3.17.43 \(\int \frac {1}{4} (-4+e^{\frac {1}{4} (e^x+4 x)} (4+e^x)-4 \log (x)) \, dx\)

Optimal. Leaf size=19 \[ e^{\frac {1}{4} \left (e^x+4 x\right )}-x \log (x) \]

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Rubi [A]  time = 0.05, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 6706, 2295} \begin {gather*} e^{\frac {1}{4} \left (4 x+e^x\right )}-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^((E^x + 4*x)/4) - x*Log[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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.89 \begin {gather*} e^{\frac {e^x}{4}+x}-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(E^x/4 + x) - x*Log[x]

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fricas [A]  time = 0.64, size = 13, normalized size = 0.68 \begin {gather*} -x \log \relax (x) + e^{\left (x + \frac {1}{4} \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(exp(x)+4)*exp(x+1/4*exp(x))-log(x)-1,x, algorithm="fricas")

[Out]

-x*log(x) + e^(x + 1/4*e^x)

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giac [A]  time = 0.15, size = 13, normalized size = 0.68 \begin {gather*} -x \log \relax (x) + e^{\left (x + \frac {1}{4} \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(exp(x)+4)*exp(x+1/4*exp(x))-log(x)-1,x, algorithm="giac")

[Out]

-x*log(x) + e^(x + 1/4*e^x)

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maple [A]  time = 0.03, size = 14, normalized size = 0.74

method	result	size
norman	\({\mathrm e}^{x +\frac {{\mathrm e}^{x}}{4}}-x \ln \relax (x )\)	\(14\)
risch	\({\mathrm e}^{x +\frac {{\mathrm e}^{x}}{4}}-x \ln \relax (x )\)	\(14\)
default	\({\mathrm e}^{x} {\mathrm e}^{\frac {{\mathrm e}^{x}}{4}}-x \ln \relax (x )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(exp(x)+4)*exp(x+1/4*exp(x))-ln(x)-1,x,method=_RETURNVERBOSE)

[Out]

exp(x+1/4*exp(x))-x*ln(x)

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maxima [A]  time = 0.39, size = 13, normalized size = 0.68 \begin {gather*} -x \log \relax (x) + e^{\left (x + \frac {1}{4} \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(exp(x)+4)*exp(x+1/4*exp(x))-log(x)-1,x, algorithm="maxima")

[Out]

-x*log(x) + e^(x + 1/4*e^x)

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mupad [B]  time = 1.14, size = 13, normalized size = 0.68 \begin {gather*} {\mathrm {e}}^{x+\frac {{\mathrm {e}}^x}{4}}-x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + exp(x)/4)*(exp(x) + 4))/4 - log(x) - 1,x)

[Out]

exp(x + exp(x)/4) - x*log(x)

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sympy [A]  time = 0.27, size = 12, normalized size = 0.63 \begin {gather*} - x \log {\relax (x )} + e^{x + \frac {e^{x}}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(exp(x)+4)*exp(x+1/4*exp(x))-ln(x)-1,x)

[Out]

-x*log(x) + exp(x + exp(x)/4)

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