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

Optimal. Leaf size=22 \[ x \left (\log (x)-\left (2+\frac {2}{e^2}\right ) (x+x \log (3)+\log (x))\right ) \]

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Rubi [B]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 2.91, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 2295} \begin {gather*} -\frac {2 x^2}{e^2}-\frac {2 \left (1+e^2\right ) x^2 \log (3)}{e^2}+\frac {\left (2+e^2\right ) x}{e^2}-\frac {2 x}{e^2}-\frac {1}{8} (4 x+1)^2-\frac {\left (2+e^2\right ) x \log (x)}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 29, normalized size = 1.32 \begin {gather*} -\frac {x \left (\left (1+e^2\right ) x (4+\log (81))+2 \left (2+e^2\right ) \log (x)\right )}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(x*((1 + E^2)*x*(4 + Log[81]) + 2*(2 + E^2)*Log[x]))/E^2

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fricas [A]  time = 0.49, size = 42, normalized size = 1.91 \begin {gather*} -{\left (2 \, x^{2} e^{2} + 2 \, x^{2} + 2 \, {\left (x^{2} e^{2} + x^{2}\right )} \log \relax (3) + {\left (x e^{2} + 2 \, x\right )} \log \relax (x)\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*x^2*e^2 + 2*x^2 + 2*(x^2*e^2 + x^2)*log(3) + (x*e^2 + 2*x)*log(x))*e^(-2)

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giac [B]  time = 0.14, size = 50, normalized size = 2.27 \begin {gather*} -{\left (2 \, x^{2} + {\left (x \log \relax (x) - x\right )} {\left (e^{2} + 2\right )} + {\left (2 \, x^{2} + x\right )} e^{2} + 2 \, {\left (x^{2} e^{2} + x^{2}\right )} \log \relax (3) + 2 \, x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*x^2 + (x*log(x) - x)*(e^2 + 2) + (2*x^2 + x)*e^2 + 2*(x^2*e^2 + x^2)*log(3) + 2*x)*e^(-2)

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maple [A]  time = 0.04, size = 35, normalized size = 1.59

method	result	size
norman	\(-2 \left ({\mathrm e}^{2} \ln \relax (3)+{\mathrm e}^{2}+\ln \relax (3)+1\right ) {\mathrm e}^{-2} x^{2}-{\mathrm e}^{-2} \left ({\mathrm e}^{2}+2\right ) x \ln \relax (x )\)	\(35\)
risch	\(-2 \,{\mathrm e}^{2} \ln \relax (3) {\mathrm e}^{-2} x^{2}-2 \,{\mathrm e}^{2} {\mathrm e}^{-2} x^{2}-2 \ln \relax (3) {\mathrm e}^{-2} x^{2}-2 x^{2} {\mathrm e}^{-2}-{\mathrm e}^{-2} \left ({\mathrm e}^{2}+2\right ) x \ln \relax (x )\)	\(49\)
default	\({\mathrm e}^{-2} \left ({\mathrm e}^{2} \left (-2 x^{2}-x \right )-2 x^{2} {\mathrm e}^{2} \ln \relax (3)-2 x^{2} \ln \relax (3)-x \,{\mathrm e}^{2} \ln \relax (x )+{\mathrm e}^{2} x -2 x \ln \relax (x )-2 x^{2}\right )\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.39, size = 50, normalized size = 2.27 \begin {gather*} -{\left (2 \, x^{2} + {\left (x \log \relax (x) - x\right )} {\left (e^{2} + 2\right )} + {\left (2 \, x^{2} + x\right )} e^{2} + 2 \, {\left (x^{2} e^{2} + x^{2}\right )} \log \relax (3) + 2 \, x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*x^2 + (x*log(x) - x)*(e^2 + 2) + (2*x^2 + x)*e^2 + 2*(x^2*e^2 + x^2)*log(3) + 2*x)*e^(-2)

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mupad [B]  time = 1.87, size = 33, normalized size = 1.50 \begin {gather*} -x^2\,\left (2\,{\mathrm {e}}^{-2}+2\,\ln \relax (3)+2\,{\mathrm {e}}^{-2}\,\ln \relax (3)+2\right )-x\,\ln \relax (x)\,\left (2\,{\mathrm {e}}^{-2}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.15, size = 44, normalized size = 2.00 \begin {gather*} \frac {x^{2} \left (- 2 e^{2} \log {\relax (3 )} - 2 e^{2} - 2 \log {\relax (3 )} - 2\right )}{e^{2}} + \frac {\left (- x e^{2} - 2 x\right ) \log {\relax (x )}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*(-2*exp(2)*log(3) - 2*exp(2) - 2*log(3) - 2)*exp(-2) + (-x*exp(2) - 2*x)*exp(-2)*log(x)

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