {e^{-x} \left (\left (2 e^{2 x} x+x\right ) \log \left (-\frac {10}{x}\right )-\left (x+e^{2 x}+\left (x^2+e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (x+e^{2 x}\right )\right )}{x \left (x+e^{2 x}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {7}{3} \int \left (\frac {e^{-x} \left (2 x \log \left (-\frac {10}{x}\right )-x \log \left (x+e^{2 x}\right ) \log \left (-\frac {10}{x}\right )-\log \left (x+e^{2 x}\right )\right )}{x}-\frac {e^{-x} (2 x-1) \log \left (-\frac {10}{x}\right )}{x+e^{2 x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {7}{3} e^{-x} \log \left (-\frac {10}{x}\right ) \log \left (x+e^{2 x}\right )\) |
Input:
Int[((7*x + 14*E^(2*x)*x)*Log[-10/x] + (-7*E^(2*x) - 7*x + (-7*E^(2*x)*x - 7*x^2)*Log[-10/x])*Log[E^(2*x) + x])/(3*E^(3*x)*x + 3*E^x*x^2),x]
Output:
(7*Log[-10/x]*Log[E^(2*x) + x])/(3*E^x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.98 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
parallelrisch | \(\frac {7 \ln \left ({\mathrm e}^{2 x}+x \right ) \ln \left (-\frac {10}{x}\right ) {\mathrm e}^{-x}}{3}\) | \(20\) |
risch | \(\frac {7 \left (-2 i \pi \operatorname {csgn}\left (\frac {i}{x}\right )^{2}+2 i \pi \operatorname {csgn}\left (\frac {i}{x}\right )^{3}+2 i \pi +2 \ln \left (2\right )+2 \ln \left (5\right )-2 \ln \left (x \right )\right ) {\mathrm e}^{-x} \ln \left ({\mathrm e}^{2 x}+x \right )}{6}\) | \(57\) |
Input:
int((((-7*x*exp(x)^2-7*x^2)*ln(-10/x)-7*exp(x)^2-7*x)*ln(exp(x)^2+x)+(14*x *exp(x)^2+7*x)*ln(-10/x))/(3*x*exp(x)^3+3*exp(x)*x^2),x,method=_RETURNVERB OSE)
Output:
7/3*ln(exp(x)^2+x)*ln(-10/x)/exp(x)
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (7 x+14 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )+\left (-7 e^{2 x}-7 x+\left (-7 e^{2 x} x-7 x^2\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (e^{2 x}+x\right )}{3 e^{3 x} x+3 e^x x^2} \, dx=\frac {7}{3} \, e^{\left (-x\right )} \log \left (x + e^{\left (2 \, x\right )}\right ) \log \left (-\frac {10}{x}\right ) \] Input:
integrate((((-7*x*exp(x)^2-7*x^2)*log(-10/x)-7*exp(x)^2-7*x)*log(exp(x)^2+ x)+(14*x*exp(x)^2+7*x)*log(-10/x))/(3*x*exp(x)^3+3*exp(x)*x^2),x, algorith m="fricas")
Output:
7/3*e^(-x)*log(x + e^(2*x))*log(-10/x)
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {\left (7 x+14 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )+\left (-7 e^{2 x}-7 x+\left (-7 e^{2 x} x-7 x^2\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (e^{2 x}+x\right )}{3 e^{3 x} x+3 e^x x^2} \, dx=\frac {7 e^{- x} \log {\left (- \frac {10}{x} \right )} \log {\left (x + e^{2 x} \right )}}{3} \] Input:
integrate((((-7*x*exp(x)**2-7*x**2)*ln(-10/x)-7*exp(x)**2-7*x)*ln(exp(x)** 2+x)+(14*x*exp(x)**2+7*x)*ln(-10/x))/(3*x*exp(x)**3+3*exp(x)*x**2),x)
Output:
7*exp(-x)*log(-10/x)*log(x + exp(2*x))/3
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\left (7 x+14 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )+\left (-7 e^{2 x}-7 x+\left (-7 e^{2 x} x-7 x^2\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (e^{2 x}+x\right )}{3 e^{3 x} x+3 e^x x^2} \, dx=\frac {7}{3} \, {\left (\log \left (5\right ) + \log \left (2\right ) - \log \left (-x\right )\right )} e^{\left (-x\right )} \log \left (x + e^{\left (2 \, x\right )}\right ) \] Input:
integrate((((-7*x*exp(x)^2-7*x^2)*log(-10/x)-7*exp(x)^2-7*x)*log(exp(x)^2+ x)+(14*x*exp(x)^2+7*x)*log(-10/x))/(3*x*exp(x)^3+3*exp(x)*x^2),x, algorith m="maxima")
Output:
7/3*(log(5) + log(2) - log(-x))*e^(-x)*log(x + e^(2*x))
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (19) = 38\).
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.96 \[ \int \frac {\left (7 x+14 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )+\left (-7 e^{2 x}-7 x+\left (-7 e^{2 x} x-7 x^2\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (e^{2 x}+x\right )}{3 e^{3 x} x+3 e^x x^2} \, dx=\frac {7}{12} \, {\left (\pi ^{2} \mathrm {sgn}\left (x + e^{\left (2 \, x\right )}\right ) \mathrm {sgn}\left (x\right ) + \pi ^{2} \mathrm {sgn}\left (x + e^{\left (2 \, x\right )}\right ) - \pi ^{2} \mathrm {sgn}\left (x\right ) - \pi ^{2} + 4 \, \log \left (10\right ) \log \left ({\left | x + e^{\left (2 \, x\right )} \right |}\right ) - 4 \, \log \left ({\left | x + e^{\left (2 \, x\right )} \right |}\right ) \log \left ({\left | x \right |}\right )\right )} e^{\left (-x\right )} \] Input:
integrate((((-7*x*exp(x)^2-7*x^2)*log(-10/x)-7*exp(x)^2-7*x)*log(exp(x)^2+ x)+(14*x*exp(x)^2+7*x)*log(-10/x))/(3*x*exp(x)^3+3*exp(x)*x^2),x, algorith m="giac")
Output:
7/12*(pi^2*sgn(x + e^(2*x))*sgn(x) + pi^2*sgn(x + e^(2*x)) - pi^2*sgn(x) - pi^2 + 4*log(10)*log(abs(x + e^(2*x))) - 4*log(abs(x + e^(2*x)))*log(abs( x)))*e^(-x)
Timed out. \[ \int \frac {\left (7 x+14 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )+\left (-7 e^{2 x}-7 x+\left (-7 e^{2 x} x-7 x^2\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (e^{2 x}+x\right )}{3 e^{3 x} x+3 e^x x^2} \, dx=\int \frac {\ln \left (-\frac {10}{x}\right )\,\left (7\,x+14\,x\,{\mathrm {e}}^{2\,x}\right )-\ln \left (x+{\mathrm {e}}^{2\,x}\right )\,\left (7\,x+7\,{\mathrm {e}}^{2\,x}+\ln \left (-\frac {10}{x}\right )\,\left (7\,x\,{\mathrm {e}}^{2\,x}+7\,x^2\right )\right )}{3\,x\,{\mathrm {e}}^{3\,x}+3\,x^2\,{\mathrm {e}}^x} \,d x \] Input:
int((log(-10/x)*(7*x + 14*x*exp(2*x)) - log(x + exp(2*x))*(7*x + 7*exp(2*x ) + log(-10/x)*(7*x*exp(2*x) + 7*x^2)))/(3*x*exp(3*x) + 3*x^2*exp(x)),x)
Output:
int((log(-10/x)*(7*x + 14*x*exp(2*x)) - log(x + exp(2*x))*(7*x + 7*exp(2*x ) + log(-10/x)*(7*x*exp(2*x) + 7*x^2)))/(3*x*exp(3*x) + 3*x^2*exp(x)), x)
Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (7 x+14 e^{2 x} x\right ) \log \left (-\frac {10}{x}\right )+\left (-7 e^{2 x}-7 x+\left (-7 e^{2 x} x-7 x^2\right ) \log \left (-\frac {10}{x}\right )\right ) \log \left (e^{2 x}+x\right )}{3 e^{3 x} x+3 e^x x^2} \, dx=\frac {7 \,\mathrm {log}\left (e^{2 x}+x \right ) \mathrm {log}\left (-\frac {10}{x}\right )}{3 e^{x}} \] Input:
int((((-7*x*exp(x)^2-7*x^2)*log(-10/x)-7*exp(x)^2-7*x)*log(exp(x)^2+x)+(14 *x*exp(x)^2+7*x)*log(-10/x))/(3*x*exp(x)^3+3*exp(x)*x^2),x)
Output:
(7*log(e**(2*x) + x)*log(( - 10)/x))/(3*e**x)