3.76.27 \(\int \frac {e^8 (5+8 x-2 x^2)+(5+8 x-2 x^2) \log (5)}{-15 x^3+3 x^4+(-30 x^2+6 x^3) \log (e^{5+x} (5 x-x^2))+(-15 x+3 x^2) \log ^2(e^{5+x} (5 x-x^2))} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^8+\log (5)}{3 \left (x+\log \left (e^{5+x} (5-x) x\right )\right )} \]

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Rubi [F]  time = 0.44, 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^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^8*(5 + 8*x - 2*x^2) + (5 + 8*x - 2*x^2)*Log[5])/(-15*x^3 + 3*x^4 + (-30*x^2 + 6*x^3)*Log[E^(5 + x)*(5*x
- x^2)] + (-15*x + 3*x^2)*Log[E^(5 + x)*(5*x - x^2)]^2),x]

[Out]

(-2*(E^8 + Log[5])*Defer[Int][(x + Log[-(E^(5 + x)*(-5 + x)*x)])^(-2), x])/3 - ((E^8 + Log[5])*Defer[Int][1/((
-5 + x)*(x + Log[-(E^(5 + x)*(-5 + x)*x)])^2), x])/3 - ((E^8 + Log[5])*Defer[Int][1/(x*(x + Log[-(E^(5 + x)*(-
5 + x)*x)])^2), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (5+8 x-2 x^2\right ) \left (e^8+\log (5)\right )}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx\\ &=\left (e^8+\log (5)\right ) \int \frac {5+8 x-2 x^2}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx\\ &=\left (e^8+\log (5)\right ) \int \frac {-5-8 x+2 x^2}{3 (5-x) x \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx\\ &=\frac {1}{3} \left (e^8+\log (5)\right ) \int \frac {-5-8 x+2 x^2}{(5-x) x \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx\\ &=\frac {1}{3} \left (e^8+\log (5)\right ) \int \left (-\frac {2}{\left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2}-\frac {1}{(-5+x) \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2}-\frac {1}{x \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2}\right ) \, dx\\ &=\frac {1}{3} \left (-e^8-\log (5)\right ) \int \frac {1}{(-5+x) \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx+\frac {1}{3} \left (-e^8-\log (5)\right ) \int \frac {1}{x \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx-\frac {1}{3} \left (2 \left (e^8+\log (5)\right )\right ) \int \frac {1}{\left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 26, normalized size = 0.96 \begin {gather*} \frac {e^8+\log (5)}{3 \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8*(5 + 8*x - 2*x^2) + (5 + 8*x - 2*x^2)*Log[5])/(-15*x^3 + 3*x^4 + (-30*x^2 + 6*x^3)*Log[E^(5 + x
)*(5*x - x^2)] + (-15*x + 3*x^2)*Log[E^(5 + x)*(5*x - x^2)]^2),x]

[Out]

(E^8 + Log[5])/(3*(x + Log[-(E^(5 + x)*(-5 + x)*x)]))

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fricas [A]  time = 0.65, size = 25, normalized size = 0.93 \begin {gather*} \frac {e^{8} + \log \relax (5)}{3 \, {\left (x + \log \left (-{\left (x^{2} - 5 \, x\right )} e^{\left (x + 5\right )}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+8*x+5)*log(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*log((-x^2+5*x)*exp(5+x))^2+(6*x^3-30*x
^2)*log((-x^2+5*x)*exp(5+x))+3*x^4-15*x^3),x, algorithm="fricas")

[Out]

1/3*(e^8 + log(5))/(x + log(-(x^2 - 5*x)*e^(x + 5)))

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giac [A]  time = 0.29, size = 24, normalized size = 0.89 \begin {gather*} \frac {e^{8} + \log \relax (5)}{3 \, {\left (2 \, x + \log \left (-x^{2} + 5 \, x\right ) + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+8*x+5)*log(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*log((-x^2+5*x)*exp(5+x))^2+(6*x^3-30*x
^2)*log((-x^2+5*x)*exp(5+x))+3*x^4-15*x^3),x, algorithm="giac")

[Out]

1/3*(e^8 + log(5))/(2*x + log(-x^2 + 5*x) + 5)

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maple [C]  time = 0.17, size = 476, normalized size = 17.63

method	result	size
risch	\(-\frac {2 i \ln \relax (5)}{3 \left (\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5+x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5+x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right ) \mathrm {csgn}\left (i \left (x -5\right )\right )-\pi \mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right )^{2} \mathrm {csgn}\left (i \left (x -5\right )\right )+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right ) \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right ) \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right ) \mathrm {csgn}\left (i x \right )+\pi \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right )^{3}-2 \pi \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right )^{2}+\pi \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right )^{2} \mathrm {csgn}\left (i x \right )+2 \pi -2 i \ln \left ({\mathrm e}^{5+x}\right )-2 i x -2 i \ln \left (x -5\right )-2 i \ln \relax (x )\right )}-\frac {2 i {\mathrm e}^{8}}{3 \left (\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5+x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5+x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right ) \mathrm {csgn}\left (i \left (x -5\right )\right )-\pi \mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right )^{2} \mathrm {csgn}\left (i \left (x -5\right )\right )+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right ) \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{5+x} \left (x -5\right )\right ) \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right ) \mathrm {csgn}\left (i x \right )+\pi \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right )^{3}-2 \pi \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right )^{2}+\pi \mathrm {csgn}\left (i x \left (x -5\right ) {\mathrm e}^{5+x}\right )^{2} \mathrm {csgn}\left (i x \right )+2 \pi -2 i \ln \left ({\mathrm e}^{5+x}\right )-2 i x -2 i \ln \left (x -5\right )-2 i \ln \relax (x )\right )}\)	\(476\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2+8*x+5)*ln(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*ln((-x^2+5*x)*exp(5+x))^2+(6*x^3-30*x^2)*ln((
-x^2+5*x)*exp(5+x))+3*x^4-15*x^3),x,method=_RETURNVERBOSE)

[Out]

-2/3*I/(Pi*csgn(I*exp(5+x))*csgn(I*exp(5+x)*(x-5))^2-Pi*csgn(I*exp(5+x))*csgn(I*exp(5+x)*(x-5))*csgn(I*(x-5))-
Pi*csgn(I*exp(5+x)*(x-5))^3+Pi*csgn(I*exp(5+x)*(x-5))^2*csgn(I*(x-5))+Pi*csgn(I*exp(5+x)*(x-5))*csgn(I*x*(x-5)
*exp(5+x))^2-Pi*csgn(I*exp(5+x)*(x-5))*csgn(I*x*(x-5)*exp(5+x))*csgn(I*x)+Pi*csgn(I*x*(x-5)*exp(5+x))^3-2*Pi*c
sgn(I*x*(x-5)*exp(5+x))^2+Pi*csgn(I*x*(x-5)*exp(5+x))^2*csgn(I*x)+2*Pi-2*I*ln(exp(5+x))-2*I*x-2*I*ln(x-5)-2*I*
ln(x))*ln(5)-2/3*I/(Pi*csgn(I*exp(5+x))*csgn(I*exp(5+x)*(x-5))^2-Pi*csgn(I*exp(5+x))*csgn(I*exp(5+x)*(x-5))*cs
gn(I*(x-5))-Pi*csgn(I*exp(5+x)*(x-5))^3+Pi*csgn(I*exp(5+x)*(x-5))^2*csgn(I*(x-5))+Pi*csgn(I*exp(5+x)*(x-5))*cs
gn(I*x*(x-5)*exp(5+x))^2-Pi*csgn(I*exp(5+x)*(x-5))*csgn(I*x*(x-5)*exp(5+x))*csgn(I*x)+Pi*csgn(I*x*(x-5)*exp(5+
x))^3-2*Pi*csgn(I*x*(x-5)*exp(5+x))^2+Pi*csgn(I*x*(x-5)*exp(5+x))^2*csgn(I*x)+2*Pi-2*I*ln(exp(5+x))-2*I*x-2*I*
ln(x-5)-2*I*ln(x))*exp(8)

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maxima [A]  time = 0.54, size = 22, normalized size = 0.81 \begin {gather*} \frac {e^{8} + \log \relax (5)}{3 \, {\left (2 \, x + \log \relax (x) + \log \left (-x + 5\right ) + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+8*x+5)*log(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*log((-x^2+5*x)*exp(5+x))^2+(6*x^3-30*x
^2)*log((-x^2+5*x)*exp(5+x))+3*x^4-15*x^3),x, algorithm="maxima")

[Out]

1/3*(e^8 + log(5))/(2*x + log(x) + log(-x + 5) + 5)

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mupad [B]  time = 5.80, size = 29, normalized size = 1.07 \begin {gather*} \frac {\frac {{\mathrm {e}}^8}{3}+\frac {\ln \relax (5)}{3}}{x+\ln \left ({\mathrm {e}}^5\,{\mathrm {e}}^x\,\left (5\,x-x^2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8)*(8*x - 2*x^2 + 5) + log(5)*(8*x - 2*x^2 + 5))/(log(exp(x + 5)*(5*x - x^2))^2*(15*x - 3*x^2) + log
(exp(x + 5)*(5*x - x^2))*(30*x^2 - 6*x^3) + 15*x^3 - 3*x^4),x)

[Out]

(exp(8)/3 + log(5)/3)/(x + log(exp(5)*exp(x)*(5*x - x^2)))

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sympy [A]  time = 0.22, size = 24, normalized size = 0.89 \begin {gather*} \frac {\log {\relax (5 )} + e^{8}}{3 x + 3 \log {\left (\left (- x^{2} + 5 x\right ) e^{x + 5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2+8*x+5)*ln(5)+(-2*x**2+8*x+5)*exp(4)**2)/((3*x**2-15*x)*ln((-x**2+5*x)*exp(5+x))**2+(6*x**3
-30*x**2)*ln((-x**2+5*x)*exp(5+x))+3*x**4-15*x**3),x)

[Out]

(log(5) + exp(8))/(3*x + 3*log((-x**2 + 5*x)*exp(x + 5)))

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