3.103.44 \(\int \frac {-64-4 x+x^2+32 \log (2)-4 \log ^2(2)+4 \log (2 e^x)}{x^2} \, dx\)

Optimal. Leaf size=25 \[ 5+x+\frac {4 \left ((4-\log (2))^2-\log \left (2 e^x\right )\right )}{x} \]

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Rubi [A]  time = 0.03, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {14, 2168, 29} \begin {gather*} x-\frac {4 \log \left (2 e^x\right )}{x}+\frac {4 (4-\log (2))^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64 - 4*x + x^2 + 32*Log[2] - 4*Log[2]^2 + 4*Log[2*E^x])/x^2,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.36 \begin {gather*} \frac {64}{x}+x-\frac {32 \log (2)}{x}+\frac {4 \log ^2(2)}{x}-\frac {4 \log \left (2 e^x\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64 - 4*x + x^2 + 32*Log[2] - 4*Log[2]^2 + 4*Log[2*E^x])/x^2,x]

[Out]

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

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fricas [A]  time = 0.57, size = 19, normalized size = 0.76 \begin {gather*} \frac {x^{2} + 4 \, \log \relax (2)^{2} - 36 \, \log \relax (2) + 64}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2*exp(x))-4*log(2)^2+32*log(2)+x^2-4*x-64)/x^2,x, algorithm="fricas")

[Out]

(x^2 + 4*log(2)^2 - 36*log(2) + 64)/x

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giac [A]  time = 0.16, size = 17, normalized size = 0.68 \begin {gather*} x + \frac {4 \, {\left (\log \relax (2)^{2} - 9 \, \log \relax (2) + 16\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2*exp(x))-4*log(2)^2+32*log(2)+x^2-4*x-64)/x^2,x, algorithm="giac")

[Out]

x + 4*(log(2)^2 - 9*log(2) + 16)/x

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maple [A]  time = 0.08, size = 29, normalized size = 1.16

method	result	size
risch	\(-\frac {4 \ln \left ({\mathrm e}^{x}\right )}{x}+\frac {64+4 \ln \relax (2)^{2}+x^{2}-36 \ln \relax (2)}{x}\)	\(29\)
norman	\(\frac {x \ln \left (2 \,{\mathrm e}^{x}\right )+64-4 \ln \left (2 \,{\mathrm e}^{x}\right )+4 \ln \relax (2)^{2}-32 \ln \relax (2)}{x}\)	\(31\)
default	\(-\frac {4 \ln \left (2 \,{\mathrm e}^{x}\right )}{x}+x +\frac {4 \ln \relax (2)^{2}}{x}-\frac {32 \ln \relax (2)}{x}+\frac {64}{x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(2*exp(x))-4*ln(2)^2+32*ln(2)+x^2-4*x-64)/x^2,x,method=_RETURNVERBOSE)

[Out]

-4/x*ln(exp(x))+(64+4*ln(2)^2+x^2-36*ln(2))/x

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maxima [A]  time = 0.35, size = 33, normalized size = 1.32 \begin {gather*} x + \frac {4 \, \log \relax (2)^{2}}{x} - \frac {32 \, \log \relax (2)}{x} - \frac {4 \, \log \left (2 \, e^{x}\right )}{x} + \frac {64}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2*exp(x))-4*log(2)^2+32*log(2)+x^2-4*x-64)/x^2,x, algorithm="maxima")

[Out]

x + 4*log(2)^2/x - 32*log(2)/x - 4*log(2*e^x)/x + 64/x

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mupad [B]  time = 6.13, size = 18, normalized size = 0.72 \begin {gather*} x+\frac {4\,{\ln \relax (2)}^2-36\,\ln \relax (2)+64}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - 32*log(2) - 4*log(2*exp(x)) + 4*log(2)^2 - x^2 + 64)/x^2,x)

[Out]

x + (4*log(2)^2 - 36*log(2) + 64)/x

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sympy [A]  time = 0.18, size = 15, normalized size = 0.60 \begin {gather*} x + \frac {- 36 \log {\relax (2 )} + 4 \log {\relax (2 )}^{2} + 64}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(2*exp(x))-4*ln(2)**2+32*ln(2)+x**2-4*x-64)/x**2,x)

[Out]

x + (-36*log(2) + 4*log(2)**2 + 64)/x

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