3.97.69 \(\int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} (15 x+3 e^2 x)+e^2 (10-6 x-4 x^2))}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx\)

Optimal. Leaf size=24 \[ e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x \]

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Rubi [F]  time = 20.11, 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 {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{25+e^4+20 x+4 x^2+e^2 (10+4 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)*(25 + E^4*(1 - x) - 5*x - 16*x^2 - 4*x^3 + E^((3*x)/(5 + E^2 + 2*x)
)*(15*x + 3*E^2*x) + E^2*(10 - 6*x - 4*x^2)))/(25 + E^4 + 20*x + 4*x^2 + E^2*(10 + 4*x)),x]

[Out]

-4*Defer[Int][E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x), x] + (5 + E^2)*Defer[Int][E^(5 + E^((3*x)/(5 + E^2 + 2*x)
) - x), x] - Defer[Int][E^(7 + E^((3*x)/(5 + E^2 + 2*x)) - x), x] - Defer[Int][E^(5 + E^((3*x)/(5 + E^2 + 2*x)
) - x)*x, x] + 25*Defer[Int][E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)/(5 + E^2 + 2*x)^2, x] + (5*(5 + E^2)*Defer[
Int][E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)/(5 + E^2 + 2*x)^2, x])/2 - 4*(5 + E^2)^2*Defer[Int][E^(5 + E^((3*x)
/(5 + E^2 + 2*x)) - x)/(5 + E^2 + 2*x)^2, x] + ((5 + E^2)^3*Defer[Int][E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)/(
5 + E^2 + 2*x)^2, x])/2 - ((7 + E^2)*Defer[Int][E^(9 + E^((3*x)/(5 + E^2 + 2*x)) - x)/(5 + E^2 + 2*x)^2, x])/2
- (3*(5 + E^2)^2*Defer[Int][E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x + (3*x)/(5 + E^2 + 2*x))/(5 + E^2 + 2*x)^2,
x])/2 - (5*Defer[Int][E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)/(5 + E^2 + 2*x), x])/2 + 8*(5 + E^2)*Defer[Int][E^
(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)/(5 + E^2 + 2*x), x] - (3*(5 + E^2)^2*Defer[Int][E^(5 + E^((3*x)/(5 + E^2 +
2*x)) - x)/(5 + E^2 + 2*x), x])/2 + (7 + 2*E^2)*Defer[Int][E^(7 + E^((3*x)/(5 + E^2 + 2*x)) - x)/(5 + E^2 + 2
*x), x] - Defer[Int][E^(9 + E^((3*x)/(5 + E^2 + 2*x)) - x)/(5 + E^2 + 2*x), x]/2 + (3*(5 + E^2)*Defer[Int][E^(
5 + E^((3*x)/(5 + E^2 + 2*x)) - x + (3*x)/(5 + E^2 + 2*x))/(5 + E^2 + 2*x), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{\left (5+e^2\right )^2+4 \left (5+e^2\right ) x+4 x^2} \, dx\\ &=\int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (25+e^4 (1-x)-5 x-16 x^2-4 x^3+e^{\frac {3 x}{5+e^2+2 x}} \left (15 x+3 e^2 x\right )+e^2 \left (10-6 x-4 x^2\right )\right )}{\left (5+e^2+2 x\right )^2} \, dx\\ &=\int \left (\frac {25 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2}-\frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x} (-1+x)}{\left (5+e^2+2 x\right )^2}-\frac {5 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x}{\left (5+e^2+2 x\right )^2}+\frac {3 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}} \left (5+e^2\right ) x}{\left (5+e^2+2 x\right )^2}-\frac {16 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x^2}{\left (5+e^2+2 x\right )^2}-\frac {4 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x^3}{\left (5+e^2+2 x\right )^2}-\frac {2 e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x} (-1+x) (5+2 x)}{\left (5+e^2+2 x\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x} (-1+x) (5+2 x)}{\left (5+e^2+2 x\right )^2} \, dx\right )-4 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x^3}{\left (5+e^2+2 x\right )^2} \, dx-5 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x}{\left (5+e^2+2 x\right )^2} \, dx-16 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x^2}{\left (5+e^2+2 x\right )^2} \, dx+25 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx+\left (3 \left (5+e^2\right )\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}} x}{\left (5+e^2+2 x\right )^2} \, dx-\int \frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x} (-1+x)}{\left (5+e^2+2 x\right )^2} \, dx\\ &=-\left (2 \int \left (\frac {1}{2} e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x}+\frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (7+e^2\right )}{2 \left (5+e^2+2 x\right )^2}+\frac {e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-7-2 e^2\right )}{2 \left (5+e^2+2 x\right )}\right ) \, dx\right )-4 \int \left (\frac {1}{4} e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-5-e^2\right )+\frac {1}{4} e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x-\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (5+e^2\right )^3}{8 \left (5+e^2+2 x\right )^2}+\frac {3 e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (5+e^2\right )^2}{8 \left (5+e^2+2 x\right )}\right ) \, dx-5 \int \left (\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-5-e^2\right )}{2 \left (5+e^2+2 x\right )^2}+\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{2 \left (5+e^2+2 x\right )}\right ) \, dx-16 \int \left (\frac {1}{4} e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}+\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (5+e^2\right )^2}{4 \left (5+e^2+2 x\right )^2}+\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-5-e^2\right )}{2 \left (5+e^2+2 x\right )}\right ) \, dx+25 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx+\left (3 \left (5+e^2\right )\right ) \int \left (\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}} \left (-5-e^2\right )}{2 \left (5+e^2+2 x\right )^2}+\frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}}}{2 \left (5+e^2+2 x\right )}\right ) \, dx-\int \left (\frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x} \left (-7-e^2\right )}{2 \left (5+e^2+2 x\right )^2}+\frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x}}{2 \left (5+e^2+2 x\right )}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx\right )-\frac {5}{2} \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx-4 \int e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \, dx+25 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx-\left (-7-2 e^2\right ) \int \frac {e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx-\frac {1}{2} \left (-7-e^2\right ) \int \frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx-\left (-5-e^2\right ) \int e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} \, dx+\frac {1}{2} \left (3 \left (5+e^2\right )\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}}}{5+e^2+2 x} \, dx+\frac {1}{2} \left (5 \left (5+e^2\right )\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx+\left (8 \left (5+e^2\right )\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx-\frac {1}{2} \left (3 \left (5+e^2\right )^2\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x+\frac {3 x}{5+e^2+2 x}}}{\left (5+e^2+2 x\right )^2} \, dx-\frac {1}{2} \left (3 \left (5+e^2\right )^2\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{5+e^2+2 x} \, dx-\left (4 \left (5+e^2\right )^2\right ) \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx+\frac {1}{2} \left (5+e^2\right )^3 \int \frac {e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx-\left (7+e^2\right ) \int \frac {e^{9+e^{\frac {3 x}{5+e^2+2 x}}-x}}{\left (5+e^2+2 x\right )^2} \, dx-\int e^{7+e^{\frac {3 x}{5+e^2+2 x}}-x} \, dx-\int e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 24, normalized size = 1.00 \begin {gather*} e^{5+e^{\frac {3 x}{5+e^2+2 x}}-x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)*(25 + E^4*(1 - x) - 5*x - 16*x^2 - 4*x^3 + E^((3*x)/(5 + E^2
+ 2*x))*(15*x + 3*E^2*x) + E^2*(10 - 6*x - 4*x^2)))/(25 + E^4 + 20*x + 4*x^2 + E^2*(10 + 4*x)),x]

[Out]

E^(5 + E^((3*x)/(5 + E^2 + 2*x)) - x)*x

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fricas [A]  time = 0.58, size = 21, normalized size = 0.88 \begin {gather*} x e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(-x+1)*exp(2)^2+(-4*x^2-6*x+10)*exp(2)-4*x^3-16*x^2-5*x+2
5)/(exp(2)^2+(4*x+10)*exp(2)+4*x^2+20*x+25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (4 \, x^{3} + 16 \, x^{2} + {\left (x - 1\right )} e^{4} + 2 \, {\left (2 \, x^{2} + 3 \, x - 5\right )} e^{2} - 3 \, {\left (x e^{2} + 5 \, x\right )} e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5 \, x - 25\right )} e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\right )}}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} e^{2} + 20 \, x + e^{4} + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(-x+1)*exp(2)^2+(-4*x^2-6*x+10)*exp(2)-4*x^3-16*x^2-5*x+2
5)/(exp(2)^2+(4*x+10)*exp(2)+4*x^2+20*x+25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x, algorithm="giac")

[Out]

integrate(-(4*x^3 + 16*x^2 + (x - 1)*e^4 + 2*(2*x^2 + 3*x - 5)*e^2 - 3*(x*e^2 + 5*x)*e^(3*x/(2*x + e^2 + 5)) +
5*x - 25)*e^(-x + e^(3*x/(2*x + e^2 + 5)) + 5)/(4*x^2 + 2*(2*x + 5)*e^2 + 20*x + e^4 + 25), x)

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maple [A]  time = 0.33, size = 22, normalized size = 0.92

method	result	size
risch	\(x \,{\mathrm e}^{{\mathrm e}^{\frac {3 x}{{\mathrm e}^{2}+5+2 x}}-x +5}\)	\(22\)
norman	\(\frac {\left (\left ({\mathrm e}^{2}+5\right ) x +2 x^{2}\right ) {\mathrm e}^{{\mathrm e}^{\frac {3 x}{{\mathrm e}^{2}+5+2 x}}-x +5}}{{\mathrm e}^{2}+5+2 x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(1-x)*exp(2)^2+(-4*x^2-6*x+10)*exp(2)-4*x^3-16*x^2-5*x+25)/(exp
(2)^2+(4*x+10)*exp(2)+4*x^2+20*x+25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x,method=_RETURNVERBOSE)

[Out]

x*exp(exp(3*x/(exp(2)+5+2*x))-x+5)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (4 \, x^{3} + 16 \, x^{2} + {\left (x - 1\right )} e^{4} + 2 \, {\left (2 \, x^{2} + 3 \, x - 5\right )} e^{2} - 3 \, {\left (x e^{2} + 5 \, x\right )} e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5 \, x - 25\right )} e^{\left (-x + e^{\left (\frac {3 \, x}{2 \, x + e^{2} + 5}\right )} + 5\right )}}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} e^{2} + 20 \, x + e^{4} + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(-x+1)*exp(2)^2+(-4*x^2-6*x+10)*exp(2)-4*x^3-16*x^2-5*x+2
5)/(exp(2)^2+(4*x+10)*exp(2)+4*x^2+20*x+25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x, algorithm="maxima")

[Out]

-integrate((4*x^3 + 16*x^2 + (x - 1)*e^4 + 2*(2*x^2 + 3*x - 5)*e^2 - 3*(x*e^2 + 5*x)*e^(3*x/(2*x + e^2 + 5)) +
5*x - 25)*e^(-x + e^(3*x/(2*x + e^2 + 5)) + 5)/(4*x^2 + 2*(2*x + 5)*e^2 + 20*x + e^4 + 25), x)

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mupad [B]  time = 8.55, size = 22, normalized size = 0.92 \begin {gather*} x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {3\,x}{2\,x+{\mathrm {e}}^2+5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp((3*x)/(2*x + exp(2) + 5)) - x + 5)*(5*x - exp((3*x)/(2*x + exp(2) + 5))*(15*x + 3*x*exp(2)) + ex
p(2)*(6*x + 4*x^2 - 10) + exp(4)*(x - 1) + 16*x^2 + 4*x^3 - 25))/(20*x + exp(4) + 4*x^2 + exp(2)*(4*x + 10) +
25),x)

[Out]

x*exp(-x)*exp(5)*exp(exp((3*x)/(2*x + exp(2) + 5)))

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sympy [A]  time = 36.81, size = 19, normalized size = 0.79 \begin {gather*} x e^{- x + e^{\frac {3 x}{2 x + 5 + e^{2}}} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(2)*x+15*x)*exp(3*x/(exp(2)+5+2*x))+(-x+1)*exp(2)**2+(-4*x**2-6*x+10)*exp(2)-4*x**3-16*x**2-5
*x+25)/(exp(2)**2+(4*x+10)*exp(2)+4*x**2+20*x+25)/exp(-exp(3*x/(exp(2)+5+2*x))+x-5),x)

[Out]

x*exp(-x + exp(3*x/(2*x + 5 + exp(2))) + 5)

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