3.26.62 \(\int e^{2 x} (-4-8 x+(-4 x-4 x^2) \log (4)) \, dx\)

Optimal. Leaf size=14 \[ -2 e^{2 x} x (2+x \log (4)) \]

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Rubi [A]  time = 0.08, antiderivative size = 21, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2196, 2194, 2176} \begin {gather*} -2 e^{2 x} x^2 \log (4)-4 e^{2 x} x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*E^(2*x)*x - 2*E^(2*x)*x^2*Log[4]

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

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Mathematica [A]  time = 0.03, size = 18, normalized size = 1.29 \begin {gather*} -4 e^{2 x} \left (x+\frac {1}{2} x^2 \log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*E^(2*x)*(x + (x^2*Log[4])/2)

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fricas [A]  time = 0.70, size = 14, normalized size = 1.00 \begin {gather*} -4 \, {\left (x^{2} \log \relax (2) + x\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(x^2*log(2) + x)*e^(2*x)

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giac [A]  time = 0.17, size = 14, normalized size = 1.00 \begin {gather*} -4 \, {\left (x^{2} \log \relax (2) + x\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(x^2*log(2) + x)*e^(2*x)

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maple [A]  time = 0.03, size = 14, normalized size = 1.00

method	result	size
gosper	\(-4 \,{\mathrm e}^{2 x} x \left (x \ln \relax (2)+1\right )\)	\(14\)
risch	\(\left (-4 x^{2} \ln \relax (2)-4 x \right ) {\mathrm e}^{2 x}\)	\(17\)
default	\(-4 x \,{\mathrm e}^{2 x}-4 x^{2} \ln \relax (2) {\mathrm e}^{2 x}\)	\(20\)
norman	\(-4 x \,{\mathrm e}^{2 x}-4 x^{2} \ln \relax (2) {\mathrm e}^{2 x}\)	\(20\)
meijerg	\(2-2 \,{\mathrm e}^{2 x}+\ln \relax (2) \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )+\frac {\left (-8 \ln \relax (2)-8\right ) \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )}{4}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 49, normalized size = 3.50 \begin {gather*} -2 \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} \log \relax (2) - 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \relax (2) - 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - 2 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*x^2 - 2*x + 1)*e^(2*x)*log(2) - 2*(2*x - 1)*e^(2*x)*log(2) - 2*(2*x - 1)*e^(2*x) - 2*e^(2*x)

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mupad [B]  time = 0.05, size = 13, normalized size = 0.93 \begin {gather*} -4\,x\,{\mathrm {e}}^{2\,x}\,\left (x\,\ln \relax (2)+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(2*x)*(8*x + 2*log(2)*(4*x + 4*x^2) + 4),x)

[Out]

-4*x*exp(2*x)*(x*log(2) + 1)

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sympy [A]  time = 0.11, size = 17, normalized size = 1.21 \begin {gather*} \left (- 4 x^{2} \log {\relax (2 )} - 4 x\right ) e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-4*x**2-4*x)*ln(2)-8*x-4)*exp(x)**2,x)

[Out]

(-4*x**2*log(2) - 4*x)*exp(2*x)

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