3.65.95 \(\int \frac {e^{-\frac {48+48 x}{-5+x}} (288 e^2+100 x^3+248 x^4+4 x^5+e (-100 x-536 x^2-4 x^3))}{25-10 x+x^2} \, dx\)

Optimal. Leaf size=22 \[ e^{-\frac {48 (1+x)}{-5+x}} \left (-e+x^2\right )^2 \]

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Rubi [F]  time = 0.91, 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 {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{25-10 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(288*E^2 + 100*x^3 + 248*x^4 + 4*x^5 + E*(-100*x - 536*x^2 - 4*x^3))/(E^((48 + 48*x)/(-5 + x))*(25 - 10*x
+ x^2)),x]

[Out]

(25 - E)^2*E^(-48 + 288/(5 - x)) - (5760*(25 - E)*ExpIntegralEi[288/(5 - x)])/E^48 + 288*(75 - 2*E)*Defer[Int]
[E^(-((48 + 48*x)/(-5 + x))), x] + 4*(720 - E)*Defer[Int][x/E^((48 + 48*x)/(-5 + x)), x] + 288*Defer[Int][x^2/
E^((48 + 48*x)/(-5 + x)), x] + 4*Defer[Int][x^3/E^((48 + 48*x)/(-5 + x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {48+48 x}{-5+x}} \left (288 e^2+100 x^3+248 x^4+4 x^5+e \left (-100 x-536 x^2-4 x^3\right )\right )}{(-5+x)^2} \, dx\\ &=\int \frac {4 e^{-\frac {48+48 x}{-5+x}} \left (72 e^2-25 e x-134 e x^2+25 \left (1-\frac {e}{25}\right ) x^3+62 x^4+x^5\right )}{(5-x)^2} \, dx\\ &=4 \int \frac {e^{-\frac {48+48 x}{-5+x}} \left (72 e^2-25 e x-134 e x^2+25 \left (1-\frac {e}{25}\right ) x^3+62 x^4+x^5\right )}{(5-x)^2} \, dx\\ &=4 \int \left (-72 e^{-\frac {48+48 x}{-5+x}} (-75+2 e)+\frac {72 (-25+e)^2 e^{-\frac {48+48 x}{-5+x}}}{(-5+x)^2}-\frac {1440 (-25+e) e^{-\frac {48+48 x}{-5+x}}}{-5+x}-(-720+e) e^{-\frac {48+48 x}{-5+x}} x+72 e^{-\frac {48+48 x}{-5+x}} x^2+e^{-\frac {48+48 x}{-5+x}} x^3\right ) \, dx\\ &=4 \int e^{-\frac {48+48 x}{-5+x}} x^3 \, dx+288 \int e^{-\frac {48+48 x}{-5+x}} x^2 \, dx+(288 (75-2 e)) \int e^{-\frac {48+48 x}{-5+x}} \, dx+(5760 (25-e)) \int \frac {e^{-\frac {48+48 x}{-5+x}}}{-5+x} \, dx+\left (288 (25-e)^2\right ) \int \frac {e^{-\frac {48+48 x}{-5+x}}}{(-5+x)^2} \, dx+(4 (720-e)) \int e^{-\frac {48+48 x}{-5+x}} x \, dx\\ &=4 \int e^{-\frac {48+48 x}{-5+x}} x^3 \, dx+288 \int e^{-\frac {48+48 x}{-5+x}} x^2 \, dx+(288 (75-2 e)) \int e^{-\frac {48+48 x}{-5+x}} \, dx+(5760 (25-e)) \int \frac {e^{-48-\frac {288}{-5+x}}}{-5+x} \, dx+\left (288 (25-e)^2\right ) \int \frac {e^{-48-\frac {288}{-5+x}}}{(-5+x)^2} \, dx+(4 (720-e)) \int e^{-\frac {48+48 x}{-5+x}} x \, dx\\ &=(25-e)^2 e^{-48+\frac {288}{5-x}}-\frac {5760 (25-e) \text {Ei}\left (\frac {288}{5-x}\right )}{e^{48}}+4 \int e^{-\frac {48+48 x}{-5+x}} x^3 \, dx+288 \int e^{-\frac {48+48 x}{-5+x}} x^2 \, dx+(288 (75-2 e)) \int e^{-\frac {48+48 x}{-5+x}} \, dx+(4 (720-e)) \int e^{-\frac {48+48 x}{-5+x}} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.61, size = 22, normalized size = 1.00 \begin {gather*} e^{-\frac {48 (1+x)}{-5+x}} \left (e-x^2\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(288*E^2 + 100*x^3 + 248*x^4 + 4*x^5 + E*(-100*x - 536*x^2 - 4*x^3))/(E^((48 + 48*x)/(-5 + x))*(25 -
10*x + x^2)),x]

[Out]

(E - x^2)^2/E^((48*(1 + x))/(-5 + x))

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fricas [A]  time = 0.63, size = 25, normalized size = 1.14 \begin {gather*} {\left (x^{4} - 2 \, x^{2} e + e^{2}\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^3)/(x^2-10*x+25)/exp((48*x+48)/(x-5)
),x, algorithm="fricas")

[Out]

(x^4 - 2*x^2*e + e^2)*e^(-48*(x + 1)/(x - 5))

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giac [B]  time = 0.38, size = 375, normalized size = 17.05 \begin {gather*} \frac {\frac {{\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{4}} - \frac {4 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{3}} + \frac {6 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{{\left (x - 5\right )}^{2}} - \frac {4 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )}}{x - 5} - \frac {50 \, {\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{4}} + \frac {80 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{3}} - \frac {12 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{{\left (x - 5\right )}^{2}} - \frac {16 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )}}{x - 5} + \frac {625 \, {\left (x + 1\right )}^{4} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{4}} + \frac {500 \, {\left (x + 1\right )}^{3} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{3}} + \frac {150 \, {\left (x + 1\right )}^{2} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{{\left (x - 5\right )}^{2}} + \frac {20 \, {\left (x + 1\right )} e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{x - 5} + e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 2\right )} - 2 \, e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5} + 1\right )} + e^{\left (-\frac {48 \, {\left (x + 1\right )}}{x - 5}\right )}}{\frac {{\left (x + 1\right )}^{4}}{{\left (x - 5\right )}^{4}} - \frac {4 \, {\left (x + 1\right )}^{3}}{{\left (x - 5\right )}^{3}} + \frac {6 \, {\left (x + 1\right )}^{2}}{{\left (x - 5\right )}^{2}} - \frac {4 \, {\left (x + 1\right )}}{x - 5} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^3)/(x^2-10*x+25)/exp((48*x+48)/(x-5)
),x, algorithm="giac")

[Out]

((x + 1)^4*e^(-48*(x + 1)/(x - 5) + 2)/(x - 5)^4 - 4*(x + 1)^3*e^(-48*(x + 1)/(x - 5) + 2)/(x - 5)^3 + 6*(x +
1)^2*e^(-48*(x + 1)/(x - 5) + 2)/(x - 5)^2 - 4*(x + 1)*e^(-48*(x + 1)/(x - 5) + 2)/(x - 5) - 50*(x + 1)^4*e^(-
48*(x + 1)/(x - 5) + 1)/(x - 5)^4 + 80*(x + 1)^3*e^(-48*(x + 1)/(x - 5) + 1)/(x - 5)^3 - 12*(x + 1)^2*e^(-48*(
x + 1)/(x - 5) + 1)/(x - 5)^2 - 16*(x + 1)*e^(-48*(x + 1)/(x - 5) + 1)/(x - 5) + 625*(x + 1)^4*e^(-48*(x + 1)/
(x - 5))/(x - 5)^4 + 500*(x + 1)^3*e^(-48*(x + 1)/(x - 5))/(x - 5)^3 + 150*(x + 1)^2*e^(-48*(x + 1)/(x - 5))/(
x - 5)^2 + 20*(x + 1)*e^(-48*(x + 1)/(x - 5))/(x - 5) + e^(-48*(x + 1)/(x - 5) + 2) - 2*e^(-48*(x + 1)/(x - 5)
+ 1) + e^(-48*(x + 1)/(x - 5)))/((x + 1)^4/(x - 5)^4 - 4*(x + 1)^3/(x - 5)^3 + 6*(x + 1)^2/(x - 5)^2 - 4*(x +
1)/(x - 5) + 1)

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maple [A]  time = 0.51, size = 26, normalized size = 1.18

method	result	size
risch	\(\left (x^{4}-2 x^{2} {\mathrm e}+{\mathrm e}^{2}\right ) {\mathrm e}^{-\frac {48 \left (x +1\right )}{x -5}}\)	\(26\)
gosper	\(\left (x^{4}-2 x^{2} {\mathrm e}+{\mathrm e}^{2}\right ) {\mathrm e}^{-\frac {48 \left (x +1\right )}{x -5}}\)	\(30\)
norman	\(\frac {\left (x^{5}+{\mathrm e}^{2} x -5 x^{4}-5 \,{\mathrm e}^{2}+10 x^{2} {\mathrm e}-2 x^{3} {\mathrm e}\right ) {\mathrm e}^{-\frac {48 x +48}{x -5}}}{x -5}\)	\(56\)
derivativedivides	\(43700 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (\left (48+\frac {288}{x -5}\right )^{3}-145 \left (48+\frac {288}{x -5}\right )^{2}+223482+\frac {2018880}{x -5}\right ) \left (x -5\right )^{4}}{6}+58 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (\left (48+\frac {288}{x -5}\right )^{2}-2302-\frac {27936}{x -5}\right ) \left (x -5\right )^{3}-3030 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (-1+\frac {288}{x -5}\right ) \left (x -5\right )^{2}+625 \,{\mathrm e}^{-48-\frac {288}{x -5}}+10368 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )}{288}+{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )-2304 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )}{288}+{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )+331776 \,{\mathrm e} \left (\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (-1+\frac {288}{x -5}\right ) \left (x -5\right )^{2}}{165888}-\frac {{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )}{2}\right )+960 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )}{6}+47 \,{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )+50 \,{\mathrm e} \left (-{\mathrm e}^{-48-\frac {288}{x -5}}-8 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )+2208 \,{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )+96 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )}{6}+47 \,{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )-{\mathrm e}^{2} \left (-{\mathrm e}^{-48-\frac {288}{x -5}}-8 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )+2208 \,{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )\)	\(412\)
default	\(43700 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (\left (48+\frac {288}{x -5}\right )^{3}-145 \left (48+\frac {288}{x -5}\right )^{2}+223482+\frac {2018880}{x -5}\right ) \left (x -5\right )^{4}}{6}+58 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (\left (48+\frac {288}{x -5}\right )^{2}-2302-\frac {27936}{x -5}\right ) \left (x -5\right )^{3}-3030 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (-1+\frac {288}{x -5}\right ) \left (x -5\right )^{2}+625 \,{\mathrm e}^{-48-\frac {288}{x -5}}+10368 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )}{288}+{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )-2304 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )}{288}+{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )+331776 \,{\mathrm e} \left (\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (-1+\frac {288}{x -5}\right ) \left (x -5\right )^{2}}{165888}-\frac {{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )}{2}\right )+960 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )}{6}+47 \,{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )+50 \,{\mathrm e} \left (-{\mathrm e}^{-48-\frac {288}{x -5}}-8 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )+2208 \,{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )+96 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )}{6}+47 \,{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )-{\mathrm e}^{2} \left (-{\mathrm e}^{-48-\frac {288}{x -5}}-8 \,{\mathrm e}^{-48-\frac {288}{x -5}} \left (x -5\right )+2208 \,{\mathrm e}^{-48} \expIntegralEi \left (1, \frac {288}{x -5}\right )\right )\)	\(412\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^3)/(x^2-10*x+25)/exp((48*x+48)/(x-5)),x,me
thod=_RETURNVERBOSE)

[Out]

(x^4-2*x^2*exp(1)+exp(2))*exp(-48*(x+1)/(x-5))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} {\left (x^{4} - 2 \, x^{2} e\right )} e^{\left (-\frac {288}{x - 5} - 48\right )} + 288 \, \int \frac {e^{\left (-\frac {288}{x - 5}\right )}}{x^{2} e^{46} - 10 \, x e^{46} + 25 \, e^{46}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((288*exp(1)^2+(-4*x^3-536*x^2-100*x)*exp(1)+4*x^5+248*x^4+100*x^3)/(x^2-10*x+25)/exp((48*x+48)/(x-5)
),x, algorithm="maxima")

[Out]

(x^4 - 2*x^2*e)*e^(-288/(x - 5) - 48) + 288*integrate(e^(-288/(x - 5))/(x^2*e^46 - 10*x*e^46 + 25*e^46), x)

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mupad [B]  time = 0.20, size = 28, normalized size = 1.27 \begin {gather*} {\mathrm {e}}^{-\frac {48\,x}{x-5}-\frac {48}{x-5}}\,{\left (\mathrm {e}-x^2\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(48*x + 48)/(x - 5))*(288*exp(2) - exp(1)*(100*x + 536*x^2 + 4*x^3) + 100*x^3 + 248*x^4 + 4*x^5))/(x
^2 - 10*x + 25),x)

[Out]

exp(- (48*x)/(x - 5) - 48/(x - 5))*(exp(1) - x^2)^2

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sympy [A]  time = 0.37, size = 24, normalized size = 1.09 \begin {gather*} \left (x^{4} - 2 e x^{2} + e^{2}\right ) e^{- \frac {48 x + 48}{x - 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((288*exp(1)**2+(-4*x**3-536*x**2-100*x)*exp(1)+4*x**5+248*x**4+100*x**3)/(x**2-10*x+25)/exp((48*x+48
)/(x-5)),x)

[Out]

(x**4 - 2*E*x**2 + exp(2))*exp(-(48*x + 48)/(x - 5))

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