3.11.25 \(\int (20-8 e^{2 x}-8 x-8 \log (\frac {4}{3})+e^x (-4+8 x+8 \log (\frac {4}{3}))) \, dx\)

Optimal. Leaf size=25 \[ 4 \left (e^x+x-\left (2+e^x-x-\log \left (\frac {4}{3}\right )\right )^2\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.80, number of steps used = 4, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2194, 2176} \begin {gather*} -4 x^2-8 e^x-4 e^{2 x}+4 x \left (5-2 \log \left (\frac {4}{3}\right )\right )-4 e^x \left (-2 x+1-2 \log \left (\frac {4}{3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[20 - 8*E^(2*x) - 8*x - 8*Log[4/3] + E^x*(-4 + 8*x + 8*Log[4/3]),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 41, normalized size = 1.64 \begin {gather*} -4 e^{2 x}+20 x-4 x^2-8 x \log \left (\frac {4}{3}\right )+e^x \left (8 x+4 \left (-3+2 \log \left (\frac {4}{3}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[20 - 8*E^(2*x) - 8*x - 8*Log[4/3] + E^x*(-4 + 8*x + 8*Log[4/3]),x]

[Out]

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

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fricas [A]  time = 0.76, size = 33, normalized size = 1.32 \begin {gather*} -4 \, x^{2} + 4 \, {\left (2 \, x - 2 \, \log \left (\frac {3}{4}\right ) - 3\right )} e^{x} + 8 \, x \log \left (\frac {3}{4}\right ) + 20 \, x - 4 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.37, size = 33, normalized size = 1.32 \begin {gather*} -4 \, x^{2} + 4 \, {\left (2 \, x - 2 \, \log \left (\frac {3}{4}\right ) - 3\right )} e^{x} + 8 \, x \log \left (\frac {3}{4}\right ) + 20 \, x - 4 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 42, normalized size = 1.68

method	result	size
default	\(20 x +8 \,{\mathrm e}^{x} x -12 \,{\mathrm e}^{x}+16 \,{\mathrm e}^{x} \ln \relax (2)-8 \ln \relax (3) {\mathrm e}^{x}-4 x^{2}-4 \,{\mathrm e}^{2 x}+8 \ln \left (\frac {3}{4}\right ) x\)	\(42\)
risch	\(-4 \,{\mathrm e}^{2 x}+\left (-12-8 \ln \relax (3)+16 \ln \relax (2)+8 x \right ) {\mathrm e}^{x}+8 x \ln \relax (3)-16 x \ln \relax (2)-4 x^{2}+20 x\)	\(42\)
norman	\(\left (-12-8 \ln \relax (3)+16 \ln \relax (2)\right ) {\mathrm e}^{x}+\left (8 \ln \relax (3)-16 \ln \relax (2)+20\right ) x -4 x^{2}-4 \,{\mathrm e}^{2 x}+8 \,{\mathrm e}^{x} x\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-8*exp(x)^2+(-8*ln(3/4)+8*x-4)*exp(x)+8*ln(3/4)-8*x+20,x,method=_RETURNVERBOSE)

[Out]

20*x+8*exp(x)*x-12*exp(x)+16*exp(x)*ln(2)-8*ln(3)*exp(x)-4*x^2-4*exp(x)^2+8*ln(3/4)*x

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maxima [A]  time = 0.53, size = 37, normalized size = 1.48 \begin {gather*} -4 \, x^{2} + 4 \, {\left (2 \, x - 2 \, \log \relax (3) + 4 \, \log \relax (2) - 3\right )} e^{x} + 8 \, x \log \left (\frac {3}{4}\right ) + 20 \, x - 4 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 35, normalized size = 1.40 \begin {gather*} x\,\left (8\,\ln \left (\frac {3}{4}\right )+20\right )-4\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (8\,\ln \left (\frac {3}{4}\right )+12\right )+8\,x\,{\mathrm {e}}^x-4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*log(3/4) - 8*exp(2*x) - 8*x - exp(x)*(8*log(3/4) - 8*x + 4) + 20,x)

[Out]

x*(8*log(3/4) + 20) - 4*exp(2*x) - exp(x)*(8*log(3/4) + 12) + 8*x*exp(x) - 4*x^2

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sympy [A]  time = 0.13, size = 42, normalized size = 1.68 \begin {gather*} - 4 x^{2} + x \left (- 16 \log {\relax (2 )} + 8 \log {\relax (3 )} + 20\right ) + \left (8 x - 12 - 8 \log {\relax (3 )} + 16 \log {\relax (2 )}\right ) e^{x} - 4 e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*exp(x)**2+(-8*ln(3/4)+8*x-4)*exp(x)+8*ln(3/4)-8*x+20,x)

[Out]

-4*x**2 + x*(-16*log(2) + 8*log(3) + 20) + (8*x - 12 - 8*log(3) + 16*log(2))*exp(x) - 4*exp(2*x)

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