3.96.49 \(\int \frac {-25-50 e^x-25 x}{16 x-8 x^2+x^3+e^x (16-8 x+x^2)+(-8 x+2 x^2+e^x (-8+2 x)) \log (e^x+x)+(e^x+x) \log ^2(e^x+x)} \, dx\)

Optimal. Leaf size=13 \[ \frac {25}{-4+x+\log \left (e^x+x\right )} \]

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Rubi [A]  time = 0.22, antiderivative size = 17, normalized size of antiderivative = 1.31, number of steps used = 3, number of rules used = 3, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {25}{-x-\log \left (x+e^x\right )+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 - 50*E^x - 25*x)/(16*x - 8*x^2 + x^3 + E^x*(16 - 8*x + x^2) + (-8*x + 2*x^2 + E^x*(-8 + 2*x))*Log[E^x
+ x] + (E^x + x)*Log[E^x + x]^2),x]

[Out]

-25/(4 - x - Log[E^x + 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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 \left (-1-2 e^x-x\right )}{\left (e^x+x\right ) \left (4-x-\log \left (e^x+x\right )\right )^2} \, dx\\ &=25 \int \frac {-1-2 e^x-x}{\left (e^x+x\right ) \left (4-x-\log \left (e^x+x\right )\right )^2} \, dx\\ &=-\frac {25}{4-x-\log \left (e^x+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 13, normalized size = 1.00 \begin {gather*} \frac {25}{-4+x+\log \left (e^x+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 50*E^x - 25*x)/(16*x - 8*x^2 + x^3 + E^x*(16 - 8*x + x^2) + (-8*x + 2*x^2 + E^x*(-8 + 2*x))*L
og[E^x + x] + (E^x + x)*Log[E^x + x]^2),x]

[Out]

25/(-4 + x + Log[E^x + x])

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fricas [A]  time = 0.58, size = 12, normalized size = 0.92 \begin {gather*} \frac {25}{x + \log \left (x + e^{x}\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(x)-25*x-25)/((exp(x)+x)*log(exp(x)+x)^2+((2*x-8)*exp(x)+2*x^2-8*x)*log(exp(x)+x)+(x^2-8*x+1
6)*exp(x)+x^3-8*x^2+16*x),x, algorithm="fricas")

[Out]

25/(x + log(x + e^x) - 4)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(x)-25*x-25)/((exp(x)+x)*log(exp(x)+x)^2+((2*x-8)*exp(x)+2*x^2-8*x)*log(exp(x)+x)+(x^2-8*x+1
6)*exp(x)+x^3-8*x^2+16*x),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.03, size = 13, normalized size = 1.00

method	result	size
risch	\(\frac {25}{\ln \left ({\mathrm e}^{x}+x \right )-4+x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-50*exp(x)-25*x-25)/((exp(x)+x)*ln(exp(x)+x)^2+((2*x-8)*exp(x)+2*x^2-8*x)*ln(exp(x)+x)+(x^2-8*x+16)*exp(x
)+x^3-8*x^2+16*x),x,method=_RETURNVERBOSE)

[Out]

25/(ln(exp(x)+x)-4+x)

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maxima [A]  time = 0.43, size = 12, normalized size = 0.92 \begin {gather*} \frac {25}{x + \log \left (x + e^{x}\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(x)-25*x-25)/((exp(x)+x)*log(exp(x)+x)^2+((2*x-8)*exp(x)+2*x^2-8*x)*log(exp(x)+x)+(x^2-8*x+1
6)*exp(x)+x^3-8*x^2+16*x),x, algorithm="maxima")

[Out]

25/(x + log(x + e^x) - 4)

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mupad [B]  time = 9.86, size = 12, normalized size = 0.92 \begin {gather*} \frac {25}{x+\ln \left (x+{\mathrm {e}}^x\right )-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x + 50*exp(x) + 25)/(16*x + exp(x)*(x^2 - 8*x + 16) + log(x + exp(x))*(exp(x)*(2*x - 8) - 8*x + 2*x^2
) + log(x + exp(x))^2*(x + exp(x)) - 8*x^2 + x^3),x)

[Out]

25/(x + log(x + exp(x)) - 4)

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sympy [A]  time = 0.16, size = 10, normalized size = 0.77 \begin {gather*} \frac {25}{x + \log {\left (x + e^{x} \right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*exp(x)-25*x-25)/((exp(x)+x)*ln(exp(x)+x)**2+((2*x-8)*exp(x)+2*x**2-8*x)*ln(exp(x)+x)+(x**2-8*x+
16)*exp(x)+x**3-8*x**2+16*x),x)

[Out]

25/(x + log(x + exp(x)) - 4)

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