3.62.85 \(\int \frac {1+e^{8/3} (-7-2 x)-7 x-2 x^2}{e^{8/3}+x} \, dx\)

Optimal. Leaf size=20 \[ 10-x-(3+x)^2+\log \left (e^{8/3}+x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 1, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {1850} \begin {gather*} -x^2-7 x+\log \left (x+e^{8/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^(8/3)*(-7 - 2*x) - 7*x - 2*x^2)/(E^(8/3) + x),x]

[Out]

-7*x - x^2 + Log[E^(8/3) + x]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-7-2 x+\frac {1}{e^{8/3}+x}\right ) \, dx\\ &=-7 x-x^2+\log \left (e^{8/3}+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.35 \begin {gather*} -7 e^{8/3}+e^{16/3}-x (7+x)+\log \left (e^{8/3}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^(8/3)*(-7 - 2*x) - 7*x - 2*x^2)/(E^(8/3) + x),x]

[Out]

-7*E^(8/3) + E^(16/3) - x*(7 + x) + Log[E^(8/3) + x]

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fricas [A]  time = 1.01, size = 14, normalized size = 0.70 \begin {gather*} -x^{2} - 7 \, x + \log \left (x + e^{\frac {8}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-7)*exp(8/3)-2*x^2-7*x+1)/(exp(8/3)+x),x, algorithm="fricas")

[Out]

-x^2 - 7*x + log(x + e^(8/3))

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giac [A]  time = 0.19, size = 15, normalized size = 0.75 \begin {gather*} -x^{2} - 7 \, x + \log \left ({\left | x + e^{\frac {8}{3}} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-7)*exp(8/3)-2*x^2-7*x+1)/(exp(8/3)+x),x, algorithm="giac")

[Out]

-x^2 - 7*x + log(abs(x + e^(8/3)))

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maple [A]  time = 0.38, size = 15, normalized size = 0.75

method	result	size
default	\(-x^{2}-7 x +\ln \left ({\mathrm e}^{\frac {8}{3}}+x \right )\)	\(15\)
norman	\(-x^{2}-7 x +\ln \left ({\mathrm e}^{\frac {8}{3}}+x \right )\)	\(15\)
risch	\(-x^{2}-7 x +\ln \left ({\mathrm e}^{\frac {8}{3}}+x \right )\)	\(15\)
meijerg	\(-7 \,{\mathrm e}^{\frac {8}{3}} \ln \left (1+x \,{\mathrm e}^{-\frac {8}{3}}\right )-2 \,{\mathrm e}^{\frac {16}{3}} \left (-\frac {x \,{\mathrm e}^{-\frac {8}{3}} \left (-3 x \,{\mathrm e}^{-\frac {8}{3}}+6\right )}{6}+\ln \left (1+x \,{\mathrm e}^{-\frac {8}{3}}\right )\right )+{\mathrm e}^{\frac {8}{3}} \left (-2 \,{\mathrm e}^{\frac {8}{3}}-7\right ) \left (x \,{\mathrm e}^{-\frac {8}{3}}-\ln \left (1+x \,{\mathrm e}^{-\frac {8}{3}}\right )\right )+\ln \left (1+x \,{\mathrm e}^{-\frac {8}{3}}\right )\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-7)*exp(8/3)-2*x^2-7*x+1)/(exp(8/3)+x),x,method=_RETURNVERBOSE)

[Out]

-x^2-7*x+ln(exp(8/3)+x)

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maxima [A]  time = 0.34, size = 14, normalized size = 0.70 \begin {gather*} -x^{2} - 7 \, x + \log \left (x + e^{\frac {8}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-7)*exp(8/3)-2*x^2-7*x+1)/(exp(8/3)+x),x, algorithm="maxima")

[Out]

-x^2 - 7*x + log(x + e^(8/3))

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mupad [B]  time = 0.07, size = 14, normalized size = 0.70 \begin {gather*} \ln \left (x+{\mathrm {e}}^{8/3}\right )-7\,x-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(7*x + 2*x^2 + exp(8/3)*(2*x + 7) - 1)/(x + exp(8/3)),x)

[Out]

log(x + exp(8/3)) - 7*x - x^2

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sympy [A]  time = 0.10, size = 14, normalized size = 0.70 \begin {gather*} - x^{2} - 7 x + \log {\left (x + e^{\frac {8}{3}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-7)*exp(8/3)-2*x**2-7*x+1)/(exp(8/3)+x),x)

[Out]

-x**2 - 7*x + log(x + exp(8/3))

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