\(\int \frac {(80-20 x) \log (\frac {1}{3} e^{-16+8 x-x^2} (-1-15 e^{16-8 x+x^2}))}{3+45 e^{16-8 x+x^2}} \, dx\) [3083]

Optimal result

Integrand size = 53, antiderivative size = 24 \[ \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3+45 e^{16-8 x+x^2}} \, dx=\frac {5}{3} \log ^2\left (-5-\frac {1}{3} e^{-(4-x)^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6873, 12, 6847, 2320, 36, 29, 31, 6820, 6818} \[ \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3+45 e^{16-8 x+x^2}} \, dx=\frac {5}{3} \log ^2\left (-\frac {1}{3} e^{-(4-x)^2}-5\right ) \]

[In]

Int[((80 - 20*x)*Log[(E^(-16 + 8*x - x^2)*(-1 - 15*E^(16 - 8*x + x^2)))/3])/(3 + 45*E^(16 - 8*x + x^2)),x]

[Out]

(5*Log[-5 - 1/(3*E^(4 - x)^2)]^2)/3

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2320

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 6818

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 6820

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

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3 \left (1+15 e^{(-4+x)^2}\right )} \, dx \\ & = \frac {1}{3} \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{1+15 e^{(-4+x)^2}} \, dx \\ & = \frac {1}{3} \int \frac {20 (4-x) \log \left (-5-\frac {1}{3} e^{-(-4+x)^2}\right )}{1+15 e^{(-4+x)^2}} \, dx \\ & = \frac {20}{3} \int \frac {(4-x) \log \left (-5-\frac {1}{3} e^{-(-4+x)^2}\right )}{1+15 e^{(-4+x)^2}} \, dx \\ & = \frac {5}{3} \log ^2\left (-5-\frac {1}{3} e^{-(4-x)^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3+45 e^{16-8 x+x^2}} \, dx=\frac {5}{3} \log ^2\left (-5-\frac {1}{3} e^{-(-4+x)^2}\right ) \]

[In]

Integrate[((80 - 20*x)*Log[(E^(-16 + 8*x - x^2)*(-1 - 15*E^(16 - 8*x + x^2)))/3])/(3 + 45*E^(16 - 8*x + x^2)),
x]

[Out]

(5*Log[-5 - 1/(3*E^(-4 + x)^2)]^2)/3

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33

method	result	size
norman	\(\frac {5 {\ln \left (\frac {\left (-15 \,{\mathrm e}^{x^{2}-8 x +16}-1\right ) {\mathrm e}^{-x^{2}+8 x -16}}{3}\right )}^{2}}{3}\)	\(32\)
parallelrisch	\(\frac {5 {\ln \left (-\frac {\left (15 \,{\mathrm e}^{x^{2}-8 x +16}+1\right ) {\mathrm e}^{-x^{2}+8 x -16}}{3}\right )}^{2}}{3}\)	\(32\)
risch	\(\text {Expression too large to display}\)	\(890\)

[In]

int((-20*x+80)*ln(1/3*(-15*exp(x^2-8*x+16)-1)/exp(x^2-8*x+16))/(45*exp(x^2-8*x+16)+3),x,method=_RETURNVERBOSE)

[Out]

5/3*ln(1/3*(-15*exp(x^2-8*x+16)-1)/exp(x^2-8*x+16))^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3+45 e^{16-8 x+x^2}} \, dx=\frac {5}{3} \, \log \left (-\frac {1}{3} \, {\left (15 \, e^{\left (x^{2} - 8 \, x + 16\right )} + 1\right )} e^{\left (-x^{2} + 8 \, x - 16\right )}\right )^{2} \]

[In]

integrate((-20*x+80)*log(1/3*(-15*exp(x^2-8*x+16)-1)/exp(x^2-8*x+16))/(45*exp(x^2-8*x+16)+3),x, algorithm="fri
cas")

[Out]

5/3*log(-1/3*(15*e^(x^2 - 8*x + 16) + 1)*e^(-x^2 + 8*x - 16))^2

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3+45 e^{16-8 x+x^2}} \, dx=\frac {5 \log {\left (\left (- 5 e^{x^{2} - 8 x + 16} - \frac {1}{3}\right ) e^{- x^{2} + 8 x - 16} \right )}^{2}}{3} \]

[In]

integrate((-20*x+80)*ln(1/3*(-15*exp(x**2-8*x+16)-1)/exp(x**2-8*x+16))/(45*exp(x**2-8*x+16)+3),x)

[Out]

5*log((-5*exp(x**2 - 8*x + 16) - 1/3)*exp(-x**2 + 8*x - 16))**2/3

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (17) = 34\).

Time = 0.35 (sec) , antiderivative size = 113, normalized size of antiderivative = 4.71 \[ \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3+45 e^{16-8 x+x^2}} \, dx=-\frac {5}{3} \, x^{4} - \frac {10}{3} \, x^{2} {\left (\log \left (5\right ) + \log \left (3\right ) + 16\right )} - \frac {10}{3} \, {\left (x^{2} - \log \left (\frac {1}{15} \, {\left (15 \, e^{\left (x^{2} + 16\right )} + e^{\left (8 \, x\right )}\right )} e^{\left (-16\right )}\right )\right )} \log \left (-\frac {1}{3} \, {\left (15 \, e^{\left (x^{2} - 8 \, x + 16\right )} + 1\right )} e^{\left (-x^{2} + 8 \, x - 16\right )}\right ) + \frac {10}{3} \, {\left (x^{2} + \log \left (5\right ) + \log \left (3\right ) + 16\right )} \log \left (15 \, e^{\left (x^{2} + 16\right )} + e^{\left (8 \, x\right )}\right ) - \frac {5}{3} \, \log \left (15 \, e^{\left (x^{2} + 16\right )} + e^{\left (8 \, x\right )}\right )^{2} \]

[In]

integrate((-20*x+80)*log(1/3*(-15*exp(x^2-8*x+16)-1)/exp(x^2-8*x+16))/(45*exp(x^2-8*x+16)+3),x, algorithm="max
ima")

[Out]

-5/3*x^4 - 10/3*x^2*(log(5) + log(3) + 16) - 10/3*(x^2 - log(1/15*(15*e^(x^2 + 16) + e^(8*x))*e^(-16)))*log(-1
/3*(15*e^(x^2 - 8*x + 16) + 1)*e^(-x^2 + 8*x - 16)) + 10/3*(x^2 + log(5) + log(3) + 16)*log(15*e^(x^2 + 16) +
e^(8*x)) - 5/3*log(15*e^(x^2 + 16) + e^(8*x))^2

Giac [F]

\[ \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3+45 e^{16-8 x+x^2}} \, dx=\int { -\frac {20 \, {\left (x - 4\right )} \log \left (-\frac {1}{3} \, {\left (15 \, e^{\left (x^{2} - 8 \, x + 16\right )} + 1\right )} e^{\left (-x^{2} + 8 \, x - 16\right )}\right )}{3 \, {\left (15 \, e^{\left (x^{2} - 8 \, x + 16\right )} + 1\right )}} \,d x } \]

[In]

integrate((-20*x+80)*log(1/3*(-15*exp(x^2-8*x+16)-1)/exp(x^2-8*x+16))/(45*exp(x^2-8*x+16)+3),x, algorithm="gia
c")

[Out]

integrate(-20/3*(x - 4)*log(-1/3*(15*e^(x^2 - 8*x + 16) + 1)*e^(-x^2 + 8*x - 16))/(15*e^(x^2 - 8*x + 16) + 1),
x)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {(80-20 x) \log \left (\frac {1}{3} e^{-16+8 x-x^2} \left (-1-15 e^{16-8 x+x^2}\right )\right )}{3+45 e^{16-8 x+x^2}} \, dx=\frac {5\,{\ln \left (-\frac {{\mathrm {e}}^{8\,x}\,{\mathrm {e}}^{-16}\,{\mathrm {e}}^{-x^2}}{3}-5\right )}^2}{3} \]

[In]

int(-(log(-exp(8*x - x^2 - 16)*(5*exp(x^2 - 8*x + 16) + 1/3))*(20*x - 80))/(45*exp(x^2 - 8*x + 16) + 3),x)

[Out]

(5*log(- (exp(8*x)*exp(-16)*exp(-x^2))/3 - 5)^2)/3