3.33.23 \(\int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ 4+\frac {1}{e^{10} (-x-\log (x))} \]

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Rubi [A]  time = 0.12, antiderivative size = 11, normalized size of antiderivative = 0.69, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {1}{e^{10} (x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x)/(E^10*x^3 + 2*E^10*x^2*Log[x] + E^10*x*Log[x]^2),x]

[Out]

-(1/(E^10*(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 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 {1+x}{e^{10} x (x+\log (x))^2} \, dx\\ &=\frac {\int \frac {1+x}{x (x+\log (x))^2} \, dx}{e^{10}}\\ &=-\frac {1}{e^{10} (x+\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 11, normalized size = 0.69 \begin {gather*} -\frac {1}{e^{10} (x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)/(E^10*x^3 + 2*E^10*x^2*Log[x] + E^10*x*Log[x]^2),x]

[Out]

-(1/(E^10*(x + Log[x])))

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fricas [A]  time = 0.47, size = 14, normalized size = 0.88 \begin {gather*} -\frac {1}{x e^{10} + e^{10} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+1)/(x*exp(10)*log(x)^2+2*x^2*exp(10)*log(x)+x^3*exp(10)),x, algorithm="fricas")

[Out]

-1/(x*e^10 + e^10*log(x))

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giac [A]  time = 0.27, size = 14, normalized size = 0.88 \begin {gather*} -\frac {1}{x e^{10} + e^{10} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+1)/(x*exp(10)*log(x)^2+2*x^2*exp(10)*log(x)+x^3*exp(10)),x, algorithm="giac")

[Out]

-1/(x*e^10 + e^10*log(x))

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maple [A]  time = 0.06, size = 11, normalized size = 0.69

method	result	size
risch	\(-\frac {{\mathrm e}^{-10}}{x +\ln \relax (x )}\)	\(11\)
norman	\(-\frac {{\mathrm e}^{-10}}{x +\ln \relax (x )}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+1)/(x*exp(10)*ln(x)^2+2*x^2*exp(10)*ln(x)+x^3*exp(10)),x,method=_RETURNVERBOSE)

[Out]

-exp(-10)/(x+ln(x))

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maxima [A]  time = 0.37, size = 14, normalized size = 0.88 \begin {gather*} -\frac {1}{x e^{10} + e^{10} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+1)/(x*exp(10)*log(x)^2+2*x^2*exp(10)*log(x)+x^3*exp(10)),x, algorithm="maxima")

[Out]

-1/(x*e^10 + e^10*log(x))

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mupad [B]  time = 2.10, size = 10, normalized size = 0.62 \begin {gather*} -\frac {{\mathrm {e}}^{-10}}{x+\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)/(x^3*exp(10) + x*exp(10)*log(x)^2 + 2*x^2*exp(10)*log(x)),x)

[Out]

-exp(-10)/(x + log(x))

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sympy [A]  time = 0.11, size = 14, normalized size = 0.88 \begin {gather*} - \frac {1}{x e^{10} + e^{10} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+1)/(x*exp(10)*ln(x)**2+2*x**2*exp(10)*ln(x)+x**3*exp(10)),x)

[Out]

-1/(x*exp(10) + exp(10)*log(x))

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