\(\int \frac {e^{1+2 x} (1+x)+(e^{2 x} (e (22 x+2 x^2)+2 x \log (5))+2 e^{1+2 x} x \log (x)) \log (e (11+x)+\log (5)+e \log (x))}{e (11 x+x^2)+x \log (5)+e x \log (x)} \, dx\) [8119]

Optimal result

Integrand size = 84, antiderivative size = 17 \[ \int \frac {e^{1+2 x} (1+x)+\left (e^{2 x} \left (e \left (22 x+2 x^2\right )+2 x \log (5)\right )+2 e^{1+2 x} x \log (x)\right ) \log (e (11+x)+\log (5)+e \log (x))}{e \left (11 x+x^2\right )+x \log (5)+e x \log (x)} \, dx=e^{2 x} \log (\log (5)+e (11+x+\log (x))) \]

[Out]

ln(exp(1)*(11+x+ln(x))+ln(5))*exp(x)^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6820, 2326} \[ \int \frac {e^{1+2 x} (1+x)+\left (e^{2 x} \left (e \left (22 x+2 x^2\right )+2 x \log (5)\right )+2 e^{1+2 x} x \log (x)\right ) \log (e (11+x)+\log (5)+e \log (x))}{e \left (11 x+x^2\right )+x \log (5)+e x \log (x)} \, dx=e^{2 x} \log (e (x+11)+e \log (x)+\log (5)) \]

[In]

Int[(E^(1 + 2*x)*(1 + x) + (E^(2*x)*(E*(22*x + 2*x^2) + 2*x*Log[5]) + 2*E^(1 + 2*x)*x*Log[x])*Log[E*(11 + x) +
Log[5] + E*Log[x]])/(E*(11*x + x^2) + x*Log[5] + E*x*Log[x]),x]

[Out]

E^(2*x)*Log[E*(11 + x) + Log[5] + E*Log[x]]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int e^{2 x} \left (\frac {e (1+x)}{x (e (11+x)+\log (5)+e \log (x))}+2 \log (e (11+x)+\log (5)+e \log (x))\right ) \, dx \\ & = e^{2 x} \log (e (11+x)+\log (5)+e \log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^{1+2 x} (1+x)+\left (e^{2 x} \left (e \left (22 x+2 x^2\right )+2 x \log (5)\right )+2 e^{1+2 x} x \log (x)\right ) \log (e (11+x)+\log (5)+e \log (x))}{e \left (11 x+x^2\right )+x \log (5)+e x \log (x)} \, dx=e^{2 x} \log (e (11+x)+\log (5)+e \log (x)) \]

[In]

Integrate[(E^(1 + 2*x)*(1 + x) + (E^(2*x)*(E*(22*x + 2*x^2) + 2*x*Log[5]) + 2*E^(1 + 2*x)*x*Log[x])*Log[E*(11
+ x) + Log[5] + E*Log[x]])/(E*(11*x + x^2) + x*Log[5] + E*x*Log[x]),x]

[Out]

E^(2*x)*Log[E*(11 + x) + Log[5] + E*Log[x]]

Maple [A] (verified)

Time = 4.91 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24

method	result	size
risch	\({\mathrm e}^{2 x} \ln \left ({\mathrm e} \ln \left (x \right )+\ln \left (5\right )+\left (11+x \right ) {\mathrm e}\right )\)	\(21\)
parallelrisch	\({\mathrm e}^{2 x} \ln \left ({\mathrm e} \ln \left (x \right )+\ln \left (5\right )+\left (11+x \right ) {\mathrm e}\right )\)	\(21\)

[In]

int(((2*x*exp(1)*exp(x)^2*ln(x)+(2*x*ln(5)+(2*x^2+22*x)*exp(1))*exp(x)^2)*ln(exp(1)*ln(x)+ln(5)+(11+x)*exp(1))
+(1+x)*exp(1)*exp(x)^2)/(x*exp(1)*ln(x)+x*ln(5)+(x^2+11*x)*exp(1)),x,method=_RETURNVERBOSE)

[Out]

exp(2*x)*ln(exp(1)*ln(x)+ln(5)+(11+x)*exp(1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {e^{1+2 x} (1+x)+\left (e^{2 x} \left (e \left (22 x+2 x^2\right )+2 x \log (5)\right )+2 e^{1+2 x} x \log (x)\right ) \log (e (11+x)+\log (5)+e \log (x))}{e \left (11 x+x^2\right )+x \log (5)+e x \log (x)} \, dx=e^{\left (2 \, x\right )} \log \left ({\left (x + 11\right )} e + e \log \left (x\right ) + \log \left (5\right )\right ) \]

[In]

integrate(((2*x*exp(1)*exp(x)^2*log(x)+(2*x*log(5)+(2*x^2+22*x)*exp(1))*exp(x)^2)*log(exp(1)*log(x)+log(5)+(11
+x)*exp(1))+(1+x)*exp(1)*exp(x)^2)/(x*exp(1)*log(x)+x*log(5)+(x^2+11*x)*exp(1)),x, algorithm="fricas")

[Out]

e^(2*x)*log((x + 11)*e + e*log(x) + log(5))

Sympy [A] (verification not implemented)

Time = 2.37 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {e^{1+2 x} (1+x)+\left (e^{2 x} \left (e \left (22 x+2 x^2\right )+2 x \log (5)\right )+2 e^{1+2 x} x \log (x)\right ) \log (e (11+x)+\log (5)+e \log (x))}{e \left (11 x+x^2\right )+x \log (5)+e x \log (x)} \, dx=e^{2 x} \log {\left (e \left (x + 11\right ) + e \log {\left (x \right )} + \log {\left (5 \right )} \right )} \]

[In]

integrate(((2*x*exp(1)*exp(x)**2*ln(x)+(2*x*ln(5)+(2*x**2+22*x)*exp(1))*exp(x)**2)*ln(exp(1)*ln(x)+ln(5)+(11+x
)*exp(1))+(1+x)*exp(1)*exp(x)**2)/(x*exp(1)*ln(x)+x*ln(5)+(x**2+11*x)*exp(1)),x)

[Out]

exp(2*x)*log(E*(x + 11) + E*log(x) + log(5))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {e^{1+2 x} (1+x)+\left (e^{2 x} \left (e \left (22 x+2 x^2\right )+2 x \log (5)\right )+2 e^{1+2 x} x \log (x)\right ) \log (e (11+x)+\log (5)+e \log (x))}{e \left (11 x+x^2\right )+x \log (5)+e x \log (x)} \, dx=e^{\left (2 \, x\right )} \log \left (x e + e \log \left (x\right ) + 11 \, e + \log \left (5\right )\right ) \]

[In]

integrate(((2*x*exp(1)*exp(x)^2*log(x)+(2*x*log(5)+(2*x^2+22*x)*exp(1))*exp(x)^2)*log(exp(1)*log(x)+log(5)+(11
+x)*exp(1))+(1+x)*exp(1)*exp(x)^2)/(x*exp(1)*log(x)+x*log(5)+(x^2+11*x)*exp(1)),x, algorithm="maxima")

[Out]

e^(2*x)*log(x*e + e*log(x) + 11*e + log(5))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {e^{1+2 x} (1+x)+\left (e^{2 x} \left (e \left (22 x+2 x^2\right )+2 x \log (5)\right )+2 e^{1+2 x} x \log (x)\right ) \log (e (11+x)+\log (5)+e \log (x))}{e \left (11 x+x^2\right )+x \log (5)+e x \log (x)} \, dx=e^{\left (2 \, x\right )} \log \left (x e + e \log \left (x\right ) + 11 \, e + \log \left (5\right )\right ) \]

[In]

integrate(((2*x*exp(1)*exp(x)^2*log(x)+(2*x*log(5)+(2*x^2+22*x)*exp(1))*exp(x)^2)*log(exp(1)*log(x)+log(5)+(11
+x)*exp(1))+(1+x)*exp(1)*exp(x)^2)/(x*exp(1)*log(x)+x*log(5)+(x^2+11*x)*exp(1)),x, algorithm="giac")

[Out]

e^(2*x)*log(x*e + e*log(x) + 11*e + log(5))

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{1+2 x} (1+x)+\left (e^{2 x} \left (e \left (22 x+2 x^2\right )+2 x \log (5)\right )+2 e^{1+2 x} x \log (x)\right ) \log (e (11+x)+\log (5)+e \log (x))}{e \left (11 x+x^2\right )+x \log (5)+e x \log (x)} \, dx=\int \frac {\ln \left (\ln \left (5\right )+\mathrm {e}\,\left (x+11\right )+\mathrm {e}\,\ln \left (x\right )\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (\mathrm {e}\,\left (2\,x^2+22\,x\right )+2\,x\,\ln \left (5\right )\right )+2\,x\,{\mathrm {e}}^{2\,x+1}\,\ln \left (x\right )\right )+{\mathrm {e}}^{2\,x+1}\,\left (x+1\right )}{x\,\ln \left (5\right )+\mathrm {e}\,\left (x^2+11\,x\right )+x\,\mathrm {e}\,\ln \left (x\right )} \,d x \]

[In]

int((log(log(5) + exp(1)*(x + 11) + exp(1)*log(x))*(exp(2*x)*(exp(1)*(22*x + 2*x^2) + 2*x*log(5)) + 2*x*exp(2*
x)*exp(1)*log(x)) + exp(2*x)*exp(1)*(x + 1))/(x*log(5) + exp(1)*(11*x + x^2) + x*exp(1)*log(x)),x)

[Out]

int((log(log(5) + exp(1)*(x + 11) + exp(1)*log(x))*(exp(2*x)*(exp(1)*(22*x + 2*x^2) + 2*x*log(5)) + 2*x*exp(2*
x + 1)*log(x)) + exp(2*x + 1)*(x + 1))/(x*log(5) + exp(1)*(11*x + x^2) + x*exp(1)*log(x)), x)