3.59.38 \(\int \frac {e^x (16-16 x-8 x^2) \log (4)+(16+8 x^2) \log ^2(4)-16 \log ^3(4)}{e^{2 x} (20+20 x+5 x^2)+(20-20 x-15 x^2+10 x^3+5 x^4) \log ^2(4)+(-40+30 x^2+10 x^3) \log ^3(4)+(20+20 x+5 x^2) \log ^4(4)+e^x ((40-30 x^2-10 x^3) \log (4)+(-40-40 x-10 x^2) \log ^2(4))} \, dx\) [5838]

Optimal. Leaf size=29 \[ \frac {8 x}{5 (2+x) \left (1-x+\frac {e^x}{\log (4)}-\log (4)\right )} \]

[Out]

8/5*x/(1/2*exp(x)/ln(2)-x+1-2*ln(2))/(2+x)

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Rubi [F]
time = 1.60, 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 {e^x \left (16-16 x-8 x^2\right ) \log (4)+\left (16+8 x^2\right ) \log ^2(4)-16 \log ^3(4)}{e^{2 x} \left (20+20 x+5 x^2\right )+\left (20-20 x-15 x^2+10 x^3+5 x^4\right ) \log ^2(4)+\left (-40+30 x^2+10 x^3\right ) \log ^3(4)+\left (20+20 x+5 x^2\right ) \log ^4(4)+e^x \left (\left (40-30 x^2-10 x^3\right ) \log (4)+\left (-40-40 x-10 x^2\right ) \log ^2(4)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(16 - 16*x - 8*x^2)*Log[4] + (16 + 8*x^2)*Log[4]^2 - 16*Log[4]^3)/(E^(2*x)*(20 + 20*x + 5*x^2) + (20
- 20*x - 15*x^2 + 10*x^3 + 5*x^4)*Log[4]^2 + (-40 + 30*x^2 + 10*x^3)*Log[4]^3 + (20 + 20*x + 5*x^2)*Log[4]^4 +
E^x*((40 - 30*x^2 - 10*x^3)*Log[4] + (-40 - 40*x - 10*x^2)*Log[4]^2)),x]

[Out]

(8*(4 - Log[4])*Log[4]^2*Defer[Int][(E^x - x*Log[4] + (1 - Log[4])*Log[4])^(-2), x])/5 - (8*Log[4]^2*Defer[Int
][x/(E^x - x*Log[4] + (1 - Log[4])*Log[4])^2, x])/5 - (16*(4 - Log[4])*Log[4]^2*Defer[Int][1/((2 + x)*(E^x - x
*Log[4] + (1 - Log[4])*Log[4])^2), x])/5 + (16*Log[4]*Defer[Int][1/((2 + x)^2*(E^x - x*Log[4] + (1 - Log[4])*L
og[4])), x])/5 + (16*Log[4]*Defer[Int][1/((2 + x)*(E^x - x*Log[4] + (1 - Log[4])*Log[4])), x])/5 + (8*Log[4]*D
efer[Int][(-E^x + Log[4]*(-1 + x + Log[4]))^(-1), x])/5

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 56, normalized size = 1.93 \begin {gather*} -\frac {8 x \left (x^2 \log (4)+2 \log ^2(4)+x \log ^2(4)-\log (256)\right )}{5 (2+x)^2 (-2+x+\log (4)) \left (-e^x+\log (4) (-1+x+\log (4))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(16 - 16*x - 8*x^2)*Log[4] + (16 + 8*x^2)*Log[4]^2 - 16*Log[4]^3)/(E^(2*x)*(20 + 20*x + 5*x^2)
+ (20 - 20*x - 15*x^2 + 10*x^3 + 5*x^4)*Log[4]^2 + (-40 + 30*x^2 + 10*x^3)*Log[4]^3 + (20 + 20*x + 5*x^2)*Log[
4]^4 + E^x*((40 - 30*x^2 - 10*x^3)*Log[4] + (-40 - 40*x - 10*x^2)*Log[4]^2)),x]

[Out]

(-8*x*(x^2*Log[4] + 2*Log[4]^2 + x*Log[4]^2 - Log[256]))/(5*(2 + x)^2*(-2 + x + Log[4])*(-E^x + Log[4]*(-1 + x
+ Log[4])))

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Maple [A]
time = 0.73, size = 33, normalized size = 1.14

method	result	size
norman	\(-\frac {16 x \ln \left (2\right )}{5 \left (2+x \right ) \left (4 \ln \left (2\right )^{2}+2 x \ln \left (2\right )-2 \ln \left (2\right )-{\mathrm e}^{x}\right )}\)	\(33\)
risch	\(-\frac {16 x \ln \left (2\right )}{5 \left (2+x \right ) \left (4 \ln \left (2\right )^{2}+2 x \ln \left (2\right )-2 \ln \left (2\right )-{\mathrm e}^{x}\right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-8*x^2-16*x+16)*ln(2)*exp(x)-128*ln(2)^3+4*(8*x^2+16)*ln(2)^2)/((5*x^2+20*x+20)*exp(x)^2+(4*(-10*x^2-4
0*x-40)*ln(2)^2+2*(-10*x^3-30*x^2+40)*ln(2))*exp(x)+16*(5*x^2+20*x+20)*ln(2)^4+8*(10*x^3+30*x^2-40)*ln(2)^3+4*
(5*x^4+10*x^3-15*x^2-20*x+20)*ln(2)^2),x,method=_RETURNVERBOSE)

[Out]

-16/5*x*ln(2)/(2+x)/(4*ln(2)^2+2*x*ln(2)-2*ln(2)-exp(x))

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Maxima [A]
time = 0.67, size = 44, normalized size = 1.52 \begin {gather*} -\frac {16 \, x \log \left (2\right )}{5 \, {\left (2 \, x^{2} \log \left (2\right ) + 2 \, {\left (2 \, \log \left (2\right )^{2} + \log \left (2\right )\right )} x - {\left (x + 2\right )} e^{x} + 8 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-8*x^2-16*x+16)*log(2)*exp(x)-128*log(2)^3+4*(8*x^2+16)*log(2)^2)/((5*x^2+20*x+20)*exp(x)^2+(4*(
-10*x^2-40*x-40)*log(2)^2+2*(-10*x^3-30*x^2+40)*log(2))*exp(x)+16*(5*x^2+20*x+20)*log(2)^4+8*(10*x^3+30*x^2-40
)*log(2)^3+4*(5*x^4+10*x^3-15*x^2-20*x+20)*log(2)^2),x, algorithm="maxima")

[Out]

-16/5*x*log(2)/(2*x^2*log(2) + 2*(2*log(2)^2 + log(2))*x - (x + 2)*e^x + 8*log(2)^2 - 4*log(2))

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Fricas [A]
time = 0.38, size = 34, normalized size = 1.17 \begin {gather*} -\frac {16 \, x \log \left (2\right )}{5 \, {\left (4 \, {\left (x + 2\right )} \log \left (2\right )^{2} - {\left (x + 2\right )} e^{x} + 2 \, {\left (x^{2} + x - 2\right )} \log \left (2\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-8*x^2-16*x+16)*log(2)*exp(x)-128*log(2)^3+4*(8*x^2+16)*log(2)^2)/((5*x^2+20*x+20)*exp(x)^2+(4*(
-10*x^2-40*x-40)*log(2)^2+2*(-10*x^3-30*x^2+40)*log(2))*exp(x)+16*(5*x^2+20*x+20)*log(2)^4+8*(10*x^3+30*x^2-40
)*log(2)^3+4*(5*x^4+10*x^3-15*x^2-20*x+20)*log(2)^2),x, algorithm="fricas")

[Out]

-16/5*x*log(2)/(4*(x + 2)*log(2)^2 - (x + 2)*e^x + 2*(x^2 + x - 2)*log(2))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).
time = 0.08, size = 49, normalized size = 1.69 \begin {gather*} \frac {16 x \log {\left (2 \right )}}{- 10 x^{2} \log {\left (2 \right )} - 20 x \log {\left (2 \right )}^{2} - 10 x \log {\left (2 \right )} + \left (5 x + 10\right ) e^{x} - 40 \log {\left (2 \right )}^{2} + 20 \log {\left (2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-8*x**2-16*x+16)*ln(2)*exp(x)-128*ln(2)**3+4*(8*x**2+16)*ln(2)**2)/((5*x**2+20*x+20)*exp(x)**2+(
4*(-10*x**2-40*x-40)*ln(2)**2+2*(-10*x**3-30*x**2+40)*ln(2))*exp(x)+16*(5*x**2+20*x+20)*ln(2)**4+8*(10*x**3+30
*x**2-40)*ln(2)**3+4*(5*x**4+10*x**3-15*x**2-20*x+20)*ln(2)**2),x)

[Out]

16*x*log(2)/(-10*x**2*log(2) - 20*x*log(2)**2 - 10*x*log(2) + (5*x + 10)*exp(x) - 40*log(2)**2 + 20*log(2))

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Giac [A]
time = 0.41, size = 46, normalized size = 1.59 \begin {gather*} -\frac {16 \, x \log \left (2\right )}{5 \, {\left (2 \, x^{2} \log \left (2\right ) + 4 \, x \log \left (2\right )^{2} - x e^{x} + 2 \, x \log \left (2\right ) + 8 \, \log \left (2\right )^{2} - 2 \, e^{x} - 4 \, \log \left (2\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-8*x^2-16*x+16)*log(2)*exp(x)-128*log(2)^3+4*(8*x^2+16)*log(2)^2)/((5*x^2+20*x+20)*exp(x)^2+(4*(
-10*x^2-40*x-40)*log(2)^2+2*(-10*x^3-30*x^2+40)*log(2))*exp(x)+16*(5*x^2+20*x+20)*log(2)^4+8*(10*x^3+30*x^2-40
)*log(2)^3+4*(5*x^4+10*x^3-15*x^2-20*x+20)*log(2)^2),x, algorithm="giac")

[Out]

-16/5*x*log(2)/(2*x^2*log(2) + 4*x*log(2)^2 - x*e^x + 2*x*log(2) + 8*log(2)^2 - 2*e^x - 4*log(2))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {128\,{\ln \left (2\right )}^3-4\,{\ln \left (2\right )}^2\,\left (8\,x^2+16\right )+2\,{\mathrm {e}}^x\,\ln \left (2\right )\,\left (8\,x^2+16\,x-16\right )}{4\,{\ln \left (2\right )}^2\,\left (5\,x^4+10\,x^3-15\,x^2-20\,x+20\right )+{\mathrm {e}}^{2\,x}\,\left (5\,x^2+20\,x+20\right )+16\,{\ln \left (2\right )}^4\,\left (5\,x^2+20\,x+20\right )-{\mathrm {e}}^x\,\left (2\,\ln \left (2\right )\,\left (10\,x^3+30\,x^2-40\right )+4\,{\ln \left (2\right )}^2\,\left (10\,x^2+40\,x+40\right )\right )+8\,{\ln \left (2\right )}^3\,\left (10\,x^3+30\,x^2-40\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(128*log(2)^3 - 4*log(2)^2*(8*x^2 + 16) + 2*exp(x)*log(2)*(16*x + 8*x^2 - 16))/(4*log(2)^2*(10*x^3 - 15*x
^2 - 20*x + 5*x^4 + 20) + exp(2*x)*(20*x + 5*x^2 + 20) + 16*log(2)^4*(20*x + 5*x^2 + 20) - exp(x)*(2*log(2)*(3
0*x^2 + 10*x^3 - 40) + 4*log(2)^2*(40*x + 10*x^2 + 40)) + 8*log(2)^3*(30*x^2 + 10*x^3 - 40)),x)

[Out]

-int((128*log(2)^3 - 4*log(2)^2*(8*x^2 + 16) + 2*exp(x)*log(2)*(16*x + 8*x^2 - 16))/(4*log(2)^2*(10*x^3 - 15*x
^2 - 20*x + 5*x^4 + 20) + exp(2*x)*(20*x + 5*x^2 + 20) + 16*log(2)^4*(20*x + 5*x^2 + 20) - exp(x)*(2*log(2)*(3
0*x^2 + 10*x^3 - 40) + 4*log(2)^2*(40*x + 10*x^2 + 40)) + 8*log(2)^3*(30*x^2 + 10*x^3 - 40)), x)

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