3.54.45 \(\int -\frac {18}{e^5+e^2 (-5-2 x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {9 \log \left (\frac {5}{2} \left (-5+e^3-2 x\right )\right )}{e^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 33, 31} \begin {gather*} \frac {9 \log \left (2 x-e^3+5\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-18/(E^5 + E^2*(-5 - 2*x)),x]

[Out]

(9*Log[5 - E^3 + 2*x])/E^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 33

Int[((a_.) + (b_.)*(u_))^(m_), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(a + b*x)^m, x], x, u], x]
/; FreeQ[{a, b, m}, x] && LinearQ[u, x] && NeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (18 \int \frac {1}{e^5+e^2 (-5-2 x)} \, dx\right )\\ &=9 \operatorname {Subst}\left (\int \frac {1}{e^5+e^2 x} \, dx,x,-5-2 x\right )\\ &=\frac {9 \log \left (5-e^3+2 x\right )}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.78 \begin {gather*} \frac {9 \log \left (-5+e^3-2 x\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-18/(E^5 + E^2*(-5 - 2*x)),x]

[Out]

(9*Log[-5 + E^3 - 2*x])/E^2

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fricas [A]  time = 0.63, size = 14, normalized size = 0.78 \begin {gather*} 9 \, e^{\left (-2\right )} \log \left (2 \, x - e^{3} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-18/(exp(2)*exp(3)+(-2*x-5)*exp(2)),x, algorithm="fricas")

[Out]

9*e^(-2)*log(2*x - e^3 + 5)

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giac [A]  time = 0.15, size = 19, normalized size = 1.06 \begin {gather*} 9 \, e^{\left (-2\right )} \log \left ({\left | {\left (2 \, x + 5\right )} e^{2} - e^{5} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-18/(exp(2)*exp(3)+(-2*x-5)*exp(2)),x, algorithm="giac")

[Out]

9*e^(-2)*log(abs((2*x + 5)*e^2 - e^5))

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maple [A]  time = 0.23, size = 15, normalized size = 0.83

method	result	size
norman	\(9 \,{\mathrm e}^{-2} \ln \left ({\mathrm e}^{3}-2 x -5\right )\)	\(15\)
risch	\(9 \,{\mathrm e}^{-2} \ln \left (2 x +5-{\mathrm e}^{3}\right )\)	\(15\)
meijerg	\(9 \,{\mathrm e}^{-2} \ln \left (1-\frac {2 x \,{\mathrm e}^{2}}{{\mathrm e}^{5}-5 \,{\mathrm e}^{2}}\right )\)	\(22\)
default	\(9 \ln \left (-2 \,{\mathrm e}^{2} x +{\mathrm e}^{2} {\mathrm e}^{3}-5 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-18/(exp(2)*exp(3)+(-2*x-5)*exp(2)),x,method=_RETURNVERBOSE)

[Out]

9/exp(2)*ln(exp(3)-2*x-5)

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maxima [A]  time = 0.36, size = 18, normalized size = 1.00 \begin {gather*} 9 \, e^{\left (-2\right )} \log \left ({\left (2 \, x + 5\right )} e^{2} - e^{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-18/(exp(2)*exp(3)+(-2*x-5)*exp(2)),x, algorithm="maxima")

[Out]

9*e^(-2)*log((2*x + 5)*e^2 - e^5)

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mupad [B]  time = 0.25, size = 14, normalized size = 0.78 \begin {gather*} 9\,{\mathrm {e}}^{-2}\,\ln \left (2\,x-{\mathrm {e}}^3+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-18/(exp(5) - exp(2)*(2*x + 5)),x)

[Out]

9*exp(-2)*log(2*x - exp(3) + 5)

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sympy [A]  time = 0.07, size = 20, normalized size = 1.11 \begin {gather*} \frac {9 \log {\left (2 x e^{2} - e^{5} + 5 e^{2} \right )}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-18/(exp(2)*exp(3)+(-2*x-5)*exp(2)),x)

[Out]

9*exp(-2)*log(2*x*exp(2) - exp(5) + 5*exp(2))

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