3.18.20 \(\int \frac {1}{4} e^{-1-3 x} (-3 x^2+3 x^3+(4-12 x) \log (2)+(-4+12 x) \log (16)) \, dx\)

Optimal. Leaf size=23 \[ e^{-1-3 x} x \left (-\frac {x^2}{4}+\log (2)-\log (16)\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 46, normalized size of antiderivative = 2.00, number of steps used = 12, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 2196, 2176, 2194} \begin {gather*} -\frac {1}{4} e^{-3 x-1} x^3-\frac {1}{3} e^{-3 x-1} \log (8)+\frac {1}{3} e^{-3 x-1} (1-3 x) \log (8) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} &=\frac {1}{4} \int e^{-1-3 x} \left (-3 x^2+3 x^3+(4-12 x) \log (2)+(-4+12 x) \log (16)\right ) \, dx\\ &=\frac {1}{4} \int \left (-3 e^{-1-3 x} x^2+3 e^{-1-3 x} x^3-4 e^{-1-3 x} (-1+3 x) \log (2) \left (1-\frac {\log (16)}{\log (2)}\right )\right ) \, dx\\ &=-\left (\frac {3}{4} \int e^{-1-3 x} x^2 \, dx\right )+\frac {3}{4} \int e^{-1-3 x} x^3 \, dx+\log (8) \int e^{-1-3 x} (-1+3 x) \, dx\\ &=\frac {1}{4} e^{-1-3 x} x^2-\frac {1}{4} e^{-1-3 x} x^3+\frac {1}{3} e^{-1-3 x} (1-3 x) \log (8)-\frac {1}{2} \int e^{-1-3 x} x \, dx+\frac {3}{4} \int e^{-1-3 x} x^2 \, dx+\log (8) \int e^{-1-3 x} \, dx\\ &=\frac {1}{6} e^{-1-3 x} x-\frac {1}{4} e^{-1-3 x} x^3-\frac {1}{3} e^{-1-3 x} \log (8)+\frac {1}{3} e^{-1-3 x} (1-3 x) \log (8)-\frac {1}{6} \int e^{-1-3 x} \, dx+\frac {1}{2} \int e^{-1-3 x} x \, dx\\ &=\frac {1}{18} e^{-1-3 x}-\frac {1}{4} e^{-1-3 x} x^3-\frac {1}{3} e^{-1-3 x} \log (8)+\frac {1}{3} e^{-1-3 x} (1-3 x) \log (8)+\frac {1}{6} \int e^{-1-3 x} \, dx\\ &=-\frac {1}{4} e^{-1-3 x} x^3-\frac {1}{3} e^{-1-3 x} \log (8)+\frac {1}{3} e^{-1-3 x} (1-3 x) \log (8)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{4} e^{-1-3 x} \left (-x^3-4 x \log (8)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(-1 - 3*x)*(-x^3 - 4*x*Log[8]))/4

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fricas [A]  time = 0.71, size = 17, normalized size = 0.74 \begin {gather*} -\frac {1}{4} \, {\left (x^{3} + 12 \, x \log \relax (2)\right )} e^{\left (-3 \, x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*(12*x-4)*log(2)+(-12*x+4)*log(2)+3*x^3-3*x^2)*exp(-3*x-1),x, algorithm="fricas")

[Out]

-1/4*(x^3 + 12*x*log(2))*e^(-3*x - 1)

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giac [A]  time = 0.15, size = 17, normalized size = 0.74 \begin {gather*} -\frac {1}{4} \, {\left (x^{3} + 12 \, x \log \relax (2)\right )} e^{\left (-3 \, x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*(12*x-4)*log(2)+(-12*x+4)*log(2)+3*x^3-3*x^2)*exp(-3*x-1),x, algorithm="giac")

[Out]

-1/4*(x^3 + 12*x*log(2))*e^(-3*x - 1)

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maple [A]  time = 0.06, size = 18, normalized size = 0.78

method	result	size
gosper	\(-\frac {{\mathrm e}^{-3 x -1} x \left (x^{2}+12 \ln \relax (2)\right )}{4}\)	\(18\)
risch	\(\frac {\left (-x^{3}-12 x \ln \relax (2)\right ) {\mathrm e}^{-3 x -1}}{4}\)	\(20\)
norman	\(-\frac {x^{3} {\mathrm e}^{-3 x -1}}{4}-3 x \ln \relax (2) {\mathrm e}^{-3 x -1}\)	\(24\)
derivativedivides	\(\frac {{\mathrm e}^{-3 x -1} \left (-3 x -1\right )}{36}+\frac {{\mathrm e}^{-3 x -1}}{108}+\frac {{\mathrm e}^{-3 x -1} \left (-3 x -1\right )^{2}}{36}+\frac {{\mathrm e}^{-3 x -1} \left (-3 x -1\right )^{3}}{108}+{\mathrm e}^{-3 x -1} \ln \relax (2)+{\mathrm e}^{-3 x -1} \ln \relax (2) \left (-3 x -1\right )\)	\(76\)
default	\(\frac {{\mathrm e}^{-3 x -1} \left (-3 x -1\right )}{36}+\frac {{\mathrm e}^{-3 x -1}}{108}+\frac {{\mathrm e}^{-3 x -1} \left (-3 x -1\right )^{2}}{36}+\frac {{\mathrm e}^{-3 x -1} \left (-3 x -1\right )^{3}}{108}+{\mathrm e}^{-3 x -1} \ln \relax (2)+{\mathrm e}^{-3 x -1} \ln \relax (2) \left (-3 x -1\right )\)	\(76\)
meijerg	\(-\ln \relax (2) {\mathrm e}^{-1} \left (1-{\mathrm e}^{-3 x}\right )+\ln \relax (2) {\mathrm e}^{-1} \left (1-\frac {\left (6 x +2\right ) {\mathrm e}^{-3 x}}{2}\right )+\frac {{\mathrm e}^{-1} \left (6-\frac {\left (108 x^{3}+108 x^{2}+72 x +24\right ) {\mathrm e}^{-3 x}}{4}\right )}{108}-\frac {{\mathrm e}^{-1} \left (2-\frac {\left (27 x^{2}+18 x +6\right ) {\mathrm e}^{-3 x}}{3}\right )}{36}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(4*(12*x-4)*ln(2)+(-12*x+4)*ln(2)+3*x^3-3*x^2)*exp(-3*x-1),x,method=_RETURNVERBOSE)

[Out]

-1/4*exp(-3*x-1)*x*(x^2+12*ln(2))

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maxima [B]  time = 0.40, size = 66, normalized size = 2.87 \begin {gather*} -{\left (3 \, x + 1\right )} e^{\left (-3 \, x - 1\right )} \log \relax (2) - \frac {1}{36} \, {\left (9 \, x^{3} + 9 \, x^{2} + 6 \, x + 2\right )} e^{\left (-3 \, x - 1\right )} + \frac {1}{36} \, {\left (9 \, x^{2} + 6 \, x + 2\right )} e^{\left (-3 \, x - 1\right )} + e^{\left (-3 \, x - 1\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*(12*x-4)*log(2)+(-12*x+4)*log(2)+3*x^3-3*x^2)*exp(-3*x-1),x, algorithm="maxima")

[Out]

-(3*x + 1)*e^(-3*x - 1)*log(2) - 1/36*(9*x^3 + 9*x^2 + 6*x + 2)*e^(-3*x - 1) + 1/36*(9*x^2 + 6*x + 2)*e^(-3*x
- 1) + e^(-3*x - 1)*log(2)

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mupad [B]  time = 1.04, size = 17, normalized size = 0.74 \begin {gather*} -\frac {x\,{\mathrm {e}}^{-3\,x-1}\,\left (x^2+12\,\ln \relax (2)\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(- 3*x - 1)*(3*log(2)*(12*x - 4) - 3*x^2 + 3*x^3))/4,x)

[Out]

-(x*exp(- 3*x - 1)*(12*log(2) + x^2))/4

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sympy [A]  time = 0.12, size = 20, normalized size = 0.87 \begin {gather*} \frac {\left (- x^{3} - 12 x \log {\relax (2 )}\right ) e^{- 3 x - 1}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*(12*x-4)*ln(2)+(-12*x+4)*ln(2)+3*x**3-3*x**2)*exp(-3*x-1),x)

[Out]

(-x**3 - 12*x*log(2))*exp(-3*x - 1)/4

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