3.48.6 \(\int \frac {-8 e^{x/2} x^2+e^{x/2} (8 x+2 x^2) \log (4 e^{5+x})}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log (4 e^{5+x})-12 e^{x/4} x \log ^2(4 e^{5+x})+\log ^3(4 e^{5+x})} \, dx\)

Optimal. Leaf size=26 \[ \frac {4}{\left (4-\frac {e^{-x/4} \log \left (4 e^{5+x}\right )}{x}\right )^2} \]

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Rubi [F]  time = 2.47, 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 {-8 e^{x/2} x^2+e^{x/2} \left (8 x+2 x^2\right ) \log \left (4 e^{5+x}\right )}{-64 e^{3 x/4} x^3+48 e^{x/2} x^2 \log \left (4 e^{5+x}\right )-12 e^{x/4} x \log ^2\left (4 e^{5+x}\right )+\log ^3\left (4 e^{5+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8*E^(x/2)*x^2 + E^(x/2)*(8*x + 2*x^2)*Log[4*E^(5 + x)])/(-64*E^((3*x)/4)*x^3 + 48*E^(x/2)*x^2*Log[4*E^(5
+ x)] - 12*E^(x/4)*x*Log[4*E^(5 + x)]^2 + Log[4*E^(5 + x)]^3),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 52, normalized size = 2.00 \begin {gather*} -\frac {\log \left (4 e^{5+x}\right ) \left (-8 e^{x/4} x+\log \left (4 e^{5+x}\right )\right )}{4 \left (-4 e^{x/4} x+\log \left (4 e^{5+x}\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*E^(x/2)*x^2 + E^(x/2)*(8*x + 2*x^2)*Log[4*E^(5 + x)])/(-64*E^((3*x)/4)*x^3 + 48*E^(x/2)*x^2*Log[
4*E^(5 + x)] - 12*E^(x/4)*x*Log[4*E^(5 + x)]^2 + Log[4*E^(5 + x)]^3),x]

[Out]

-1/4*(Log[4*E^(5 + x)]*(-8*E^(x/4)*x + Log[4*E^(5 + x)]))/(-4*E^(x/4)*x + Log[4*E^(5 + x)])^2

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fricas [B]  time = 0.78, size = 91, normalized size = 3.50 \begin {gather*} -\frac {x^{2} - 8 \, {\left (x^{2} + 2 \, x \log \relax (2) + 5 \, x\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, {\left (x + 5\right )} \log \relax (2) + 4 \, \log \relax (2)^{2} + 10 \, x + 25}{4 \, {\left (16 \, x^{2} e^{\left (\frac {1}{2} \, x\right )} + x^{2} - 8 \, {\left (x^{2} + 2 \, x \log \relax (2) + 5 \, x\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, {\left (x + 5\right )} \log \relax (2) + 4 \, \log \relax (2)^{2} + 10 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x)*exp(1/4*x)^2*log(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(log(4*exp(5+x))^3-12*x*exp(1/4*x)*log
(4*exp(5+x))^2+48*x^2*exp(1/4*x)^2*log(4*exp(5+x))-64*x^3*exp(1/4*x)^3),x, algorithm="fricas")

[Out]

-1/4*(x^2 - 8*(x^2 + 2*x*log(2) + 5*x)*e^(1/4*x) + 4*(x + 5)*log(2) + 4*log(2)^2 + 10*x + 25)/(16*x^2*e^(1/2*x
) + x^2 - 8*(x^2 + 2*x*log(2) + 5*x)*e^(1/4*x) + 4*(x + 5)*log(2) + 4*log(2)^2 + 10*x + 25)

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giac [B]  time = 2.62, size = 111, normalized size = 4.27 \begin {gather*} \frac {8 \, x^{2} e^{\left (\frac {1}{4} \, x\right )} + 16 \, x e^{\left (\frac {1}{4} \, x\right )} \log \relax (2) - x^{2} + 40 \, x e^{\left (\frac {1}{4} \, x\right )} - 4 \, x \log \relax (2) - 4 \, \log \relax (2)^{2} - 10 \, x - 20 \, \log \relax (2) - 25}{4 \, {\left (16 \, x^{2} e^{\left (\frac {1}{2} \, x\right )} - 8 \, x^{2} e^{\left (\frac {1}{4} \, x\right )} - 16 \, x e^{\left (\frac {1}{4} \, x\right )} \log \relax (2) + x^{2} - 40 \, x e^{\left (\frac {1}{4} \, x\right )} + 4 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} + 10 \, x + 20 \, \log \relax (2) + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x)*exp(1/4*x)^2*log(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(log(4*exp(5+x))^3-12*x*exp(1/4*x)*log
(4*exp(5+x))^2+48*x^2*exp(1/4*x)^2*log(4*exp(5+x))-64*x^3*exp(1/4*x)^3),x, algorithm="giac")

[Out]

1/4*(8*x^2*e^(1/4*x) + 16*x*e^(1/4*x)*log(2) - x^2 + 40*x*e^(1/4*x) - 4*x*log(2) - 4*log(2)^2 - 10*x - 20*log(
2) - 25)/(16*x^2*e^(1/2*x) - 8*x^2*e^(1/4*x) - 16*x*e^(1/4*x)*log(2) + x^2 - 40*x*e^(1/4*x) + 4*x*log(2) + 4*l
og(2)^2 + 10*x + 20*log(2) + 25)

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maple [A]  time = 5.88, size = 31, normalized size = 1.19

method	result	size
default	\(\frac {4 x^{2} {\mathrm e}^{\frac {x}{2}}}{\left (4 x \,{\mathrm e}^{\frac {x}{4}}-\ln \left (4 \,{\mathrm e}^{5+x}\right )\right )^{2}}\)	\(31\)
risch	\(\frac {16 x^{2} {\mathrm e}^{\frac {x}{2}}}{\left (i \pi \mathrm {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right )^{3}-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right )^{2}-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right )^{2}+i \pi \mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right )^{3}-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}+i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{3}-2 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right )^{2}+i \pi \mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right )+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right )+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{\frac {x}{4}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {3 x}{4}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )-10+8 x \,{\mathrm e}^{\frac {x}{4}}-4 \ln \relax (2)-8 \ln \left ({\mathrm e}^{\frac {x}{4}}\right )\right )^{2}}\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+8*x)*exp(1/4*x)^2*ln(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(ln(4*exp(5+x))^3-12*x*exp(1/4*x)*ln(4*exp(5+
x))^2+48*x^2*exp(1/4*x)^2*ln(4*exp(5+x))-64*x^3*exp(1/4*x)^3),x,method=_RETURNVERBOSE)

[Out]

4*x^2*exp(1/4*x)^2/(4*x*exp(1/4*x)-ln(4*exp(5+x)))^2

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maxima [B]  time = 0.54, size = 97, normalized size = 3.73 \begin {gather*} -\frac {x^{2} + 2 \, x {\left (2 \, \log \relax (2) + 5\right )} - 8 \, {\left (x^{2} + x {\left (2 \, \log \relax (2) + 5\right )}\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, \log \relax (2)^{2} + 20 \, \log \relax (2) + 25}{4 \, {\left (16 \, x^{2} e^{\left (\frac {1}{2} \, x\right )} + x^{2} + 2 \, x {\left (2 \, \log \relax (2) + 5\right )} - 8 \, {\left (x^{2} + x {\left (2 \, \log \relax (2) + 5\right )}\right )} e^{\left (\frac {1}{4} \, x\right )} + 4 \, \log \relax (2)^{2} + 20 \, \log \relax (2) + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+8*x)*exp(1/4*x)^2*log(4*exp(5+x))-8*x^2*exp(1/4*x)^2)/(log(4*exp(5+x))^3-12*x*exp(1/4*x)*log
(4*exp(5+x))^2+48*x^2*exp(1/4*x)^2*log(4*exp(5+x))-64*x^3*exp(1/4*x)^3),x, algorithm="maxima")

[Out]

-1/4*(x^2 + 2*x*(2*log(2) + 5) - 8*(x^2 + x*(2*log(2) + 5))*e^(1/4*x) + 4*log(2)^2 + 20*log(2) + 25)/(16*x^2*e
^(1/2*x) + x^2 + 2*x*(2*log(2) + 5) - 8*(x^2 + x*(2*log(2) + 5))*e^(1/4*x) + 4*log(2)^2 + 20*log(2) + 25)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {8\,x^2\,{\mathrm {e}}^{x/2}-\ln \left (4\,{\mathrm {e}}^{x+5}\right )\,{\mathrm {e}}^{x/2}\,\left (2\,x^2+8\,x\right )}{{\ln \left (4\,{\mathrm {e}}^{x+5}\right )}^3-64\,x^3\,{\mathrm {e}}^{\frac {3\,x}{4}}+48\,x^2\,\ln \left (4\,{\mathrm {e}}^{x+5}\right )\,{\mathrm {e}}^{x/2}-12\,x\,{\ln \left (4\,{\mathrm {e}}^{x+5}\right )}^2\,{\mathrm {e}}^{x/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x^2*exp(x/2) - log(4*exp(x + 5))*exp(x/2)*(8*x + 2*x^2))/(log(4*exp(x + 5))^3 - 64*x^3*exp((3*x)/4) +
48*x^2*log(4*exp(x + 5))*exp(x/2) - 12*x*log(4*exp(x + 5))^2*exp(x/4)),x)

[Out]

int(-(8*x^2*exp(x/2) - log(4*exp(x + 5))*exp(x/2)*(8*x + 2*x^2))/(log(4*exp(x + 5))^3 - 64*x^3*exp((3*x)/4) +
48*x^2*log(4*exp(x + 5))*exp(x/2) - 12*x*log(4*exp(x + 5))^2*exp(x/4)), x)

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sympy [B]  time = 0.44, size = 107, normalized size = 4.12 \begin {gather*} \frac {- \frac {x^{2}}{4} - \frac {5 x}{2} - x \log {\relax (2 )} + \left (2 x^{2} + 4 x \log {\relax (2 )} + 10 x\right ) e^{\frac {x}{4}} - \frac {25}{4} - 5 \log {\relax (2 )} - \log {\relax (2 )}^{2}}{16 x^{2} e^{\frac {x}{2}} + x^{2} + 4 x \log {\relax (2 )} + 10 x + \left (- 8 x^{2} - 40 x - 16 x \log {\relax (2 )}\right ) e^{\frac {x}{4}} + 4 \log {\relax (2 )}^{2} + 20 \log {\relax (2 )} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+8*x)*exp(1/4*x)**2*ln(4*exp(5+x))-8*x**2*exp(1/4*x)**2)/(ln(4*exp(5+x))**3-12*x*exp(1/4*x)*
ln(4*exp(5+x))**2+48*x**2*exp(1/4*x)**2*ln(4*exp(5+x))-64*x**3*exp(1/4*x)**3),x)

[Out]

(-x**2/4 - 5*x/2 - x*log(2) + (2*x**2 + 4*x*log(2) + 10*x)*exp(x/4) - 25/4 - 5*log(2) - log(2)**2)/(16*x**2*ex
p(x/2) + x**2 + 4*x*log(2) + 10*x + (-8*x**2 - 40*x - 16*x*log(2))*exp(x/4) + 4*log(2)**2 + 20*log(2) + 25)

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