3.103.8 \(\int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} (-2 e^2 x-32 x^2) \log (4)}{e^2 \log (4)} \, dx\)

Optimal. Leaf size=24 \[ x-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)} \]

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Rubi [A]  time = 0.09, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} -e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^2*x + E^2*Log[4] + E^((32*x)/E^2)*(-2*E^2*x - 32*x^2)*Log[4])/(E^2*Log[4]),x]

[Out]

x - E^((32*x)/E^2)*x^2 - x^2/Log[4]

Rule 12

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 {\int \left (-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)\right ) \, dx}{e^2 \log (4)}\\ &=x-\frac {x^2}{\log (4)}+\frac {\int e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \, dx}{e^2}\\ &=x-\frac {x^2}{\log (4)}+\frac {\int e^{\frac {32 x}{e^2}} \left (-2 e^2-32 x\right ) x \, dx}{e^2}\\ &=x-\frac {x^2}{\log (4)}+\frac {\int \left (-2 e^{2+\frac {32 x}{e^2}} x-32 e^{\frac {32 x}{e^2}} x^2\right ) \, dx}{e^2}\\ &=x-\frac {x^2}{\log (4)}-\frac {2 \int e^{2+\frac {32 x}{e^2}} x \, dx}{e^2}-\frac {32 \int e^{\frac {32 x}{e^2}} x^2 \, dx}{e^2}\\ &=x-\frac {1}{16} e^{2+\frac {32 x}{e^2}} x-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)}+\frac {1}{16} \int e^{2+\frac {32 x}{e^2}} \, dx+2 \int e^{\frac {32 x}{e^2}} x \, dx\\ &=\frac {1}{512} e^{4+\frac {32 x}{e^2}}+x-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)}-\frac {1}{16} e^2 \int e^{\frac {32 x}{e^2}} \, dx\\ &=x-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 22, normalized size = 0.92 \begin {gather*} x \left (1-e^{\frac {32 x}{e^2}} x-\frac {x}{\log (4)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^2*x + E^2*Log[4] + E^((32*x)/E^2)*(-2*E^2*x - 32*x^2)*Log[4])/(E^2*Log[4]),x]

[Out]

x*(1 - E^((32*x)/E^2)*x - x/Log[4])

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fricas [A]  time = 0.71, size = 28, normalized size = 1.17 \begin {gather*} -\frac {2 \, x^{2} e^{\left (32 \, x e^{\left (-2\right )}\right )} \log \relax (2) + x^{2} - 2 \, x \log \relax (2)}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-2*x*exp(1)^2-32*x^2)*log(2)*exp(16*x/exp(1)^2)^2+2*exp(1)^2*log(2)-2*x*exp(1)^2)/exp(1)^2/l
og(2),x, algorithm="fricas")

[Out]

-1/2*(2*x^2*e^(32*x*e^(-2))*log(2) + x^2 - 2*x*log(2))/log(2)

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giac [B]  time = 0.16, size = 71, normalized size = 2.96 \begin {gather*} -\frac {{\left (256 \, x^{2} e^{2} - 512 \, x e^{2} \log \relax (2) + {\left ({\left (32 \, x e^{2} - e^{4}\right )} e^{\left (2 \, {\left (16 \, x + e^{2}\right )} e^{\left (-2\right )}\right )} + {\left (512 \, x^{2} e^{2} - 32 \, x e^{4} + e^{6}\right )} e^{\left (32 \, x e^{\left (-2\right )}\right )}\right )} \log \relax (2)\right )} e^{\left (-2\right )}}{512 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-2*x*exp(1)^2-32*x^2)*log(2)*exp(16*x/exp(1)^2)^2+2*exp(1)^2*log(2)-2*x*exp(1)^2)/exp(1)^2/l
og(2),x, algorithm="giac")

[Out]

-1/512*(256*x^2*e^2 - 512*x*e^2*log(2) + ((32*x*e^2 - e^4)*e^(2*(16*x + e^2)*e^(-2)) + (512*x^2*e^2 - 32*x*e^4
+ e^6)*e^(32*x*e^(-2)))*log(2))*e^(-2)/log(2)

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maple [A]  time = 0.06, size = 23, normalized size = 0.96

method	result	size
risch	\(x -\frac {x^{2}}{2 \ln \relax (2)}-{\mathrm e}^{32 x \,{\mathrm e}^{-2}} x^{2}\)	\(23\)
derivativedivides	\(\frac {-128 x^{2}+256 x \ln \relax (2)-256 \ln \relax (2) {\mathrm e}^{32 x \,{\mathrm e}^{-2}} x^{2}}{256 \ln \relax (2)}\)	\(35\)
norman	\(\left (x \,{\mathrm e}-x^{2} {\mathrm e} \,{\mathrm e}^{32 x \,{\mathrm e}^{-2}}-\frac {{\mathrm e} x^{2}}{2 \ln \relax (2)}\right ) {\mathrm e}^{-1}\)	\(39\)
default	\(\frac {{\mathrm e}^{-2} \left (x \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} \ln \relax (2) {\mathrm e}^{32 x \,{\mathrm e}^{-2}} x^{2}-\frac {x^{2} {\mathrm e}^{2}}{2}\right )}{\ln \relax (2)}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*(-2*x*exp(1)^2-32*x^2)*ln(2)*exp(16*x/exp(1)^2)^2+2*exp(1)^2*ln(2)-2*x*exp(1)^2)/exp(1)^2/ln(2),x,m
ethod=_RETURNVERBOSE)

[Out]

x-1/2*x^2/ln(2)-exp(32*x*exp(-2))*x^2

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maxima [A]  time = 0.35, size = 37, normalized size = 1.54 \begin {gather*} -\frac {{\left (2 \, x^{2} e^{\left (32 \, x e^{\left (-2\right )} + 2\right )} \log \relax (2) + x^{2} e^{2} - 2 \, x e^{2} \log \relax (2)\right )} e^{\left (-2\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-2*x*exp(1)^2-32*x^2)*log(2)*exp(16*x/exp(1)^2)^2+2*exp(1)^2*log(2)-2*x*exp(1)^2)/exp(1)^2/l
og(2),x, algorithm="maxima")

[Out]

-1/2*(2*x^2*e^(32*x*e^(-2) + 2)*log(2) + x^2*e^2 - 2*x*e^2*log(2))*e^(-2)/log(2)

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mupad [B]  time = 0.11, size = 22, normalized size = 0.92 \begin {gather*} x-\frac {x^2}{2\,\ln \relax (2)}-x^2\,{\mathrm {e}}^{32\,x\,{\mathrm {e}}^{-2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*(x*exp(2) - exp(2)*log(2) + exp(32*x*exp(-2))*log(2)*(2*x*exp(2) + 32*x^2)))/log(2),x)

[Out]

x - x^2/(2*log(2)) - x^2*exp(32*x*exp(-2))

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sympy [A]  time = 0.14, size = 20, normalized size = 0.83 \begin {gather*} - x^{2} e^{\frac {32 x}{e^{2}}} - \frac {x^{2}}{2 \log {\relax (2 )}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-2*x*exp(1)**2-32*x**2)*ln(2)*exp(16*x/exp(1)**2)**2+2*exp(1)**2*ln(2)-2*x*exp(1)**2)/exp(1)
**2/ln(2),x)

[Out]

-x**2*exp(32*x*exp(-2)) - x**2/(2*log(2)) + x

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