3.102.58 \(\int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log (\frac {\log (2)}{4})}{4 x^3} \, dx\)

Optimal. Leaf size=20 \[ \frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \]

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Rubi [A]  time = 0.16, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2177, 2178} \begin {gather*} \frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x/4)*(-128 + 16*x) + E^(x/4)*(-8 + x)*Log[Log[2]/4])/(4*x^3),x]

[Out]

(E^(x/4)*(16 + Log[Log[2]/4]))/x^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 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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{x^3} \, dx\\ &=\frac {1}{4} \int \left (-\frac {128 e^{x/4} \left (1+\frac {1}{16} \log \left (\frac {\log (2)}{4}\right )\right )}{x^3}+\frac {16 e^{x/4} \left (1+\frac {1}{16} \log \left (\frac {\log (2)}{4}\right )\right )}{x^2}\right ) \, dx\\ &=\frac {1}{4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4}}{x^2} \, dx-\left (2 \left (16+\log \left (\frac {\log (2)}{4}\right )\right )\right ) \int \frac {e^{x/4}}{x^3} \, dx\\ &=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2}-\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{4 x}+\frac {1}{16} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4}}{x} \, dx-\frac {1}{4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4}}{x^2} \, dx\\ &=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2}+\frac {1}{16} \text {Ei}\left (\frac {x}{4}\right ) \left (16+\log \left (\frac {\log (2)}{4}\right )\right )-\frac {1}{16} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4}}{x} \, dx\\ &=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} \frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/4)*(-128 + 16*x) + E^(x/4)*(-8 + x)*Log[Log[2]/4])/(4*x^3),x]

[Out]

(E^(x/4)*(16 + Log[Log[2]/4]))/x^2

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fricas [A]  time = 0.56, size = 21, normalized size = 1.05 \begin {gather*} \frac {e^{\left (\frac {1}{4} \, x\right )} \log \left (\frac {1}{4} \, \log \relax (2)\right ) + 16 \, e^{\left (\frac {1}{4} \, x\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8+x)*exp(1/4*x)*log(1/4*log(2))+(16*x-128)*exp(1/4*x))/x^3,x, algorithm="fricas")

[Out]

(e^(1/4*x)*log(1/4*log(2)) + 16*e^(1/4*x))/x^2

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giac [A]  time = 0.15, size = 29, normalized size = 1.45 \begin {gather*} -\frac {2 \, e^{\left (\frac {1}{4} \, x\right )} \log \relax (2) - e^{\left (\frac {1}{4} \, x\right )} \log \left (\log \relax (2)\right ) - 16 \, e^{\left (\frac {1}{4} \, x\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8+x)*exp(1/4*x)*log(1/4*log(2))+(16*x-128)*exp(1/4*x))/x^3,x, algorithm="giac")

[Out]

-(2*e^(1/4*x)*log(2) - e^(1/4*x)*log(log(2)) - 16*e^(1/4*x))/x^2

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maple [A]  time = 0.11, size = 16, normalized size = 0.80

method	result	size
gosper	\(\frac {\left (16+\ln \left (\frac {\ln \relax (2)}{4}\right )\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\)	\(16\)
norman	\(\frac {\left (-2 \ln \relax (2)+\ln \left (\ln \relax (2)\right )+16\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\)	\(18\)
risch	\(-\frac {\left (2 \ln \relax (2)-\ln \left (\ln \relax (2)\right )-16\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\)	\(21\)
derivativedivides	\(\frac {16 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}} \ln \relax (2)}{x^{2}}+\frac {{\mathrm e}^{\frac {x}{4}} \ln \left (\ln \relax (2)\right )}{x^{2}}\)	\(33\)
default	\(\frac {16 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}} \ln \relax (2)}{x^{2}}+\frac {{\mathrm e}^{\frac {x}{4}} \ln \left (\ln \relax (2)\right )}{x^{2}}\)	\(33\)
meijerg	\(-4 \left (\frac {1}{4}+\frac {\ln \left (\frac {\ln \relax (2)}{4}\right )}{64}\right ) \left (-\frac {2 \left (2+\frac {x}{2}\right )}{x}+\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}+\ln \left (-\frac {x}{4}\right )+\expIntegralEi \left (1, -\frac {x}{4}\right )+1-\ln \relax (x )+2 \ln \relax (2)-i \pi +\frac {4}{x}\right )-4 \left (\frac {\ln \left (\frac {\ln \relax (2)}{4}\right )}{32}+\frac {1}{2}\right ) \left (\frac {\frac {3}{4} x^{2}+4 x +8}{x^{2}}-\frac {8 \left (3+\frac {3 x}{4}\right ) {\mathrm e}^{\frac {x}{4}}}{3 x^{2}}-\frac {\ln \left (-\frac {x}{4}\right )}{2}-\frac {\expIntegralEi \left (1, -\frac {x}{4}\right )}{2}-\frac {3}{4}+\frac {\ln \relax (x )}{2}-\ln \relax (2)+\frac {i \pi }{2}-\frac {8}{x^{2}}-\frac {4}{x}\right )\)	\(137\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-8+x)*exp(1/4*x)*ln(1/4*ln(2))+(16*x-128)*exp(1/4*x))/x^3,x,method=_RETURNVERBOSE)

[Out]

(16+ln(1/4*ln(2)))/x^2*exp(1/4*x)

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maxima [C]  time = 0.37, size = 37, normalized size = 1.85 \begin {gather*} \frac {1}{16} \, \Gamma \left (-1, -\frac {1}{4} \, x\right ) \log \left (\frac {1}{4} \, \log \relax (2)\right ) + \frac {1}{8} \, \Gamma \left (-2, -\frac {1}{4} \, x\right ) \log \left (\frac {1}{4} \, \log \relax (2)\right ) + \Gamma \left (-1, -\frac {1}{4} \, x\right ) + 2 \, \Gamma \left (-2, -\frac {1}{4} \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8+x)*exp(1/4*x)*log(1/4*log(2))+(16*x-128)*exp(1/4*x))/x^3,x, algorithm="maxima")

[Out]

1/16*gamma(-1, -1/4*x)*log(1/4*log(2)) + 1/8*gamma(-2, -1/4*x)*log(1/4*log(2)) + gamma(-1, -1/4*x) + 2*gamma(-
2, -1/4*x)

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mupad [B]  time = 0.07, size = 15, normalized size = 0.75 \begin {gather*} \frac {{\mathrm {e}}^{x/4}\,\left (\ln \left (\frac {\ln \relax (2)}{4}\right )+16\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x/4)*(16*x - 128))/4 + (log(log(2)/4)*exp(x/4)*(x - 8))/4)/x^3,x)

[Out]

(exp(x/4)*(log(log(2)/4) + 16))/x^2

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sympy [A]  time = 0.14, size = 19, normalized size = 0.95 \begin {gather*} \frac {\left (- 2 \log {\relax (2 )} + \log {\left (\log {\relax (2 )} \right )} + 16\right ) e^{\frac {x}{4}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-8+x)*exp(1/4*x)*ln(1/4*ln(2))+(16*x-128)*exp(1/4*x))/x**3,x)

[Out]

(-2*log(2) + log(log(2)) + 16)*exp(x/4)/x**2

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