\(\int \frac {-6 x+e^{3+x^2} (-1-10 x^2)+(-x-2 e^{3+x^2} x^2) \log (x)+(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x)) \log (5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x))}{(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x)) \log ^2(5 e^{3+x^2}+5 x+(e^{3+x^2}+x) \log (x))} \, dx\) [3159]

Optimal result

Integrand size = 146, antiderivative size = 21 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=-5+\frac {x}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx \]

[In]

Int[(-6*x + E^(3 + x^2)*(-1 - 10*x^2) + (-x - 2*E^(3 + x^2)*x^2)*Log[x] + (5*E^(3 + x^2) + 5*x + (E^(3 + x^2)
+ x)*Log[x])*Log[5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Log[x]])/((5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Lo
g[x])*Log[5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Log[x]]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (e^{3+x^2}+x\right ) (5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx \\ & = \int \frac {-6 x-e^{3+x^2} \left (1+10 x^2\right )-x \left (1+2 e^{3+x^2} x\right ) \log (x)+\left (e^{3+x^2}+x\right ) (5+\log (x)) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{\left (e^{3+x^2}+x\right ) (5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx \\ & = \int \left (\frac {x \left (-1+2 x^2\right )}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {-1-10 x^2-2 x^2 \log (x)+5 \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )+\log (x) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx \\ & = \int \frac {x \left (-1+2 x^2\right )}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {-1-10 x^2-2 x^2 \log (x)+5 \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )+\log (x) \log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx \\ & = \int \left (-\frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {2 x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx+\int \left (\frac {-1-10 x^2-2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}+\frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx \\ & = 2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {-1-10 x^2-2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx \\ & = 2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \left (-\frac {1}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}-\frac {10 x^2}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}-\frac {2 x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )}\right ) \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx \\ & = 2 \int \frac {x^3}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-2 \int \frac {x^2 \log (x)}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-10 \int \frac {x^2}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {x}{\left (e^{3+x^2}+x\right ) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx-\int \frac {1}{(5+\log (x)) \log ^2\left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx+\int \frac {1}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log \left (\left (e^{3+x^2}+x\right ) (5+\log (x))\right )} \]

[In]

Integrate[(-6*x + E^(3 + x^2)*(-1 - 10*x^2) + (-x - 2*E^(3 + x^2)*x^2)*Log[x] + (5*E^(3 + x^2) + 5*x + (E^(3 +
x^2) + x)*Log[x])*Log[5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Log[x]])/((5*E^(3 + x^2) + 5*x + (E^(3 + x^2) +
x)*Log[x])*Log[5*E^(3 + x^2) + 5*x + (E^(3 + x^2) + x)*Log[x]]^2),x]

[Out]

x/Log[(E^(3 + x^2) + x)*(5 + Log[x])]

Maple [A] (verified)

Time = 1.95 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38

method	result	size
parallelrisch	\(\frac {x}{\ln \left (\left ({\mathrm e}^{x^{2}+3}+x \right ) \ln \left (x \right )+5 \,{\mathrm e}^{x^{2}+3}+5 x \right )}\)	\(29\)
risch	\(\frac {2 i x}{\pi \,\operatorname {csgn}\left (i \left (5+\ln \left (x \right )\right )\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right )\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \left (x \right )\right )\right )-\pi \,\operatorname {csgn}\left (i \left (5+\ln \left (x \right )\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \left (x \right )\right )\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \left (x \right )\right )\right )}^{2}+\pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}+3}+x \right ) \left (5+\ln \left (x \right )\right )\right )}^{3}+2 i \ln \left (5+\ln \left (x \right )\right )+2 i \ln \left ({\mathrm e}^{x^{2}+3}+x \right )}\)	\(148\)

[In]

int((((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+5*x)*ln((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+5*x)+(-2*x^2*exp(x^2+3)-x)*l
n(x)+(-10*x^2-1)*exp(x^2+3)-6*x)/((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+5*x)/ln((exp(x^2+3)+x)*ln(x)+5*exp(x^2+3)+
5*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log \left ({\left (x + e^{\left (x^{2} + 3\right )}\right )} \log \left (x\right ) + 5 \, x + 5 \, e^{\left (x^{2} + 3\right )}\right )} \]

[In]

integrate((((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)*log((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)+(-2*x^2*exp(x^
2+3)-x)*log(x)+(-10*x^2-1)*exp(x^2+3)-6*x)/((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)/log((exp(x^2+3)+x)*log(x)+
5*exp(x^2+3)+5*x)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log {\left (5 x + \left (x + e^{x^{2} + 3}\right ) \log {\left (x \right )} + 5 e^{x^{2} + 3} \right )}} \]

[In]

integrate((((exp(x**2+3)+x)*ln(x)+5*exp(x**2+3)+5*x)*ln((exp(x**2+3)+x)*ln(x)+5*exp(x**2+3)+5*x)+(-2*x**2*exp(
x**2+3)-x)*ln(x)+(-10*x**2-1)*exp(x**2+3)-6*x)/((exp(x**2+3)+x)*ln(x)+5*exp(x**2+3)+5*x)/ln((exp(x**2+3)+x)*ln
(x)+5*exp(x**2+3)+5*x)**2,x)

[Out]

x/log(5*x + (x + exp(x**2 + 3))*log(x) + 5*exp(x**2 + 3))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log \left (x + e^{\left (x^{2} + 3\right )}\right ) + \log \left (\log \left (x\right ) + 5\right )} \]

[In]

integrate((((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)*log((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)+(-2*x^2*exp(x^
2+3)-x)*log(x)+(-10*x^2-1)*exp(x^2+3)-6*x)/((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)/log((exp(x^2+3)+x)*log(x)+
5*exp(x^2+3)+5*x)^2,x, algorithm="maxima")

[Out]

x/(log(x + e^(x^2 + 3)) + log(log(x) + 5))

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\frac {x}{\log \left (x \log \left (x\right ) + e^{\left (x^{2} + 3\right )} \log \left (x\right ) + 5 \, x + 5 \, e^{\left (x^{2} + 3\right )}\right )} \]

[In]

integrate((((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)*log((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)+(-2*x^2*exp(x^
2+3)-x)*log(x)+(-10*x^2-1)*exp(x^2+3)-6*x)/((exp(x^2+3)+x)*log(x)+5*exp(x^2+3)+5*x)/log((exp(x^2+3)+x)*log(x)+
5*exp(x^2+3)+5*x)^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-6 x+e^{3+x^2} \left (-1-10 x^2\right )+\left (-x-2 e^{3+x^2} x^2\right ) \log (x)+\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log \left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )}{\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right ) \log ^2\left (5 e^{3+x^2}+5 x+\left (e^{3+x^2}+x\right ) \log (x)\right )} \, dx=\int -\frac {6\,x+{\mathrm {e}}^{x^2+3}\,\left (10\,x^2+1\right )-\ln \left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )\,\left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )+\ln \left (x\right )\,\left (x+2\,x^2\,{\mathrm {e}}^{x^2+3}\right )}{{\ln \left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )}^2\,\left (5\,x+5\,{\mathrm {e}}^{x^2+3}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^{x^2+3}\right )\right )} \,d x \]

[In]

int(-(6*x + exp(x^2 + 3)*(10*x^2 + 1) - log(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))*(5*x + 5*exp(x^2
+ 3) + log(x)*(x + exp(x^2 + 3))) + log(x)*(x + 2*x^2*exp(x^2 + 3)))/(log(5*x + 5*exp(x^2 + 3) + log(x)*(x +
exp(x^2 + 3)))^2*(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))),x)

[Out]

int(-(6*x + exp(x^2 + 3)*(10*x^2 + 1) - log(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))*(5*x + 5*exp(x^2
+ 3) + log(x)*(x + exp(x^2 + 3))) + log(x)*(x + 2*x^2*exp(x^2 + 3)))/(log(5*x + 5*exp(x^2 + 3) + log(x)*(x +
exp(x^2 + 3)))^2*(5*x + 5*exp(x^2 + 3) + log(x)*(x + exp(x^2 + 3)))), x)