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

Optimal. Leaf size=17 \[ -1+\log \left (e^{5+\frac {1}{15} x (11+x)}+x\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 1, number of rules used = 1, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6684} \begin {gather*} \log \left (e^{\frac {x^2}{15}+\frac {11 x}{15}+5}+x\right ) \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]  time = 0.16, size = 19, normalized size = 1.12 \begin {gather*} \log \left (e^{5+\frac {11 x}{15}+\frac {x^2}{15}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[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^(5 + (11*x)/15 + x^2/15) + x]

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fricas [A]  time = 0.61, size = 14, normalized size = 0.82 \begin {gather*} \log \left (x + e^{\left (\frac {1}{15} \, x^{2} + \frac {11}{15} \, x + 5\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.25, size = 14, normalized size = 0.82 \begin {gather*} \log \left (x + e^{\left (\frac {1}{15} \, x^{2} + \frac {11}{15} \, x + 5\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.06, size = 17, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maxima [A]  time = 0.38, size = 25, normalized size = 1.47 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.16, size = 14, normalized size = 0.82 \begin {gather*} \ln \left (x+{\mathrm {e}}^{\frac {x^2}{15}+\frac {11\,x}{15}+5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.15, size = 29, normalized size = 1.71 \begin {gather*} \frac {14 x^{2}}{225} + \frac {154 x}{225} + \frac {\log {\left (x + e^{\frac {x^{2}}{15} + \frac {11 x}{15} + 5} \right )}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

14*x**2/225 + 154*x/225 + log(x + exp(x**2/15 + 11*x/15 + 5))/15

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