3.10.81 \(\int \frac {e^{4/x} (-144-27 x)+e^{4/x} (36+9 x) \log (\frac {\log (4)}{x})}{8 x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac {9 e^{4/x} \left (4-\log \left (\frac {\log (4)}{x}\right )\right )}{8 x} \]

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Rubi [A]  time = 0.28, antiderivative size = 36, normalized size of antiderivative = 1.44, number of steps used = 15, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {12, 14, 2212, 2209, 2554, 6742, 2210} \begin {gather*} \frac {9 e^{4/x}}{2 x}-\frac {9 e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4/x)*(-144 - 27*x) + E^(4/x)*(36 + 9*x)*Log[Log[4]/x])/(8*x^3),x]

[Out]

(9*E^(4/x))/(2*x) - (9*E^(4/x)*Log[Log[4]/x])/(8*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \frac {e^{4/x} (-144-27 x)+e^{4/x} (36+9 x) \log \left (\frac {\log (4)}{x}\right )}{x^3} \, dx\\ &=\frac {1}{8} \int \left (-\frac {144 e^{4/x}}{x^3}-\frac {27 e^{4/x}}{x^2}+\frac {36 e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{x^3}+\frac {9 e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{x^2}\right ) \, dx\\ &=\frac {9}{8} \int \frac {e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{x^2} \, dx-\frac {27}{8} \int \frac {e^{4/x}}{x^2} \, dx+\frac {9}{2} \int \frac {e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{x^3} \, dx-18 \int \frac {e^{4/x}}{x^3} \, dx\\ &=\frac {27 e^{4/x}}{32}+\frac {9 e^{4/x}}{2 x}-\frac {9 e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{8 x}-\frac {9}{8} \int \frac {e^{4/x}}{4 x} \, dx+\frac {9}{2} \int \frac {e^{4/x}}{x^2} \, dx-\frac {9}{2} \int \frac {e^{4/x} (4-x)}{16 x^2} \, dx\\ &=-\frac {9 e^{4/x}}{32}+\frac {9 e^{4/x}}{2 x}-\frac {9 e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{8 x}-\frac {9}{32} \int \frac {e^{4/x} (4-x)}{x^2} \, dx-\frac {9}{32} \int \frac {e^{4/x}}{x} \, dx\\ &=-\frac {9 e^{4/x}}{32}+\frac {9 e^{4/x}}{2 x}+\frac {9 \text {Ei}\left (\frac {4}{x}\right )}{32}-\frac {9 e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{8 x}-\frac {9}{32} \int \left (\frac {4 e^{4/x}}{x^2}-\frac {e^{4/x}}{x}\right ) \, dx\\ &=-\frac {9 e^{4/x}}{32}+\frac {9 e^{4/x}}{2 x}+\frac {9 \text {Ei}\left (\frac {4}{x}\right )}{32}-\frac {9 e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{8 x}+\frac {9}{32} \int \frac {e^{4/x}}{x} \, dx-\frac {9}{8} \int \frac {e^{4/x}}{x^2} \, dx\\ &=\frac {9 e^{4/x}}{2 x}-\frac {9 e^{4/x} \log \left (\frac {\log (4)}{x}\right )}{8 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 23, normalized size = 0.92 \begin {gather*} -\frac {9 e^{4/x} \left (-4+\log \left (\frac {\log (4)}{x}\right )\right )}{8 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4/x)*(-144 - 27*x) + E^(4/x)*(36 + 9*x)*Log[Log[4]/x])/(8*x^3),x]

[Out]

(-9*E^(4/x)*(-4 + Log[Log[4]/x]))/(8*x)

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fricas [A]  time = 0.60, size = 29, normalized size = 1.16 \begin {gather*} -\frac {9 \, {\left (e^{\frac {4}{x}} \log \left (\frac {2 \, \log \relax (2)}{x}\right ) - 4 \, e^{\frac {4}{x}}\right )}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((9*x+36)*exp(4/x)*log(2*log(2)/x)+(-27*x-144)*exp(4/x))/x^3,x, algorithm="fricas")

[Out]

-9/8*(e^(4/x)*log(2*log(2)/x) - 4*e^(4/x))/x

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giac [B]  time = 0.38, size = 43, normalized size = 1.72 \begin {gather*} -\frac {9 \, {\left (e^{\frac {4}{x}} \log \relax (2) - e^{\frac {4}{x}} \log \relax (x) + e^{\frac {4}{x}} \log \left (\log \relax (2)\right ) - 4 \, e^{\frac {4}{x}}\right )}}{8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((9*x+36)*exp(4/x)*log(2*log(2)/x)+(-27*x-144)*exp(4/x))/x^3,x, algorithm="giac")

[Out]

-9/8*(e^(4/x)*log(2) - e^(4/x)*log(x) + e^(4/x)*log(log(2)) - 4*e^(4/x))/x

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maple [A]  time = 0.10, size = 32, normalized size = 1.28

method	result	size
norman	\(\frac {\frac {9 x \,{\mathrm e}^{\frac {4}{x}}}{2}-\frac {9 x \,{\mathrm e}^{\frac {4}{x}} \ln \left (\frac {2 \ln \relax (2)}{x}\right )}{8}}{x^{2}}\)	\(32\)
risch	\(\frac {9 \,{\mathrm e}^{\frac {4}{x}} \ln \relax (x )}{8 x}-\frac {9 \left (-8+2 \ln \relax (2)+2 \ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{\frac {4}{x}}}{16 x}\)	\(37\)
default	\(\frac {\left (36-9 \ln \left (\frac {2 \ln \relax (2)}{x}\right )-9 \ln \relax (x )\right ) x \,{\mathrm e}^{\frac {4}{x}}+9 x \,{\mathrm e}^{\frac {4}{x}} \ln \relax (x )}{8 x^{2}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*((9*x+36)*exp(4/x)*ln(2*ln(2)/x)+(-27*x-144)*exp(4/x))/x^3,x,method=_RETURNVERBOSE)

[Out]

(9/2*x*exp(4/x)-9/8*x*exp(4/x)*ln(2*ln(2)/x))/x^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {9}{8} \, \Gamma \left (2, -\frac {4}{x}\right ) - \frac {9}{8} \, \int -\frac {{\left (x {\left (\log \relax (2) + \log \left (\log \relax (2)\right ) - 3\right )} - {\left (x + 4\right )} \log \relax (x) + 4 \, \log \relax (2) + 4 \, \log \left (\log \relax (2)\right )\right )} e^{\frac {4}{x}}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((9*x+36)*exp(4/x)*log(2*log(2)/x)+(-27*x-144)*exp(4/x))/x^3,x, algorithm="maxima")

[Out]

-9/8*gamma(2, -4/x) - 9/8*integrate(-(x*(log(2) + log(log(2)) - 3) - (x + 4)*log(x) + 4*log(2) + 4*log(log(2))
)*e^(4/x)/x^3, x)

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mupad [B]  time = 0.85, size = 21, normalized size = 0.84 \begin {gather*} -\frac {9\,{\mathrm {e}}^{4/x}\,\left (\ln \left (\frac {2\,\ln \relax (2)}{x}\right )-4\right )}{8\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(4/x)*(27*x + 144))/8 - (log((2*log(2))/x)*exp(4/x)*(9*x + 36))/8)/x^3,x)

[Out]

-(9*exp(4/x)*(log((2*log(2))/x) - 4))/(8*x)

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sympy [A]  time = 0.31, size = 19, normalized size = 0.76 \begin {gather*} \frac {\left (36 - 9 \log {\left (\frac {2 \log {\relax (2 )}}{x} \right )}\right ) e^{\frac {4}{x}}}{8 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((9*x+36)*exp(4/x)*ln(2*ln(2)/x)+(-27*x-144)*exp(4/x))/x**3,x)

[Out]

(36 - 9*log(2*log(2)/x))*exp(4/x)/(8*x)

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