3.103.72 \(\int \frac {6 x^3-4 x^2 \log (2)+e^x (-16 x^5+(-16 x^3+32 x^4) \log (2)+(20 x^2-24 x^3) \log ^2(2)+(-8 x+8 x^2) \log ^3(2)+(1-x) \log ^4(2))+e^{2 x} (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2))}{x^2+e^x (8 x^3-8 x^2 \log (2)+2 x \log ^2(2))+e^{2 x} (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2))} \, dx\)

Optimal. Leaf size=26 \[ -x^2+\frac {x}{e^x+\frac {x}{(2 x-\log (2))^2}} \]

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Rubi [F]  time = 3.17, 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 {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(6*x^3 - 4*x^2*Log[2] + E^x*(-16*x^5 + (-16*x^3 + 32*x^4)*Log[2] + (20*x^2 - 24*x^3)*Log[2]^2 + (-8*x + 8*
x^2)*Log[2]^3 + (1 - x)*Log[2]^4) + E^(2*x)*(-32*x^5 + 64*x^4*Log[2] - 48*x^3*Log[2]^2 + 16*x^2*Log[2]^3 - 2*x
*Log[2]^4))/(x^2 + E^x*(8*x^3 - 8*x^2*Log[2] + 2*x*Log[2]^2) + E^(2*x)*(16*x^4 - 32*x^3*Log[2] + 24*x^2*Log[2]
^2 - 8*x*Log[2]^3 + Log[2]^4)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2} \, dx\\ &=\int \left (-2 x-\frac {(-1+x) (2 x-\log (2))^2}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)}+\frac {x \left (4 x^3+4 x^2 (1-\log (2))-\log ^2(2)+x \log ^2(2)\right )}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2}\right ) \, dx\\ &=-x^2-\int \frac {(-1+x) (2 x-\log (2))^2}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)} \, dx+\int \frac {x \left (4 x^3+4 x^2 (1-\log (2))-\log ^2(2)+x \log ^2(2)\right )}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2} \, dx\\ &=-x^2+\int \left (\frac {4 x^4}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2}-\frac {4 x^3 (-1+\log (2))}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2}-\frac {x \log ^2(2)}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2}+\frac {x^2 \log ^2(2)}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2}\right ) \, dx-\int \left (\frac {4 x^3}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)}-\frac {\log ^2(2)}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)}-\frac {4 x^2 (1+\log (2))}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)}+\frac {x \log (2) (4+\log (2))}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)}\right ) \, dx\\ &=-x^2+4 \int \frac {x^4}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2} \, dx-4 \int \frac {x^3}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)} \, dx-(4 (-1+\log (2))) \int \frac {x^3}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2} \, dx-\log ^2(2) \int \frac {x}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2} \, dx+\log ^2(2) \int \frac {x^2}{\left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )^2} \, dx+\log ^2(2) \int \frac {1}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)} \, dx+(4 (1+\log (2))) \int \frac {x^2}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)} \, dx-(\log (2) (4+\log (2))) \int \frac {x}{x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.44, size = 154, normalized size = 5.92 \begin {gather*} x \left (-x+\frac {64 x^7-\log ^6(2)+x \log ^5(2) (8+\log (2))-64 x^6 (-1+\log (8))-x^2 \log ^3(2) \left (4 \log ^2(2)+\log (4)+2 \log (2) (9+\log (16))\right )-16 x^4 \log ^2(2) (-5+\log (1024))+16 x^5 \log (2) (-8+\log (32768))+4 x^3 \log ^3(2) \log (32768)}{(2 x-\log (2))^3 \left (2 x^2-x (-2+\log (2))+\log (2)\right ) \left (x+4 e^x x^2-4 e^x x \log (2)+e^x \log ^2(2)\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*x^3 - 4*x^2*Log[2] + E^x*(-16*x^5 + (-16*x^3 + 32*x^4)*Log[2] + (20*x^2 - 24*x^3)*Log[2]^2 + (-8*
x + 8*x^2)*Log[2]^3 + (1 - x)*Log[2]^4) + E^(2*x)*(-32*x^5 + 64*x^4*Log[2] - 48*x^3*Log[2]^2 + 16*x^2*Log[2]^3
- 2*x*Log[2]^4))/(x^2 + E^x*(8*x^3 - 8*x^2*Log[2] + 2*x*Log[2]^2) + E^(2*x)*(16*x^4 - 32*x^3*Log[2] + 24*x^2*
Log[2]^2 - 8*x*Log[2]^3 + Log[2]^4)),x]

[Out]

x*(-x + (64*x^7 - Log[2]^6 + x*Log[2]^5*(8 + Log[2]) - 64*x^6*(-1 + Log[8]) - x^2*Log[2]^3*(4*Log[2]^2 + Log[4
] + 2*Log[2]*(9 + Log[16])) - 16*x^4*Log[2]^2*(-5 + Log[1024]) + 16*x^5*Log[2]*(-8 + Log[32768]) + 4*x^3*Log[2
]^3*Log[32768])/((2*x - Log[2])^3*(2*x^2 - x*(-2 + Log[2]) + Log[2])*(x + 4*E^x*x^2 - 4*E^x*x*Log[2] + E^x*Log
[2]^2)))

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fricas [B]  time = 0.63, size = 67, normalized size = 2.58 \begin {gather*} \frac {3 \, x^{3} - 4 \, x^{2} \log \relax (2) + x \log \relax (2)^{2} - {\left (4 \, x^{4} - 4 \, x^{3} \log \relax (2) + x^{2} \log \relax (2)^{2}\right )} e^{x}}{{\left (4 \, x^{2} - 4 \, x \log \relax (2) + \log \relax (2)^{2}\right )} e^{x} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(2)^4+16*x^2*log(2)^3-48*x^3*log(2)^2+64*x^4*log(2)-32*x^5)*exp(x)^2+((-x+1)*log(2)^4+(8*x
^2-8*x)*log(2)^3+(-24*x^3+20*x^2)*log(2)^2+(32*x^4-16*x^3)*log(2)-16*x^5)*exp(x)-4*x^2*log(2)+6*x^3)/((log(2)^
4-8*x*log(2)^3+24*x^2*log(2)^2-32*x^3*log(2)+16*x^4)*exp(x)^2+(2*x*log(2)^2-8*x^2*log(2)+8*x^3)*exp(x)+x^2),x,
algorithm="fricas")

[Out]

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

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giac [B]  time = 0.20, size = 73, normalized size = 2.81 \begin {gather*} -\frac {4 \, x^{4} e^{x} - 4 \, x^{3} e^{x} \log \relax (2) + x^{2} e^{x} \log \relax (2)^{2} - 3 \, x^{3} + 4 \, x^{2} \log \relax (2) - x \log \relax (2)^{2}}{4 \, x^{2} e^{x} - 4 \, x e^{x} \log \relax (2) + e^{x} \log \relax (2)^{2} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(2)^4+16*x^2*log(2)^3-48*x^3*log(2)^2+64*x^4*log(2)-32*x^5)*exp(x)^2+((-x+1)*log(2)^4+(8*x
^2-8*x)*log(2)^3+(-24*x^3+20*x^2)*log(2)^2+(32*x^4-16*x^3)*log(2)-16*x^5)*exp(x)-4*x^2*log(2)+6*x^3)/((log(2)^
4-8*x*log(2)^3+24*x^2*log(2)^2-32*x^3*log(2)+16*x^4)*exp(x)^2+(2*x*log(2)^2-8*x^2*log(2)+8*x^3)*exp(x)+x^2),x,
algorithm="giac")

[Out]

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

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maple [A]  time = 0.24, size = 42, normalized size = 1.62

method	result	size
risch	\(-x^{2}+\frac {x \left (\ln \relax (2)-2 x \right )^{2}}{\ln \relax (2)^{2} {\mathrm e}^{x}-4 x \ln \relax (2) {\mathrm e}^{x}+4 \,{\mathrm e}^{x} x^{2}+x}\)	\(42\)
norman	\(\frac {\frac {5 x \ln \relax (2)^{2}}{4}+\frac {\ln \relax (2)^{4} {\mathrm e}^{x}}{4}-\ln \relax (2)^{3} {\mathrm e}^{x} x +3 x^{3}-4 x^{2} \ln \relax (2)-4 \,{\mathrm e}^{x} x^{4}+4 x^{3} \ln \relax (2) {\mathrm e}^{x}}{\ln \relax (2)^{2} {\mathrm e}^{x}-4 x \ln \relax (2) {\mathrm e}^{x}+4 \,{\mathrm e}^{x} x^{2}+x}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*ln(2)^4+16*x^2*ln(2)^3-48*x^3*ln(2)^2+64*x^4*ln(2)-32*x^5)*exp(x)^2+((1-x)*ln(2)^4+(8*x^2-8*x)*ln(2
)^3+(-24*x^3+20*x^2)*ln(2)^2+(32*x^4-16*x^3)*ln(2)-16*x^5)*exp(x)-4*x^2*ln(2)+6*x^3)/((ln(2)^4-8*x*ln(2)^3+24*
x^2*ln(2)^2-32*x^3*ln(2)+16*x^4)*exp(x)^2+(2*x*ln(2)^2-8*x^2*ln(2)+8*x^3)*exp(x)+x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.53, size = 67, normalized size = 2.58 \begin {gather*} \frac {3 \, x^{3} - 4 \, x^{2} \log \relax (2) + x \log \relax (2)^{2} - {\left (4 \, x^{4} - 4 \, x^{3} \log \relax (2) + x^{2} \log \relax (2)^{2}\right )} e^{x}}{{\left (4 \, x^{2} - 4 \, x \log \relax (2) + \log \relax (2)^{2}\right )} e^{x} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(2)^4+16*x^2*log(2)^3-48*x^3*log(2)^2+64*x^4*log(2)-32*x^5)*exp(x)^2+((-x+1)*log(2)^4+(8*x
^2-8*x)*log(2)^3+(-24*x^3+20*x^2)*log(2)^2+(32*x^4-16*x^3)*log(2)-16*x^5)*exp(x)-4*x^2*log(2)+6*x^3)/((log(2)^
4-8*x*log(2)^3+24*x^2*log(2)^2-32*x^3*log(2)+16*x^4)*exp(x)^2+(2*x*log(2)^2-8*x^2*log(2)+8*x^3)*exp(x)+x^2),x,
algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^x\,\left ({\ln \relax (2)}^4\,\left (x-1\right )+{\ln \relax (2)}^3\,\left (8\,x-8\,x^2\right )+\ln \relax (2)\,\left (16\,x^3-32\,x^4\right )+16\,x^5-{\ln \relax (2)}^2\,\left (20\,x^2-24\,x^3\right )\right )+{\mathrm {e}}^{2\,x}\,\left (32\,x^5-64\,\ln \relax (2)\,x^4+48\,{\ln \relax (2)}^2\,x^3-16\,{\ln \relax (2)}^3\,x^2+2\,{\ln \relax (2)}^4\,x\right )+4\,x^2\,\ln \relax (2)-6\,x^3}{{\mathrm {e}}^x\,\left (8\,x^3-8\,\ln \relax (2)\,x^2+2\,{\ln \relax (2)}^2\,x\right )+{\mathrm {e}}^{2\,x}\,\left (16\,x^4-32\,\ln \relax (2)\,x^3+24\,{\ln \relax (2)}^2\,x^2-8\,{\ln \relax (2)}^3\,x+{\ln \relax (2)}^4\right )+x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(log(2)^4*(x - 1) + log(2)^3*(8*x - 8*x^2) + log(2)*(16*x^3 - 32*x^4) + 16*x^5 - log(2)^2*(20*x^2
- 24*x^3)) + exp(2*x)*(48*x^3*log(2)^2 - 16*x^2*log(2)^3 + 2*x*log(2)^4 - 64*x^4*log(2) + 32*x^5) + 4*x^2*log
(2) - 6*x^3)/(exp(x)*(2*x*log(2)^2 - 8*x^2*log(2) + 8*x^3) + exp(2*x)*(24*x^2*log(2)^2 - 8*x*log(2)^3 - 32*x^3
*log(2) + log(2)^4 + 16*x^4) + x^2),x)

[Out]

int(-(exp(x)*(log(2)^4*(x - 1) + log(2)^3*(8*x - 8*x^2) + log(2)*(16*x^3 - 32*x^4) + 16*x^5 - log(2)^2*(20*x^2
- 24*x^3)) + exp(2*x)*(48*x^3*log(2)^2 - 16*x^2*log(2)^3 + 2*x*log(2)^4 - 64*x^4*log(2) + 32*x^5) + 4*x^2*log
(2) - 6*x^3)/(exp(x)*(2*x*log(2)^2 - 8*x^2*log(2) + 8*x^3) + exp(2*x)*(24*x^2*log(2)^2 - 8*x*log(2)^3 - 32*x^3
*log(2) + log(2)^4 + 16*x^4) + x^2), x)

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sympy [A]  time = 0.40, size = 44, normalized size = 1.69 \begin {gather*} - x^{2} + \frac {4 x^{3} - 4 x^{2} \log {\relax (2 )} + x \log {\relax (2 )}^{2}}{x + \left (4 x^{2} - 4 x \log {\relax (2 )} + \log {\relax (2 )}^{2}\right ) e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*ln(2)**4+16*x**2*ln(2)**3-48*x**3*ln(2)**2+64*x**4*ln(2)-32*x**5)*exp(x)**2+((-x+1)*ln(2)**4+
(8*x**2-8*x)*ln(2)**3+(-24*x**3+20*x**2)*ln(2)**2+(32*x**4-16*x**3)*ln(2)-16*x**5)*exp(x)-4*x**2*ln(2)+6*x**3)
/((ln(2)**4-8*x*ln(2)**3+24*x**2*ln(2)**2-32*x**3*ln(2)+16*x**4)*exp(x)**2+(2*x*ln(2)**2-8*x**2*ln(2)+8*x**3)*
exp(x)+x**2),x)

[Out]

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

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