\(\int \frac {1}{2} (e^x (16+e^4 (6+x^4))+e^{4+x} (6+6 x+5 x^4+x^5) \log (x)) \, dx\) [4160]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 41, antiderivative size = 22 \[ \int \frac {1}{2} \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx=2 e^x \left (4+\frac {1}{4} e^4 x \left (6+x^4\right ) \log (x)\right ) \]

[Out]

2*exp(x)*(1/4*(x^4+6)*x*exp(4)*ln(x)+4)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 22, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {12, 2227, 2225, 2207, 2634} \[ \int \frac {1}{2} \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx=\frac {1}{2} e^{x+4} x^5 \log (x)+8 e^x+3 e^{x+4} x \log (x) \]

[In]

Int[(E^x*(16 + E^4*(6 + x^4)) + E^(4 + x)*(6 + 6*x + 5*x^4 + x^5)*Log[x])/2,x]

[Out]

8*E^x + 3*E^(4 + x)*x*Log[x] + (E^(4 + x)*x^5*Log[x])/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !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 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx \\ & = \frac {1}{2} \int e^x \left (16+e^4 \left (6+x^4\right )\right ) \, dx+\frac {1}{2} \int e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x) \, dx \\ & = 3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x)-\frac {1}{2} \int e^{4+x} \left (6+x^4\right ) \, dx+\frac {1}{2} \int \left (16 e^x+e^{4+x} \left (6+x^4\right )\right ) \, dx \\ & = 3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x)+\frac {1}{2} \int e^{4+x} \left (6+x^4\right ) \, dx-\frac {1}{2} \int \left (6 e^{4+x}+e^{4+x} x^4\right ) \, dx+8 \int e^x \, dx \\ & = 8 e^x+3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x)-\frac {1}{2} \int e^{4+x} x^4 \, dx+\frac {1}{2} \int \left (6 e^{4+x}+e^{4+x} x^4\right ) \, dx-3 \int e^{4+x} \, dx \\ & = 8 e^x-3 e^{4+x}-\frac {1}{2} e^{4+x} x^4+3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x)+\frac {1}{2} \int e^{4+x} x^4 \, dx+2 \int e^{4+x} x^3 \, dx+3 \int e^{4+x} \, dx \\ & = 8 e^x+2 e^{4+x} x^3+3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x)-2 \int e^{4+x} x^3 \, dx-6 \int e^{4+x} x^2 \, dx \\ & = 8 e^x-6 e^{4+x} x^2+3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x)+6 \int e^{4+x} x^2 \, dx+12 \int e^{4+x} x \, dx \\ & = 8 e^x+12 e^{4+x} x+3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x)-12 \int e^{4+x} \, dx-12 \int e^{4+x} x \, dx \\ & = 8 e^x-12 e^{4+x}+3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x)+12 \int e^{4+x} \, dx \\ & = 8 e^x+3 e^{4+x} x \log (x)+\frac {1}{2} e^{4+x} x^5 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1}{2} \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx=\frac {1}{2} e^x \left (16+e^4 x \left (6+x^4\right ) \log (x)\right ) \]

[In]

Integrate[(E^x*(16 + E^4*(6 + x^4)) + E^(4 + x)*(6 + 6*x + 5*x^4 + x^5)*Log[x])/2,x]

[Out]

(E^x*(16 + E^4*x*(6 + x^4)*Log[x]))/2

Maple [A] (verified)

Time = 3.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18

method result size
risch \(\frac {\ln \left (x \right ) {\mathrm e}^{4+x} x^{5}}{2}+3 \ln \left (x \right ) {\mathrm e}^{4+x} x +8 \,{\mathrm e}^{x}\) \(26\)
parallelrisch \(\frac {\ln \left (x \right ) {\mathrm e}^{4} {\mathrm e}^{x} x^{5}}{2}+3 \ln \left (x \right ) {\mathrm e}^{4} {\mathrm e}^{x} x +8 \,{\mathrm e}^{x}\) \(26\)

[In]

int(1/2*(x^5+5*x^4+6*x+6)*exp(4)*exp(x)*ln(x)+1/2*((x^4+6)*exp(4)+16)*exp(x),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(x)*exp(4+x)*x^5+3*ln(x)*exp(4+x)*x+8*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {1}{2} \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx=\frac {1}{2} \, {\left ({\left (x^{5} + 6 \, x\right )} e^{\left (x + 8\right )} \log \left (x\right ) + 16 \, e^{\left (x + 4\right )}\right )} e^{\left (-4\right )} \]

[In]

integrate(1/2*(x^5+5*x^4+6*x+6)*exp(4)*exp(x)*log(x)+1/2*((x^4+6)*exp(4)+16)*exp(x),x, algorithm="fricas")

[Out]

1/2*((x^5 + 6*x)*e^(x + 8)*log(x) + 16*e^(x + 4))*e^(-4)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{2} \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx=\frac {\left (x^{5} e^{4} \log {\left (x \right )} + 6 x e^{4} \log {\left (x \right )} + 16\right ) e^{x}}{2} \]

[In]

integrate(1/2*(x**5+5*x**4+6*x+6)*exp(4)*exp(x)*ln(x)+1/2*((x**4+6)*exp(4)+16)*exp(x),x)

[Out]

(x**5*exp(4)*log(x) + 6*x*exp(4)*log(x) + 16)*exp(x)/2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (17) = 34\).

Time = 0.20 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.41 \[ \int \frac {1}{2} \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx=\frac {1}{2} \, {\left (x^{5} e^{4} + 6 \, x e^{4}\right )} e^{x} \log \left (x\right ) - \frac {1}{2} \, {\left (x^{4} e^{4} - 4 \, x^{3} e^{4} + 12 \, x^{2} e^{4} - 24 \, x e^{4} + 30 \, e^{4}\right )} e^{x} + \frac {1}{2} \, {\left (x^{4} e^{4} - 4 \, x^{3} e^{4} + 12 \, x^{2} e^{4} - 24 \, x e^{4} + 24 \, e^{4}\right )} e^{x} + 3 \, e^{\left (x + 4\right )} + 8 \, e^{x} \]

[In]

integrate(1/2*(x^5+5*x^4+6*x+6)*exp(4)*exp(x)*log(x)+1/2*((x^4+6)*exp(4)+16)*exp(x),x, algorithm="maxima")

[Out]

1/2*(x^5*e^4 + 6*x*e^4)*e^x*log(x) - 1/2*(x^4*e^4 - 4*x^3*e^4 + 12*x^2*e^4 - 24*x*e^4 + 30*e^4)*e^x + 1/2*(x^4
*e^4 - 4*x^3*e^4 + 12*x^2*e^4 - 24*x*e^4 + 24*e^4)*e^x + 3*e^(x + 4) + 8*e^x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (17) = 34\).

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.36 \[ \int \frac {1}{2} \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx=\frac {1}{2} \, {\left (x^{5} + 6 \, x\right )} e^{\left (x + 4\right )} \log \left (x\right ) + \frac {1}{2} \, {\left (x^{4} - 4 \, x^{3} + 12 \, x^{2} - 24 \, x + 30\right )} e^{\left (x + 4\right )} - \frac {1}{2} \, {\left (x^{4} - 4 \, x^{3} + 12 \, x^{2} - 24 \, x + 24\right )} e^{\left (x + 4\right )} - 3 \, e^{\left (x + 4\right )} + 8 \, e^{x} \]

[In]

integrate(1/2*(x^5+5*x^4+6*x+6)*exp(4)*exp(x)*log(x)+1/2*((x^4+6)*exp(4)+16)*exp(x),x, algorithm="giac")

[Out]

1/2*(x^5 + 6*x)*e^(x + 4)*log(x) + 1/2*(x^4 - 4*x^3 + 12*x^2 - 24*x + 30)*e^(x + 4) - 1/2*(x^4 - 4*x^3 + 12*x^
2 - 24*x + 24)*e^(x + 4) - 3*e^(x + 4) + 8*e^x

Mupad [B] (verification not implemented)

Time = 10.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {1}{2} \left (e^x \left (16+e^4 \left (6+x^4\right )\right )+e^{4+x} \left (6+6 x+5 x^4+x^5\right ) \log (x)\right ) \, dx=8\,{\mathrm {e}}^x+{\mathrm {e}}^x\,\ln \left (x\right )\,\left (\frac {{\mathrm {e}}^4\,x^5}{2}+3\,{\mathrm {e}}^4\,x\right ) \]

[In]

int((exp(x)*(exp(4)*(x^4 + 6) + 16))/2 + (exp(4)*exp(x)*log(x)*(6*x + 5*x^4 + x^5 + 6))/2,x)

[Out]

8*exp(x) + exp(x)*log(x)*(3*x*exp(4) + (x^5*exp(4))/2)