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

Optimal. Leaf size=17 \[ e^{2 x} \log (\log (5)+e (11+x+\log (x))) \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 2, number of rules used = 2, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6688, 2288} \begin {gather*} e^{2 x} \log (e (x+11)+e \log (x)+\log (5)) \end {gather*}

Antiderivative was successfully verified.

[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 2288

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 19, normalized size = 1.12 \begin {gather*} e^{2 x} \log (e (11+x)+\log (5)+e \log (x)) \end {gather*}

Antiderivative was successfully verified.

[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]]

________________________________________________________________________________________

fricas [A]  time = 1.11, size = 20, normalized size = 1.18 \begin {gather*} e^{\left (2 \, x\right )} \log \left ({\left (x + 11\right )} e + e \log \relax (x) + \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))+(x+1)*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))

________________________________________________________________________________________

giac [A]  time = 0.23, size = 22, normalized size = 1.29 \begin {gather*} e^{\left (2 \, x\right )} \log \left (x e + e \log \relax (x) + 11 \, e + \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))+(x+1)*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))

________________________________________________________________________________________

maple [A]  time = 0.10, size = 21, normalized size = 1.24

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

Verification of antiderivative is not currently implemented for this CAS.

[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))
+(x+1)*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))

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 22, normalized size = 1.29 \begin {gather*} e^{\left (2 \, x\right )} \log \left (x e + e \log \relax (x) + 11 \, e + \log \relax (5)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))+(x+1)*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))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {\ln \left (\ln \relax (5)+\mathrm {e}\,\left (x+11\right )+\mathrm {e}\,\ln \relax (x)\right )\,\left ({\mathrm {e}}^{2\,x}\,\left (\mathrm {e}\,\left (2\,x^2+22\,x\right )+2\,x\,\ln \relax (5)\right )+2\,x\,{\mathrm {e}}^{2\,x+1}\,\ln \relax (x)\right )+{\mathrm {e}}^{2\,x+1}\,\left (x+1\right )}{x\,\ln \relax (5)+\mathrm {e}\,\left (x^2+11\,x\right )+x\,\mathrm {e}\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 4.59, size = 22, normalized size = 1.29 \begin {gather*} e^{2 x} \log {\left (e \left (x + 11\right ) + e \log {\relax (x )} + \log {\relax (5 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))+(x+1)*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))

________________________________________________________________________________________