3.36.94 \(\int \frac {x-\frac {8 e^{5-e^3}}{x^2 \log (4)}}{4 x} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{4} (-5+x)+\frac {e^{5-e^3}}{x^2 \log (4)} \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 14} \begin {gather*} \frac {e^{5-e^3}}{x^2 \log (4)}+\frac {x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x - (8*E^(5 - E^3))/(x^2*Log[4]))/(4*x),x]

[Out]

x/4 + E^(5 - E^3)/(x^2*Log[4])

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {x-\frac {8 e^{5-e^3}}{x^2 \log (4)}}{x} \, dx\\ &=\frac {1}{4} \int \left (1-\frac {8 e^{5-e^3}}{x^3 \log (4)}\right ) \, dx\\ &=\frac {x}{4}+\frac {e^{5-e^3}}{x^2 \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.92 \begin {gather*} \frac {x}{4}+\frac {e^{5-e^3}}{x^2 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - (8*E^(5 - E^3))/(x^2*Log[4]))/(4*x),x]

[Out]

x/4 + E^(5 - E^3)/(x^2*Log[4])

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fricas [A]  time = 0.95, size = 29, normalized size = 1.16 \begin {gather*} \frac {{\left (x^{3} e^{\left (e^{3} - 5\right )} \log \relax (2) + 2\right )} e^{\left (-e^{3} + 5\right )}}{4 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-8*exp(-log(2*x^2*log(2))-exp(3)+5)+x)/x,x, algorithm="fricas")

[Out]

1/4*(x^3*e^(e^3 - 5)*log(2) + 2)*e^(-e^3 + 5)/(x^2*log(2))

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giac [A]  time = 0.13, size = 20, normalized size = 0.80 \begin {gather*} \frac {1}{4} \, x + \frac {e^{\left (-e^{3} + 5\right )}}{2 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-8*exp(-log(2*x^2*log(2))-exp(3)+5)+x)/x,x, algorithm="giac")

[Out]

1/4*x + 1/2*e^(-e^3 + 5)/(x^2*log(2))

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maple [A]  time = 0.05, size = 21, normalized size = 0.84

method	result	size
risch	\(\frac {x}{4}+\frac {{\mathrm e}^{-{\mathrm e}^{3}+5}}{2 x^{2} \ln \relax (2)}\)	\(21\)
default	\(\frac {x}{4}+{\mathrm e}^{-\ln \left (2 x^{2} \ln \relax (2)\right )-{\mathrm e}^{3}+5}\)	\(22\)
norman	\(\frac {\frac {x^{3}}{4}+\frac {{\mathrm e}^{-{\mathrm e}^{3}} {\mathrm e}^{5}}{2 \ln \relax (2)}}{x^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-8*exp(-ln(2*x^2*ln(2))-exp(3)+5)+x)/x,x,method=_RETURNVERBOSE)

[Out]

1/4*x+1/2*exp(-exp(3)+5)/x^2/ln(2)

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maxima [A]  time = 0.49, size = 20, normalized size = 0.80 \begin {gather*} \frac {1}{4} \, x + \frac {e^{\left (-e^{3} + 5\right )}}{2 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-8*exp(-log(2*x^2*log(2))-exp(3)+5)+x)/x,x, algorithm="maxima")

[Out]

1/4*x + 1/2*e^(-e^3 + 5)/(x^2*log(2))

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mupad [B]  time = 2.18, size = 20, normalized size = 0.80 \begin {gather*} \frac {x}{4}+\frac {{\mathrm {e}}^{5-{\mathrm {e}}^3}}{2\,x^2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x/4 - 2*exp(5 - exp(3) - log(2*x^2*log(2))))/x,x)

[Out]

x/4 + exp(5 - exp(3))/(2*x^2*log(2))

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sympy [A]  time = 0.11, size = 27, normalized size = 1.08 \begin {gather*} \frac {x e^{e^{3}} \log {\relax (2 )} + \frac {2 e^{5}}{x^{2}}}{4 e^{e^{3}} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-8*exp(-ln(2*x**2*ln(2))-exp(3)+5)+x)/x,x)

[Out]

(x*exp(exp(3))*log(2) + 2*exp(5)/x**2)*exp(-exp(3))/(4*log(2))

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