3.73.74 \(\int \frac {-e^x x+(3-e^x) \log (\frac {-25-e^2}{-15+5 e^x})}{3 x^2-e^x x^2+(-3 x+e^x x) \log (\frac {-25-e^2}{-15+5 e^x})} \, dx\)

Optimal. Leaf size=32 \[ \log \left (1-\frac {\log \left (\frac {\left (5+\frac {e^2}{5}\right ) x}{3 x-e^x x}\right )}{x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.97, antiderivative size = 24, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 3, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6741, 6712, 31} \begin {gather*} \log \left (1-\frac {\log \left (\frac {25+e^2}{15-5 e^x}\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^x*x) + (3 - E^x)*Log[(-25 - E^2)/(-15 + 5*E^x)])/(3*x^2 - E^x*x^2 + (-3*x + E^x*x)*Log[(-25 - E^2)/(-
15 + 5*E^x)]),x]

[Out]

Log[1 - Log[(25 + E^2)/(15 - 5*E^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 6712

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, -Dist[c*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ
[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

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^x x+\left (3-e^x\right ) \log \left (\frac {-25-e^2}{-15+5 e^x}\right )}{\left (3-e^x\right ) x \left (x-\log \left (\frac {25+e^2}{15-5 e^x}\right )\right )} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {\log \left (\frac {25+e^2}{15-5 e^x}\right )}{x}\right )\\ &=\log \left (1-\frac {\log \left (\frac {25+e^2}{15-5 e^x}\right )}{x}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 26, normalized size = 0.81 \begin {gather*} -\log (x)+\log \left (x-\log \left (\frac {25+e^2}{15-5 e^x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^x*x) + (3 - E^x)*Log[(-25 - E^2)/(-15 + 5*E^x)])/(3*x^2 - E^x*x^2 + (-3*x + E^x*x)*Log[(-25 - E
^2)/(-15 + 5*E^x)]),x]

[Out]

-Log[x] + Log[x - Log[(25 + E^2)/(15 - 5*E^x)]]

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 23, normalized size = 0.72 \begin {gather*} -\log \relax (x) + \log \left (-x + \log \left (-\frac {e^{2} + 25}{5 \, {\left (e^{x} - 3\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)+3)*log((-exp(2)-25)/(5*exp(x)-15))-exp(x)*x)/((exp(x)*x-3*x)*log((-exp(2)-25)/(5*exp(x)-15
))-exp(x)*x^2+3*x^2),x, algorithm="fricas")

[Out]

-log(x) + log(-x + log(-1/5*(e^2 + 25)/(e^x - 3)))

________________________________________________________________________________________

giac [A]  time = 0.17, size = 23, normalized size = 0.72 \begin {gather*} -\log \relax (x) + \log \left (-x + \log \left (-\frac {e^{2} + 25}{5 \, {\left (e^{x} - 3\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)+3)*log((-exp(2)-25)/(5*exp(x)-15))-exp(x)*x)/((exp(x)*x-3*x)*log((-exp(2)-25)/(5*exp(x)-15
))-exp(x)*x^2+3*x^2),x, algorithm="giac")

[Out]

-log(x) + log(-x + log(-1/5*(e^2 + 25)/(e^x - 3)))

________________________________________________________________________________________

maple [A]  time = 0.18, size = 27, normalized size = 0.84

method	result	size
default	\(-\ln \relax (x )+\ln \left (x -\ln \left (\frac {-{\mathrm e}^{2}-25}{5 \,{\mathrm e}^{x}-15}\right )\right )\)	\(27\)
norman	\(-\ln \relax (x )+\ln \left (x -\ln \left (\frac {-{\mathrm e}^{2}-25}{5 \,{\mathrm e}^{x}-15}\right )\right )\)	\(27\)
risch	\(-\ln \relax (x )+\ln \left (\ln \left ({\mathrm e}^{x}-3\right )+\frac {i \left (2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}-3}\right )^{2}-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}-3}\right )^{3}-2 i \ln \relax (5)+2 i \ln \left (-{\mathrm e}^{2}-25\right )-2 i x \right )}{2}\right )\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)+3)*ln((-exp(2)-25)/(5*exp(x)-15))-exp(x)*x)/((exp(x)*x-3*x)*ln((-exp(2)-25)/(5*exp(x)-15))-exp(x
)*x^2+3*x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(x-ln((-exp(2)-25)/(5*exp(x)-15)))

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 24, normalized size = 0.75 \begin {gather*} \log \left (x + \log \relax (5) - \log \left (e^{2} + 25\right ) + \log \left (-e^{x} + 3\right )\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)+3)*log((-exp(2)-25)/(5*exp(x)-15))-exp(x)*x)/((exp(x)*x-3*x)*log((-exp(2)-25)/(5*exp(x)-15
))-exp(x)*x^2+3*x^2),x, algorithm="maxima")

[Out]

log(x + log(5) - log(e^2 + 25) + log(-e^x + 3)) - log(x)

________________________________________________________________________________________

mupad [B]  time = 4.60, size = 28, normalized size = 0.88 \begin {gather*} \ln \left (x-\ln \left (-\frac {1}{{\mathrm {e}}^x-3}\right )-\ln \left (\frac {{\mathrm {e}}^2}{5}+5\right )\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-(exp(2) + 25)/(5*exp(x) - 15))*(exp(x) - 3) + x*exp(x))/(x^2*exp(x) - 3*x^2 + log(-(exp(2) + 25)/(5*
exp(x) - 15))*(3*x - x*exp(x))),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.24, size = 20, normalized size = 0.62 \begin {gather*} - \log {\relax (x )} + \log {\left (- x + \log {\left (\frac {-25 - e^{2}}{5 e^{x} - 15} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)+3)*ln((-exp(2)-25)/(5*exp(x)-15))-exp(x)*x)/((exp(x)*x-3*x)*ln((-exp(2)-25)/(5*exp(x)-15))
-exp(x)*x**2+3*x**2),x)

[Out]

-log(x) + log(-x + log((-25 - exp(2))/(5*exp(x) - 15)))

________________________________________________________________________________________