3.47.49 \(\int (20+2 x+3 x^2-8 x^3-5 x^4+(8-16 x-24 x^2) \log (2)-16 \log ^2(2)+e^{2 x} (-1-2 x) \log ^2(2)+e^x ((-2+2 x+8 x^2+2 x^3) \log (2)+(8+8 x) \log ^2(2))) \, dx\)

Optimal. Leaf size=25 \[ x-x \left (-20+x+\left (-1+x+x^2-\left (-4+e^x\right ) \log (2)\right )^2\right ) \]

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Rubi [B]  time = 0.16, antiderivative size = 127, normalized size of antiderivative = 5.08, number of steps used = 20, number of rules used = 3, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2176, 2194, 2196} \begin {gather*} -x^5-2 x^4+x^3+2 e^x x^3 \log (2)-8 x^3 \log (2)+x^2+2 e^x x^2 \log (2)-8 x^2 \log (2)+4 x \left (5-4 \log ^2(2)\right )-8 e^x \log ^2(2)+\frac {1}{2} e^{2 x} \log ^2(2)+8 e^x (x+1) \log ^2(2)-\frac {1}{2} e^{2 x} (2 x+1) \log ^2(2)-2 e^x x \log (2)+8 x \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^2 + x^3 - 2*x^4 - x^5 + 8*x*Log[2] - 2*E^x*x*Log[2] - 8*x^2*Log[2] + 2*E^x*x^2*Log[2] - 8*x^3*Log[2] + 2*E^x
*x^3*Log[2] - 8*E^x*Log[2]^2 + (E^(2*x)*Log[2]^2)/2 + 8*E^x*(1 + x)*Log[2]^2 - (E^(2*x)*(1 + 2*x)*Log[2]^2)/2
+ 4*x*(5 - 4*Log[2]^2)

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x^2+x^3-2 x^4-x^5+4 x \left (5-4 \log ^2(2)\right )+\log (2) \int \left (8-16 x-24 x^2\right ) \, dx+\log ^2(2) \int e^{2 x} (-1-2 x) \, dx+\int e^x \left (\left (-2+2 x+8 x^2+2 x^3\right ) \log (2)+(8+8 x) \log ^2(2)\right ) \, dx\\ &=x^2+x^3-2 x^4-x^5+8 x \log (2)-8 x^2 \log (2)-8 x^3 \log (2)-\frac {1}{2} e^{2 x} (1+2 x) \log ^2(2)+4 x \left (5-4 \log ^2(2)\right )+\log ^2(2) \int e^{2 x} \, dx+\int \left (2 e^x \left (-1+x+4 x^2+x^3\right ) \log (2)+8 e^x (1+x) \log ^2(2)\right ) \, dx\\ &=x^2+x^3-2 x^4-x^5+8 x \log (2)-8 x^2 \log (2)-8 x^3 \log (2)+\frac {1}{2} e^{2 x} \log ^2(2)-\frac {1}{2} e^{2 x} (1+2 x) \log ^2(2)+4 x \left (5-4 \log ^2(2)\right )+(2 \log (2)) \int e^x \left (-1+x+4 x^2+x^3\right ) \, dx+\left (8 \log ^2(2)\right ) \int e^x (1+x) \, dx\\ &=x^2+x^3-2 x^4-x^5+8 x \log (2)-8 x^2 \log (2)-8 x^3 \log (2)+\frac {1}{2} e^{2 x} \log ^2(2)+8 e^x (1+x) \log ^2(2)-\frac {1}{2} e^{2 x} (1+2 x) \log ^2(2)+4 x \left (5-4 \log ^2(2)\right )+(2 \log (2)) \int \left (-e^x+e^x x+4 e^x x^2+e^x x^3\right ) \, dx-\left (8 \log ^2(2)\right ) \int e^x \, dx\\ &=x^2+x^3-2 x^4-x^5+8 x \log (2)-8 x^2 \log (2)-8 x^3 \log (2)-8 e^x \log ^2(2)+\frac {1}{2} e^{2 x} \log ^2(2)+8 e^x (1+x) \log ^2(2)-\frac {1}{2} e^{2 x} (1+2 x) \log ^2(2)+4 x \left (5-4 \log ^2(2)\right )-(2 \log (2)) \int e^x \, dx+(2 \log (2)) \int e^x x \, dx+(2 \log (2)) \int e^x x^3 \, dx+(8 \log (2)) \int e^x x^2 \, dx\\ &=x^2+x^3-2 x^4-x^5-2 e^x \log (2)+8 x \log (2)+2 e^x x \log (2)-8 x^2 \log (2)+8 e^x x^2 \log (2)-8 x^3 \log (2)+2 e^x x^3 \log (2)-8 e^x \log ^2(2)+\frac {1}{2} e^{2 x} \log ^2(2)+8 e^x (1+x) \log ^2(2)-\frac {1}{2} e^{2 x} (1+2 x) \log ^2(2)+4 x \left (5-4 \log ^2(2)\right )-(2 \log (2)) \int e^x \, dx-(6 \log (2)) \int e^x x^2 \, dx-(16 \log (2)) \int e^x x \, dx\\ &=x^2+x^3-2 x^4-x^5-4 e^x \log (2)+8 x \log (2)-14 e^x x \log (2)-8 x^2 \log (2)+2 e^x x^2 \log (2)-8 x^3 \log (2)+2 e^x x^3 \log (2)-8 e^x \log ^2(2)+\frac {1}{2} e^{2 x} \log ^2(2)+8 e^x (1+x) \log ^2(2)-\frac {1}{2} e^{2 x} (1+2 x) \log ^2(2)+4 x \left (5-4 \log ^2(2)\right )+(12 \log (2)) \int e^x x \, dx+(16 \log (2)) \int e^x \, dx\\ &=x^2+x^3-2 x^4-x^5+12 e^x \log (2)+8 x \log (2)-2 e^x x \log (2)-8 x^2 \log (2)+2 e^x x^2 \log (2)-8 x^3 \log (2)+2 e^x x^3 \log (2)-8 e^x \log ^2(2)+\frac {1}{2} e^{2 x} \log ^2(2)+8 e^x (1+x) \log ^2(2)-\frac {1}{2} e^{2 x} (1+2 x) \log ^2(2)+4 x \left (5-4 \log ^2(2)\right )-(12 \log (2)) \int e^x \, dx\\ &=x^2+x^3-2 x^4-x^5+8 x \log (2)-2 e^x x \log (2)-8 x^2 \log (2)+2 e^x x^2 \log (2)-8 x^3 \log (2)+2 e^x x^3 \log (2)-8 e^x \log ^2(2)+\frac {1}{2} e^{2 x} \log ^2(2)+8 e^x (1+x) \log ^2(2)-\frac {1}{2} e^{2 x} (1+2 x) \log ^2(2)+4 x \left (5-4 \log ^2(2)\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 78, normalized size = 3.12 \begin {gather*} 20 x+x^2+x^3-2 x^4-x^5+8 x \log (2)-8 x^2 \log (2)-8 x^3 \log (2)-16 x \log ^2(2)-e^{2 x} x \log ^2(2)+2 e^x \log (2) \left (x^2+x^3+x (-1+\log (16))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

20*x + x^2 + x^3 - 2*x^4 - x^5 + 8*x*Log[2] - 8*x^2*Log[2] - 8*x^3*Log[2] - 16*x*Log[2]^2 - E^(2*x)*x*Log[2]^2
+ 2*E^x*Log[2]*(x^2 + x^3 + x*(-1 + Log[16]))

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fricas [B]  time = 0.69, size = 77, normalized size = 3.08 \begin {gather*} -x^{5} - 2 \, x^{4} - x e^{\left (2 \, x\right )} \log \relax (2)^{2} + x^{3} - 16 \, x \log \relax (2)^{2} + x^{2} + 2 \, {\left (4 \, x \log \relax (2)^{2} + {\left (x^{3} + x^{2} - x\right )} \log \relax (2)\right )} e^{x} - 8 \, {\left (x^{3} + x^{2} - x\right )} \log \relax (2) + 20 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^5 - 2*x^4 - x*e^(2*x)*log(2)^2 + x^3 - 16*x*log(2)^2 + x^2 + 2*(4*x*log(2)^2 + (x^3 + x^2 - x)*log(2))*e^x
- 8*(x^3 + x^2 - x)*log(2) + 20*x

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giac [B]  time = 0.13, size = 81, normalized size = 3.24 \begin {gather*} -x^{5} - 2 \, x^{4} - x e^{\left (2 \, x\right )} \log \relax (2)^{2} + x^{3} - 16 \, x \log \relax (2)^{2} + x^{2} + 2 \, {\left (x^{3} \log \relax (2) + x^{2} \log \relax (2) + 4 \, x \log \relax (2)^{2} - x \log \relax (2)\right )} e^{x} - 8 \, {\left (x^{3} + x^{2} - x\right )} \log \relax (2) + 20 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^5 - 2*x^4 - x*e^(2*x)*log(2)^2 + x^3 - 16*x*log(2)^2 + x^2 + 2*(x^3*log(2) + x^2*log(2) + 4*x*log(2)^2 - x*
log(2))*e^x - 8*(x^3 + x^2 - x)*log(2) + 20*x

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maple [B]  time = 0.04, size = 88, normalized size = 3.52

method	result	size
risch	\(-x \ln \relax (2)^{2} {\mathrm e}^{2 x}+\left (2 x^{3} \ln \relax (2)+8 x \ln \relax (2)^{2}+2 x^{2} \ln \relax (2)-2 x \ln \relax (2)\right ) {\mathrm e}^{x}-16 x \ln \relax (2)^{2}-8 x^{3} \ln \relax (2)-8 x^{2} \ln \relax (2)+8 x \ln \relax (2)-x^{5}-2 x^{4}+x^{3}+x^{2}+20 x\)	\(88\)
norman	\(\left (-8 \ln \relax (2)+1\right ) x^{2}+\left (-8 \ln \relax (2)+1\right ) x^{3}+\left (-16 \ln \relax (2)^{2}+8 \ln \relax (2)+20\right ) x +\left (8 \ln \relax (2)^{2}-2 \ln \relax (2)\right ) x \,{\mathrm e}^{x}-2 x^{4}-x^{5}-x \ln \relax (2)^{2} {\mathrm e}^{2 x}+2 x^{2} \ln \relax (2) {\mathrm e}^{x}+2 x^{3} \ln \relax (2) {\mathrm e}^{x}\)	\(90\)
default	\(20 x +2 x^{2} \ln \relax (2) {\mathrm e}^{x}-2 x \ln \relax (2) {\mathrm e}^{x}+2 x^{3} \ln \relax (2) {\mathrm e}^{x}+8 x \ln \relax (2)^{2} {\mathrm e}^{x}-8 x^{3} \ln \relax (2)-8 x^{2} \ln \relax (2)+8 x \ln \relax (2)-x \ln \relax (2)^{2} {\mathrm e}^{2 x}+x^{2}+x^{3}-2 x^{4}-x^{5}-16 x \ln \relax (2)^{2}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x-1)*ln(2)^2*exp(x)^2+((8*x+8)*ln(2)^2+(2*x^3+8*x^2+2*x-2)*ln(2))*exp(x)-16*ln(2)^2+(-24*x^2-16*x+8)*l
n(2)-5*x^4-8*x^3+3*x^2+2*x+20,x,method=_RETURNVERBOSE)

[Out]

-x*ln(2)^2*exp(2*x)+(2*x^3*ln(2)+8*x*ln(2)^2+2*x^2*ln(2)-2*x*ln(2))*exp(x)-16*x*ln(2)^2-8*x^3*ln(2)-8*x^2*ln(2
)+8*x*ln(2)-x^5-2*x^4+x^3+x^2+20*x

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maxima [B]  time = 0.45, size = 82, normalized size = 3.28 \begin {gather*} -x^{5} - 2 \, x^{4} - x e^{\left (2 \, x\right )} \log \relax (2)^{2} + x^{3} - 16 \, x \log \relax (2)^{2} + x^{2} + 2 \, {\left (x^{3} \log \relax (2) + x^{2} \log \relax (2) + {\left (4 \, \log \relax (2)^{2} - \log \relax (2)\right )} x\right )} e^{x} - 8 \, {\left (x^{3} + x^{2} - x\right )} \log \relax (2) + 20 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^5 - 2*x^4 - x*e^(2*x)*log(2)^2 + x^3 - 16*x*log(2)^2 + x^2 + 2*(x^3*log(2) + x^2*log(2) + (4*log(2)^2 - log
(2))*x)*e^x - 8*(x^3 + x^2 - x)*log(2) + 20*x

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mupad [B]  time = 0.11, size = 82, normalized size = 3.28 \begin {gather*} x\,\left (\ln \left (256\right )-16\,{\ln \relax (2)}^2+20\right )-x^3\,\left (\ln \left (256\right )-1\right )-x^2\,\left (\ln \left (256\right )-1\right )-2\,x^4-x^5+x^2\,{\mathrm {e}}^x\,\ln \relax (4)+x^3\,{\mathrm {e}}^x\,\ln \relax (4)-x\,{\mathrm {e}}^x\,\left (\ln \relax (4)-8\,{\ln \relax (2)}^2\right )-x\,{\mathrm {e}}^{2\,x}\,{\ln \relax (2)}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(log(256) - 16*log(2)^2 + 20) - x^3*(log(256) - 1) - x^2*(log(256) - 1) - 2*x^4 - x^5 + x^2*exp(x)*log(4) +
x^3*exp(x)*log(4) - x*exp(x)*(log(4) - 8*log(2)^2) - x*exp(2*x)*log(2)^2

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sympy [B]  time = 0.18, size = 90, normalized size = 3.60 \begin {gather*} - x^{5} - 2 x^{4} + x^{3} \left (1 - 8 \log {\relax (2 )}\right ) + x^{2} \left (1 - 8 \log {\relax (2 )}\right ) - x e^{2 x} \log {\relax (2 )}^{2} + x \left (- 16 \log {\relax (2 )}^{2} + 8 \log {\relax (2 )} + 20\right ) + \left (2 x^{3} \log {\relax (2 )} + 2 x^{2} \log {\relax (2 )} - 2 x \log {\relax (2 )} + 8 x \log {\relax (2 )}^{2}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-1)*ln(2)**2*exp(x)**2+((8*x+8)*ln(2)**2+(2*x**3+8*x**2+2*x-2)*ln(2))*exp(x)-16*ln(2)**2+(-24*x
**2-16*x+8)*ln(2)-5*x**4-8*x**3+3*x**2+2*x+20,x)

[Out]

-x**5 - 2*x**4 + x**3*(1 - 8*log(2)) + x**2*(1 - 8*log(2)) - x*exp(2*x)*log(2)**2 + x*(-16*log(2)**2 + 8*log(2
) + 20) + (2*x**3*log(2) + 2*x**2*log(2) - 2*x*log(2) + 8*x*log(2)**2)*exp(x)

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