3.11.39 \(\int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x (-5-5 x+e^2 (1+x))}{-15 x-5 e x-4 x^2+e^x (-5 x+e^2 x)+e^2 (3 x+e x+x^2)} \, dx\)

Optimal. Leaf size=26 \[ e^3+\log (x)+\log \left (3+e+e^x+x-\frac {x}{5-e^2}\right ) \]

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Rubi [F]  time = 0.86, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{-15 x-5 e x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-15 - 5*E - 8*x + E^2*(3 + E + 2*x) + E^x*(-5 - 5*x + E^2*(1 + x)))/(-15*x - 5*E*x - 4*x^2 + E^x*(-5*x +
E^2*x) + E^2*(3*x + E*x + x^2)),x]

[Out]

x + Log[x] + (11 + E*(5 - 2*E - E^2))*Defer[Int][(-5*E^x*(1 - E^2/5) - 15*(1 - (E*(-5 + 3*E + E^2))/15) - 4*(1
- E^2/4)*x)^(-1), x] - (2 - E)*(2 + E)*Defer[Int][x/(5*E^x*(1 - E^2/5) + 15*(1 - (E*(-5 + 3*E + E^2))/15) + 4
*(1 - E^2/4)*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-15-5 e-8 x+e^2 (3+e+2 x)+e^x \left (-5-5 x+e^2 (1+x)\right )}{(-15-5 e) x-4 x^2+e^x \left (-5 x+e^2 x\right )+e^2 \left (3 x+e x+x^2\right )} \, dx\\ &=\int \frac {15 \left (1+\frac {e}{3}\right )+8 x-e^2 (3+e+2 x)-e^x \left (-5-5 x+e^2 (1+x)\right )}{-((-15-5 e) x)+4 x^2-e^x \left (-5 x+e^2 x\right )-e^2 \left (3 x+e x+x^2\right )} \, dx\\ &=\int \left (\frac {1+x}{x}+\frac {-11-5 e+2 e^2+e^3-\left (4-e^2\right ) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x}\right ) \, dx\\ &=\int \frac {1+x}{x} \, dx+\int \frac {-11-5 e+2 e^2+e^3-\left (4-e^2\right ) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x} \, dx\\ &=\int \left (1+\frac {1}{x}\right ) \, dx+\int \left (\frac {11 \left (1-\frac {1}{11} e \left (-5+2 e+e^2\right )\right )}{-5 e^x \left (1-\frac {e^2}{5}\right )-15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )-4 \left (1-\frac {e^2}{4}\right ) x}+\frac {(-2-e) (2-e) x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x}\right ) \, dx\\ &=x+\log (x)+((-2-e) (2-e)) \int \frac {x}{5 e^x \left (1-\frac {e^2}{5}\right )+15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )+4 \left (1-\frac {e^2}{4}\right ) x} \, dx+\left (11+e \left (5-2 e-e^2\right )\right ) \int \frac {1}{-5 e^x \left (1-\frac {e^2}{5}\right )-15 \left (1-\frac {1}{15} e \left (-5+3 e+e^2\right )\right )-4 \left (1-\frac {e^2}{4}\right ) x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.39, size = 40, normalized size = 1.54 \begin {gather*} \log (x)+\log \left (15+5 e-3 e^2-e^3+5 e^x-e^{2+x}+4 x-e^2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 - 5*E - 8*x + E^2*(3 + E + 2*x) + E^x*(-5 - 5*x + E^2*(1 + x)))/(-15*x - 5*E*x - 4*x^2 + E^x*(-
5*x + E^2*x) + E^2*(3*x + E*x + x^2)),x]

[Out]

Log[x] + Log[15 + 5*E - 3*E^2 - E^3 + 5*E^x - E^(2 + x) + 4*x - E^2*x]

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fricas [A]  time = 0.83, size = 28, normalized size = 1.08 \begin {gather*} \log \left ({\left (x + 3\right )} e^{2} + {\left (e^{2} - 5\right )} e^{x} - 4 \, x + e^{3} - 5 \, e - 15\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x-15)/((exp(2)*x-5*x)*exp(x)+(x*exp(1)
+x^2+3*x)*exp(2)-5*x*exp(1)-4*x^2-15*x),x, algorithm="fricas")

[Out]

log((x + 3)*e^2 + (e^2 - 5)*e^x - 4*x + e^3 - 5*e - 15) + log(x)

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giac [A]  time = 0.38, size = 36, normalized size = 1.38 \begin {gather*} \log \left (-x e^{2} + 4 \, x - e^{3} - 3 \, e^{2} + 5 \, e - e^{\left (x + 2\right )} + 5 \, e^{x} + 15\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x-15)/((exp(2)*x-5*x)*exp(x)+(x*exp(1)
+x^2+3*x)*exp(2)-5*x*exp(1)-4*x^2-15*x),x, algorithm="giac")

[Out]

log(-x*e^2 + 4*x - e^3 - 3*e^2 + 5*e - e^(x + 2) + 5*e^x + 15) + log(x)

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maple [A]  time = 0.07, size = 34, normalized size = 1.31

method	result	size
risch	\(\ln \relax (x )+\ln \left ({\mathrm e}^{x}+\frac {{\mathrm e}^{3}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-4 x -15}{{\mathrm e}^{2}-5}\right )\)	\(34\)
norman	\(\ln \relax (x )+\ln \left ({\mathrm e} \,{\mathrm e}^{2}+{\mathrm e}^{2} {\mathrm e}^{x}+{\mathrm e}^{2} x +3 \,{\mathrm e}^{2}-5 \,{\mathrm e}-5 \,{\mathrm e}^{x}-4 x -15\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x-15)/((exp(2)*x-5*x)*exp(x)+(x*exp(1)+x^2+3
*x)*exp(2)-5*x*exp(1)-4*x^2-15*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(exp(x)+(exp(3)+exp(2)*x+3*exp(2)-5*exp(1)-4*x-15)/(exp(2)-5))

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maxima [A]  time = 0.50, size = 36, normalized size = 1.38 \begin {gather*} \log \relax (x) + \log \left (\frac {x {\left (e^{2} - 4\right )} + {\left (e^{2} - 5\right )} e^{x} + e^{3} + 3 \, e^{2} - 5 \, e - 15}{e^{2} - 5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x-15)/((exp(2)*x-5*x)*exp(x)+(x*exp(1)
+x^2+3*x)*exp(2)-5*x*exp(1)-4*x^2-15*x),x, algorithm="maxima")

[Out]

log(x) + log((x*(e^2 - 4) + (e^2 - 5)*e^x + e^3 + 3*e^2 - 5*e - 15)/(e^2 - 5))

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mupad [B]  time = 0.57, size = 36, normalized size = 1.38 \begin {gather*} \ln \left (4\,x-{\mathrm {e}}^{x+2}+5\,\mathrm {e}-3\,{\mathrm {e}}^2-{\mathrm {e}}^3+5\,{\mathrm {e}}^x-x\,{\mathrm {e}}^2+15\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 5*exp(1) + exp(x)*(5*x - exp(2)*(x + 1) + 5) - exp(2)*(2*x + exp(1) + 3) + 15)/(15*x - exp(2)*(3*x
+ x*exp(1) + x^2) + 5*x*exp(1) + exp(x)*(5*x - x*exp(2)) + 4*x^2),x)

[Out]

log(4*x - exp(x + 2) + 5*exp(1) - 3*exp(2) - exp(3) + 5*exp(x) - x*exp(2) + 15) + log(x)

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sympy [A]  time = 0.27, size = 36, normalized size = 1.38 \begin {gather*} \log {\relax (x )} + \log {\left (\frac {- 4 x + x e^{2} - 15 - 5 e + e^{3} + 3 e^{2}}{-5 + e^{2}} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(2)-5*x-5)*exp(x)+(exp(1)+2*x+3)*exp(2)-5*exp(1)-8*x-15)/((exp(2)*x-5*x)*exp(x)+(x*exp(1)
+x**2+3*x)*exp(2)-5*x*exp(1)-4*x**2-15*x),x)

[Out]

log(x) + log((-4*x + x*exp(2) - 15 - 5*E + exp(3) + 3*exp(2))/(-5 + exp(2)) + exp(x))

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