\(\int \frac {e^x (4+x)+(3 e^x+e^x (12+3 x) \log (4+x)) \log (2 \log (5))}{4+x} \, dx\) [130]

Optimal result

Integrand size = 39, antiderivative size = 17 \[ \int \frac {e^x (4+x)+\left (3 e^x+e^x (12+3 x) \log (4+x)\right ) \log (2 \log (5))}{4+x} \, dx=e^x (1+3 \log (4+x) \log (2 \log (5))) \]

[Out]

(1+3*ln(2*ln(5))*ln(4+x))*exp(x)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6873, 6874, 2225, 2634, 2209, 2230} \[ \int \frac {e^x (4+x)+\left (3 e^x+e^x (12+3 x) \log (4+x)\right ) \log (2 \log (5))}{4+x} \, dx=e^x+3 e^x \log (\log (25)) \log (x+4) \]

[In]

Int[(E^x*(4 + x) + (3*E^x + E^x*(12 + 3*x)*Log[4 + x])*Log[2*Log[5]])/(4 + x),x]

[Out]

E^x + 3*E^x*Log[4 + x]*Log[Log[25]]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (x+4 \left (1+\frac {3}{4} \log (\log (25))\right )+12 \log (4+x) \log (\log (25))+3 x \log (4+x) \log (\log (25))\right )}{4+x} \, dx \\ & = \int \left (3 e^x \log (4+x) \log (\log (25))+\frac {e^x (4+x+3 \log (\log (25)))}{4+x}\right ) \, dx \\ & = (3 \log (\log (25))) \int e^x \log (4+x) \, dx+\int \frac {e^x (4+x+3 \log (\log (25)))}{4+x} \, dx \\ & = 3 e^x \log (4+x) \log (\log (25))-(3 \log (\log (25))) \int \frac {e^x}{4+x} \, dx+\int \left (e^x+\frac {3 e^x \log (\log (25))}{4+x}\right ) \, dx \\ & = -\frac {3 \operatorname {ExpIntegralEi}(4+x) \log (\log (25))}{e^4}+3 e^x \log (4+x) \log (\log (25))+(3 \log (\log (25))) \int \frac {e^x}{4+x} \, dx+\int e^x \, dx \\ & = e^x+3 e^x \log (4+x) \log (\log (25)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^x (4+x)+\left (3 e^x+e^x (12+3 x) \log (4+x)\right ) \log (2 \log (5))}{4+x} \, dx=e^x (1+3 \log (4+x) \log (\log (25))) \]

[In]

Integrate[(E^x*(4 + x) + (3*E^x + E^x*(12 + 3*x)*Log[4 + x])*Log[2*Log[5]])/(4 + x),x]

[Out]

E^x*(1 + 3*Log[4 + x]*Log[Log[25]])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
default	\(3 \,{\mathrm e}^{x} \ln \left (4+x \right ) \ln \left (2 \ln \left (5\right )\right )+{\mathrm e}^{x}\)	\(17\)
parallelrisch	\(3 \,{\mathrm e}^{x} \ln \left (4+x \right ) \ln \left (2 \ln \left (5\right )\right )+{\mathrm e}^{x}\)	\(17\)
risch	\(3 \left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )\right ) {\mathrm e}^{x} \ln \left (4+x \right )+{\mathrm e}^{x}\)	\(18\)
norman	\(\left (3 \ln \left (2\right )+3 \ln \left (\ln \left (5\right )\right )\right ) {\mathrm e}^{x} \ln \left (4+x \right )+{\mathrm e}^{x}\)	\(21\)

[In]

int((((3*x+12)*exp(x)*ln(4+x)+3*exp(x))*ln(2*ln(5))+(4+x)*exp(x))/(4+x),x,method=_RETURNVERBOSE)

[Out]

3*exp(x)*ln(4+x)*ln(2*ln(5))+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {e^x (4+x)+\left (3 e^x+e^x (12+3 x) \log (4+x)\right ) \log (2 \log (5))}{4+x} \, dx=3 \, e^{x} \log \left (x + 4\right ) \log \left (2 \, \log \left (5\right )\right ) + e^{x} \]

[In]

integrate((((3*x+12)*exp(x)*log(4+x)+3*exp(x))*log(2*log(5))+(4+x)*exp(x))/(4+x),x, algorithm="fricas")

[Out]

3*e^x*log(x + 4)*log(2*log(5)) + e^x

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {e^x (4+x)+\left (3 e^x+e^x (12+3 x) \log (4+x)\right ) \log (2 \log (5))}{4+x} \, dx=\left (3 \log {\left (x + 4 \right )} \log {\left (\log {\left (5 \right )} \right )} + 3 \log {\left (2 \right )} \log {\left (x + 4 \right )} + 1\right ) e^{x} \]

[In]

integrate((((3*x+12)*exp(x)*ln(4+x)+3*exp(x))*ln(2*ln(5))+(4+x)*exp(x))/(4+x),x)

[Out]

(3*log(x + 4)*log(log(5)) + 3*log(2)*log(x + 4) + 1)*exp(x)

Maxima [F]

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

[In]

integrate((((3*x+12)*exp(x)*log(4+x)+3*exp(x))*log(2*log(5))+(4+x)*exp(x))/(4+x),x, algorithm="maxima")

[Out]

-3*e^(-4)*exp_integral_e(1, -x - 4)*log(2*log(5)) - 4*e^(-4)*exp_integral_e(1, -x - 4) + (3*(x*(log(2) + log(l
og(5))) + 4*log(2) + 4*log(log(5)))*e^x*log(x + 4) + x*e^x)/(x + 4) - integrate((3*x*(log(2) + log(log(5))) +
12*log(2) + 12*log(log(5)) + 4)*e^x/(x^2 + 8*x + 16), x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {e^x (4+x)+\left (3 e^x+e^x (12+3 x) \log (4+x)\right ) \log (2 \log (5))}{4+x} \, dx=3 \, e^{x} \log \left (x + 4\right ) \log \left (2 \, \log \left (5\right )\right ) + e^{x} \]

[In]

integrate((((3*x+12)*exp(x)*log(4+x)+3*exp(x))*log(2*log(5))+(4+x)*exp(x))/(4+x),x, algorithm="giac")

[Out]

3*e^x*log(x + 4)*log(2*log(5)) + e^x

Mupad [B] (verification not implemented)

Time = 8.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^x (4+x)+\left (3 e^x+e^x (12+3 x) \log (4+x)\right ) \log (2 \log (5))}{4+x} \, dx={\mathrm {e}}^x\,\left (3\,\ln \left (x+4\right )\,\ln \left (\ln \left (25\right )\right )+1\right ) \]

[In]

int((exp(x)*(x + 4) + log(2*log(5))*(3*exp(x) + log(x + 4)*exp(x)*(3*x + 12)))/(x + 4),x)

[Out]

exp(x)*(3*log(x + 4)*log(log(25)) + 1)