\(\int \frac {e^5 (-10-2 x+10 x^2+2 x^3)+e^{10} (10-8 x-2 x^2) \log ^2(5+x)+\log (\frac {3 e^x}{x}) (e^5 (-10 x-2 x^2)+4 e^{10} x \log (5+x))}{5 x+11 x^2+7 x^3+x^4+e^5 (-10 x-12 x^2-2 x^3) \log ^2(5+x)+e^{10} (5 x+x^2) \log ^4(5+x)} \, dx\) [6869]

Optimal result

Integrand size = 135, antiderivative size = 29 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=\frac {2 \log \left (\frac {3 e^x}{x}\right )}{\frac {1+x}{e^5}-\log ^2(5+x)} \]

[Out]

2*ln(exp(ln(3)+x)/x)/((1+x)/exp(5)-ln(5+x)^2)

Rubi [F]

\[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=\int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx \]

[In]

Int[(E^5*(-10 - 2*x + 10*x^2 + 2*x^3) + E^10*(10 - 8*x - 2*x^2)*Log[5 + x]^2 + Log[(3*E^x)/x]*(E^5*(-10*x - 2*
x^2) + 4*E^10*x*Log[5 + x]))/(5*x + 11*x^2 + 7*x^3 + x^4 + E^5*(-10*x - 12*x^2 - 2*x^3)*Log[5 + x]^2 + E^10*(5
*x + x^2)*Log[5 + x]^4),x]

[Out]

-2*E^5*Defer[Int][Log[(3*E^x)/x]/(1 + x - E^5*Log[5 + x]^2)^2, x] + 4*E^10*Defer[Int][(Log[(3*E^x)/x]*Log[5 +
x])/((5 + x)*(1 + x - E^5*Log[5 + x]^2)^2), x] + 2*E^5*Defer[Int][(1 + x - E^5*Log[5 + x]^2)^(-1), x] - 2*E^5*
Defer[Int][1/(x*(1 + x - E^5*Log[5 + x]^2)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^5 \left (-x \log \left (\frac {3 e^x}{x}\right ) \left (5+x-2 e^5 \log (5+x)\right )+\left (-5+4 x+x^2\right ) \left (1+x-e^5 \log ^2(5+x)\right )\right )}{x (5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx \\ & = \left (2 e^5\right ) \int \frac {-x \log \left (\frac {3 e^x}{x}\right ) \left (5+x-2 e^5 \log (5+x)\right )+\left (-5+4 x+x^2\right ) \left (1+x-e^5 \log ^2(5+x)\right )}{x (5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx \\ & = \left (2 e^5\right ) \int \left (-\frac {\log \left (\frac {3 e^x}{x}\right ) \left (5+x-2 e^5 \log (5+x)\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}+\frac {-1+x}{x \left (1+x-e^5 \log ^2(5+x)\right )}\right ) \, dx \\ & = -\left (\left (2 e^5\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right ) \left (5+x-2 e^5 \log (5+x)\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\right )+\left (2 e^5\right ) \int \frac {-1+x}{x \left (1+x-e^5 \log ^2(5+x)\right )} \, dx \\ & = -\left (\left (2 e^5\right ) \int \left (\frac {5 \log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}+\frac {x \log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}-\frac {2 e^5 \log \left (\frac {3 e^x}{x}\right ) \log (5+x)}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}\right ) \, dx\right )+\left (2 e^5\right ) \int \left (\frac {1}{1+x-e^5 \log ^2(5+x)}-\frac {1}{x \left (1+x-e^5 \log ^2(5+x)\right )}\right ) \, dx \\ & = -\left (\left (2 e^5\right ) \int \frac {x \log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\right )+\left (2 e^5\right ) \int \frac {1}{1+x-e^5 \log ^2(5+x)} \, dx-\left (2 e^5\right ) \int \frac {1}{x \left (1+x-e^5 \log ^2(5+x)\right )} \, dx-\left (10 e^5\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx+\left (4 e^{10}\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right ) \log (5+x)}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx \\ & = \left (2 e^5\right ) \int \frac {1}{1+x-e^5 \log ^2(5+x)} \, dx-\left (2 e^5\right ) \int \frac {1}{x \left (1+x-e^5 \log ^2(5+x)\right )} \, dx-\left (2 e^5\right ) \int \left (\frac {\log \left (\frac {3 e^x}{x}\right )}{\left (1+x-e^5 \log ^2(5+x)\right )^2}-\frac {5 \log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}\right ) \, dx-\left (10 e^5\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx+\left (4 e^{10}\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right ) \log (5+x)}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx \\ & = -\left (\left (2 e^5\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right )}{\left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\right )+\left (2 e^5\right ) \int \frac {1}{1+x-e^5 \log ^2(5+x)} \, dx-\left (2 e^5\right ) \int \frac {1}{x \left (1+x-e^5 \log ^2(5+x)\right )} \, dx+\left (4 e^{10}\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right ) \log (5+x)}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=-\frac {2 e^5 \log \left (\frac {3 e^x}{x}\right )}{-1-x+e^5 \log ^2(5+x)} \]

[In]

Integrate[(E^5*(-10 - 2*x + 10*x^2 + 2*x^3) + E^10*(10 - 8*x - 2*x^2)*Log[5 + x]^2 + Log[(3*E^x)/x]*(E^5*(-10*
x - 2*x^2) + 4*E^10*x*Log[5 + x]))/(5*x + 11*x^2 + 7*x^3 + x^4 + E^5*(-10*x - 12*x^2 - 2*x^3)*Log[5 + x]^2 + E
^10*(5*x + x^2)*Log[5 + x]^4),x]

[Out]

(-2*E^5*Log[(3*E^x)/x])/(-1 - x + E^5*Log[5 + x]^2)

Maple [A] (verified)

Time = 23.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(-\frac {2 \ln \left (\frac {{\mathrm e}^{\ln \left (3\right )+x}}{x}\right ) {\mathrm e}^{5}}{{\mathrm e}^{5} \ln \left (5+x \right )^{2}-x -1}\)	\(31\)
risch	\(-\frac {2 \,{\mathrm e}^{5} \ln \left (3 \,{\mathrm e}^{x}\right )}{{\mathrm e}^{5} \ln \left (5+x \right )^{2}-x -1}+\frac {{\mathrm e}^{5} \left (-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2}+i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{3}-i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 \ln \left (x \right )\right )}{{\mathrm e}^{5} \ln \left (5+x \right )^{2}-x -1}\)	\(135\)

[In]

int(((4*x*exp(5)^2*ln(5+x)+(-2*x^2-10*x)*exp(5))*ln(exp(ln(3)+x)/x)+(-2*x^2-8*x+10)*exp(5)^2*ln(5+x)^2+(2*x^3+
10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp(5)^2*ln(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*ln(5+x)^2+x^4+7*x^3+11*x^2+5*
x),x,method=_RETURNVERBOSE)

[Out]

-2*ln(exp(ln(3)+x)/x)*exp(5)/(exp(5)*ln(5+x)^2-x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=-\frac {2 \, e^{5} \log \left (\frac {e^{\left (x + \log \left (3\right )\right )}}{x}\right )}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \]

[In]

integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)
^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x
^3+11*x^2+5*x),x, algorithm="fricas")

[Out]

-2*e^5*log(e^(x + log(3))/x)/(e^5*log(x + 5)^2 - x - 1)

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((4*x*exp(5)**2*ln(5+x)+(-2*x**2-10*x)*exp(5))*ln(exp(ln(3)+x)/x)+(-2*x**2-8*x+10)*exp(5)**2*ln(5+x)
**2+(2*x**3+10*x**2-2*x-10)*exp(5))/((x**2+5*x)*exp(5)**2*ln(5+x)**4+(-2*x**3-12*x**2-10*x)*exp(5)*ln(5+x)**2+
x**4+7*x**3+11*x**2+5*x),x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=-\frac {2 \, {\left (x e^{5} + e^{5} \log \left (3\right ) - e^{5} \log \left (x\right )\right )}}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \]

[In]

integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)
^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x
^3+11*x^2+5*x),x, algorithm="maxima")

[Out]

-2*(x*e^5 + e^5*log(3) - e^5*log(x))/(e^5*log(x + 5)^2 - x - 1)

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=-\frac {4 \, {\left (x e^{5} + e^{5} \log \left (3\right ) - e^{5} \log \left (x\right )\right )}}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \]

[In]

integrate(((4*x*exp(5)^2*log(5+x)+(-2*x^2-10*x)*exp(5))*log(exp(log(3)+x)/x)+(-2*x^2-8*x+10)*exp(5)^2*log(5+x)
^2+(2*x^3+10*x^2-2*x-10)*exp(5))/((x^2+5*x)*exp(5)^2*log(5+x)^4+(-2*x^3-12*x^2-10*x)*exp(5)*log(5+x)^2+x^4+7*x
^3+11*x^2+5*x),x, algorithm="giac")

[Out]

-4*(x*e^5 + e^5*log(3) - e^5*log(x))/(e^5*log(x + 5)^2 - x - 1)

Mupad [B] (verification not implemented)

Time = 12.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {e^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+x)} \, dx=\frac {2\,{\mathrm {e}}^5\,\ln \left (\frac {3\,{\mathrm {e}}^x}{x}\right )}{-{\mathrm {e}}^5\,{\ln \left (x+5\right )}^2+x+1} \]

[In]

int(-(exp(5)*(2*x - 10*x^2 - 2*x^3 + 10) + log(exp(x + log(3))/x)*(exp(5)*(10*x + 2*x^2) - 4*x*log(x + 5)*exp(
10)) + log(x + 5)^2*exp(10)*(8*x + 2*x^2 - 10))/(5*x + 11*x^2 + 7*x^3 + x^4 - log(x + 5)^2*exp(5)*(10*x + 12*x
^2 + 2*x^3) + log(x + 5)^4*exp(10)*(5*x + x^2)),x)

[Out]

(2*exp(5)*log((3*exp(x))/x))/(x - log(x + 5)^2*exp(5) + 1)