3.1.54 \(\int \frac {5+e^{2 e^{2 x} (1-e^3)+2 x-2 e^3 x} (2-2 e^3+e^{2 x} (4-4 e^3))}{-9+e^{2 e^{2 x} (1-e^3)+2 x-2 e^3 x}+5 x} \, dx\)

Optimal. Leaf size=30 \[ \log \left (4-e^{2 \left (1-e^3\right ) \left (e^{2 x}+x\right )}+5 (1-x)\right ) \]

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Rubi [A]  time = 0.42, antiderivative size = 34, normalized size of antiderivative = 1.13, number of steps used = 1, number of rules used = 1, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6684} \begin {gather*} \log \left (-5 x-e^{-2 e^3 x+2 x+2 \left (1-e^3\right ) e^{2 x}}+9\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + E^(2*E^(2*x)*(1 - E^3) + 2*x - 2*E^3*x)*(2 - 2*E^3 + E^(2*x)*(4 - 4*E^3)))/(-9 + E^(2*E^(2*x)*(1 - E^
3) + 2*x - 2*E^3*x) + 5*x),x]

[Out]

Log[9 - E^(2*E^(2*x)*(1 - E^3) + 2*x - 2*E^3*x) - 5*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 (9-e^{2 e^{2 x} \left (1-e^3\right )+2 x-2 e^3 x}-5 x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 22, normalized size = 0.73 \begin {gather*} \log \left (-9+e^{-2 \left (-1+e^3\right ) \left (e^{2 x}+x\right )}+5 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E^(2*E^(2*x)*(1 - E^3) + 2*x - 2*E^3*x)*(2 - 2*E^3 + E^(2*x)*(4 - 4*E^3)))/(-9 + E^(2*E^(2*x)*(
1 - E^3) + 2*x - 2*E^3*x) + 5*x),x]

[Out]

Log[-9 + E^(-2*(-1 + E^3)*(E^(2*x) + x)) + 5*x]

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fricas [A]  time = 0.72, size = 26, normalized size = 0.87 \begin {gather*} \log \left (5 \, x + e^{\left (-2 \, x e^{3} - 2 \, {\left (e^{3} - 1\right )} e^{\left (2 \, x\right )} + 2 \, x\right )} - 9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(3)+4)*exp(2*x)-2*exp(3)+2)*exp((-exp(3)+1)*exp(2*x)-x*exp(3)+x)^2+5)/(exp((-exp(3)+1)*exp(
2*x)-x*exp(3)+x)^2+5*x-9),x, algorithm="fricas")

[Out]

log(5*x + e^(-2*x*e^3 - 2*(e^3 - 1)*e^(2*x) + 2*x) - 9)

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giac [A]  time = 0.54, size = 30, normalized size = 1.00 \begin {gather*} \log \left (5 \, x + e^{\left (-2 \, x e^{3} + 2 \, x + 2 \, e^{\left (2 \, x\right )} - 2 \, e^{\left (2 \, x + 3\right )}\right )} - 9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(3)+4)*exp(2*x)-2*exp(3)+2)*exp((-exp(3)+1)*exp(2*x)-x*exp(3)+x)^2+5)/(exp((-exp(3)+1)*exp(
2*x)-x*exp(3)+x)^2+5*x-9),x, algorithm="giac")

[Out]

log(5*x + e^(-2*x*e^3 + 2*x + 2*e^(2*x) - 2*e^(2*x + 3)) - 9)

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maple [A]  time = 0.10, size = 20, normalized size = 0.67

method	result	size
risch	\(\ln \left ({\mathrm e}^{-2 \left ({\mathrm e}^{2 x}+x \right ) \left ({\mathrm e}^{3}-1\right )}+5 x -9\right )\)	\(20\)
norman	\(\ln \left ({\mathrm e}^{2 \left (-{\mathrm e}^{3}+1\right ) {\mathrm e}^{2 x}-2 x \,{\mathrm e}^{3}+2 x}+5 x -9\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*exp(3)+4)*exp(2*x)-2*exp(3)+2)*exp((-exp(3)+1)*exp(2*x)-x*exp(3)+x)^2+5)/(exp((-exp(3)+1)*exp(2*x)-x
*exp(3)+x)^2+5*x-9),x,method=_RETURNVERBOSE)

[Out]

ln(exp(-2*(exp(2*x)+x)*(exp(3)-1))+5*x-9)

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maxima [B]  time = 0.94, size = 63, normalized size = 2.10 \begin {gather*} -2 \, e^{\left (2 \, x + 3\right )} + \log \left (5 \, x - 9\right ) + \log \left (\frac {{\left ({\left (5 \, x - 9\right )} e^{\left (2 \, x e^{3} + 2 \, e^{\left (2 \, x + 3\right )}\right )} + e^{\left (2 \, x + 2 \, e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x e^{3}\right )}}{5 \, x - 9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(3)+4)*exp(2*x)-2*exp(3)+2)*exp((-exp(3)+1)*exp(2*x)-x*exp(3)+x)^2+5)/(exp((-exp(3)+1)*exp(
2*x)-x*exp(3)+x)^2+5*x-9),x, algorithm="maxima")

[Out]

-2*e^(2*x + 3) + log(5*x - 9) + log(((5*x - 9)*e^(2*x*e^3 + 2*e^(2*x + 3)) + e^(2*x + 2*e^(2*x)))*e^(-2*x*e^3)
/(5*x - 9))

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mupad [B]  time = 0.56, size = 33, normalized size = 1.10 \begin {gather*} \ln \left (5\,x+{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^3}-9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x - 2*x*exp(3) - 2*exp(2*x)*(exp(3) - 1))*(2*exp(3) + exp(2*x)*(4*exp(3) - 4) - 2) - 5)/(5*x + exp
(2*x - 2*x*exp(3) - 2*exp(2*x)*(exp(3) - 1)) - 9),x)

[Out]

log(5*x + exp(2*exp(2*x))*exp(2*x)*exp(-2*exp(2*x)*exp(3))*exp(-2*x*exp(3)) - 9)

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sympy [A]  time = 0.24, size = 29, normalized size = 0.97 \begin {gather*} \log {\left (5 x + e^{- 2 x e^{3} + 2 x + 2 \left (1 - e^{3}\right ) e^{2 x}} - 9 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(3)+4)*exp(2*x)-2*exp(3)+2)*exp((-exp(3)+1)*exp(2*x)-x*exp(3)+x)**2+5)/(exp((-exp(3)+1)*exp
(2*x)-x*exp(3)+x)**2+5*x-9),x)

[Out]

log(5*x + exp(-2*x*exp(3) + 2*x + 2*(1 - exp(3))*exp(2*x)) - 9)

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