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

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

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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} $$\begin{array}{c}\log (1-\frac{\log \left(\frac{25+e^2}{15-5e^x}\right)}{x})\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-e^{x}x+(3-e^x)\log \left(\frac{-25-e^2}{-15+5e^x}\right)}{(3-e^x)x(x-\log \left(\frac{25+e^2}{15-5e^x}\right))}dx\\ & =-\mathrm{Subst}(\int \frac{1}{1-x}dx,x,\frac{\log \left(\frac{25+e^2}{15-5e^x}\right)}{x})\\ & =\log (1-\frac{\log \left(\frac{25+e^2}{15-5e^x}\right)}{x})\end{array}\end{array}$$

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

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)]]

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fricas [A]  time = 0.55, size = 23, normalized size = 0.72 $$\begin{array}{c}-\log (x)+\log (-x+\log (-\frac{e^2+25}{5(e^x-3)}))\end{array}$$

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)))

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giac [A]  time = 0.17, size = 23, normalized size = 0.72 $$\begin{array}{c}-\log (x)+\log (-x+\log (-\frac{e^2+25}{5(e^x-3)}))\end{array}$$

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)))

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

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

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)))

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maxima [A]  time = 0.49, size = 24, normalized size = 0.75 $$\begin{array}{c}\log (x+\log (5)-\log (e^2+25)+\log (-e^x+3))-\log (x)\end{array}$$

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)

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mupad [B]  time = 4.60, size = 28, normalized size = 0.88 $$\begin{array}{c}\ln (x-\ln (-\frac{1}{{\mathrm{e}}^x-3})-\ln (\frac{{\mathrm{e}}^2}{5}+5))-\ln (x)\end{array}$$

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)

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

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)))

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