3.47.23 \(\int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} (x+x^2+(-2-3 x-2 x^2+2 x^3) \log (2))}{(-2 x-3 x^2+x^4) \log (2)} \, dx\)

Optimal. Leaf size=32 \[ \frac {e^{x \left (2+\frac {\log (2 x)+\frac {\log (-4+2 x)}{\log (2)}}{x}\right )}}{1+x} \]

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Rubi [F]  time = 2.19, 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 {\exp \left (\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}\right ) \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\left (-2 x-3 x^2+x^4\right ) \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(2*E^4*(-4 + 2*x)^Log[2]^(-1)*Gamma[1 + Log[2]^(-1), 4 - 2*x])/(4 - 2*x)^Log[2]^(-1) - (2*E^4*(-4 + 2*x)^Log[2
]^(-1)*Gamma[Log[2]^(-1), 4 - 2*x]*(1 - Log[4]))/((4 - 2*x)^Log[2]^(-1)*Log[2]) - (4*Log[8]*Defer[Int][(E^(2*x
)*(-4 + 2*x)^(-1 + Log[2]^(-1)))/(1 + x)^2, x])/Log[2] - (4*(1 - Log[128])*Defer[Int][(E^(2*x)*(-4 + 2*x)^(-1
+ Log[2]^(-1)))/(1 + x), x])/Log[2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\exp \left (\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}\right ) \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2 x-3 x^2+x^4} \, dx}{\log (2)}\\ &=\frac {\int \frac {\exp \left (\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}\right ) \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{x \left (-2-3 x+x^3\right )} \, dx}{\log (2)}\\ &=\frac {\int \frac {2 e^{\frac {2 x \log (2)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2-3 x+x^3} \, dx}{\log (2)}\\ &=\frac {2 \int \frac {e^{\frac {2 x \log (2)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2-3 x+x^3} \, dx}{\log (2)}\\ &=\frac {2 \int \frac {e^{2 x} (-4+2 x)^{\frac {1}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2-3 x+x^3} \, dx}{\log (2)}\\ &=\frac {2 \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{\frac {1}{2}+x+\frac {x^2}{2}} \, dx}{\log (2)}\\ &=\frac {2 \int \frac {2 e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{(1+x)^2} \, dx}{\log (2)}\\ &=\frac {4 \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{(1+x)^2} \, dx}{\log (2)}\\ &=\frac {4 \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \left (2 x^3 \log (2)+x^2 (1-\log (4))-\log (4)+x (1-\log (8))\right )}{(1+x)^2} \, dx}{\log (2)}\\ &=\frac {4 \int \left (e^{2 x} (-4+2 x)^{\frac {1}{\log (2)}} \log (2)+e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} (1-\log (4))-\frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \log (8)}{(1+x)^2}+\frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} (-1+\log (128))}{1+x}\right ) \, dx}{\log (2)}\\ &=4 \int e^{2 x} (-4+2 x)^{\frac {1}{\log (2)}} \, dx+\frac {(4 (1-\log (4))) \int e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}} \, dx}{\log (2)}-\frac {(4 \log (8)) \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}}}{(1+x)^2} \, dx}{\log (2)}-\frac {(4 (1-\log (128))) \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}}}{1+x} \, dx}{\log (2)}\\ &=2 e^4 (4-2 x)^{-\frac {1}{\log (2)}} (-4+2 x)^{\frac {1}{\log (2)}} \Gamma \left (1+\frac {1}{\log (2)},4-2 x\right )-\frac {2 e^4 (4-2 x)^{-\frac {1}{\log (2)}} (-4+2 x)^{\frac {1}{\log (2)}} \Gamma \left (\frac {1}{\log (2)},4-2 x\right ) (1-\log (4))}{\log (2)}-\frac {(4 \log (8)) \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}}}{(1+x)^2} \, dx}{\log (2)}-\frac {(4 (1-\log (128))) \int \frac {e^{2 x} (-4+2 x)^{-1+\frac {1}{\log (2)}}}{1+x} \, dx}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 3.68, size = 71, normalized size = 2.22 \begin {gather*} \frac {\int \frac {e^{\frac {2 x \log (2)+\log (2) \log (2 x)+\log (-4+2 x)}{\log (2)}} \left (x+x^2+\left (-2-3 x-2 x^2+2 x^3\right ) \log (2)\right )}{-2 x-3 x^2+x^4} \, dx}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 31, normalized size = 0.97 \begin {gather*} \frac {e^{\left (\frac {2 \, x \log \relax (2) + \log \relax (2) \log \left (2 \, x\right ) + \log \left (2 \, x - 4\right )}{\log \relax (2)}\right )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 29, normalized size = 0.91 \begin {gather*} \frac {x e^{\left (\frac {2 \, x \log \relax (2) + \log \relax (2)^{2} + \log \relax (2) + \log \left (x - 2\right )}{\log \relax (2)}\right )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 23, normalized size = 0.72

method	result	size
risch	\(\frac {2 x \left (2 x -4\right )^{\frac {1}{\ln \relax (2)}} {\mathrm e}^{2 x}}{x +1}\)	\(23\)
gosper	\(\frac {{\mathrm e}^{\frac {\ln \relax (2) \ln \left (2 x \right )+\ln \left (2 x -4\right )+2 x \ln \relax (2)}{\ln \relax (2)}}}{x +1}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-2*x^2-3*x-2)*ln(2)+x^2+x)*exp((ln(2)*ln(2*x)+ln(2*x-4)+2*x*ln(2))/ln(2))/(x^4-3*x^2-2*x)/ln(2),x,m
ethod=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.56, size = 23, normalized size = 0.72 \begin {gather*} \frac {2 \, x e^{\left (2 \, x + \frac {\log \left (x - 2\right )}{\log \relax (2)} + 1\right )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {\ln \left (2\,x-4\right )+\ln \left (2\,x\right )\,\ln \relax (2)+2\,x\,\ln \relax (2)}{\ln \relax (2)}}\,\left (x-\ln \relax (2)\,\left (-2\,x^3+2\,x^2+3\,x+2\right )+x^2\right )}{\ln \relax (2)\,\left (-x^4+3\,x^2+2\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((log(2*x - 4) + log(2*x)*log(2) + 2*x*log(2))/log(2))*(x - log(2)*(3*x + 2*x^2 - 2*x^3 + 2) + x^2))/
(log(2)*(2*x + 3*x^2 - x^4)),x)

[Out]

int(-(exp((log(2*x - 4) + log(2*x)*log(2) + 2*x*log(2))/log(2))*(x - log(2)*(3*x + 2*x^2 - 2*x^3 + 2) + x^2))/
(log(2)*(2*x + 3*x^2 - x^4)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-2*x**2-3*x-2)*ln(2)+x**2+x)*exp((ln(2)*ln(2*x)+ln(2*x-4)+2*x*ln(2))/ln(2))/(x**4-3*x**2-2*x
)/ln(2),x)

[Out]

Timed out

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