\(\int \frac {15+e^{\frac {1}{15} (75+11 x+x^2)} (11+2 x)}{15 e^{\frac {1}{15} (75+11 x+x^2)}+15 x} \, dx\) [6713]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 45, antiderivative size = 17 \[ \int \frac {15+e^{\frac {1}{15} \left (75+11 x+x^2\right )} (11+2 x)}{15 e^{\frac {1}{15} \left (75+11 x+x^2\right )}+15 x} \, dx=-1+\log \left (e^{5+\frac {1}{15} x (11+x)}+x\right ) \]

[Out]

ln(exp(1/15*(11+x)*x+5)+x)-1

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6816} \[ \int \frac {15+e^{\frac {1}{15} \left (75+11 x+x^2\right )} (11+2 x)}{15 e^{\frac {1}{15} \left (75+11 x+x^2\right )}+15 x} \, dx=\log \left (e^{\frac {x^2}{15}+\frac {11 x}{15}+5}+x\right ) \]

[In]

Int[(15 + E^((75 + 11*x + x^2)/15)*(11 + 2*x))/(15*E^((75 + 11*x + x^2)/15) + 15*x),x]

[Out]

Log[E^(5 + (11*x)/15 + x^2/15) + x]

Rule 6816

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

Rubi steps \begin{align*} \text {integral}& = \log \left (e^{5+\frac {11 x}{15}+\frac {x^2}{15}}+x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {15+e^{\frac {1}{15} \left (75+11 x+x^2\right )} (11+2 x)}{15 e^{\frac {1}{15} \left (75+11 x+x^2\right )}+15 x} \, dx=\log \left (e^{\frac {1}{15} \left (75+11 x+x^2\right )}+x\right ) \]

[In]

Integrate[(15 + E^((75 + 11*x + x^2)/15)*(11 + 2*x))/(15*E^((75 + 11*x + x^2)/15) + 15*x),x]

[Out]

Log[E^((75 + 11*x + x^2)/15) + x]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method result size
parallelrisch \(\ln \left ({\mathrm e}^{\frac {1}{15} x^{2}+\frac {11}{15} x +5}+x \right )\) \(15\)
risch \(-5+\ln \left ({\mathrm e}^{\frac {1}{15} x^{2}+\frac {11}{15} x +5}+x \right )\) \(17\)
norman \(\ln \left (15 \,{\mathrm e}^{\frac {1}{15} x^{2}+\frac {11}{15} x +5}+15 x \right )\) \(19\)

[In]

int(((2*x+11)*exp(1/15*x^2+11/15*x+5)+15)/(15*exp(1/15*x^2+11/15*x+5)+15*x),x,method=_RETURNVERBOSE)

[Out]

ln(exp(1/15*x^2+11/15*x+5)+x)

Fricas [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {15+e^{\frac {1}{15} \left (75+11 x+x^2\right )} (11+2 x)}{15 e^{\frac {1}{15} \left (75+11 x+x^2\right )}+15 x} \, dx=\log \left (x + e^{\left (\frac {1}{15} \, x^{2} + \frac {11}{15} \, x + 5\right )}\right ) \]

[In]

integrate(((2*x+11)*exp(1/15*x^2+11/15*x+5)+15)/(15*exp(1/15*x^2+11/15*x+5)+15*x),x, algorithm="fricas")

[Out]

log(x + e^(1/15*x^2 + 11/15*x + 5))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {15+e^{\frac {1}{15} \left (75+11 x+x^2\right )} (11+2 x)}{15 e^{\frac {1}{15} \left (75+11 x+x^2\right )}+15 x} \, dx=\log {\left (x + e^{\frac {x^{2}}{15} + \frac {11 x}{15} + 5} \right )} \]

[In]

integrate(((2*x+11)*exp(1/15*x**2+11/15*x+5)+15)/(15*exp(1/15*x**2+11/15*x+5)+15*x),x)

[Out]

log(x + exp(x**2/15 + 11*x/15 + 5))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {15+e^{\frac {1}{15} \left (75+11 x+x^2\right )} (11+2 x)}{15 e^{\frac {1}{15} \left (75+11 x+x^2\right )}+15 x} \, dx=\frac {11}{15} \, x + \log \left ({\left (x + e^{\left (\frac {1}{15} \, x^{2} + \frac {11}{15} \, x + 5\right )}\right )} e^{\left (-\frac {11}{15} \, x - 5\right )}\right ) \]

[In]

integrate(((2*x+11)*exp(1/15*x^2+11/15*x+5)+15)/(15*exp(1/15*x^2+11/15*x+5)+15*x),x, algorithm="maxima")

[Out]

11/15*x + log((x + e^(1/15*x^2 + 11/15*x + 5))*e^(-11/15*x - 5))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {15+e^{\frac {1}{15} \left (75+11 x+x^2\right )} (11+2 x)}{15 e^{\frac {1}{15} \left (75+11 x+x^2\right )}+15 x} \, dx=\log \left (x + e^{\left (\frac {1}{15} \, x^{2} + \frac {11}{15} \, x + 5\right )}\right ) \]

[In]

integrate(((2*x+11)*exp(1/15*x^2+11/15*x+5)+15)/(15*exp(1/15*x^2+11/15*x+5)+15*x),x, algorithm="giac")

[Out]

log(x + e^(1/15*x^2 + 11/15*x + 5))

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {15+e^{\frac {1}{15} \left (75+11 x+x^2\right )} (11+2 x)}{15 e^{\frac {1}{15} \left (75+11 x+x^2\right )}+15 x} \, dx=\ln \left (x+{\mathrm {e}}^{\frac {x^2}{15}+\frac {11\,x}{15}+5}\right ) \]

[In]

int((exp((11*x)/15 + x^2/15 + 5)*(2*x + 11) + 15)/(15*x + 15*exp((11*x)/15 + x^2/15 + 5)),x)

[Out]

log(x + exp((11*x)/15 + x^2/15 + 5))