3.34.57 \(\int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} (250 x-100 x^2+10 x^3))}{125-50 x+5 x^2} \, dx\)

Optimal. Leaf size=24 \[ e^{-\frac {e^{2 x}}{5 (-5+x)}} (13+e)+x^2 \]

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Rubi [F]  time = 1.76, 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^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(143 + E*(11 - 2*x) - 26*x) + E^(E^(2*x)/(-25 + 5*x))*(250*x - 100*x^2 + 10*x^3))/(E^(E^(2*x)/(-2
5 + 5*x))*(125 - 50*x + 5*x^2)),x]

[Out]

x^2 + ((13 + E)*Defer[Int][E^(2*x - E^(2*x)/(-25 + 5*x))/(-5 + x)^2, x])/5 - (2*(13 + E)*Defer[Int][E^(2*x - E
^(2*x)/(-25 + 5*x))/(-5 + x), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{5 (-5+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{(-5+x)^2} \, dx\\ &=\frac {1}{5} \int \left (10 x-\frac {e^{2 x-\frac {e^{2 x}}{-25+5 x}} (13+e) (-11+2 x)}{(-5+x)^2}\right ) \, dx\\ &=x^2+\frac {1}{5} (-13-e) \int \frac {e^{2 x-\frac {e^{2 x}}{-25+5 x}} (-11+2 x)}{(-5+x)^2} \, dx\\ &=x^2+\frac {1}{5} (-13-e) \int \left (-\frac {e^{2 x-\frac {e^{2 x}}{-25+5 x}}}{(-5+x)^2}+\frac {2 e^{2 x-\frac {e^{2 x}}{-25+5 x}}}{-5+x}\right ) \, dx\\ &=x^2+\frac {1}{5} (13+e) \int \frac {e^{2 x-\frac {e^{2 x}}{-25+5 x}}}{(-5+x)^2} \, dx-\frac {1}{5} (2 (13+e)) \int \frac {e^{2 x-\frac {e^{2 x}}{-25+5 x}}}{-5+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 39, normalized size = 1.62 \begin {gather*} -25+e^{1+\frac {e^{2 x}}{25-5 x}}+13 e^{\frac {e^{2 x}}{25-5 x}}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(143 + E*(11 - 2*x) - 26*x) + E^(E^(2*x)/(-25 + 5*x))*(250*x - 100*x^2 + 10*x^3))/(E^(E^(2*
x)/(-25 + 5*x))*(125 - 50*x + 5*x^2)),x]

[Out]

-25 + E^(1 + E^(2*x)/(25 - 5*x)) + 13*E^(E^(2*x)/(25 - 5*x)) + x^2

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fricas [A]  time = 0.57, size = 33, normalized size = 1.38 \begin {gather*} {\left (x^{2} e^{\left (\frac {e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )} + e + 13\right )} e^{\left (-\frac {e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1)-26*x+143)*exp(x)^2)/(5*x^2-50*x+125
)/exp(exp(x)^2/(5*x-25)),x, algorithm="fricas")

[Out]

(x^2*e^(1/5*e^(2*x)/(x - 5)) + e + 13)*e^(-1/5*e^(2*x)/(x - 5))

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giac [B]  time = 0.25, size = 64, normalized size = 2.67 \begin {gather*} {\left (x^{2} e^{\left (2 \, x\right )} + e^{\left (\frac {10 \, x^{2} - 50 \, x - e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}} + 1\right )} + 13 \, e^{\left (\frac {10 \, x^{2} - 50 \, x - e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1)-26*x+143)*exp(x)^2)/(5*x^2-50*x+125
)/exp(exp(x)^2/(5*x-25)),x, algorithm="giac")

[Out]

(x^2*e^(2*x) + e^(1/5*(10*x^2 - 50*x - e^(2*x))/(x - 5) + 1) + 13*e^(1/5*(10*x^2 - 50*x - e^(2*x))/(x - 5)))*e
^(-2*x)

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maple [A]  time = 0.50, size = 34, normalized size = 1.42

method	result	size
risch	\(x^{2}+{\mathrm e}^{-\frac {{\mathrm e}^{2 x}}{5 \left (x -5\right )}} {\mathrm e}+13 \,{\mathrm e}^{-\frac {{\mathrm e}^{2 x}}{5 \left (x -5\right )}}\)	\(34\)
norman	\(\frac {\left (\left ({\mathrm e}+13\right ) x +{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{5 x -25}} x^{3}-65-5 \,{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{5 x -25}} x^{2}-5 \,{\mathrm e}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{2 x}}{5 x -25}}}{x -5}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1)-26*x+143)*exp(x)^2)/(5*x^2-50*x+125)/exp(
exp(x)^2/(5*x-25)),x,method=_RETURNVERBOSE)

[Out]

x^2+exp(-1/5*exp(2*x)/(x-5))*exp(1)+13*exp(-1/5*exp(2*x)/(x-5))

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maxima [A]  time = 0.60, size = 33, normalized size = 1.38 \begin {gather*} {\left (x^{2} e^{\left (\frac {e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )} + e + 13\right )} e^{\left (-\frac {e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1)-26*x+143)*exp(x)^2)/(5*x^2-50*x+125
)/exp(exp(x)^2/(5*x-25)),x, algorithm="maxima")

[Out]

(x^2*e^(1/5*e^(2*x)/(x - 5)) + e + 13)*e^(-1/5*e^(2*x)/(x - 5))

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mupad [B]  time = 2.26, size = 23, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{-\frac {{\mathrm {e}}^{2\,x}}{5\,x-25}}\,\left (\mathrm {e}+13\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(2*x)/(5*x - 25))*(exp(exp(2*x)/(5*x - 25))*(250*x - 100*x^2 + 10*x^3) - exp(2*x)*(26*x + exp(1)*
(2*x - 11) - 143)))/(5*x^2 - 50*x + 125),x)

[Out]

exp(-exp(2*x)/(5*x - 25))*(exp(1) + 13) + x^2

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sympy [A]  time = 0.31, size = 19, normalized size = 0.79 \begin {gather*} x^{2} + \left (e + 13\right ) e^{- \frac {e^{2 x}}{5 x - 25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**3-100*x**2+250*x)*exp(exp(x)**2/(5*x-25))+((-2*x+11)*exp(1)-26*x+143)*exp(x)**2)/(5*x**2-50*
x+125)/exp(exp(x)**2/(5*x-25)),x)

[Out]

x**2 + (E + 13)*exp(-exp(2*x)/(5*x - 25))

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