\(\int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx\) [3223]

Optimal result

Integrand size = 33, antiderivative size = 16 \[ \int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx=4+\frac {1}{e^{10} (-x-\log (x))} \]

[Out]

exp(-10)/(-x-ln(x))+4

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6820, 12, 6818} \[ \int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx=-\frac {1}{e^{10} (x+\log (x))} \]

[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 6818

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 6820

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx=-\frac {1}{e^{10} (x+\log (x))} \]

[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])))

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

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

[In]

int((1+x)/(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))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx=-\frac {1}{x e^{10} + e^{10} \log \left (x\right )} \]

[In]

integrate((1+x)/(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))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx=- \frac {1}{x e^{10} + e^{10} \log {\left (x \right )}} \]

[In]

integrate((1+x)/(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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx=-\frac {1}{x e^{10} + e^{10} \log \left (x\right )} \]

[In]

integrate((1+x)/(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))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx=-\frac {1}{x e^{10} + e^{10} \log \left (x\right )} \]

[In]

integrate((1+x)/(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))

Mupad [B] (verification not implemented)

Time = 9.51 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {1+x}{e^{10} x^3+2 e^{10} x^2 \log (x)+e^{10} x \log ^2(x)} \, dx=-\frac {{\mathrm {e}}^{-10}}{x+\ln \left (x\right )} \]

[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))