∫ −36exx3+(−1500−1500x+ex(500+500x)) log(3−ex)+(900+ex(−300−300x)+900x) log(3−ex) log(log(3−ex))+(−180−180x+ex(60+60x)) log(3−ex) log2(log(3−ex))+(12+ex(−4−4x)+12x) log(3−ex) log3(log(3−ex))________________________________________________________________________________________________________________________________________________________ (375x3−125exx3) log(3−ex)+(−225x3+75exx3) log(3−ex) log(log(3−ex))+(45x3−15exx3) log(3−ex) log2(log(3−ex))+(−3x3+exx3) log(3−ex) log3(log(3−ex)) dx

3.1.3 \(\int \frac {-36 e^x x^3+(-1500-1500 x+e^x (500+500 x)) \log (3-e^x)+(900+e^x (-300-300 x)+900 x) \log (3-e^x) \log (\log (3-e^x))+(-180-180 x+e^x (60+60 x)) \log (3-e^x) \log ^2(\log (3-e^x))+(12+e^x (-4-4 x)+12 x) \log (3-e^x) \log ^3(\log (3-e^x))}{(375 x^3-125 e^x x^3) \log (3-e^x)+(-225 x^3+75 e^x x^3) \log (3-e^x) \log (\log (3-e^x))+(45 x^3-15 e^x x^3) \log (3-e^x) \log ^2(\log (3-e^x))+(-3 x^3+e^x x^3) \log (3-e^x) \log ^3(\log (3-e^x))} \, dx\)

Optimal. Leaf size=33 \[ 3+\frac {1}{2} \left (\left (2+\frac {2}{x}\right )^2+\frac {36}{\left (5-\log \left (\log \left (3-e^x\right )\right )\right )^2}\right ) \]

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Rubi [A] time = 0.48, antiderivative size = 28, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, integrand size = 258, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6688, 37, 2282, 6686} \begin {gather*} \frac {2 (x+1)^2}{x^2}+\frac {18}{\left (5-\log \left (\log \left (3-e^x\right )\right )\right )^2} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(-36*E^x*x^3 + (-1500 - 1500*x + E^x*(500 + 500*x))*Log[3 - E^x] + (900 + E^x*(-300 - 300*x) + 900*x)*Log[

3 - E^x]*Log[Log[3 - E^x]] + (-180 - 180*x + E^x*(60 + 60*x))*Log[3 - E^x]*Log[Log[3 - E^x]]^2 + (12 + E^x*(-4

- 4*x) + 12*x)*Log[3 - E^x]*Log[Log[3 - E^x]]^3)/((375*x^3 - 125*E^x*x^3)*Log[3 - E^x] + (-225*x^3 + 75*E^x*x

^3)*Log[3 - E^x]*Log[Log[3 - E^x]] + (45*x^3 - 15*E^x*x^3)*Log[3 - E^x]*Log[Log[3 - E^x]]^2 + (-3*x^3 + E^x*x^

3)*Log[3 - E^x]*Log[Log[3 - E^x]]^3),x]

[Out]

(2*(1 + x)^2)/x^2 + 18/(5 - Log[Log[3 - E^x]])^2

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +

1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ

[m, -1]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu

nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F

reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x

] && InverseFunctionQ[F[x]]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /; !F

alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {4 (1+x)}{x^3}-\frac {36 e^x}{\left (-3+e^x\right ) \log \left (3-e^x\right ) \left (-5+\log \left (\log \left (3-e^x\right )\right )\right )^3}\right ) \, dx\\ &=-\left (4 \int \frac {1+x}{x^3} \, dx\right )-36 \int \frac {e^x}{\left (-3+e^x\right ) \log \left (3-e^x\right ) \left (-5+\log \left (\log \left (3-e^x\right )\right )\right )^3} \, dx\\ &=\frac {2 (1+x)^2}{x^2}-36 \operatorname {Subst}\left (\int \frac {1}{(-3+x) \log (3-x) (-5+\log (\log (3-x)))^3} \, dx,x,e^x\right )\\ &=\frac {2 (1+x)^2}{x^2}+\frac {18}{\left (5-\log \left (\log \left (3-e^x\right )\right )\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.06, size = 26, normalized size = 0.79 \begin {gather*} \frac {2}{x^2}+\frac {4}{x}+\frac {18}{\left (-5+\log \left (\log \left (3-e^x\right )\right )\right )^2} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-36*E^x*x^3 + (-1500 - 1500*x + E^x*(500 + 500*x))*Log[3 - E^x] + (900 + E^x*(-300 - 300*x) + 900*x

)*Log[3 - E^x]*Log[Log[3 - E^x]] + (-180 - 180*x + E^x*(60 + 60*x))*Log[3 - E^x]*Log[Log[3 - E^x]]^2 + (12 + E

^x*(-4 - 4*x) + 12*x)*Log[3 - E^x]*Log[Log[3 - E^x]]^3)/((375*x^3 - 125*E^x*x^3)*Log[3 - E^x] + (-225*x^3 + 75

*E^x*x^3)*Log[3 - E^x]*Log[Log[3 - E^x]] + (45*x^3 - 15*E^x*x^3)*Log[3 - E^x]*Log[Log[3 - E^x]]^2 + (-3*x^3 +

E^x*x^3)*Log[3 - E^x]*Log[Log[3 - E^x]]^3),x]

[Out]

2/x^2 + 4/x + 18/(-5 + Log[Log[3 - E^x]])^2

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fricas [B] time = 0.49, size = 78, normalized size = 2.36 \begin {gather*} \frac {2 \, {\left ({\left (2 \, x + 1\right )} \log \left (\log \left (-e^{x} + 3\right )\right )^{2} + 9 \, x^{2} - 10 \, {\left (2 \, x + 1\right )} \log \left (\log \left (-e^{x} + 3\right )\right ) + 50 \, x + 25\right )}}{x^{2} \log \left (\log \left (-e^{x} + 3\right )\right )^{2} - 10 \, x^{2} \log \left (\log \left (-e^{x} + 3\right )\right ) + 25 \, x^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-4)*exp(x)+12*x+12)*log(-exp(x)+3)*log(log(-exp(x)+3))^3+((60*x+60)*exp(x)-180*x-180)*log(-ex

p(x)+3)*log(log(-exp(x)+3))^2+((-300*x-300)*exp(x)+900*x+900)*log(-exp(x)+3)*log(log(-exp(x)+3))+((500*x+500)*

exp(x)-1500*x-1500)*log(-exp(x)+3)-36*exp(x)*x^3)/((exp(x)*x^3-3*x^3)*log(-exp(x)+3)*log(log(-exp(x)+3))^3+(-1

5*exp(x)*x^3+45*x^3)*log(-exp(x)+3)*log(log(-exp(x)+3))^2+(75*exp(x)*x^3-225*x^3)*log(-exp(x)+3)*log(log(-exp(

x)+3))+(-125*exp(x)*x^3+375*x^3)*log(-exp(x)+3)),x, algorithm="fricas")

[Out]

2*((2*x + 1)*log(log(-e^x + 3))^2 + 9*x^2 - 10*(2*x + 1)*log(log(-e^x + 3)) + 50*x + 25)/(x^2*log(log(-e^x + 3

))^2 - 10*x^2*log(log(-e^x + 3)) + 25*x^2)

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giac [B] time = 0.78, size = 91, normalized size = 2.76 \begin {gather*} \frac {2 \, {\left (2 \, x \log \left (\log \left (-e^{x} + 3\right )\right )^{2} + 9 \, x^{2} - 20 \, x \log \left (\log \left (-e^{x} + 3\right )\right ) + \log \left (\log \left (-e^{x} + 3\right )\right )^{2} + 50 \, x - 10 \, \log \left (\log \left (-e^{x} + 3\right )\right ) + 25\right )}}{x^{2} \log \left (\log \left (-e^{x} + 3\right )\right )^{2} - 10 \, x^{2} \log \left (\log \left (-e^{x} + 3\right )\right ) + 25 \, x^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-4)*exp(x)+12*x+12)*log(-exp(x)+3)*log(log(-exp(x)+3))^3+((60*x+60)*exp(x)-180*x-180)*log(-ex

p(x)+3)*log(log(-exp(x)+3))^2+((-300*x-300)*exp(x)+900*x+900)*log(-exp(x)+3)*log(log(-exp(x)+3))+((500*x+500)*

exp(x)-1500*x-1500)*log(-exp(x)+3)-36*exp(x)*x^3)/((exp(x)*x^3-3*x^3)*log(-exp(x)+3)*log(log(-exp(x)+3))^3+(-1

5*exp(x)*x^3+45*x^3)*log(-exp(x)+3)*log(log(-exp(x)+3))^2+(75*exp(x)*x^3-225*x^3)*log(-exp(x)+3)*log(log(-exp(

x)+3))+(-125*exp(x)*x^3+375*x^3)*log(-exp(x)+3)),x, algorithm="giac")

[Out]

2*(2*x*log(log(-e^x + 3))^2 + 9*x^2 - 20*x*log(log(-e^x + 3)) + log(log(-e^x + 3))^2 + 50*x - 10*log(log(-e^x

+ 3)) + 25)/(x^2*log(log(-e^x + 3))^2 - 10*x^2*log(log(-e^x + 3)) + 25*x^2)

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maple [A] time = 0.06, size = 26, normalized size = 0.79


method	result	size

risch	\(\frac {4 x +2}{x^{2}}+\frac {18}{\left (\ln \left (\ln \left (-{\mathrm e}^{x}+3\right )\right )-5\right )^{2}}\)	\(26\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x-4)*exp(x)+12*x+12)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))^3+((60*x+60)*exp(x)-180*x-180)*ln(-exp(x)+3)*ln

(ln(-exp(x)+3))^2+((-300*x-300)*exp(x)+900*x+900)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))+((500*x+500)*exp(x)-1500*x-1

500)*ln(-exp(x)+3)-36*exp(x)*x^3)/((exp(x)*x^3-3*x^3)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))^3+(-15*exp(x)*x^3+45*x^3

)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))^2+(75*exp(x)*x^3-225*x^3)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))+(-125*exp(x)*x^3+3

75*x^3)*ln(-exp(x)+3)),x,method=_RETURNVERBOSE)

[Out]

2*(2*x+1)/x^2+18/(ln(ln(-exp(x)+3))-5)^2

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maxima [A] time = 0.86, size = 37, normalized size = 1.12 \begin {gather*} \frac {18}{\log \left (\log \left (-e^{x} + 3\right )\right )^{2} - 10 \, \log \left (\log \left (-e^{x} + 3\right )\right ) + 25} + \frac {2 \, {\left (2 \, x + 1\right )}}{x^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-4)*exp(x)+12*x+12)*log(-exp(x)+3)*log(log(-exp(x)+3))^3+((60*x+60)*exp(x)-180*x-180)*log(-ex

p(x)+3)*log(log(-exp(x)+3))^2+((-300*x-300)*exp(x)+900*x+900)*log(-exp(x)+3)*log(log(-exp(x)+3))+((500*x+500)*

exp(x)-1500*x-1500)*log(-exp(x)+3)-36*exp(x)*x^3)/((exp(x)*x^3-3*x^3)*log(-exp(x)+3)*log(log(-exp(x)+3))^3+(-1

5*exp(x)*x^3+45*x^3)*log(-exp(x)+3)*log(log(-exp(x)+3))^2+(75*exp(x)*x^3-225*x^3)*log(-exp(x)+3)*log(log(-exp(

x)+3))+(-125*exp(x)*x^3+375*x^3)*log(-exp(x)+3)),x, algorithm="maxima")

[Out]

18/(log(log(-e^x + 3))^2 - 10*log(log(-e^x + 3)) + 25) + 2*(2*x + 1)/x^2

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mupad [B] time = 0.48, size = 36, normalized size = 1.09 \begin {gather*} \frac {18}{{\ln \left (\ln \left (3-{\mathrm {e}}^x\right )\right )}^2-10\,\ln \left (\ln \left (3-{\mathrm {e}}^x\right )\right )+25}+\frac {4\,x+2}{x^2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((log(3 - exp(x))*(1500*x - exp(x)*(500*x + 500) + 1500) + 36*x^3*exp(x) - log(3 - exp(x))*log(log(3 - exp(

x)))*(900*x - exp(x)*(300*x + 300) + 900) - log(3 - exp(x))*log(log(3 - exp(x)))^3*(12*x - exp(x)*(4*x + 4) +

12) + log(3 - exp(x))*log(log(3 - exp(x)))^2*(180*x - exp(x)*(60*x + 60) + 180))/(log(3 - exp(x))*(125*x^3*exp

(x) - 375*x^3) - log(3 - exp(x))*log(log(3 - exp(x)))*(75*x^3*exp(x) - 225*x^3) - log(3 - exp(x))*log(log(3 -

exp(x)))^3*(x^3*exp(x) - 3*x^3) + log(3 - exp(x))*log(log(3 - exp(x)))^2*(15*x^3*exp(x) - 45*x^3)),x)

[Out]

18/(log(log(3 - exp(x)))^2 - 10*log(log(3 - exp(x))) + 25) + (4*x + 2)/x^2

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sympy [A] time = 0.34, size = 32, normalized size = 0.97 \begin {gather*} \frac {18}{\log {\left (\log {\left (3 - e^{x} \right )} \right )}^{2} - 10 \log {\left (\log {\left (3 - e^{x} \right )} \right )} + 25} - \frac {- 4 x - 2}{x^{2}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x-4)*exp(x)+12*x+12)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))**3+((60*x+60)*exp(x)-180*x-180)*ln(-exp(x

)+3)*ln(ln(-exp(x)+3))**2+((-300*x-300)*exp(x)+900*x+900)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))+((500*x+500)*exp(x)-

1500*x-1500)*ln(-exp(x)+3)-36*exp(x)*x**3)/((exp(x)*x**3-3*x**3)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))**3+(-15*exp(x

)*x**3+45*x**3)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))**2+(75*exp(x)*x**3-225*x**3)*ln(-exp(x)+3)*ln(ln(-exp(x)+3))+(

-125*exp(x)*x**3+375*x**3)*ln(-exp(x)+3)),x)

[Out]

18/(log(log(3 - exp(x)))**2 - 10*log(log(3 - exp(x))) + 25) - (-4*x - 2)/x**2

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