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

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

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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 {gather*} 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 ) \end {gather*}

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

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

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

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

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

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giac [C]  time = 0.27, size = 128, normalized size = 4.92 \begin {gather*} -\frac {4}{3} \, {\rm Ei}\left (x + \log \relax (3) - \log \relax (2) - 3\right ) e^{\left (-\log \relax (3) + \log \relax (2) + 3\right )} \log \relax (3) + \frac {4}{3} \, {\rm Ei}\left (x + \log \relax (3) - \log \relax (2) - 3\right ) e^{\left (-\log \relax (3) + \log \relax (2) + 3\right )} \log \relax (2) - \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 \relax (3) - \log \relax (2) - 3\right ) e^{\left (-\log \relax (3) + \log \relax (2) + 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 \end {gather*}

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

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




method result size



risch \(\frac {4 \left (2+x \right ) {\mathrm e}^{x} \ln \left (\ln \relax (2)-\ln \relax (3)+3-x \right )}{3}+2 x\) \(24\)
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 \(2 x +\frac {8 \,{\mathrm e}^{x} \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}+\frac {4 \,{\mathrm e}^{x} x \ln \left (\ln \left (\frac {2}{3}\right )+3-x \right )}{3}\) \(30\)



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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4}{3} \, {\left (x + 2\right )} e^{x} \log \left (-x - \log \relax (3) + \log \relax (2) + 3\right ) - \frac {16}{9} \, e^{3} E_{1}\left (-x + \log \left (\frac {2}{3}\right ) + 3\right ) + 2 \, {\left (\log \left (\frac {2}{3}\right ) + 3\right )} \log \left (x - \log \left (\frac {2}{3}\right ) - 3\right ) - 2 \, \log \left (\frac {2}{3}\right ) \log \left (x - \log \left (\frac {2}{3}\right ) - 3\right ) + 2 \, x - \frac {8}{3} \, \int \frac {e^{x}}{x + \log \relax (3) - \log \relax (2) - 3}\,{d x} - 6 \, \log \left (x - \log \left (\frac {2}{3}\right ) - 3\right ) \end {gather*}

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \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 \end {gather*}

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)

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sympy [A]  time = 0.77, size = 32, normalized size = 1.23 \begin {gather*} 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} \end {gather*}

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

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