\(\int \frac {18+e^x (-8-4 x)-6 x-6 \log (\frac {3}{2})+e^x (36-4 x^2-(12+4 x) \log (\frac {3}{2})) \log (3-x-\log (\frac {3}{2}))}{9-3 x-3 \log (\frac {3}{2})} \, dx\) [6898]

Optimal result

Integrand size = 68, antiderivative size = 26 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=2 x+\frac {4}{3} e^x (2+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right ) \]

[Out]

2*x+4/3*exp(x)*(2+x)*ln(ln(2/3)+3-x)

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6873, 6874, 6820, 2230, 2225, 2209, 2207, 2634} \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=2 x-\frac {4}{3} e^x \log \left (-x+3-\log \left (\frac {3}{2}\right )\right )+\frac {4}{3} e^x (x+3) \log \left (-x+3-\log \left (\frac {3}{2}\right )\right ) \]

[In]

Int[(18 + E^x*(-8 - 4*x) - 6*x - 6*Log[3/2] + E^x*(36 - 4*x^2 - (12 + 4*x)*Log[3/2])*Log[3 - x - Log[3/2]])/(9
- 3*x - 3*Log[3/2]),x]

[Out]

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

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 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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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 (-8-4 x)-6 x+18 \left (1-\frac {1}{3} \log \left (\frac {3}{2}\right )\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx \\ & = \int \left (2+\frac {4 e^x \left (-2-x-x^2 \log \left (3-x-\log \left (\frac {3}{2}\right )\right )+9 \left (1-\frac {1}{3} \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )-x \log \left (\frac {3}{2}\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )\right )}{3 \left (3-x-\log \left (\frac {3}{2}\right )\right )}\right ) \, dx \\ & = 2 x+\frac {4}{3} \int \frac {e^x \left (-2-x-x^2 \log \left (3-x-\log \left (\frac {3}{2}\right )\right )+9 \left (1-\frac {1}{3} \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )-x \log \left (\frac {3}{2}\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )\right )}{3-x-\log \left (\frac {3}{2}\right )} \, dx \\ & = 2 x+\frac {4}{3} \int \frac {e^x \left (-2-x-(3+x) \left (-3+x+\log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )\right )}{3-x-\log \left (\frac {3}{2}\right )} \, dx \\ & = 2 x+\frac {4}{3} \int \left (\frac {e^x (2+x)}{-3+x+\log \left (\frac {3}{2}\right )}+e^x (3+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )\right ) \, dx \\ & = 2 x+\frac {4}{3} \int \frac {e^x (2+x)}{-3+x+\log \left (\frac {3}{2}\right )} \, dx+\frac {4}{3} \int e^x (3+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right ) \, dx \\ & = 2 x-\frac {4}{3} e^x \log \left (3-x-\log \left (\frac {3}{2}\right )\right )+\frac {4}{3} e^x (3+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )-\frac {4}{3} \int \frac {e^x (-2-x)}{3-x-\log \left (\frac {3}{2}\right )} \, dx+\frac {4}{3} \int \left (e^x+\frac {e^x \left (5-\log \left (\frac {3}{2}\right )\right )}{-3+x+\log \left (\frac {3}{2}\right )}\right ) \, dx \\ & = 2 x-\frac {4}{3} e^x \log \left (3-x-\log \left (\frac {3}{2}\right )\right )+\frac {4}{3} e^x (3+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )+\frac {4 \int e^x \, dx}{3}-\frac {4}{3} \int \left (e^x+\frac {e^x \left (5-\log \left (\frac {3}{2}\right )\right )}{-3+x+\log \left (\frac {3}{2}\right )}\right ) \, dx+\frac {1}{3} \left (4 \left (5-\log \left (\frac {3}{2}\right )\right )\right ) \int \frac {e^x}{-3+x+\log \left (\frac {3}{2}\right )} \, dx \\ & = \frac {4 e^x}{3}+2 x+\frac {8}{9} e^3 \text {Ei}\left (-3+x+\log \left (\frac {3}{2}\right )\right ) \left (5-\log \left (\frac {3}{2}\right )\right )-\frac {4}{3} e^x \log \left (3-x-\log \left (\frac {3}{2}\right )\right )+\frac {4}{3} e^x (3+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )-\frac {4 \int e^x \, dx}{3}-\frac {1}{3} \left (4 \left (5-\log \left (\frac {3}{2}\right )\right )\right ) \int \frac {e^x}{-3+x+\log \left (\frac {3}{2}\right )} \, dx \\ & = 2 x-\frac {4}{3} e^x \log \left (3-x-\log \left (\frac {3}{2}\right )\right )+\frac {4}{3} e^x (3+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 6.45 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\frac {2}{3} \left (3 x+2 e^x (2+x) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )\right ) \]

[In]

Integrate[(18 + E^x*(-8 - 4*x) - 6*x - 6*Log[3/2] + E^x*(36 - 4*x^2 - (12 + 4*x)*Log[3/2])*Log[3 - x - Log[3/2
]])/(9 - 3*x - 3*Log[3/2]),x]

[Out]

(2*(3*x + 2*E^x*(2 + x)*Log[3 - x - Log[3/2]]))/3

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {4 \left (2+x \right ) {\mathrm e}^{x} \ln \left (\ln \left (2\right )-\ln \left (3\right )+3-x \right )}{3}+2 x\)	\(24\)
default	\(2 x +\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}\)	\(30\)
norman	\(2 x +\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}\)	\(30\)
parts	\(2 x +\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}\)	\(30\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+12+\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+4 \ln \left (\frac {2}{3}\right )+2 x\)	\(35\)

[In]

int((((4*x+12)*ln(2/3)-4*x^2+36)*exp(x)*ln(ln(2/3)+3-x)+(-4*x-8)*exp(x)+6*ln(2/3)-6*x+18)/(3*ln(2/3)-3*x+9),x,
method=_RETURNVERBOSE)

[Out]

4/3*(2+x)*exp(x)*ln(ln(2)-ln(3)+3-x)+2*x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\frac {4}{3} \, {\left (x + 2\right )} e^{x} \log \left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + 2 \, x \]

[In]

integrate((((4*x+12)*log(2/3)-4*x^2+36)*exp(x)*log(log(2/3)+3-x)+(-4*x-8)*exp(x)+6*log(2/3)-6*x+18)/(3*log(2/3
)-3*x+9),x, algorithm="fricas")

[Out]

4/3*(x + 2)*e^x*log(-x + log(2/3) + 3) + 2*x

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=2 x + \frac {\left (4 x \log {\left (- x + \log {\left (\frac {2}{3} \right )} + 3 \right )} + 8 \log {\left (- x + \log {\left (\frac {2}{3} \right )} + 3 \right )}\right ) e^{x}}{3} \]

[In]

integrate((((4*x+12)*ln(2/3)-4*x**2+36)*exp(x)*ln(ln(2/3)+3-x)+(-4*x-8)*exp(x)+6*ln(2/3)-6*x+18)/(3*ln(2/3)-3*
x+9),x)

[Out]

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

Maxima [F]

\[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\int { \frac {2 \, {\left (2 \, {\left (x^{2} - {\left (x + 3\right )} \log \left (\frac {2}{3}\right ) - 9\right )} e^{x} \log \left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + 2 \, {\left (x + 2\right )} e^{x} + 3 \, x - 3 \, \log \left (\frac {2}{3}\right ) - 9\right )}}{3 \, {\left (x - \log \left (\frac {2}{3}\right ) - 3\right )}} \,d x } \]

[In]

integrate((((4*x+12)*log(2/3)-4*x^2+36)*exp(x)*log(log(2/3)+3-x)+(-4*x-8)*exp(x)+6*log(2/3)-6*x+18)/(3*log(2/3
)-3*x+9),x, algorithm="maxima")

[Out]

4/3*(x + 2)*e^x*log(-x - log(3) + log(2) + 3) - 16/9*e^3*exp_integral_e(1, -x + log(2/3) + 3) + 2*(log(2/3) +
3)*log(x - log(2/3) - 3) - 2*log(2/3)*log(x - log(2/3) - 3) + 2*x - 8/3*integrate(e^x/(x + log(3) - log(2) - 3
), x) - 6*log(x - log(2/3) - 3)

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 4.92 \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=-\frac {4}{3} \, {\rm Ei}\left (x + \log \left (3\right ) - \log \left (2\right ) - 3\right ) e^{\left (-\log \left (3\right ) + \log \left (2\right ) + 3\right )} \log \left (3\right ) + \frac {4}{3} \, {\rm Ei}\left (x + \log \left (3\right ) - \log \left (2\right ) - 3\right ) e^{\left (-\log \left (3\right ) + \log \left (2\right ) + 3\right )} \log \left (2\right ) - \frac {4}{3} \, {\rm Ei}\left (x - \log \left (\frac {2}{3}\right ) - 3\right ) e^{\left (\log \left (\frac {2}{3}\right ) + 3\right )} \log \left (\frac {2}{3}\right ) + \frac {4}{3} \, x e^{x} \log \left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + \frac {20}{3} \, {\rm Ei}\left (x + \log \left (3\right ) - \log \left (2\right ) - 3\right ) e^{\left (-\log \left (3\right ) + \log \left (2\right ) + 3\right )} - \frac {20}{3} \, {\rm Ei}\left (x - \log \left (\frac {2}{3}\right ) - 3\right ) e^{\left (\log \left (\frac {2}{3}\right ) + 3\right )} + \frac {8}{3} \, e^{x} \log \left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + 2 \, x \]

[In]

integrate((((4*x+12)*log(2/3)-4*x^2+36)*exp(x)*log(log(2/3)+3-x)+(-4*x-8)*exp(x)+6*log(2/3)-6*x+18)/(3*log(2/3
)-3*x+9),x, algorithm="giac")

[Out]

-4/3*Ei(x + log(3) - log(2) - 3)*e^(-log(3) + log(2) + 3)*log(3) + 4/3*Ei(x + log(3) - log(2) - 3)*e^(-log(3)
+ log(2) + 3)*log(2) - 4/3*Ei(x - log(2/3) - 3)*e^(log(2/3) + 3)*log(2/3) + 4/3*x*e^x*log(-x + log(2/3) + 3) +
20/3*Ei(x + log(3) - log(2) - 3)*e^(-log(3) + log(2) + 3) - 20/3*Ei(x - log(2/3) - 3)*e^(log(2/3) + 3) + 8/3*
e^x*log(-x + log(2/3) + 3) + 2*x

Mupad [F(-1)]

Timed out. \[ \int \frac {18+e^x (-8-4 x)-6 x-6 \log \left (\frac {3}{2}\right )+e^x \left (36-4 x^2-(12+4 x) \log \left (\frac {3}{2}\right )\right ) \log \left (3-x-\log \left (\frac {3}{2}\right )\right )}{9-3 x-3 \log \left (\frac {3}{2}\right )} \, dx=\int \frac {6\,\ln \left (\frac {2}{3}\right )-6\,x-{\mathrm {e}}^x\,\left (4\,x+8\right )+\ln \left (\ln \left (\frac {2}{3}\right )-x+3\right )\,{\mathrm {e}}^x\,\left (\ln \left (\frac {2}{3}\right )\,\left (4\,x+12\right )-4\,x^2+36\right )+18}{3\,\ln \left (\frac {2}{3}\right )-3\,x+9} \,d x \]

[In]

int((6*log(2/3) - 6*x - exp(x)*(4*x + 8) + log(log(2/3) - x + 3)*exp(x)*(log(2/3)*(4*x + 12) - 4*x^2 + 36) + 1
8)/(3*log(2/3) - 3*x + 9),x)

[Out]

int((6*log(2/3) - 6*x - exp(x)*(4*x + 8) + log(log(2/3) - x + 3)*exp(x)*(log(2/3)*(4*x + 12) - 4*x^2 + 36) + 1
8)/(3*log(2/3) - 3*x + 9), x)