3.56.45 \(\int \frac {3 e^5+e^x (-16 x+8 x^2)}{(-e^5 x+8 e^x x^2+32 x^4) \log (\frac {-e^5+8 e^x x+32 x^3}{8 x^3})} \, dx\) [5545]

Optimal. Leaf size=24 \[ \log \left (20 \log \left (4+\frac {e^x-\frac {e^5}{8 x}}{x^2}\right )\right ) \]

[Out]

ln(20*ln(1/x^2*(-1/8*exp(5)/x+exp(x))+4))

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Rubi [A]
time = 0.11, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6816} \begin {gather*} \log \left (\log \left (-\frac {-32 x^3-8 e^x x+e^5}{8 x^3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*E^5 + E^x*(-16*x + 8*x^2))/((-(E^5*x) + 8*E^x*x^2 + 32*x^4)*Log[(-E^5 + 8*E^x*x + 32*x^3)/(8*x^3)]),x]

[Out]

Log[Log[-1/8*(E^5 - 8*E^x*x - 32*x^3)/x^3]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (\log \left (-\frac {e^5-8 e^x x-32 x^3}{8 x^3}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.09, size = 21, normalized size = 0.88 \begin {gather*} \log \left (\log \left (4-\frac {e^5}{8 x^3}+\frac {e^x}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*E^5 + E^x*(-16*x + 8*x^2))/((-(E^5*x) + 8*E^x*x^2 + 32*x^4)*Log[(-E^5 + 8*E^x*x + 32*x^3)/(8*x^3)
]),x]

[Out]

Log[Log[4 - E^5/(8*x^3) + E^x/x^2]]

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Maple [A]
time = 0.56, size = 23, normalized size = 0.96

method	result	size
norman	\(\ln \left (\ln \left (\frac {8 \,{\mathrm e}^{x} x -{\mathrm e}^{5}+32 x^{3}}{8 x^{3}}\right )\right )\)	\(23\)
risch	\(\ln \left (-3 \ln \left (2\right )+\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )^{2}}{2}-3 \ln \left (x \right )+\ln \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )+\frac {i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )}{2}+\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )^{3}}{2}-\frac {i \pi \mathrm {csgn}\left (i x^{3}\right )^{2} \mathrm {csgn}\left (i x \right )}{2}+\frac {i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi -i \pi \mathrm {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )^{2}-\frac {i \pi \mathrm {csgn}\left (i x^{3}\right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{x^{3}}\right ) \mathrm {csgn}\left (i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-32 x^{3}-8 \,{\mathrm e}^{x} x +{\mathrm e}^{5}\right )}{x^{3}}\right )}{2}\right )\)	\(328\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2-16*x)*exp(x)+3*exp(5))/(8*exp(x)*x^2-x*exp(5)+32*x^4)/ln(1/8*(8*exp(x)*x-exp(5)+32*x^3)/x^3),x,met
hod=_RETURNVERBOSE)

[Out]

ln(ln(1/8*(8*exp(x)*x-exp(5)+32*x^3)/x^3))

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Maxima [A]
time = 0.53, size = 26, normalized size = 1.08 \begin {gather*} \log \left (-3 \, \log \left (2\right ) + \log \left (32 \, x^{3} + 8 \, x e^{x} - e^{5}\right ) - 3 \, \log \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-16*x)*exp(x)+3*exp(5))/(8*exp(x)*x^2-x*exp(5)+32*x^4)/log(1/8*(8*exp(x)*x-exp(5)+32*x^3)/x^3
),x, algorithm="maxima")

[Out]

log(-3*log(2) + log(32*x^3 + 8*x*e^x - e^5) - 3*log(x))

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Fricas [A]
time = 0.35, size = 22, normalized size = 0.92 \begin {gather*} \log \left (\log \left (\frac {32 \, x^{3} + 8 \, x e^{x} - e^{5}}{8 \, x^{3}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-16*x)*exp(x)+3*exp(5))/(8*exp(x)*x^2-x*exp(5)+32*x^4)/log(1/8*(8*exp(x)*x-exp(5)+32*x^3)/x^3
),x, algorithm="fricas")

[Out]

log(log(1/8*(32*x^3 + 8*x*e^x - e^5)/x^3))

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Sympy [A]
time = 0.21, size = 20, normalized size = 0.83 \begin {gather*} \log {\left (\log {\left (\frac {4 x^{3} + x e^{x} - \frac {e^{5}}{8}}{x^{3}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2-16*x)*exp(x)+3*exp(5))/(8*exp(x)*x**2-x*exp(5)+32*x**4)/ln(1/8*(8*exp(x)*x-exp(5)+32*x**3)/
x**3),x)

[Out]

log(log((4*x**3 + x*exp(x) - exp(5)/8)/x**3))

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Giac [A]
time = 0.42, size = 22, normalized size = 0.92 \begin {gather*} \log \left (\log \left (\frac {32 \, x^{3} + 8 \, x e^{x} - e^{5}}{8 \, x^{3}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-16*x)*exp(x)+3*exp(5))/(8*exp(x)*x^2-x*exp(5)+32*x^4)/log(1/8*(8*exp(x)*x-exp(5)+32*x^3)/x^3
),x, algorithm="giac")

[Out]

log(log(1/8*(32*x^3 + 8*x*e^x - e^5)/x^3))

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Mupad [B]
time = 4.29, size = 22, normalized size = 0.92 \begin {gather*} \ln \left (\ln \left (\frac {8\,x\,{\mathrm {e}}^x-{\mathrm {e}}^5+32\,x^3}{8\,x^3}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(5) - exp(x)*(16*x - 8*x^2))/(log((x*exp(x) - exp(5)/8 + 4*x^3)/x^3)*(8*x^2*exp(x) - x*exp(5) + 32*x
^4)),x)

[Out]

log(log((8*x*exp(x) - exp(5) + 32*x^3)/(8*x^3)))

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