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

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

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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 {gather*} \int \frac {e^{2 x} (-20+4 x) \log \left (\frac {1}{3} (-5+x)\right )+\log \left (3+e^x\right ) \left (-12 e^x-4 e^{2 x}+\left (e^{2 x} (-20+4 x)+e^x (-60+12 x)\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{\left (-15+e^x (-5+x)+3 x\right ) \log ^2\left (\frac {1}{3} (-5+x)\right )} \, dx \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^x \left (\frac {e^x \log \left (\frac {1}{3} (-5+x)\right )}{3+e^x}+\frac {\log \left (3+e^x\right ) \left (-1+(-5+x) \log \left (\frac {1}{3} (-5+x)\right )\right )}{-5+x}\right )}{\log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx\\ &=4 \int \frac {e^x \left (\frac {e^x \log \left (\frac {1}{3} (-5+x)\right )}{3+e^x}+\frac {\log \left (3+e^x\right ) \left (-1+(-5+x) \log \left (\frac {1}{3} (-5+x)\right )\right )}{-5+x}\right )}{\log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx\\ &=4 \int \left (\frac {e^x \left (\log \left (3+e^x\right )+5 \log \left (\frac {1}{3} (-5+x)\right )-x \log \left (\frac {1}{3} (-5+x)\right )+5 \log \left (3+e^x\right ) \log \left (\frac {1}{3} (-5+x)\right )-x \log \left (3+e^x\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}+\frac {e^x (\log (27)-3 \log (-5+x))}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}\right ) \, dx\\ &=4 \int \frac {e^x \left (\log \left (3+e^x\right )+5 \log \left (\frac {1}{3} (-5+x)\right )-x \log \left (\frac {1}{3} (-5+x)\right )+5 \log \left (3+e^x\right ) \log \left (\frac {1}{3} (-5+x)\right )-x \log \left (3+e^x\right ) \log \left (\frac {1}{3} (-5+x)\right )\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \frac {e^x (\log (27)-3 \log (-5+x))}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx\\ &=4 \int \frac {e^x \left (-\left ((-5+x) \log \left (\frac {1}{3} (-5+x)\right )\right )-\log \left (3+e^x\right ) \left (-1+(-5+x) \log \left (\frac {1}{3} (-5+x)\right )\right )\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \left (\frac {e^x \log (27)}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}+\frac {3 e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}\right ) \, dx\\ &=4 \int \left (\frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )}+\frac {e^x \left (1+\log \left (3+e^x\right )\right )}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )}\right ) \, dx+12 \int \frac {e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+(4 \log (27)) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx\\ &=4 \int \frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \frac {e^x \left (1+\log \left (3+e^x\right )\right )}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+12 \int \frac {e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+(4 \log (27)) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx\\ &=4 \int \left (\frac {e^x}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )}+\frac {e^x \log \left (3+e^x\right )}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )}\right ) \, dx+4 \int \frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+12 \int \frac {e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+(4 \log (27)) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx\\ &=4 \int \frac {e^x \log \left (3+e^x\right )}{(5-x) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \frac {e^x}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+4 \int \frac {e^x \log \left (3+e^x\right )}{\log \left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+12 \int \frac {e^x \log (-5+x)}{\left (-3-e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx+(4 \log (27)) \int \frac {e^x}{\left (3+e^x\right ) \log ^2\left (-\frac {5}{3}+\frac {x}{3}\right )} \, dx\\ \end {aligned} \end {gather*}

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

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 {gather*} \frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \left (\frac {1}{3} \, x - \frac {5}{3}\right )} \end {gather*}

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 {gather*} -\frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \relax (3) - \log \left (x - 5\right )} \end {gather*}

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	\(\frac {4 \ln \left (3+{\mathrm e}^{x}\right ) {\mathrm e}^{x}}{\ln \left (\frac {x}{3}-\frac {5}{3}\right )}\)	\(18\)

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 {gather*} -\frac {4 \, e^{x} \log \left (e^{x} + 3\right )}{\log \relax (3) - \log \left (x - 5\right )} \end {gather*}

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 {gather*} \frac {4\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^x+3\right )}{\ln \left (\frac {x}{3}-\frac {5}{3}\right )} \end {gather*}

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 {gather*} \text {Timed out} \end {gather*}

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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