\(\int \frac {e^2 (2 x+2 x^2)+e (52 x+4 x^2) \log (x)+(50 x+2 x^2) \log ^2(x)+(e (96+48 x)+e^2 (2 x+x^2)+e (4 x+2 x^2) \log (x)+(2 x+x^2) \log ^2(x)) \log (2+x)}{e^2 (2 x+x^2)+e (4 x+2 x^2) \log (x)+(2 x+x^2) \log ^2(x)} \, dx\) [7935]

Optimal result

Integrand size = 130, antiderivative size = 22 \[ \int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=x-\left (-x-\frac {48 \log (x)}{e+\log (x)}\right ) \log (2+x) \]

[Out]

x-ln(2+x)*(-x-48/(exp(1)+ln(x))*ln(x))

Rubi [F]

\[ \int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx \]

[In]

Int[(E^2*(2*x + 2*x^2) + E*(52*x + 4*x^2)*Log[x] + (50*x + 2*x^2)*Log[x]^2 + (E*(96 + 48*x) + E^2*(2*x + x^2)
+ E*(4*x + 2*x^2)*Log[x] + (2*x + x^2)*Log[x]^2)*Log[2 + x])/(E^2*(2*x + x^2) + E*(4*x + 2*x^2)*Log[x] + (2*x
+ x^2)*Log[x]^2),x]

[Out]

2*x + 46*Log[2 + x] + 2*E^2*Defer[Int][(1 + x)/((2 + x)*(E + Log[x])^2), x] - 4*E^2*Defer[Int][(13 + x)/((2 +
x)*(E + Log[x])^2), x] + 2*E^2*Defer[Int][(25 + x)/((2 + x)*(E + Log[x])^2), x] + 4*E*Defer[Int][(13 + x)/((2
+ x)*(E + Log[x])), x] - 4*E*Defer[Int][(25 + x)/((2 + x)*(E + Log[x])), x] + E^2*Defer[Int][Log[2 + x]/(E + L
og[x])^2, x] + 48*E*Defer[Int][Log[2 + x]/(x*(E + Log[x])^2), x] + 2*E*Defer[Int][(Log[x]*Log[2 + x])/(E + Log
[x])^2, x] + Defer[Int][(Log[x]^2*Log[2 + x])/(E + Log[x])^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e x \log (x) (2 (13+x)+(2+x) \log (2+x))+x \log ^2(x) (2 (25+x)+(2+x) \log (2+x))+e (2 e x (1+x)+(2+x) (48+e x) \log (2+x))}{x (2+x) (e+\log (x))^2} \, dx \\ & = \int \left (\frac {2 e^2 (1+x)}{(2+x) (e+\log (x))^2}+\frac {4 e (13+x) \log (x)}{(2+x) (e+\log (x))^2}+\frac {2 (25+x) \log ^2(x)}{(2+x) (e+\log (x))^2}+\frac {\left (48 e+e^2 x+2 e x \log (x)+x \log ^2(x)\right ) \log (2+x)}{x (e+\log (x))^2}\right ) \, dx \\ & = 2 \int \frac {(25+x) \log ^2(x)}{(2+x) (e+\log (x))^2} \, dx+(4 e) \int \frac {(13+x) \log (x)}{(2+x) (e+\log (x))^2} \, dx+\left (2 e^2\right ) \int \frac {1+x}{(2+x) (e+\log (x))^2} \, dx+\int \frac {\left (48 e+e^2 x+2 e x \log (x)+x \log ^2(x)\right ) \log (2+x)}{x (e+\log (x))^2} \, dx \\ & = 2 \int \left (\frac {25+x}{2+x}+\frac {e^2 (25+x)}{(2+x) (e+\log (x))^2}-\frac {2 e (25+x)}{(2+x) (e+\log (x))}\right ) \, dx+(4 e) \int \left (-\frac {e (13+x)}{(2+x) (e+\log (x))^2}+\frac {13+x}{(2+x) (e+\log (x))}\right ) \, dx+\left (2 e^2\right ) \int \frac {1+x}{(2+x) (e+\log (x))^2} \, dx+\int \left (\frac {e^2 \log (2+x)}{(e+\log (x))^2}+\frac {48 e \log (2+x)}{x (e+\log (x))^2}+\frac {2 e \log (x) \log (2+x)}{(e+\log (x))^2}+\frac {\log ^2(x) \log (2+x)}{(e+\log (x))^2}\right ) \, dx \\ & = 2 \int \frac {25+x}{2+x} \, dx+(2 e) \int \frac {\log (x) \log (2+x)}{(e+\log (x))^2} \, dx+(4 e) \int \frac {13+x}{(2+x) (e+\log (x))} \, dx-(4 e) \int \frac {25+x}{(2+x) (e+\log (x))} \, dx+(48 e) \int \frac {\log (2+x)}{x (e+\log (x))^2} \, dx+e^2 \int \frac {\log (2+x)}{(e+\log (x))^2} \, dx+\left (2 e^2\right ) \int \frac {1+x}{(2+x) (e+\log (x))^2} \, dx+\left (2 e^2\right ) \int \frac {25+x}{(2+x) (e+\log (x))^2} \, dx-\left (4 e^2\right ) \int \frac {13+x}{(2+x) (e+\log (x))^2} \, dx+\int \frac {\log ^2(x) \log (2+x)}{(e+\log (x))^2} \, dx \\ & = 2 \int \left (1+\frac {23}{2+x}\right ) \, dx+(2 e) \int \frac {\log (x) \log (2+x)}{(e+\log (x))^2} \, dx+(4 e) \int \frac {13+x}{(2+x) (e+\log (x))} \, dx-(4 e) \int \frac {25+x}{(2+x) (e+\log (x))} \, dx+(48 e) \int \frac {\log (2+x)}{x (e+\log (x))^2} \, dx+e^2 \int \frac {\log (2+x)}{(e+\log (x))^2} \, dx+\left (2 e^2\right ) \int \frac {1+x}{(2+x) (e+\log (x))^2} \, dx+\left (2 e^2\right ) \int \frac {25+x}{(2+x) (e+\log (x))^2} \, dx-\left (4 e^2\right ) \int \frac {13+x}{(2+x) (e+\log (x))^2} \, dx+\int \frac {\log ^2(x) \log (2+x)}{(e+\log (x))^2} \, dx \\ & = 2 x+46 \log (2+x)+(2 e) \int \frac {\log (x) \log (2+x)}{(e+\log (x))^2} \, dx+(4 e) \int \frac {13+x}{(2+x) (e+\log (x))} \, dx-(4 e) \int \frac {25+x}{(2+x) (e+\log (x))} \, dx+(48 e) \int \frac {\log (2+x)}{x (e+\log (x))^2} \, dx+e^2 \int \frac {\log (2+x)}{(e+\log (x))^2} \, dx+\left (2 e^2\right ) \int \frac {1+x}{(2+x) (e+\log (x))^2} \, dx+\left (2 e^2\right ) \int \frac {25+x}{(2+x) (e+\log (x))^2} \, dx-\left (4 e^2\right ) \int \frac {13+x}{(2+x) (e+\log (x))^2} \, dx+\int \frac {\log ^2(x) \log (2+x)}{(e+\log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=x+48 \log (2+x)+\frac {(-48 e+e x+x \log (x)) \log (2+x)}{e+\log (x)} \]

[In]

Integrate[(E^2*(2*x + 2*x^2) + E*(52*x + 4*x^2)*Log[x] + (50*x + 2*x^2)*Log[x]^2 + (E*(96 + 48*x) + E^2*(2*x +
x^2) + E*(4*x + 2*x^2)*Log[x] + (2*x + x^2)*Log[x]^2)*Log[2 + x])/(E^2*(2*x + x^2) + E*(4*x + 2*x^2)*Log[x] +
(2*x + x^2)*Log[x]^2),x]

[Out]

x + 48*Log[2 + x] + ((-48*E + E*x + x*Log[x])*Log[2 + x])/(E + Log[x])

Maple [A] (verified)

Time = 6.89 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55

method	result	size
risch	\(\frac {\left (x \,{\mathrm e}+x \ln \left (x \right )-48 \,{\mathrm e}\right ) \ln \left (2+x \right )}{{\mathrm e}+\ln \left (x \right )}+x +48 \ln \left (2+x \right )\)	\(34\)
parallelrisch	\(\frac {{\mathrm e} \ln \left (2+x \right ) x +\ln \left (x \right ) \ln \left (2+x \right ) x +x \,{\mathrm e}+x \ln \left (x \right )+48 \ln \left (2+x \right ) \ln \left (x \right )-4 \,{\mathrm e}-4 \ln \left (x \right )}{{\mathrm e}+\ln \left (x \right )}\)	\(50\)

[In]

int((((x^2+2*x)*ln(x)^2+(2*x^2+4*x)*exp(1)*ln(x)+(x^2+2*x)*exp(1)^2+(48*x+96)*exp(1))*ln(2+x)+(2*x^2+50*x)*ln(
x)^2+(4*x^2+52*x)*exp(1)*ln(x)+(2*x^2+2*x)*exp(1)^2)/((x^2+2*x)*ln(x)^2+(2*x^2+4*x)*exp(1)*ln(x)+(x^2+2*x)*exp
(1)^2),x,method=_RETURNVERBOSE)

[Out]

(x*exp(1)+x*ln(x)-48*exp(1))/(exp(1)+ln(x))*ln(2+x)+x+48*ln(2+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\frac {x e + {\left (x e + {\left (x + 48\right )} \log \left (x\right )\right )} \log \left (x + 2\right ) + x \log \left (x\right )}{e + \log \left (x\right )} \]

[In]

integrate((((x^2+2*x)*log(x)^2+(2*x^2+4*x)*exp(1)*log(x)+(x^2+2*x)*exp(1)^2+(48*x+96)*exp(1))*log(2+x)+(2*x^2+
50*x)*log(x)^2+(4*x^2+52*x)*exp(1)*log(x)+(2*x^2+2*x)*exp(1)^2)/((x^2+2*x)*log(x)^2+(2*x^2+4*x)*exp(1)*log(x)+
(x^2+2*x)*exp(1)^2),x, algorithm="fricas")

[Out]

(x*e + (x*e + (x + 48)*log(x))*log(x + 2) + x*log(x))/(e + log(x))

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((((x**2+2*x)*ln(x)**2+(2*x**2+4*x)*exp(1)*ln(x)+(x**2+2*x)*exp(1)**2+(48*x+96)*exp(1))*ln(2+x)+(2*x*
*2+50*x)*ln(x)**2+(4*x**2+52*x)*exp(1)*ln(x)+(2*x**2+2*x)*exp(1)**2)/((x**2+2*x)*ln(x)**2+(2*x**2+4*x)*exp(1)*
ln(x)+(x**2+2*x)*exp(1)**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\frac {x e + {\left (x e + {\left (x + 48\right )} \log \left (x\right )\right )} \log \left (x + 2\right ) + x \log \left (x\right )}{e + \log \left (x\right )} \]

[In]

integrate((((x^2+2*x)*log(x)^2+(2*x^2+4*x)*exp(1)*log(x)+(x^2+2*x)*exp(1)^2+(48*x+96)*exp(1))*log(2+x)+(2*x^2+
50*x)*log(x)^2+(4*x^2+52*x)*exp(1)*log(x)+(2*x^2+2*x)*exp(1)^2)/((x^2+2*x)*log(x)^2+(2*x^2+4*x)*exp(1)*log(x)+
(x^2+2*x)*exp(1)^2),x, algorithm="maxima")

[Out]

(x*e + (x*e + (x + 48)*log(x))*log(x + 2) + x*log(x))/(e + log(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\frac {x e \log \left (x + 2\right ) + x \log \left (x + 2\right ) \log \left (x\right ) + x e + x \log \left (x\right ) + 48 \, \log \left (x + 2\right ) \log \left (x\right )}{e + \log \left (x\right )} \]

[In]

integrate((((x^2+2*x)*log(x)^2+(2*x^2+4*x)*exp(1)*log(x)+(x^2+2*x)*exp(1)^2+(48*x+96)*exp(1))*log(2+x)+(2*x^2+
50*x)*log(x)^2+(4*x^2+52*x)*exp(1)*log(x)+(2*x^2+2*x)*exp(1)^2)/((x^2+2*x)*log(x)^2+(2*x^2+4*x)*exp(1)*log(x)+
(x^2+2*x)*exp(1)^2),x, algorithm="giac")

[Out]

(x*e*log(x + 2) + x*log(x + 2)*log(x) + x*e + x*log(x) + 48*log(x + 2)*log(x))/(e + log(x))

Mupad [B] (verification not implemented)

Time = 13.98 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {e^2 \left (2 x+2 x^2\right )+e \left (52 x+4 x^2\right ) \log (x)+\left (50 x+2 x^2\right ) \log ^2(x)+\left (e (96+48 x)+e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)\right ) \log (2+x)}{e^2 \left (2 x+x^2\right )+e \left (4 x+2 x^2\right ) \log (x)+\left (2 x+x^2\right ) \log ^2(x)} \, dx=\frac {x\,\mathrm {e}+48\,\ln \left (x+2\right )\,\ln \left (x\right )+x\,\ln \left (x\right )+x\,\ln \left (x+2\right )\,\mathrm {e}+x\,\ln \left (x+2\right )\,\ln \left (x\right )}{\mathrm {e}+\ln \left (x\right )} \]

[In]

int((log(x)^2*(50*x + 2*x^2) + exp(2)*(2*x + 2*x^2) + log(x + 2)*(log(x)^2*(2*x + x^2) + exp(2)*(2*x + x^2) +
exp(1)*(48*x + 96) + exp(1)*log(x)*(4*x + 2*x^2)) + exp(1)*log(x)*(52*x + 4*x^2))/(log(x)^2*(2*x + x^2) + exp(
2)*(2*x + x^2) + exp(1)*log(x)*(4*x + 2*x^2)),x)

[Out]

(x*exp(1) + 48*log(x + 2)*log(x) + x*log(x) + x*log(x + 2)*exp(1) + x*log(x + 2)*log(x))/(exp(1) + log(x))