3.69.9 $\int \frac{e^{2x}(-20+4x)\log (\frac{1}{3}(-5+x))+\log (3+e^x)(-12e^x-4e^{2x}+(e^{2x}(-20+4x)+e^x(-60+12x))\log (\frac{1}{3}(-5+x)))}{(-15+e^x(-5+x)+3x){\log }^2(\frac{1}{3}(-5+x))}dx$

Optimal. Leaf size=23 $$-5+\frac{4e^x\log (3+e^x)}{\log (\frac{1}{3}(-5+x))}$$

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Rubi [F]  time = 3.23, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.000, Rules used = {} $$\begin{array}{c}\int \frac{e^{2x}(-20+4x)\log (\frac{1}{3}(-5+x))+\log (3+e^x)(-12e^x-4e^{2x}+(e^{2x}(-20+4x)+e^x(-60+12x))\log (\frac{1}{3}(-5+x)))}{(-15+e^x(-5+x)+3x){\log }^2(\frac{1}{3}(-5+x))}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(-20 + 4*x)*Log[(-5 + x)/3] + Log[3 + E^x]*(-12*E^x - 4*E^(2*x) + (E^(2*x)*(-20 + 4*x) + E^x*(-60
+ 12*x))*Log[(-5 + x)/3]))/((-15 + E^x*(-5 + x) + 3*x)*Log[(-5 + x)/3]^2),x]

[Out]

4*Log[27]*Defer[Int][E^x/((3 + E^x)*Log[-5/3 + x/3]^2), x] + 4*Defer[Int][(E^x*Log[3 + E^x])/((5 - x)*Log[-5/3
+ x/3]^2), x] + 4*Defer[Int][E^x/Log[-5/3 + x/3], x] + 4*Defer[Int][(E^x*Log[3 + E^x])/Log[-5/3 + x/3], x] +
12*Defer[Int][(E^x*Log[-5 + x])/((-3 - E^x)*Log[-5/3 + x/3]^2), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{4e^x(\frac{e^x\log (\frac{1}{3}(-5+x))}{3+e^x}+\frac{\log (3+e^x)(-1+(-5+x)\log (\frac{1}{3}(-5+x)))}{-5+x})}{{\log }^2(-\frac{5}{3}+\frac{x}{3})}dx\\ & =4\int \frac{e^x(\frac{e^x\log (\frac{1}{3}(-5+x))}{3+e^x}+\frac{\log (3+e^x)(-1+(-5+x)\log (\frac{1}{3}(-5+x)))}{-5+x})}{{\log }^2(-\frac{5}{3}+\frac{x}{3})}dx\\ & =4\int (\frac{e^x(\log (3+e^x)+5\log (\frac{1}{3}(-5+x))-x\log (\frac{1}{3}(-5+x))+5\log (3+e^x)\log (\frac{1}{3}(-5+x))-x\log (3+e^x)\log (\frac{1}{3}(-5+x)))}{(5-x){\log }^2(-\frac{5}{3}+\frac{x}{3})}+\frac{e^x(\log (27)-3\log (-5+x))}{(3+e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})})dx\\ & =4\int \frac{e^x(\log (3+e^x)+5\log (\frac{1}{3}(-5+x))-x\log (\frac{1}{3}(-5+x))+5\log (3+e^x)\log (\frac{1}{3}(-5+x))-x\log (3+e^x)\log (\frac{1}{3}(-5+x)))}{(5-x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+4\int \frac{e^x(\log (27)-3\log (-5+x))}{(3+e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx\\ & =4\int \frac{e^x(-((-5+x)\log (\frac{1}{3}(-5+x)))-\log (3+e^x)(-1+(-5+x)\log (\frac{1}{3}(-5+x))))}{(5-x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+4\int (\frac{e^x\log (27)}{(3+e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}+\frac{3e^x\log (-5+x)}{(-3-e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})})dx\\ & =4\int (\frac{e^x\log (3+e^x)}{(5-x){\log }^2(-\frac{5}{3}+\frac{x}{3})}+\frac{e^x(1+\log (3+e^x))}{\log (-\frac{5}{3}+\frac{x}{3})})dx+12\int \frac{e^x\log (-5+x)}{(-3-e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+(4\log (27))\int \frac{e^x}{(3+e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx\\ & =4\int \frac{e^x\log (3+e^x)}{(5-x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+4\int \frac{e^x(1+\log (3+e^x))}{\log (-\frac{5}{3}+\frac{x}{3})}dx+12\int \frac{e^x\log (-5+x)}{(-3-e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+(4\log (27))\int \frac{e^x}{(3+e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx\\ & =4\int (\frac{e^x}{\log (-\frac{5}{3}+\frac{x}{3})}+\frac{e^x\log (3+e^x)}{\log (-\frac{5}{3}+\frac{x}{3})})dx+4\int \frac{e^x\log (3+e^x)}{(5-x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+12\int \frac{e^x\log (-5+x)}{(-3-e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+(4\log (27))\int \frac{e^x}{(3+e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx\\ & =4\int \frac{e^x\log (3+e^x)}{(5-x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+4\int \frac{e^x}{\log (-\frac{5}{3}+\frac{x}{3})}dx+4\int \frac{e^x\log (3+e^x)}{\log (-\frac{5}{3}+\frac{x}{3})}dx+12\int \frac{e^x\log (-5+x)}{(-3-e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx+(4\log (27))\int \frac{e^x}{(3+e^x){\log }^2(-\frac{5}{3}+\frac{x}{3})}dx\end{array}\end{array}$$

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Mathematica [A]  time = 1.56, size = 21, normalized size = 0.91 $$\begin{array}{c}\frac{4e^x\log (3+e^x)}{\log (\frac{1}{3}(-5+x))}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(-20 + 4*x)*Log[(-5 + x)/3] + Log[3 + E^x]*(-12*E^x - 4*E^(2*x) + (E^(2*x)*(-20 + 4*x) + E^
x*(-60 + 12*x))*Log[(-5 + x)/3]))/((-15 + E^x*(-5 + x) + 3*x)*Log[(-5 + x)/3]^2),x]

[Out]

(4*E^x*Log[3 + E^x])/Log[(-5 + x)/3]

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fricas [A]  time = 0.57, size = 17, normalized size = 0.74 $$\begin{array}{c}\frac{4e^x\log (e^x+3)}{\log (\frac{1}{3}x-\frac{5}{3})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*log(1/3*x-5/3)-4*exp(x)^2-12*exp(x))*log(3+exp(x))+(4*x-20)*e
xp(x)^2*log(1/3*x-5/3))/((x-5)*exp(x)+3*x-15)/log(1/3*x-5/3)^2,x, algorithm="fricas")

[Out]

4*e^x*log(e^x + 3)/log(1/3*x - 5/3)

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giac [A]  time = 0.16, size = 20, normalized size = 0.87 $$\begin{array}{c}-\frac{4e^x\log (e^x+3)}{\log (3)-\log (x-5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*log(1/3*x-5/3)-4*exp(x)^2-12*exp(x))*log(3+exp(x))+(4*x-20)*e
xp(x)^2*log(1/3*x-5/3))/((x-5)*exp(x)+3*x-15)/log(1/3*x-5/3)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 18, normalized size = 0.78

method	result	size
risch	$\frac{4\ln (3+{\mathrm{e}}^x){\mathrm{e}}^x}{\ln (\frac{x}{3}-\frac{5}{3})}$	$18$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*ln(1/3*x-5/3)-4*exp(x)^2-12*exp(x))*ln(3+exp(x))+(4*x-20)*exp(x)^2*
ln(1/3*x-5/3))/((x-5)*exp(x)+3*x-15)/ln(1/3*x-5/3)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.54, size = 20, normalized size = 0.87 $$\begin{array}{c}-\frac{4e^x\log (e^x+3)}{\log (3)-\log (x-5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x-20)*exp(x)^2+(12*x-60)*exp(x))*log(1/3*x-5/3)-4*exp(x)^2-12*exp(x))*log(3+exp(x))+(4*x-20)*e
xp(x)^2*log(1/3*x-5/3))/((x-5)*exp(x)+3*x-15)/log(1/3*x-5/3)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.27, size = 17, normalized size = 0.74 $$\begin{array}{c}\frac{4{\mathrm{e}}^x\ln ({\mathrm{e}}^x+3)}{\ln (\frac{x}{3}-\frac{5}{3})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp(x) + 3)*(4*exp(2*x) + 12*exp(x) - log(x/3 - 5/3)*(exp(x)*(12*x - 60) + exp(2*x)*(4*x - 20))) - e
xp(2*x)*log(x/3 - 5/3)*(4*x - 20))/(log(x/3 - 5/3)^2*(3*x + exp(x)*(x - 5) - 15)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Timed out}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x-20)*exp(x)**2+(12*x-60)*exp(x))*ln(1/3*x-5/3)-4*exp(x)**2-12*exp(x))*ln(3+exp(x))+(4*x-20)*e
xp(x)**2*ln(1/3*x-5/3))/((x-5)*exp(x)+3*x-15)/ln(1/3*x-5/3)**2,x)

[Out]

Timed out

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