3.35.56 \(\int \frac {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+(-24 x^2+16 x^3) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x ((96-64 x) \log (2)-8 x^2 \log ^2(2))} \, dx\)

Optimal. Leaf size=25 \[ \frac {x}{e^x-\frac {x^2}{4}+\frac {3-2 x}{\log (2)}} \]

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Rubi [F]  time = 1.46, 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 {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+\left (-24 x^2+16 x^3\right ) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(48*Log[2] + E^x*(16 - 16*x)*Log[2]^2 + 4*x^2*Log[2]^2)/(144 - 192*x + 64*x^2 + (-24*x^2 + 16*x^3)*Log[2]
+ 16*E^(2*x)*Log[2]^2 + x^4*Log[2]^2 + E^x*((96 - 64*x)*Log[2] - 8*x^2*Log[2]^2)),x]

[Out]

80*Log[2]*Defer[Int][x/(-12 + 8*x - 4*E^x*Log[2] + x^2*Log[2])^2, x] - 8*(4 - Log[2])*Log[2]*Defer[Int][x^2/(-
12 + 8*x - 4*E^x*Log[2] + x^2*Log[2])^2, x] - 4*Log[2]^2*Defer[Int][x^3/(-12 + 8*x - 4*E^x*Log[2] + x^2*Log[2]
)^2, x] + 4*Log[2]*Defer[Int][x/(-12 + 8*x - 4*E^x*Log[2] + x^2*Log[2]), x] - 4*Log[2]*Defer[Int][(-12 + 8*x +
x^2*Log[2] - E^x*Log[16])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \log (2) \left (12-4 e^x (-1+x) \log (2)+x^2 \log (2)\right )}{\left (8 x+x^2 \log (2)-4 \left (3+e^x \log (2)\right )\right )^2} \, dx\\ &=(4 \log (2)) \int \frac {12-4 e^x (-1+x) \log (2)+x^2 \log (2)}{\left (8 x+x^2 \log (2)-4 \left (3+e^x \log (2)\right )\right )^2} \, dx\\ &=(4 \log (2)) \int \left (\frac {x \left (20-2 x (4-\log (2))-x^2 \log (2)\right )}{\left (12-8 x+4 e^x \log (2)-x^2 \log (2)\right )^2}+\frac {-1+x}{-12+8 x-4 e^x \log (2)+x^2 \log (2)}\right ) \, dx\\ &=(4 \log (2)) \int \frac {x \left (20-2 x (4-\log (2))-x^2 \log (2)\right )}{\left (12-8 x+4 e^x \log (2)-x^2 \log (2)\right )^2} \, dx+(4 \log (2)) \int \frac {-1+x}{-12+8 x-4 e^x \log (2)+x^2 \log (2)} \, dx\\ &=(4 \log (2)) \int \left (\frac {20 x}{\left (-12+8 x-4 e^x \log (2)+x^2 \log (2)\right )^2}+\frac {2 x^2 (-4+\log (2))}{\left (-12+8 x-4 e^x \log (2)+x^2 \log (2)\right )^2}-\frac {x^3 \log (2)}{\left (-12+8 x-4 e^x \log (2)+x^2 \log (2)\right )^2}\right ) \, dx+(4 \log (2)) \int \left (\frac {x}{-12+8 x-4 e^x \log (2)+x^2 \log (2)}-\frac {1}{-12+8 x+x^2 \log (2)-e^x \log (16)}\right ) \, dx\\ &=(4 \log (2)) \int \frac {x}{-12+8 x-4 e^x \log (2)+x^2 \log (2)} \, dx-(4 \log (2)) \int \frac {1}{-12+8 x+x^2 \log (2)-e^x \log (16)} \, dx+(80 \log (2)) \int \frac {x}{\left (-12+8 x-4 e^x \log (2)+x^2 \log (2)\right )^2} \, dx-(8 (4-\log (2)) \log (2)) \int \frac {x^2}{\left (-12+8 x-4 e^x \log (2)+x^2 \log (2)\right )^2} \, dx-\left (4 \log ^2(2)\right ) \int \frac {x^3}{\left (-12+8 x-4 e^x \log (2)+x^2 \log (2)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.71, size = 26, normalized size = 1.04 \begin {gather*} \frac {4 x \log (2)}{12-8 x+4 e^x \log (2)-x^2 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48*Log[2] + E^x*(16 - 16*x)*Log[2]^2 + 4*x^2*Log[2]^2)/(144 - 192*x + 64*x^2 + (-24*x^2 + 16*x^3)*L
og[2] + 16*E^(2*x)*Log[2]^2 + x^4*Log[2]^2 + E^x*((96 - 64*x)*Log[2] - 8*x^2*Log[2]^2)),x]

[Out]

(4*x*Log[2])/(12 - 8*x + 4*E^x*Log[2] - x^2*Log[2])

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fricas [A]  time = 0.96, size = 24, normalized size = 0.96 \begin {gather*} -\frac {4 \, x \log \relax (2)}{x^{2} \log \relax (2) - 4 \, e^{x} \log \relax (2) + 8 \, x - 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x+16)*log(2)^2*exp(x)+4*x^2*log(2)^2+48*log(2))/(16*log(2)^2*exp(x)^2+(-8*x^2*log(2)^2+(-64*x+
96)*log(2))*exp(x)+x^4*log(2)^2+(16*x^3-24*x^2)*log(2)+64*x^2-192*x+144),x, algorithm="fricas")

[Out]

-4*x*log(2)/(x^2*log(2) - 4*e^x*log(2) + 8*x - 12)

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giac [A]  time = 0.17, size = 24, normalized size = 0.96 \begin {gather*} -\frac {4 \, x \log \relax (2)}{x^{2} \log \relax (2) - 4 \, e^{x} \log \relax (2) + 8 \, x - 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x+16)*log(2)^2*exp(x)+4*x^2*log(2)^2+48*log(2))/(16*log(2)^2*exp(x)^2+(-8*x^2*log(2)^2+(-64*x+
96)*log(2))*exp(x)+x^4*log(2)^2+(16*x^3-24*x^2)*log(2)+64*x^2-192*x+144),x, algorithm="giac")

[Out]

-4*x*log(2)/(x^2*log(2) - 4*e^x*log(2) + 8*x - 12)

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maple [A]  time = 0.20, size = 25, normalized size = 1.00

method	result	size
risch	\(-\frac {4 x \ln \relax (2)}{x^{2} \ln \relax (2)-4 \,{\mathrm e}^{x} \ln \relax (2)+8 x -12}\)	\(25\)
norman	\(\frac {-2 \ln \relax (2)^{2} {\mathrm e}^{x}+\frac {x^{2} \ln \relax (2)^{2}}{2}-6 \ln \relax (2)}{x^{2} \ln \relax (2)-4 \,{\mathrm e}^{x} \ln \relax (2)+8 x -12}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x+16)*ln(2)^2*exp(x)+4*x^2*ln(2)^2+48*ln(2))/(16*ln(2)^2*exp(x)^2+(-8*x^2*ln(2)^2+(-64*x+96)*ln(2))*
exp(x)+x^4*ln(2)^2+(16*x^3-24*x^2)*ln(2)+64*x^2-192*x+144),x,method=_RETURNVERBOSE)

[Out]

-4*x*ln(2)/(x^2*ln(2)-4*exp(x)*ln(2)+8*x-12)

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maxima [A]  time = 0.57, size = 24, normalized size = 0.96 \begin {gather*} -\frac {4 \, x \log \relax (2)}{x^{2} \log \relax (2) - 4 \, e^{x} \log \relax (2) + 8 \, x - 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x+16)*log(2)^2*exp(x)+4*x^2*log(2)^2+48*log(2))/(16*log(2)^2*exp(x)^2+(-8*x^2*log(2)^2+(-64*x+
96)*log(2))*exp(x)+x^4*log(2)^2+(16*x^3-24*x^2)*log(2)+64*x^2-192*x+144),x, algorithm="maxima")

[Out]

-4*x*log(2)/(x^2*log(2) - 4*e^x*log(2) + 8*x - 12)

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mupad [B]  time = 3.23, size = 24, normalized size = 0.96 \begin {gather*} -\frac {4\,x\,\ln \relax (2)}{8\,x+x^2\,\ln \relax (2)-4\,{\mathrm {e}}^x\,\ln \relax (2)-12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((48*log(2) + 4*x^2*log(2)^2 - exp(x)*log(2)^2*(16*x - 16))/(x^4*log(2)^2 - 192*x + 16*exp(2*x)*log(2)^2 -
log(2)*(24*x^2 - 16*x^3) - exp(x)*(log(2)*(64*x - 96) + 8*x^2*log(2)^2) + 64*x^2 + 144),x)

[Out]

-(4*x*log(2))/(8*x + x^2*log(2) - 4*exp(x)*log(2) - 12)

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sympy [A]  time = 0.17, size = 26, normalized size = 1.04 \begin {gather*} \frac {4 x \log {\relax (2 )}}{- x^{2} \log {\relax (2 )} - 8 x + 4 e^{x} \log {\relax (2 )} + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x+16)*ln(2)**2*exp(x)+4*x**2*ln(2)**2+48*ln(2))/(16*ln(2)**2*exp(x)**2+(-8*x**2*ln(2)**2+(-64*
x+96)*ln(2))*exp(x)+x**4*ln(2)**2+(16*x**3-24*x**2)*ln(2)+64*x**2-192*x+144),x)

[Out]

4*x*log(2)/(-x**2*log(2) - 8*x + 4*exp(x)*log(2) + 12)

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