3.82.58 \(\int \frac {3 x+4 e^x (-20 x+5 x^2)+(-12+4 x) \log (2)+e^x (16 x-4 x^2) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x (-10 x^2-40 e^x x \log (2))+(160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {3-x}{20 e^x-x-4 \log (2) \left (e^x-\log (x)\right )} \]

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Rubi [F]  time = 3.59, 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 {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{400 e^{2 x} x+x^3+8 e^x x^2 \log (2)+16 e^{2 x} x \log ^2(2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3*x + 4*E^x*(-20*x + 5*x^2) + (-12 + 4*x)*Log[2] + E^x*(16*x - 4*x^2)*Log[2] - 4*x*Log[2]*Log[x])/(400*E^
(2*x)*x + x^3 + 8*E^x*x^2*Log[2] + 16*E^(2*x)*x*Log[2]^2 + 4*E^x*(-10*x^2 - 40*E^x*x*Log[2]) + (160*E^x*x*Log[
2] - 8*x^2*Log[2] - 32*E^x*x*Log[2]^2)*Log[x] + 16*x*Log[2]^2*Log[x]^2),x]

[Out]

(3 + Log[16])*Defer[Int][(x - 20*E^x*(1 - Log[2]/5) - 4*Log[2]*Log[x])^(-2), x] - 12*Log[2]*Defer[Int][1/(x*(x
- 20*E^x*(1 - Log[2]/5) - 4*Log[2]*Log[x])^2), x] - 4*Defer[Int][x/(x - 20*E^x*(1 - Log[2]/5) - 4*Log[2]*Log[
x])^2, x] + Defer[Int][x^2/(x - 20*E^x*(1 - Log[2]/5) - 4*Log[2]*Log[x])^2, x] + 12*Log[2]*Defer[Int][Log[x]/(
x - 20*E^x*(1 - Log[2]/5) - 4*Log[2]*Log[x])^2, x] - 4*Log[2]*Defer[Int][(x*Log[x])/(x - 20*E^x*(1 - Log[2]/5)
- 4*Log[2]*Log[x])^2, x] + 4*Defer[Int][(x - 20*E^x*(1 - Log[2]/5) - 4*Log[2]*Log[x])^(-1), x] + Defer[Int][x
/(-x + 20*E^x*(1 - Log[2]/5) + 4*Log[2]*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 x+4 e^x \left (-20 x+5 x^2\right )+(-12+4 x) \log (2)+e^x \left (16 x-4 x^2\right ) \log (2)-4 x \log (2) \log (x)}{x^3+8 e^x x^2 \log (2)+4 e^x \left (-10 x^2-40 e^x x \log (2)\right )+e^{2 x} x \left (400+16 \log ^2(2)\right )+\left (160 e^x x \log (2)-8 x^2 \log (2)-32 e^x x \log ^2(2)\right ) \log (x)+16 x \log ^2(2) \log ^2(x)} \, dx\\ &=\int \frac {-4 e^x x^2 (-5+\log (2))-12 \log (2)+x \left (3+16 e^x (-5+\log (2))+\log (16)\right )-4 x \log (2) \log (x)}{x \left (x+4 e^x (-5+\log (2))-4 \log (2) \log (x)\right )^2} \, dx\\ &=\int \left (\frac {4-x}{x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)}+\frac {-4 x^2+x^3-12 \log (2)+3 x \left (1+\frac {\log (16)}{3}\right )+12 x \log (2) \log (x)-4 x^2 \log (2) \log (x)}{x \left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}\right ) \, dx\\ &=\int \frac {4-x}{x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)} \, dx+\int \frac {-4 x^2+x^3-12 \log (2)+3 x \left (1+\frac {\log (16)}{3}\right )+12 x \log (2) \log (x)-4 x^2 \log (2) \log (x)}{x \left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx\\ &=\int \left (-\frac {4 x}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}+\frac {x^2}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}-\frac {12 \log (2)}{x \left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}+\frac {3+\log (16)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}+\frac {12 \log (2) \log (x)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}-\frac {4 x \log (2) \log (x)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2}\right ) \, dx+\int \left (\frac {4}{x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)}+\frac {x}{-x+20 e^x \left (1-\frac {\log (2)}{5}\right )+4 \log (2) \log (x)}\right ) \, dx\\ &=-\left (4 \int \frac {x}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx\right )+4 \int \frac {1}{x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)} \, dx-(4 \log (2)) \int \frac {x \log (x)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx-(12 \log (2)) \int \frac {1}{x \left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx+(12 \log (2)) \int \frac {\log (x)}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx+(3+\log (16)) \int \frac {1}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx+\int \frac {x^2}{\left (x-20 e^x \left (1-\frac {\log (2)}{5}\right )-4 \log (2) \log (x)\right )^2} \, dx+\int \frac {x}{-x+20 e^x \left (1-\frac {\log (2)}{5}\right )+4 \log (2) \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.98, size = 72, normalized size = 2.48 \begin {gather*} \frac {x^2 \left (1+4 e^x (-5+\log (2))\right )-x \left (3+12 e^x (-5+\log (2))+\log (16)\right )+\log (4096)}{\left (x+4 e^x x (-5+\log (2))-4 \log (2)\right ) \left (x+4 e^x (-5+\log (2))-4 \log (2) \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*x + 4*E^x*(-20*x + 5*x^2) + (-12 + 4*x)*Log[2] + E^x*(16*x - 4*x^2)*Log[2] - 4*x*Log[2]*Log[x])/(
400*E^(2*x)*x + x^3 + 8*E^x*x^2*Log[2] + 16*E^(2*x)*x*Log[2]^2 + 4*E^x*(-10*x^2 - 40*E^x*x*Log[2]) + (160*E^x*
x*Log[2] - 8*x^2*Log[2] - 32*E^x*x*Log[2]^2)*Log[x] + 16*x*Log[2]^2*Log[x]^2),x]

[Out]

(x^2*(1 + 4*E^x*(-5 + Log[2])) - x*(3 + 12*E^x*(-5 + Log[2]) + Log[16]) + Log[4096])/((x + 4*E^x*x*(-5 + Log[2
]) - 4*Log[2])*(x + 4*E^x*(-5 + Log[2]) - 4*Log[2]*Log[x]))

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fricas [A]  time = 0.64, size = 26, normalized size = 0.90 \begin {gather*} \frac {x - 3}{{\left (\log \relax (2) - 5\right )} e^{\left (x + 2 \, \log \relax (2)\right )} - 4 \, \log \relax (2) \log \relax (x) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(2)*log(x)+(5*x^2-20*x)*exp(x+2*log(2))+(-4*x^2+16*x)*log(2)*exp(x)+(4*x-12)*log(2)+3*x)/(1
6*x*log(2)^2*log(x)^2+(40*x*log(2)*exp(x+2*log(2))-32*x*log(2)^2*exp(x)-8*x^2*log(2))*log(x)+25*x*exp(x+2*log(
2))^2+(-40*x*log(2)*exp(x)-10*x^2)*exp(x+2*log(2))+16*x*log(2)^2*exp(x)^2+8*x^2*log(2)*exp(x)+x^3),x, algorith
m="fricas")

[Out]

(x - 3)/((log(2) - 5)*e^(x + 2*log(2)) - 4*log(2)*log(x) + x)

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giac [A]  time = 0.34, size = 24, normalized size = 0.83 \begin {gather*} \frac {x - 3}{4 \, e^{x} \log \relax (2) - 4 \, \log \relax (2) \log \relax (x) + x - 20 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(2)*log(x)+(5*x^2-20*x)*exp(x+2*log(2))+(-4*x^2+16*x)*log(2)*exp(x)+(4*x-12)*log(2)+3*x)/(1
6*x*log(2)^2*log(x)^2+(40*x*log(2)*exp(x+2*log(2))-32*x*log(2)^2*exp(x)-8*x^2*log(2))*log(x)+25*x*exp(x+2*log(
2))^2+(-40*x*log(2)*exp(x)-10*x^2)*exp(x+2*log(2))+16*x*log(2)^2*exp(x)^2+8*x^2*log(2)*exp(x)+x^3),x, algorith
m="giac")

[Out]

(x - 3)/(4*e^x*log(2) - 4*log(2)*log(x) + x - 20*e^x)

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maple [A]  time = 0.06, size = 25, normalized size = 0.86

method	result	size
risch	\(\frac {x -3}{4 \,{\mathrm e}^{x} \ln \relax (2)-4 \ln \relax (2) \ln \relax (x )+x -20 \,{\mathrm e}^{x}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x*ln(2)*ln(x)+(5*x^2-20*x)*exp(x+2*ln(2))+(-4*x^2+16*x)*ln(2)*exp(x)+(4*x-12)*ln(2)+3*x)/(16*x*ln(2)^2
*ln(x)^2+(40*x*ln(2)*exp(x+2*ln(2))-32*x*ln(2)^2*exp(x)-8*x^2*ln(2))*ln(x)+25*x*exp(x+2*ln(2))^2+(-40*x*ln(2)*
exp(x)-10*x^2)*exp(x+2*ln(2))+16*x*ln(2)^2*exp(x)^2+8*x^2*ln(2)*exp(x)+x^3),x,method=_RETURNVERBOSE)

[Out]

(x-3)/(4*exp(x)*ln(2)-4*ln(2)*ln(x)+x-20*exp(x))

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maxima [A]  time = 0.52, size = 22, normalized size = 0.76 \begin {gather*} \frac {x - 3}{4 \, {\left (\log \relax (2) - 5\right )} e^{x} - 4 \, \log \relax (2) \log \relax (x) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(2)*log(x)+(5*x^2-20*x)*exp(x+2*log(2))+(-4*x^2+16*x)*log(2)*exp(x)+(4*x-12)*log(2)+3*x)/(1
6*x*log(2)^2*log(x)^2+(40*x*log(2)*exp(x+2*log(2))-32*x*log(2)^2*exp(x)-8*x^2*log(2))*log(x)+25*x*exp(x+2*log(
2))^2+(-40*x*log(2)*exp(x)-10*x^2)*exp(x+2*log(2))+16*x*log(2)^2*exp(x)^2+8*x^2*log(2)*exp(x)+x^3),x, algorith
m="maxima")

[Out]

(x - 3)/(4*(log(2) - 5)*e^x - 4*log(2)*log(x) + x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {3\,x+\ln \relax (2)\,\left (4\,x-12\right )-{\mathrm {e}}^{x+2\,\ln \relax (2)}\,\left (20\,x-5\,x^2\right )+{\mathrm {e}}^x\,\ln \relax (2)\,\left (16\,x-4\,x^2\right )-4\,x\,\ln \relax (2)\,\ln \relax (x)}{25\,x\,{\mathrm {e}}^{2\,x+4\,\ln \relax (2)}-{\mathrm {e}}^{x+2\,\ln \relax (2)}\,\left (10\,x^2+40\,x\,{\mathrm {e}}^x\,\ln \relax (2)\right )-\ln \relax (x)\,\left (8\,x^2\,\ln \relax (2)-40\,x\,{\mathrm {e}}^{x+2\,\ln \relax (2)}\,\ln \relax (2)+32\,x\,{\mathrm {e}}^x\,{\ln \relax (2)}^2\right )+x^3+8\,x^2\,{\mathrm {e}}^x\,\ln \relax (2)+16\,x\,{\mathrm {e}}^{2\,x}\,{\ln \relax (2)}^2+16\,x\,{\ln \relax (2)}^2\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + log(2)*(4*x - 12) - exp(x + 2*log(2))*(20*x - 5*x^2) + exp(x)*log(2)*(16*x - 4*x^2) - 4*x*log(2)*lo
g(x))/(25*x*exp(2*x + 4*log(2)) - exp(x + 2*log(2))*(10*x^2 + 40*x*exp(x)*log(2)) - log(x)*(8*x^2*log(2) - 40*
x*exp(x + 2*log(2))*log(2) + 32*x*exp(x)*log(2)^2) + x^3 + 8*x^2*exp(x)*log(2) + 16*x*exp(2*x)*log(2)^2 + 16*x
*log(2)^2*log(x)^2),x)

[Out]

int((3*x + log(2)*(4*x - 12) - exp(x + 2*log(2))*(20*x - 5*x^2) + exp(x)*log(2)*(16*x - 4*x^2) - 4*x*log(2)*lo
g(x))/(25*x*exp(2*x + 4*log(2)) - exp(x + 2*log(2))*(10*x^2 + 40*x*exp(x)*log(2)) - log(x)*(8*x^2*log(2) - 40*
x*exp(x + 2*log(2))*log(2) + 32*x*exp(x)*log(2)^2) + x^3 + 8*x^2*exp(x)*log(2) + 16*x*exp(2*x)*log(2)^2 + 16*x
*log(2)^2*log(x)^2), x)

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sympy [A]  time = 0.44, size = 22, normalized size = 0.76 \begin {gather*} \frac {x - 3}{x + \left (-20 + 4 \log {\relax (2 )}\right ) e^{x} - 4 \log {\relax (2 )} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*ln(2)*ln(x)+(5*x**2-20*x)*exp(x+2*ln(2))+(-4*x**2+16*x)*ln(2)*exp(x)+(4*x-12)*ln(2)+3*x)/(16*x
*ln(2)**2*ln(x)**2+(40*x*ln(2)*exp(x+2*ln(2))-32*x*ln(2)**2*exp(x)-8*x**2*ln(2))*ln(x)+25*x*exp(x+2*ln(2))**2+
(-40*x*ln(2)*exp(x)-10*x**2)*exp(x+2*ln(2))+16*x*ln(2)**2*exp(x)**2+8*x**2*ln(2)*exp(x)+x**3),x)

[Out]

(x - 3)/(x + (-20 + 4*log(2))*exp(x) - 4*log(2)*log(x))

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