3.69.98 $\int \frac{18+e^x(-8-4x)-6x-6\log (\frac{3}{2})+e^x(36-4x^2-(12+4x)\log (\frac{3}{2}))\log (3-x-\log (\frac{3}{2}))}{9-3x-3\log (\frac{3}{2})}dx$

Optimal. Leaf size=26 $$2x+\frac{4}{3}{e}^x(2+x)\log (3-x-\log \left(\frac{3}{2}\right))$$

________________________________________________________________________________________

Rubi [A]  time = 1.72, antiderivative size = 45, normalized size of antiderivative = 1.73, number of steps used = 15, number of rules used = 8, integrand size = 68, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.118, Rules used = {6741, 6742, 6688, 2199, 2194, 2178, 2176, 2554} $$\begin{array}{c}2x-\frac{4}{3}{e}^x\log (-x+3-\log \left(\frac{3}{2}\right))+\frac{4}{3}{e}^x(x+3)\log (-x+3-\log \left(\frac{3}{2}\right))\end{array}$$

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

Rule 2178

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

Rule 2194

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 2199

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] &&  !$UseGamma === True

Rule 2554

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 6688

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

Rule 6741

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^x(-8-4x)-6x+18(1-\frac{1}{3}\log \left(\frac{3}{2}\right))+e^x(36-4x^2-(12+4x)\log \left(\frac{3}{2}\right))\log (3-x-\log \left(\frac{3}{2}\right))}{9-3x-3\log \left(\frac{3}{2}\right)}dx\\ & =\int (2+\frac{4e^x(-2-x-x^2\log (3-x-\log \left(\frac{3}{2}\right))+9(1-\frac{1}{3}\log \left(\frac{3}{2}\right))\log (3-x-\log \left(\frac{3}{2}\right))-x\log \left(\frac{3}{2}\right)\log (3-x-\log \left(\frac{3}{2}\right)))}{3(3-x-\log \left(\frac{3}{2}\right))})dx\\ & =2x+\frac{4}{3}\int \frac{e^x(-2-x-x^2\log (3-x-\log \left(\frac{3}{2}\right))+9(1-\frac{1}{3}\log \left(\frac{3}{2}\right))\log (3-x-\log \left(\frac{3}{2}\right))-x\log \left(\frac{3}{2}\right)\log (3-x-\log \left(\frac{3}{2}\right)))}{3-x-\log \left(\frac{3}{2}\right)}dx\\ & =2x+\frac{4}{3}\int \frac{e^x(-2-x-(3+x)(-3+x+\log \left(\frac{3}{2}\right))\log (3-x-\log \left(\frac{3}{2}\right)))}{3-x-\log \left(\frac{3}{2}\right)}dx\\ & =2x+\frac{4}{3}\int (\frac{e^x(2+x)}{-3+x+\log \left(\frac{3}{2}\right)}+e^x(3+x)\log (3-x-\log \left(\frac{3}{2}\right)))dx\\ & =2x+\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 (3-x-\log \left(\frac{3}{2}\right))dx\\ & =2x-\frac{4}{3}{e}^x\log (3-x-\log \left(\frac{3}{2}\right))+\frac{4}{3}{e}^x(3+x)\log (3-x-\log \left(\frac{3}{2}\right))-\frac{4}{3}\int \frac{e^x(-2-x)}{3-x-\log \left(\frac{3}{2}\right)}dx+\frac{4}{3}\int (e^x+\frac{e^x(5-\log \left(\frac{3}{2}\right))}{-3+x+\log \left(\frac{3}{2}\right)})dx\\ & =2x-\frac{4}{3}{e}^x\log (3-x-\log \left(\frac{3}{2}\right))+\frac{4}{3}{e}^x(3+x)\log (3-x-\log \left(\frac{3}{2}\right))+\frac{4\int e^{x}dx}{3}-\frac{4}{3}\int (e^x+\frac{e^x(5-\log \left(\frac{3}{2}\right))}{-3+x+\log \left(\frac{3}{2}\right)})dx+\frac{1}{3}\left(4(5-\log \left(\frac{3}{2}\right))\right)\int \frac{e^x}{-3+x+\log \left(\frac{3}{2}\right)}dx\\ & =\frac{4e^x}{3}+2x+\frac{8}{9}{e}^3\text{Ei}(-3+x+\log \left(\frac{3}{2}\right))(5-\log \left(\frac{3}{2}\right))-\frac{4}{3}{e}^x\log (3-x-\log \left(\frac{3}{2}\right))+\frac{4}{3}{e}^x(3+x)\log (3-x-\log \left(\frac{3}{2}\right))-\frac{4\int e^{x}dx}{3}-\frac{1}{3}\left(4(5-\log \left(\frac{3}{2}\right))\right)\int \frac{e^x}{-3+x+\log \left(\frac{3}{2}\right)}dx\\ & =2x-\frac{4}{3}{e}^x\log (3-x-\log \left(\frac{3}{2}\right))+\frac{4}{3}{e}^x(3+x)\log (3-x-\log \left(\frac{3}{2}\right))\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [A]  time = 2.84, size = 28, normalized size = 1.08 $$\begin{array}{c}\frac{2}{3}(3x+2e^x(2+x)\log (3-x-\log \left(\frac{3}{2}\right)))\end{array}$$

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 19, normalized size = 0.73 $$\begin{array}{c}\frac{4}{3}(x+2)e^x\log (-x+\log \left(\frac{2}{3}\right)+3)+2x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [C]  time = 0.27, size = 128, normalized size = 4.92 $$\begin{array}{c}-\frac{4}{3}\mathrm{E}\mathrm{i}(x+\log (3)-\log (2)-3)e^{(-\log (3)+\log (2)+3)}\log (3)+\frac{4}{3}\mathrm{E}\mathrm{i}(x+\log (3)-\log (2)-3)e^{(-\log (3)+\log (2)+3)}\log (2)-\frac{4}{3}\mathrm{E}\mathrm{i}(x-\log \left(\frac{2}{3}\right)-3)e^{(\log \left(\frac{2}{3}\right)+3)}\log \left(\frac{2}{3}\right)+\frac{4}{3}x{e}^x\log (-x+\log \left(\frac{2}{3}\right)+3)+\frac{20}{3}\mathrm{E}\mathrm{i}(x+\log (3)-\log (2)-3)e^{(-\log (3)+\log (2)+3)}-\frac{20}{3}\mathrm{E}\mathrm{i}(x-\log \left(\frac{2}{3}\right)-3)e^{(\log \left(\frac{2}{3}\right)+3)}+\frac{8}{3}{e}^x\log (-x+\log \left(\frac{2}{3}\right)+3)+2x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maple [A]  time = 0.19, size = 24, normalized size = 0.92

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

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\frac{4}{3}(x+2)e^x\log (-x-\log (3)+\log (2)+3)-\frac{16}{9}{e}^3{E}_1(-x+\log \left(\frac{2}{3}\right)+3)+2(\log \left(\frac{2}{3}\right)+3)\log (x-\log \left(\frac{2}{3}\right)-3)-2\log \left(\frac{2}{3}\right)\log (x-\log \left(\frac{2}{3}\right)-3)+2x-\frac{8}{3}\int \frac{e^x}{x+\log (3)-\log (2)-3}dx-6\log (x-\log \left(\frac{2}{3}\right)-3)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\int \frac{6\ln \left(\frac{2}{3}\right)-6x-{\mathrm{e}}^x(4x+8)+\ln (\ln \left(\frac{2}{3}\right)-x+3){\mathrm{e}}^x(\ln \left(\frac{2}{3}\right)(4x+12)-4x^2+36)+18}{3\ln \left(\frac{2}{3}\right)-3x+9}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 0.77, size = 32, normalized size = 1.23 $$\begin{array}{c}2x+\frac{(4x\log (-x+\log \left(\frac{2}{3}\right)+3)+8\log (-x+\log \left(\frac{2}{3}\right)+3))e^x}{3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________