3.62.17 \(\int \frac {e^{\frac {-x+(3-e-e^4) \log ^2(3)+(-6+2 e+2 e^4) \log (3) \log (-5+x)+(3-e-e^4) \log ^2(-5+x)}{-3+e+e^4}} (5-x+(-6+2 e+2 e^4) \log (3)+(6-2 e-2 e^4) \log (-5+x))}{15+e (-5+x)+e^4 (-5+x)-3 x} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {x}{3-e-e^4}-(\log (3)-\log (-5+x))^2} \]

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Rubi [B]  time = 1.74, antiderivative size = 109, normalized size of antiderivative = 3.63, number of steps used = 5, number of rules used = 5, integrand size = 117, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6, 2274, 1586, 12, 2288} \begin {gather*} -\frac {(x-5)^{\log (9)-1} \left (-x+2 \left (3-e-e^4\right ) \log (x-5)+5\right ) \exp \left (\frac {x-\left (\left (3-e-e^4\right ) \log ^2(x-5)\right )-\left (3-e-e^4\right ) \log ^2(3)}{3-e-e^4}\right )}{\frac {2 \left (3-e-e^4\right ) \log (x-5)}{5-x}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-x + (3 - E - E^4)*Log[3]^2 + (-6 + 2*E + 2*E^4)*Log[3]*Log[-5 + x] + (3 - E - E^4)*Log[-5 + x]^2)/(-
3 + E + E^4))*(5 - x + (-6 + 2*E + 2*E^4)*Log[3] + (6 - 2*E - 2*E^4)*Log[-5 + x]))/(15 + E*(-5 + x) + E^4*(-5
+ x) - 3*x),x]

[Out]

-((E^((x - (3 - E - E^4)*Log[3]^2 - (3 - E - E^4)*Log[-5 + x]^2)/(3 - E - E^4))*(-5 + x)^(-1 + Log[9])*(5 - x
+ 2*(3 - E - E^4)*Log[-5 + x]))/(1 + (2*(3 - E - E^4)*Log[-5 + x])/(5 - x)))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-x+\left (3-e-e^4\right ) \log ^2(3)+\left (-6+2 e+2 e^4\right ) \log (3) \log (-5+x)+\left (3-e-e^4\right ) \log ^2(-5+x)}{-3+e+e^4}\right ) \left (5-x+\left (-6+2 e+2 e^4\right ) \log (3)+\left (6-2 e-2 e^4\right ) \log (-5+x)\right )}{15+\left (e+e^4\right ) (-5+x)-3 x} \, dx\\ &=\int \frac {\exp \left (\frac {-x+\left (3-e-e^4\right ) \log ^2(3)+\left (3-e-e^4\right ) \log ^2(-5+x)}{-3+e+e^4}\right ) (-5+x)^{\frac {\left (-6+2 e+2 e^4\right ) \log (3)}{-3+e+e^4}} \left (5-x+\left (-6+2 e+2 e^4\right ) \log (3)+\left (6-2 e-2 e^4\right ) \log (-5+x)\right )}{15+\left (e+e^4\right ) (-5+x)-3 x} \, dx\\ &=\int \frac {\exp \left (\frac {-x+\left (3-e-e^4\right ) \log ^2(3)+\left (3-e-e^4\right ) \log ^2(-5+x)}{-3+e+e^4}\right ) (-5+x)^{-1+\frac {\left (-6+2 e+2 e^4\right ) \log (3)}{-3+e+e^4}} \left (5-x+\left (-6+2 e+2 e^4\right ) \log (3)+\left (6-2 e-2 e^4\right ) \log (-5+x)\right )}{-3+e+e^4} \, dx\\ &=\frac {\int \exp \left (\frac {-x+\left (3-e-e^4\right ) \log ^2(3)+\left (3-e-e^4\right ) \log ^2(-5+x)}{-3+e+e^4}\right ) (-5+x)^{-1+\frac {\left (-6+2 e+2 e^4\right ) \log (3)}{-3+e+e^4}} \left (5-x+\left (-6+2 e+2 e^4\right ) \log (3)+\left (6-2 e-2 e^4\right ) \log (-5+x)\right ) \, dx}{-3+e+e^4}\\ &=-\frac {\exp \left (\frac {x-\left (3-e-e^4\right ) \log ^2(3)-\left (3-e-e^4\right ) \log ^2(-5+x)}{3-e-e^4}\right ) (-5+x)^{-1+\log (9)} \left (5-x+2 \left (3-e-e^4\right ) \log (-5+x)\right )}{1+\frac {2 \left (3-e-e^4\right ) \log (-5+x)}{5-x}}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.28, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-x+\left (3-e-e^4\right ) \log ^2(3)+\left (-6+2 e+2 e^4\right ) \log (3) \log (-5+x)+\left (3-e-e^4\right ) \log ^2(-5+x)}{-3+e+e^4}} \left (5-x+\left (-6+2 e+2 e^4\right ) \log (3)+\left (6-2 e-2 e^4\right ) \log (-5+x)\right )}{15+e (-5+x)+e^4 (-5+x)-3 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((-x + (3 - E - E^4)*Log[3]^2 + (-6 + 2*E + 2*E^4)*Log[3]*Log[-5 + x] + (3 - E - E^4)*Log[-5 + x]
^2)/(-3 + E + E^4))*(5 - x + (-6 + 2*E + 2*E^4)*Log[3] + (6 - 2*E - 2*E^4)*Log[-5 + x]))/(15 + E*(-5 + x) + E^
4*(-5 + x) - 3*x),x]

[Out]

Integrate[(E^((-x + (3 - E - E^4)*Log[3]^2 + (-6 + 2*E + 2*E^4)*Log[3]*Log[-5 + x] + (3 - E - E^4)*Log[-5 + x]
^2)/(-3 + E + E^4))*(5 - x + (-6 + 2*E + 2*E^4)*Log[3] + (6 - 2*E - 2*E^4)*Log[-5 + x]))/(15 + E*(-5 + x) + E^
4*(-5 + x) - 3*x), x]

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fricas [A]  time = 0.64, size = 51, normalized size = 1.70 \begin {gather*} e^{\left (-\frac {{\left (e^{4} + e - 3\right )} \log \relax (3)^{2} - 2 \, {\left (e^{4} + e - 3\right )} \log \relax (3) \log \left (x - 5\right ) + {\left (e^{4} + e - 3\right )} \log \left (x - 5\right )^{2} + x}{e^{4} + e - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4)-2*exp(1)+6)*log(x-5)+(2*exp(4)+2*exp(1)-6)*log(3)+5-x)*exp(((-exp(4)+3-exp(1))*log(x-5)^
2+(2*exp(4)+2*exp(1)-6)*log(3)*log(x-5)+(-exp(4)+3-exp(1))*log(3)^2-x)/(exp(4)+exp(1)-3))/((x-5)*exp(4)+(x-5)*
exp(1)+15-3*x),x, algorithm="fricas")

[Out]

e^(-((e^4 + e - 3)*log(3)^2 - 2*(e^4 + e - 3)*log(3)*log(x - 5) + (e^4 + e - 3)*log(x - 5)^2 + x)/(e^4 + e - 3
))

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giac [B]  time = 1.02, size = 163, normalized size = 5.43 \begin {gather*} e^{\left (-\frac {e^{4} \log \relax (3)^{2}}{e^{4} + e - 3} - \frac {e \log \relax (3)^{2}}{e^{4} + e - 3} + \frac {2 \, e^{4} \log \relax (3) \log \left (x - 5\right )}{e^{4} + e - 3} + \frac {2 \, e \log \relax (3) \log \left (x - 5\right )}{e^{4} + e - 3} - \frac {e^{4} \log \left (x - 5\right )^{2}}{e^{4} + e - 3} - \frac {e \log \left (x - 5\right )^{2}}{e^{4} + e - 3} + \frac {3 \, \log \relax (3)^{2}}{e^{4} + e - 3} - \frac {6 \, \log \relax (3) \log \left (x - 5\right )}{e^{4} + e - 3} + \frac {3 \, \log \left (x - 5\right )^{2}}{e^{4} + e - 3} - \frac {x}{e^{4} + e - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4)-2*exp(1)+6)*log(x-5)+(2*exp(4)+2*exp(1)-6)*log(3)+5-x)*exp(((-exp(4)+3-exp(1))*log(x-5)^
2+(2*exp(4)+2*exp(1)-6)*log(3)*log(x-5)+(-exp(4)+3-exp(1))*log(3)^2-x)/(exp(4)+exp(1)-3))/((x-5)*exp(4)+(x-5)*
exp(1)+15-3*x),x, algorithm="giac")

[Out]

e^(-e^4*log(3)^2/(e^4 + e - 3) - e*log(3)^2/(e^4 + e - 3) + 2*e^4*log(3)*log(x - 5)/(e^4 + e - 3) + 2*e*log(3)
*log(x - 5)/(e^4 + e - 3) - e^4*log(x - 5)^2/(e^4 + e - 3) - e*log(x - 5)^2/(e^4 + e - 3) + 3*log(3)^2/(e^4 +
e - 3) - 6*log(3)*log(x - 5)/(e^4 + e - 3) + 3*log(x - 5)^2/(e^4 + e - 3) - x/(e^4 + e - 3))

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maple [B]  time = 0.37, size = 64, normalized size = 2.13

method	result	size
norman	\({\mathrm e}^{\frac {\left (-{\mathrm e}^{4}+3-{\mathrm e}\right ) \ln \left (x -5\right )^{2}+\left (2 \,{\mathrm e}^{4}+2 \,{\mathrm e}-6\right ) \ln \relax (3) \ln \left (x -5\right )+\left (-{\mathrm e}^{4}+3-{\mathrm e}\right ) \ln \relax (3)^{2}-x}{{\mathrm e}^{4}+{\mathrm e}-3}}\)	\(64\)
risch	\({\mathrm e}^{-\frac {{\mathrm e} \ln \relax (3)^{2}+{\mathrm e}^{4} \ln \relax (3)^{2}-2 \ln \relax (3) \ln \left (x -5\right ) {\mathrm e}-2 \ln \relax (3) \ln \left (x -5\right ) {\mathrm e}^{4}+\ln \left (x -5\right )^{2} {\mathrm e}+\ln \left (x -5\right )^{2} {\mathrm e}^{4}-3 \ln \relax (3)^{2}+6 \ln \relax (3) \ln \left (x -5\right )-3 \ln \left (x -5\right )^{2}+x}{{\mathrm e}^{4}+{\mathrm e}-3}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(4)-2*exp(1)+6)*ln(x-5)+(2*exp(4)+2*exp(1)-6)*ln(3)+5-x)*exp(((-exp(4)+3-exp(1))*ln(x-5)^2+(2*exp(
4)+2*exp(1)-6)*ln(3)*ln(x-5)+(-exp(4)+3-exp(1))*ln(3)^2-x)/(exp(4)+exp(1)-3))/((x-5)*exp(4)+(x-5)*exp(1)+15-3*
x),x,method=_RETURNVERBOSE)

[Out]

exp(((-exp(4)+3-exp(1))*ln(x-5)^2+(2*exp(4)+2*exp(1)-6)*ln(3)*ln(x-5)+(-exp(4)+3-exp(1))*ln(3)^2-x)/(exp(4)+ex
p(1)-3))

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maxima [B]  time = 0.70, size = 163, normalized size = 5.43 \begin {gather*} e^{\left (-\frac {e^{4} \log \relax (3)^{2}}{e^{4} + e - 3} - \frac {e \log \relax (3)^{2}}{e^{4} + e - 3} + \frac {2 \, e^{4} \log \relax (3) \log \left (x - 5\right )}{e^{4} + e - 3} + \frac {2 \, e \log \relax (3) \log \left (x - 5\right )}{e^{4} + e - 3} - \frac {e^{4} \log \left (x - 5\right )^{2}}{e^{4} + e - 3} - \frac {e \log \left (x - 5\right )^{2}}{e^{4} + e - 3} + \frac {3 \, \log \relax (3)^{2}}{e^{4} + e - 3} - \frac {6 \, \log \relax (3) \log \left (x - 5\right )}{e^{4} + e - 3} + \frac {3 \, \log \left (x - 5\right )^{2}}{e^{4} + e - 3} - \frac {x}{e^{4} + e - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4)-2*exp(1)+6)*log(x-5)+(2*exp(4)+2*exp(1)-6)*log(3)+5-x)*exp(((-exp(4)+3-exp(1))*log(x-5)^
2+(2*exp(4)+2*exp(1)-6)*log(3)*log(x-5)+(-exp(4)+3-exp(1))*log(3)^2-x)/(exp(4)+exp(1)-3))/((x-5)*exp(4)+(x-5)*
exp(1)+15-3*x),x, algorithm="maxima")

[Out]

e^(-e^4*log(3)^2/(e^4 + e - 3) - e*log(3)^2/(e^4 + e - 3) + 2*e^4*log(3)*log(x - 5)/(e^4 + e - 3) + 2*e*log(3)
*log(x - 5)/(e^4 + e - 3) - e^4*log(x - 5)^2/(e^4 + e - 3) - e*log(x - 5)^2/(e^4 + e - 3) + 3*log(3)^2/(e^4 +
e - 3) - 6*log(3)*log(x - 5)/(e^4 + e - 3) + 3*log(x - 5)^2/(e^4 + e - 3) - x/(e^4 + e - 3))

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mupad [B]  time = 1.95, size = 125, normalized size = 4.17 \begin {gather*} {\mathrm {e}}^{\frac {3\,{\ln \relax (3)}^2}{\mathrm {e}+{\mathrm {e}}^4-3}}\,{\mathrm {e}}^{-\frac {{\ln \left (x-5\right )}^2\,\mathrm {e}}{\mathrm {e}+{\mathrm {e}}^4-3}}\,{\mathrm {e}}^{-\frac {{\ln \left (x-5\right )}^2\,{\mathrm {e}}^4}{\mathrm {e}+{\mathrm {e}}^4-3}}\,{\mathrm {e}}^{-\frac {\mathrm {e}\,{\ln \relax (3)}^2}{\mathrm {e}+{\mathrm {e}}^4-3}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^4\,{\ln \relax (3)}^2}{\mathrm {e}+{\mathrm {e}}^4-3}}\,{\mathrm {e}}^{-\frac {x}{\mathrm {e}+{\mathrm {e}}^4-3}}\,{\mathrm {e}}^{\frac {3\,{\ln \left (x-5\right )}^2}{\mathrm {e}+{\mathrm {e}}^4-3}}\,{\left (x-5\right )}^{2\,\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x + log(x - 5)^2*(exp(1) + exp(4) - 3) + log(3)^2*(exp(1) + exp(4) - 3) - log(x - 5)*log(3)*(2*exp
(1) + 2*exp(4) - 6))/(exp(1) + exp(4) - 3))*(x + log(x - 5)*(2*exp(1) + 2*exp(4) - 6) - log(3)*(2*exp(1) + 2*e
xp(4) - 6) - 5))/(exp(1)*(x - 5) - 3*x + exp(4)*(x - 5) + 15),x)

[Out]

exp((3*log(3)^2)/(exp(1) + exp(4) - 3))*exp(-(log(x - 5)^2*exp(1))/(exp(1) + exp(4) - 3))*exp(-(log(x - 5)^2*e
xp(4))/(exp(1) + exp(4) - 3))*exp(-(exp(1)*log(3)^2)/(exp(1) + exp(4) - 3))*exp(-(exp(4)*log(3)^2)/(exp(1) + e
xp(4) - 3))*exp(-x/(exp(1) + exp(4) - 3))*exp((3*log(x - 5)^2)/(exp(1) + exp(4) - 3))*(x - 5)^(2*log(3))

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sympy [B]  time = 0.79, size = 60, normalized size = 2.00 \begin {gather*} e^{\frac {- x + \left (- e^{4} - e + 3\right ) \log {\left (x - 5 \right )}^{2} + \left (-6 + 2 e + 2 e^{4}\right ) \log {\relax (3 )} \log {\left (x - 5 \right )} + \left (- e^{4} - e + 3\right ) \log {\relax (3 )}^{2}}{-3 + e + e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4)-2*exp(1)+6)*ln(x-5)+(2*exp(4)+2*exp(1)-6)*ln(3)+5-x)*exp(((-exp(4)+3-exp(1))*ln(x-5)**2+
(2*exp(4)+2*exp(1)-6)*ln(3)*ln(x-5)+(-exp(4)+3-exp(1))*ln(3)**2-x)/(exp(4)+exp(1)-3))/((x-5)*exp(4)+(x-5)*exp(
1)+15-3*x),x)

[Out]

exp((-x + (-exp(4) - E + 3)*log(x - 5)**2 + (-6 + 2*E + 2*exp(4))*log(3)*log(x - 5) + (-exp(4) - E + 3)*log(3)
**2)/(-3 + E + exp(4)))

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