3.64.11 \(\int \frac {-6+e^x (-2+2 x)+8 x \log (16 e^{2 x})-2 \log ^2(16 e^{2 x})}{x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {2 \left (3+e^x+\log ^2\left (16 e^{2 x}\right )\right )}{x} \]

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Rubi [A]  time = 0.09, antiderivative size = 29, normalized size of antiderivative = 1.45, number of steps used = 10, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {14, 2197, 2158, 29, 2168} \begin {gather*} \frac {2 e^x}{x}+\frac {6}{x}+\frac {2 \log ^2\left (16 e^{2 x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6 + E^x*(-2 + 2*x) + 8*x*Log[16*E^(2*x)] - 2*Log[16*E^(2*x)]^2)/x^2,x]

[Out]

6/x + (2*E^x)/x + (2*Log[16*E^(2*x)]^2)/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 2158

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(b*x)/a, x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, 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]))

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 29, normalized size = 1.45 \begin {gather*} \frac {2 \left (3+e^x+4 x^2+\left (-2 x+\log \left (16 e^{2 x}\right )\right )^2\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 + E^x*(-2 + 2*x) + 8*x*Log[16*E^(2*x)] - 2*Log[16*E^(2*x)]^2)/x^2,x]

[Out]

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

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fricas [A]  time = 0.77, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (4 \, x^{2} + 16 \, \log \relax (2)^{2} + e^{x} + 3\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(16*exp(x)^2)^2+8*x*log(16*exp(x)^2)+(2*x-2)*exp(x)-6)/x^2,x, algorithm="fricas")

[Out]

2*(4*x^2 + 16*log(2)^2 + e^x + 3)/x

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giac [A]  time = 0.16, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (4 \, x^{2} + 16 \, \log \relax (2)^{2} + e^{x} + 3\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(16*exp(x)^2)^2+8*x*log(16*exp(x)^2)+(2*x-2)*exp(x)-6)/x^2,x, algorithm="giac")

[Out]

2*(4*x^2 + 16*log(2)^2 + e^x + 3)/x

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maple [A]  time = 0.14, size = 22, normalized size = 1.10

method	result	size
norman	\(\frac {6+2 \ln \left (16 \,{\mathrm e}^{2 x}\right )^{2}+2 \,{\mathrm e}^{x}}{x}\)	\(22\)
default	\(\frac {6}{x}+\frac {2 \ln \left (16 \,{\mathrm e}^{2 x}\right )^{2}}{x}+\frac {2 \,{\mathrm e}^{x}}{x}\)	\(28\)
risch	\(\frac {8 \ln \left ({\mathrm e}^{x}\right )^{2}}{x}+\frac {4 \left (-i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+2 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+8 \ln \relax (2)\right ) \ln \left ({\mathrm e}^{x}\right )}{x}+\frac {12+64 \ln \relax (2)^{2}+4 \,{\mathrm e}^{x}-16 i \ln \relax (2) \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-\pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{4} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+4 \pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{3} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-6 \pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{4}+4 \pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{5}-\pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{6}-16 i \ln \relax (2) \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+32 i \ln \relax (2) \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}}{2 x}\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(16*exp(x)^2)^2+8*x*ln(16*exp(x)^2)+(2*x-2)*exp(x)-6)/x^2,x,method=_RETURNVERBOSE)

[Out]

(6+2*ln(16*exp(x)^2)^2+2*exp(x))/x

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maxima [C]  time = 0.39, size = 31, normalized size = 1.55 \begin {gather*} \frac {2 \, \log \left (16 \, e^{\left (2 \, x\right )}\right )^{2}}{x} + \frac {6}{x} + 2 \, {\rm Ei}\relax (x) - 2 \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(16*exp(x)^2)^2+8*x*log(16*exp(x)^2)+(2*x-2)*exp(x)-6)/x^2,x, algorithm="maxima")

[Out]

2*log(16*e^(2*x))^2/x + 6/x + 2*Ei(x) - 2*gamma(-1, -x)

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mupad [B]  time = 0.08, size = 20, normalized size = 1.00 \begin {gather*} 8\,x+\frac {2\,{\mathrm {e}}^x+32\,{\ln \relax (2)}^2+6}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(16*exp(2*x))^2 - exp(x)*(2*x - 2) - 8*x*log(16*exp(2*x)) + 6)/x^2,x)

[Out]

8*x + (2*exp(x) + 32*log(2)^2 + 6)/x

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sympy [A]  time = 0.15, size = 19, normalized size = 0.95 \begin {gather*} 8 x + \frac {2 e^{x}}{x} + \frac {6 + 32 \log {\relax (2 )}^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(16*exp(x)**2)**2+8*x*ln(16*exp(x)**2)+(2*x-2)*exp(x)-6)/x**2,x)

[Out]

8*x + 2*exp(x)/x + (6 + 32*log(2)**2)/x

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