3.58.34 \(\int \frac {e^8 (25 x^2-10 x^3+x^4)+e^{\frac {-x+e^8 (-10 x+2 x^2)}{e^8 (-5+x)}} (5 x+e^8 (-25+60 x-21 x^2+2 x^3))}{e^8 (25 x^2-10 x^3+x^4)} \, dx\)

Optimal. Leaf size=25 \[ -1+\frac {e^{2 x+\frac {x}{e^8 (5-x)}}}{x}+x \]

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Rubi [F]  time = 2.01, 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^8 \left (25 x^2-10 x^3+x^4\right )+\exp \left (\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}\right ) \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{e^8 \left (25 x^2-10 x^3+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^8*(25*x^2 - 10*x^3 + x^4) + E^((-x + E^8*(-10*x + 2*x^2))/(E^8*(-5 + x)))*(5*x + E^8*(-25 + 60*x - 21*x
^2 + 2*x^3)))/(E^8*(25*x^2 - 10*x^3 + x^4)),x]

[Out]

x + Defer[Int][E^(2*x - x/(E^8*(-5 + x)))/(-5 + x)^2, x]/E^8 - Defer[Int][E^(2*x - x/(E^8*(-5 + x)))/(-5 + x),
x]/(5*E^8) - Defer[Int][E^(8 + 2*x - x/(E^8*(-5 + x)))/x^2, x]/E^8 + ((10 + E^(-8))*Defer[Int][E^(2*x - x/(E^
8*(-5 + x)))/x, x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+\exp \left (\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}\right ) \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{25 x^2-10 x^3+x^4} \, dx}{e^8}\\ &=\frac {\int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+\exp \left (\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}\right ) \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{x^2 \left (25-10 x+x^2\right )} \, dx}{e^8}\\ &=\frac {\int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+\exp \left (\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}\right ) \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{(-5+x)^2 x^2} \, dx}{e^8}\\ &=\frac {\int \left (e^8+\frac {e^{2 x-\frac {x}{e^8 (-5+x)}} \left (-25 e^8+5 \left (1+12 e^8\right ) x-21 e^8 x^2+2 e^8 x^3\right )}{(5-x)^2 x^2}\right ) \, dx}{e^8}\\ &=x+\frac {\int \frac {e^{2 x-\frac {x}{e^8 (-5+x)}} \left (-25 e^8+5 \left (1+12 e^8\right ) x-21 e^8 x^2+2 e^8 x^3\right )}{(5-x)^2 x^2} \, dx}{e^8}\\ &=x+\frac {\int \left (\frac {e^{2 x-\frac {x}{e^8 (-5+x)}}}{(-5+x)^2}-\frac {e^{2 x-\frac {x}{e^8 (-5+x)}}}{5 (-5+x)}-\frac {e^{8+2 x-\frac {x}{e^8 (-5+x)}}}{x^2}+\frac {e^{2 x-\frac {x}{e^8 (-5+x)}} \left (1+10 e^8\right )}{5 x}\right ) \, dx}{e^8}\\ &=x+\frac {1}{5} \left (10+\frac {1}{e^8}\right ) \int \frac {e^{2 x-\frac {x}{e^8 (-5+x)}}}{x} \, dx-\frac {\int \frac {e^{2 x-\frac {x}{e^8 (-5+x)}}}{-5+x} \, dx}{5 e^8}+\frac {\int \frac {e^{2 x-\frac {x}{e^8 (-5+x)}}}{(-5+x)^2} \, dx}{e^8}-\frac {\int \frac {e^{8+2 x-\frac {x}{e^8 (-5+x)}}}{x^2} \, dx}{e^8}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 22, normalized size = 0.88 \begin {gather*} \frac {e^{\left (2-\frac {1}{e^8 (-5+x)}\right ) x}}{x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8*(25*x^2 - 10*x^3 + x^4) + E^((-x + E^8*(-10*x + 2*x^2))/(E^8*(-5 + x)))*(5*x + E^8*(-25 + 60*x
- 21*x^2 + 2*x^3)))/(E^8*(25*x^2 - 10*x^3 + x^4)),x]

[Out]

E^((2 - 1/(E^8*(-5 + x)))*x)/x + x

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fricas [A]  time = 0.71, size = 32, normalized size = 1.28 \begin {gather*} \frac {x^{2} + e^{\left (\frac {{\left (2 \, {\left (x^{2} - 5 \, x\right )} e^{8} - x\right )} e^{\left (-8\right )}}{x - 5}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-21*x^2+60*x-25)*exp(4)^2+5*x)*exp(((2*x^2-10*x)*exp(4)^2-x)/(x-5)/exp(4)^2)+(x^4-10*x^3+25*
x^2)*exp(4)^2)/(x^4-10*x^3+25*x^2)/exp(4)^2,x, algorithm="fricas")

[Out]

(x^2 + e^((2*(x^2 - 5*x)*e^8 - x)*e^(-8)/(x - 5)))/x

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giac [A]  time = 0.38, size = 44, normalized size = 1.76 \begin {gather*} \frac {{\left (x^{2} e^{8} + e^{\left (\frac {2 \, x^{2} e^{8} - 10 \, x e^{8} - x}{x e^{8} - 5 \, e^{8}} + 8\right )}\right )} e^{\left (-8\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-21*x^2+60*x-25)*exp(4)^2+5*x)*exp(((2*x^2-10*x)*exp(4)^2-x)/(x-5)/exp(4)^2)+(x^4-10*x^3+25*
x^2)*exp(4)^2)/(x^4-10*x^3+25*x^2)/exp(4)^2,x, algorithm="giac")

[Out]

(x^2*e^8 + e^((2*x^2*e^8 - 10*x*e^8 - x)/(x*e^8 - 5*e^8) + 8))*e^(-8)/x

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maple [A]  time = 0.23, size = 28, normalized size = 1.12

method	result	size
risch	\(x +\frac {{\mathrm e}^{\frac {x \left (2 x \,{\mathrm e}^{8}-10 \,{\mathrm e}^{8}-1\right ) {\mathrm e}^{-8}}{x -5}}}{x}\)	\(28\)
norman	\(\frac {\left (x^{3} {\mathrm e}^{4}-25 x \,{\mathrm e}^{4}+x \,{\mathrm e}^{4} {\mathrm e}^{\frac {\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{8}-x \right ) {\mathrm e}^{-8}}{x -5}}-5 \,{\mathrm e}^{4} {\mathrm e}^{\frac {\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{8}-x \right ) {\mathrm e}^{-8}}{x -5}}\right ) {\mathrm e}^{-4}}{x \left (x -5\right )}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3-21*x^2+60*x-25)*exp(4)^2+5*x)*exp(((2*x^2-10*x)*exp(4)^2-x)/(x-5)/exp(4)^2)+(x^4-10*x^3+25*x^2)*e
xp(4)^2)/(x^4-10*x^3+25*x^2)/exp(4)^2,x,method=_RETURNVERBOSE)

[Out]

x+1/x*exp(x*(2*x*exp(8)-10*exp(8)-1)*exp(-8)/(x-5))

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maxima [B]  time = 0.61, size = 76, normalized size = 3.04 \begin {gather*} {\left ({\left (x - \frac {25}{x - 5} + 10 \, \log \left (x - 5\right )\right )} e^{8} + 10 \, {\left (\frac {5}{x - 5} - \log \left (x - 5\right )\right )} e^{8} - \frac {25 \, e^{8}}{x - 5} + \frac {e^{\left (2 \, x - \frac {5}{x e^{8} - 5 \, e^{8}} - e^{\left (-8\right )} + 8\right )}}{x}\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-21*x^2+60*x-25)*exp(4)^2+5*x)*exp(((2*x^2-10*x)*exp(4)^2-x)/(x-5)/exp(4)^2)+(x^4-10*x^3+25*
x^2)*exp(4)^2)/(x^4-10*x^3+25*x^2)/exp(4)^2,x, algorithm="maxima")

[Out]

((x - 25/(x - 5) + 10*log(x - 5))*e^8 + 10*(5/(x - 5) - log(x - 5))*e^8 - 25*e^8/(x - 5) + e^(2*x - 5/(x*e^8 -
5*e^8) - e^(-8) + 8)/x)*e^(-8)

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mupad [B]  time = 0.39, size = 37, normalized size = 1.48 \begin {gather*} x+\frac {{\mathrm {e}}^{-\frac {10\,x}{x-5}}\,{\mathrm {e}}^{\frac {2\,x^2}{x-5}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{-8}}{x-5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-8)*(exp(8)*(25*x^2 - 10*x^3 + x^4) + exp(-(exp(-8)*(x + exp(8)*(10*x - 2*x^2)))/(x - 5))*(5*x + exp(
8)*(60*x - 21*x^2 + 2*x^3 - 25))))/(25*x^2 - 10*x^3 + x^4),x)

[Out]

x + (exp(-(10*x)/(x - 5))*exp((2*x^2)/(x - 5))*exp(-(x*exp(-8))/(x - 5)))/x

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sympy [A]  time = 0.23, size = 24, normalized size = 0.96 \begin {gather*} x + \frac {e^{\frac {- x + \left (2 x^{2} - 10 x\right ) e^{8}}{\left (x - 5\right ) e^{8}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3-21*x**2+60*x-25)*exp(4)**2+5*x)*exp(((2*x**2-10*x)*exp(4)**2-x)/(x-5)/exp(4)**2)+(x**4-10*
x**3+25*x**2)*exp(4)**2)/(x**4-10*x**3+25*x**2)/exp(4)**2,x)

[Out]

x + exp((-x + (2*x**2 - 10*x)*exp(8))*exp(-8)/(x - 5))/x

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