3.80.44 \(\int \frac {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+(256-192 x+48 x^2-4 x^3) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+(256 x-192 x^2+48 x^3-4 x^4) \log (16)} \, dx\)

Optimal. Leaf size=24 \[ \log \left (\frac {e^{2 x}}{(-4+x)^2}-4 \left (x-\frac {x}{\log (16)}\right )\right ) \]

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Rubi [F]  time = 2.22, 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 {-256+192 x-48 x^2+4 x^3+e^{2 x} (-10+2 x) \log (16)+\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{-256 x+192 x^2-48 x^3+4 x^4+e^{2 x} (-4+x) \log (16)+\left (256 x-192 x^2+48 x^3-4 x^4\right ) \log (16)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-256 + 192*x - 48*x^2 + 4*x^3 + E^(2*x)*(-10 + 2*x)*Log[16] + (256 - 192*x + 48*x^2 - 4*x^3)*Log[16])/(-2
56*x + 192*x^2 - 48*x^3 + 4*x^4 + E^(2*x)*(-4 + x)*Log[16] + (256*x - 192*x^2 + 48*x^3 - 4*x^4)*Log[16]),x]

[Out]

2*x - 2*Log[4 - x] + 192*(1 - Log[16])*Defer[Int][x/(-64*x*(1 - Log[16]) + 32*x^2*(1 - Log[16]) - 4*x^3*(1 - L
og[16]) - E^(2*x)*Log[16]), x] + 8*(1 - Log[16])*Defer[Int][x^3/(-64*x*(1 - Log[16]) + 32*x^2*(1 - Log[16]) -
4*x^3*(1 - Log[16]) - E^(2*x)*Log[16]), x] + 64*(1 - Log[16])*Defer[Int][(64*x*(1 - Log[16]) - 32*x^2*(1 - Log
[16]) + 4*x^3*(1 - Log[16]) + E^(2*x)*Log[16])^(-1), x] + 76*(1 - Log[16])*Defer[Int][x^2/(64*x*(1 - Log[16])
- 32*x^2*(1 - Log[16]) + 4*x^3*(1 - Log[16]) + E^(2*x)*Log[16]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {256-192 x+48 x^2-4 x^3-e^{2 x} (-10+2 x) \log (16)-\left (256-192 x+48 x^2-4 x^3\right ) \log (16)}{(4-x) \left (64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)\right )} \, dx\\ &=\int \left (\frac {2 (-5+x)}{-4+x}+\frac {4 \left (16-48 x+19 x^2-2 x^3\right ) (1-\log (16))}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)}\right ) \, dx\\ &=2 \int \frac {-5+x}{-4+x} \, dx+(4 (1-\log (16))) \int \frac {16-48 x+19 x^2-2 x^3}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)} \, dx\\ &=2 \int \left (1+\frac {1}{4-x}\right ) \, dx+(4 (1-\log (16))) \int \left (\frac {48 x}{-64 x (1-\log (16))+32 x^2 (1-\log (16))-4 x^3 (1-\log (16))-e^{2 x} \log (16)}+\frac {2 x^3}{-64 x (1-\log (16))+32 x^2 (1-\log (16))-4 x^3 (1-\log (16))-e^{2 x} \log (16)}+\frac {16}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)}+\frac {19 x^2}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)}\right ) \, dx\\ &=2 x-2 \log (4-x)+(8 (1-\log (16))) \int \frac {x^3}{-64 x (1-\log (16))+32 x^2 (1-\log (16))-4 x^3 (1-\log (16))-e^{2 x} \log (16)} \, dx+(64 (1-\log (16))) \int \frac {1}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)} \, dx+(76 (1-\log (16))) \int \frac {x^2}{64 x (1-\log (16))-32 x^2 (1-\log (16))+4 x^3 (1-\log (16))+e^{2 x} \log (16)} \, dx+(192 (1-\log (16))) \int \frac {x}{-64 x (1-\log (16))+32 x^2 (1-\log (16))-4 x^3 (1-\log (16))-e^{2 x} \log (16)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 52, normalized size = 2.17 \begin {gather*} -2 \log (4-x)+\log \left (-64 x+32 x^2-4 x^3-e^{2 x} \log (16)+64 x \log (16)-32 x^2 \log (16)+4 x^3 \log (16)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-256 + 192*x - 48*x^2 + 4*x^3 + E^(2*x)*(-10 + 2*x)*Log[16] + (256 - 192*x + 48*x^2 - 4*x^3)*Log[16
])/(-256*x + 192*x^2 - 48*x^3 + 4*x^4 + E^(2*x)*(-4 + x)*Log[16] + (256*x - 192*x^2 + 48*x^3 - 4*x^4)*Log[16])
,x]

[Out]

-2*Log[4 - x] + Log[-64*x + 32*x^2 - 4*x^3 - E^(2*x)*Log[16] + 64*x*Log[16] - 32*x^2*Log[16] + 4*x^3*Log[16]]

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fricas [B]  time = 0.59, size = 43, normalized size = 1.79 \begin {gather*} \log \left (x^{3} - 8 \, x^{2} - 4 \, {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \relax (2) + e^{\left (2 \, x\right )} \log \relax (2) + 16 \, x\right ) - 2 \, \log \left (x - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(2*x-10)*log(2)*exp(x)^2+4*(-4*x^3+48*x^2-192*x+256)*log(2)+4*x^3-48*x^2+192*x-256)/(4*(x-4)*log(
2)*exp(x)^2+4*(-4*x^4+48*x^3-192*x^2+256*x)*log(2)+4*x^4-48*x^3+192*x^2-256*x),x, algorithm="fricas")

[Out]

log(x^3 - 8*x^2 - 4*(x^3 - 8*x^2 + 16*x)*log(2) + e^(2*x)*log(2) + 16*x) - 2*log(x - 4)

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giac [B]  time = 0.23, size = 46, normalized size = 1.92 \begin {gather*} \log \left (-4 \, x^{3} \log \relax (2) + x^{3} + 32 \, x^{2} \log \relax (2) - 8 \, x^{2} - 64 \, x \log \relax (2) + e^{\left (2 \, x\right )} \log \relax (2) + 16 \, x\right ) - 2 \, \log \left (x - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(2*x-10)*log(2)*exp(x)^2+4*(-4*x^3+48*x^2-192*x+256)*log(2)+4*x^3-48*x^2+192*x-256)/(4*(x-4)*log(
2)*exp(x)^2+4*(-4*x^4+48*x^3-192*x^2+256*x)*log(2)+4*x^4-48*x^3+192*x^2-256*x),x, algorithm="giac")

[Out]

log(-4*x^3*log(2) + x^3 + 32*x^2*log(2) - 8*x^2 - 64*x*log(2) + e^(2*x)*log(2) + 16*x) - 2*log(x - 4)

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maple [B]  time = 0.23, size = 47, normalized size = 1.96

method	result	size
risch	\(-2 \ln \left (x -4\right )+\ln \left ({\mathrm e}^{2 x}-\frac {x \left (4 x^{2} \ln \relax (2)-32 x \ln \relax (2)-x^{2}+64 \ln \relax (2)+8 x -16\right )}{\ln \relax (2)}\right )\)	\(47\)
norman	\(-2 \ln \left (x -4\right )+\ln \left (4 x^{3} \ln \relax (2)-32 x^{2} \ln \relax (2)-\ln \relax (2) {\mathrm e}^{2 x}-x^{3}+64 x \ln \relax (2)+8 x^{2}-16 x \right )\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*(2*x-10)*ln(2)*exp(x)^2+4*(-4*x^3+48*x^2-192*x+256)*ln(2)+4*x^3-48*x^2+192*x-256)/(4*(x-4)*ln(2)*exp(x)
^2+4*(-4*x^4+48*x^3-192*x^2+256*x)*ln(2)+4*x^4-48*x^3+192*x^2-256*x),x,method=_RETURNVERBOSE)

[Out]

-2*ln(x-4)+ln(exp(2*x)-x*(4*x^2*ln(2)-32*x*ln(2)-x^2+64*ln(2)+8*x-16)/ln(2))

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maxima [B]  time = 0.48, size = 53, normalized size = 2.21 \begin {gather*} -2 \, \log \left (x - 4\right ) + \log \left (-\frac {x^{3} {\left (4 \, \log \relax (2) - 1\right )} - 8 \, x^{2} {\left (4 \, \log \relax (2) - 1\right )} + 16 \, x {\left (4 \, \log \relax (2) - 1\right )} - e^{\left (2 \, x\right )} \log \relax (2)}{\log \relax (2)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(2*x-10)*log(2)*exp(x)^2+4*(-4*x^3+48*x^2-192*x+256)*log(2)+4*x^3-48*x^2+192*x-256)/(4*(x-4)*log(
2)*exp(x)^2+4*(-4*x^4+48*x^3-192*x^2+256*x)*log(2)+4*x^4-48*x^3+192*x^2-256*x),x, algorithm="maxima")

[Out]

-2*log(x - 4) + log(-(x^3*(4*log(2) - 1) - 8*x^2*(4*log(2) - 1) + 16*x*(4*log(2) - 1) - e^(2*x)*log(2))/log(2)
)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {192\,x-4\,\ln \relax (2)\,\left (4\,x^3-48\,x^2+192\,x-256\right )-48\,x^2+4\,x^3+4\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)\,\left (2\,x-10\right )-256}{4\,\ln \relax (2)\,\left (-4\,x^4+48\,x^3-192\,x^2+256\,x\right )-256\,x+192\,x^2-48\,x^3+4\,x^4+4\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)\,\left (x-4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((192*x - 4*log(2)*(192*x - 48*x^2 + 4*x^3 - 256) - 48*x^2 + 4*x^3 + 4*exp(2*x)*log(2)*(2*x - 10) - 256)/(4
*log(2)*(256*x - 192*x^2 + 48*x^3 - 4*x^4) - 256*x + 192*x^2 - 48*x^3 + 4*x^4 + 4*exp(2*x)*log(2)*(x - 4)),x)

[Out]

int((192*x - 4*log(2)*(192*x - 48*x^2 + 4*x^3 - 256) - 48*x^2 + 4*x^3 + 4*exp(2*x)*log(2)*(2*x - 10) - 256)/(4
*log(2)*(256*x - 192*x^2 + 48*x^3 - 4*x^4) - 256*x + 192*x^2 - 48*x^3 + 4*x^4 + 4*exp(2*x)*log(2)*(x - 4)), x)

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sympy [B]  time = 0.34, size = 51, normalized size = 2.12 \begin {gather*} - 2 \log {\left (x - 4 \right )} + \log {\left (\frac {- 4 x^{3} \log {\relax (2 )} + x^{3} - 8 x^{2} + 32 x^{2} \log {\relax (2 )} - 64 x \log {\relax (2 )} + 16 x}{\log {\relax (2 )}} + e^{2 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(2*x-10)*ln(2)*exp(x)**2+4*(-4*x**3+48*x**2-192*x+256)*ln(2)+4*x**3-48*x**2+192*x-256)/(4*(x-4)*l
n(2)*exp(x)**2+4*(-4*x**4+48*x**3-192*x**2+256*x)*ln(2)+4*x**4-48*x**3+192*x**2-256*x),x)

[Out]

-2*log(x - 4) + log((-4*x**3*log(2) + x**3 - 8*x**2 + 32*x**2*log(2) - 64*x*log(2) + 16*x)/log(2) + exp(2*x))

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