3.89.75 $\int \frac{-10+5\log (3)+e^{e^x+x}(-10+5\log (3))}{e^{e^x}+x}dx$

Optimal. Leaf size=24 $$(-10+5\log (3))(2+\log \left(\frac{1}{3}{e}^5(e^{e^x}+x)\right))$$

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Rubi [A]  time = 0.06, antiderivative size = 16, normalized size of antiderivative = 0.67, number of steps used = 1, number of rules used = 1, integrand size = 30, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.033, Rules used = {6684} $$\begin{array}{c}-5(2-\log (3))\log (x+e^{e^x})\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-10 + 5*Log[3] + E^(E^x + x)*(-10 + 5*Log[3]))/(E^E^x + x),x]

[Out]

-5*(2 - Log[3])*Log[E^E^x + x]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =-5(2-\log (3))\log (e^{e^x}+x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.07, size = 14, normalized size = 0.58 $$\begin{array}{c}5(-2+\log (3))\log (e^{e^x}+x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-10 + 5*Log[3] + E^(E^x + x)*(-10 + 5*Log[3]))/(E^E^x + x),x]

[Out]

5*(-2 + Log[3])*Log[E^E^x + x]

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fricas [A]  time = 0.45, size = 26, normalized size = 1.08 $$\begin{array}{c}-5x\log (3)+5(\log (3)-2)\log (x{e}^x+e^{(x+e^x)})+10x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(3)-10)*exp(x)*exp(exp(x))+5*log(3)-10)/(x+exp(exp(x))),x, algorithm="fricas")

[Out]

-5*x*log(3) + 5*(log(3) - 2)*log(x*e^x + e^(x + e^x)) + 10*x

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giac [B]  time = 0.22, size = 37, normalized size = 1.54 $$\begin{array}{c}-5x\log (3)+5\log (3)\log (x{e}^x+e^{(x+e^x)})+10x-10\log (x{e}^x+e^{(x+e^x)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(3)-10)*exp(x)*exp(exp(x))+5*log(3)-10)/(x+exp(exp(x))),x, algorithm="giac")

[Out]

-5*x*log(3) + 5*log(3)*log(x*e^x + e^(x + e^x)) + 10*x - 10*log(x*e^x + e^(x + e^x))

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maple [A]  time = 0.05, size = 14, normalized size = 0.58

method	result	size
norman	$(5\ln (3)-10)\ln (x+{\mathrm{e}}^{{\mathrm{e}}^x})$	$14$
risch	$5\ln (x+{\mathrm{e}}^{{\mathrm{e}}^x})\ln (3)-10\ln (x+{\mathrm{e}}^{{\mathrm{e}}^x})$	$20$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*ln(3)-10)*exp(x)*exp(exp(x))+5*ln(3)-10)/(x+exp(exp(x))),x,method=_RETURNVERBOSE)

[Out]

(5*ln(3)-10)*ln(x+exp(exp(x)))

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maxima [A]  time = 0.49, size = 12, normalized size = 0.50 $$\begin{array}{c}5(\log (3)-2)\log (x+e^{\left(e^x\right)})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*log(3)-10)*exp(x)*exp(exp(x))+5*log(3)-10)/(x+exp(exp(x))),x, algorithm="maxima")

[Out]

5*(log(3) - 2)*log(x + e^(e^x))

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mupad [B]  time = 5.16, size = 13, normalized size = 0.54 $$\begin{array}{c}\ln (x+{\mathrm{e}}^{{\mathrm{e}}^x})(5\ln (3)-10)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*log(3) + exp(exp(x))*exp(x)*(5*log(3) - 10) - 10)/(x + exp(exp(x))),x)

[Out]

log(x + exp(exp(x)))*(5*log(3) - 10)

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sympy [A]  time = 0.14, size = 14, normalized size = 0.58 $$\begin{array}{c}5(-2+\log (3))\log (x+e^{e^x})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*ln(3)-10)*exp(x)*exp(exp(x))+5*ln(3)-10)/(x+exp(exp(x))),x)

[Out]

5*(-2 + log(3))*log(x + exp(exp(x)))

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