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

Optimal. Leaf size=24 \[ e^5 \log \left (\log \left (\frac {5+2 e^{-x/3}-x}{x}\right )\right ) \]

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Rubi [A]  time = 1.06, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {6741, 12, 6688, 6684} \begin {gather*} e^5 \log \left (\log \left (\frac {2 e^{-x/3}}{x}+\frac {5}{x}-1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^5*Log[Log[-1 + 5/x + 2/(E^(x/3)*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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 24, normalized size = 1.00 \begin {gather*} e^5 \log \left (\log \left (\frac {5+2 e^{-x/3}-x}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^5*Log[Log[(5 + 2/E^(x/3) - x)/x]]

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fricas [A]  time = 0.74, size = 31, normalized size = 1.29 \begin {gather*} e^{5} \log \left (\log \left (-\frac {{\left ({\left (x - 5\right )} e^{\left (\frac {1}{3} \, x + 5\right )} - 2 \, e^{5}\right )} e^{\left (-\frac {1}{3} \, x - 5\right )}}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 28, normalized size = 1.17 \begin {gather*} e^{5} \log \left (x - 3 \, \log \left (-x e^{\left (\frac {1}{3} \, x\right )} + 5 \, e^{\left (\frac {1}{3} \, x\right )} + 2\right ) + 3 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.23, size = 28, normalized size = 1.17

method	result	size
norman	\({\mathrm e}^{5} \ln \left (\ln \left (\frac {\left (\left (5-x \right ) {\mathrm e}^{\frac {x}{3}}+2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )\right )\)	\(28\)
risch	\({\mathrm e}^{5} \ln \left (\ln \left ({\mathrm e}^{\frac {x}{3}}\right )-\frac {i \left (-2 \pi \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{3}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right )+\pi \,\mathrm {csgn}\left (i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right )^{2}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{3}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right )^{3}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-\frac {x}{3}} \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right ) {\mathrm e}^{-\frac {x}{3}}}{x}\right )^{3}+2 i \ln \relax (x )-2 i \ln \left (x \,{\mathrm e}^{\frac {x}{3}}-5 \,{\mathrm e}^{\frac {x}{3}}-2\right )+2 \pi \right )}{2}\right )\)	\(400\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 24, normalized size = 1.00 \begin {gather*} e^{5} \log \left (-\frac {1}{3} \, x + \log \left (-{\left (x - 5\right )} e^{\left (\frac {1}{3} \, x\right )} + 2\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.61, size = 20, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^5\,\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{-\frac {x}{3}}-x+5}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.45, size = 22, normalized size = 0.92 \begin {gather*} e^{5} \log {\left (\log {\left (\frac {\left (\left (5 - x\right ) e^{\frac {x}{3}} + 2\right ) e^{- \frac {x}{3}}}{x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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