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

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

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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} $$\begin{array}{c}{e}^{\frac{1}{4}(4x+e^x)}-x\log (x)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{4}\int (-4+e^{\frac{1}{4}(e^x+4x)}(4+e^x)-4\log (x))dx\\ & =-x+\frac{1}{4}\int e^{\frac{1}{4}(e^x+4x)}(4+e^x)dx-\int \log (x)dx\\ & =e^{\frac{1}{4}(e^x+4x)}-x\log (x)\end{array}\end{array}$$

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

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 $$\begin{array}{c}-x\log (x)+e^{(x+\frac{1}{4}{e}^x)}\end{array}$$

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 $$\begin{array}{c}-x\log (x)+e^{(x+\frac{1}{4}{e}^x)}\end{array}$$

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	${\mathrm{e}}^{x+\frac{{\mathrm{e}}^x}{4}}-x\ln (x)$	$14$
risch	${\mathrm{e}}^{x+\frac{{\mathrm{e}}^x}{4}}-x\ln (x)$	$14$
default	${\mathrm{e}}^x{\mathrm{e}}^{\frac{{\mathrm{e}}^x}{4}}-x\ln (x)$	$15$

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 $$\begin{array}{c}-x\log (x)+e^{(x+\frac{1}{4}{e}^x)}\end{array}$$

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 $$\begin{array}{c}{\mathrm{e}}^{x+\frac{{\mathrm{e}}^x}{4}}-x\ln (x)\end{array}$$

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 $$\begin{array}{c}-x\log (x)+e^{x+\frac{e^x}{4}}\end{array}$$

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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