3.97.84 \(\int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} (626-250 x-25 x^2)+e^x (-125-175 x-55 x^2-5 x^3)}{125+50 x+5 x^2} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {-\frac {1}{5}+5 (5-x) x}{5+x}}-e^x x \]

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Rubi [F]  time = 0.72, 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 {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-1 + 125*x - 25*x^2)/(25 + 5*x))*(626 - 250*x - 25*x^2) + E^x*(-125 - 175*x - 55*x^2 - 5*x^3))/(125 +
50*x + 5*x^2),x]

[Out]

E^x - E^x*(1 + x) - 5*Defer[Int][E^((-1 + 125*x - 25*x^2)/(5*(5 + x))), x] + (1251*Defer[Int][E^((-1 + 125*x -
25*x^2)/(5*(5 + x)))/(5 + x)^2, x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{5 (5+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{(5+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-5 e^x (1+x)+\frac {e^{\frac {-1+125 x-25 x^2}{5 (5+x)}} \left (626-250 x-25 x^2\right )}{(5+x)^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {-1+125 x-25 x^2}{5 (5+x)}} \left (626-250 x-25 x^2\right )}{(5+x)^2} \, dx-\int e^x (1+x) \, dx\\ &=-e^x (1+x)+\frac {1}{5} \int \left (-25 e^{\frac {-1+125 x-25 x^2}{5 (5+x)}}+\frac {1251 e^{\frac {-1+125 x-25 x^2}{5 (5+x)}}}{(5+x)^2}\right ) \, dx+\int e^x \, dx\\ &=e^x-e^x (1+x)-5 \int e^{\frac {-1+125 x-25 x^2}{5 (5+x)}} \, dx+\frac {1251}{5} \int \frac {e^{\frac {-1+125 x-25 x^2}{5 (5+x)}}}{(5+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 25, normalized size = 0.93 \begin {gather*} e^{75-\frac {1251}{5 (5+x)}-5 (5+x)}-e^x x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-1 + 125*x - 25*x^2)/(25 + 5*x))*(626 - 250*x - 25*x^2) + E^x*(-125 - 175*x - 55*x^2 - 5*x^3))/
(125 + 50*x + 5*x^2),x]

[Out]

E^(75 - 1251/(5*(5 + x)) - 5*(5 + x)) - E^x*x

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fricas [A]  time = 0.54, size = 24, normalized size = 0.89 \begin {gather*} -x e^{x} + e^{\left (-\frac {25 \, x^{2} - 125 \, x + 1}{5 \, {\left (x + 5\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x^2+125*x-1)/(25+5*x)))/(5*x^2+50*x+1
25),x, algorithm="fricas")

[Out]

-x*e^x + e^(-1/5*(25*x^2 - 125*x + 1)/(x + 5))

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giac [A]  time = 0.23, size = 25, normalized size = 0.93 \begin {gather*} -x e^{x} + e^{\left (-\frac {125 \, x^{2} - 626 \, x}{25 \, {\left (x + 5\right )}} - \frac {1}{25}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x^2+125*x-1)/(25+5*x)))/(5*x^2+50*x+1
25),x, algorithm="giac")

[Out]

-x*e^x + e^(-1/25*(125*x^2 - 626*x)/(x + 5) - 1/25)

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maple [A]  time = 0.15, size = 25, normalized size = 0.93

method	result	size
risch	\(-{\mathrm e}^{x} x +{\mathrm e}^{-\frac {25 x^{2}-125 x +1}{5 \left (5+x \right )}}\)	\(25\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}-5 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}}{5+x}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x^2+125*x-1)/(25+5*x)))/(5*x^2+50*x+125),x,
method=_RETURNVERBOSE)

[Out]

-exp(x)*x+exp(-1/5*(25*x^2-125*x+1)/(5+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -{\left (x e^{\left (6 \, x\right )} - e^{\left (-\frac {1251}{5 \, {\left (x + 5\right )}} + 50\right )}\right )} e^{\left (-5 \, x\right )} + \frac {25 \, e^{\left (-5\right )} E_{2}\left (-x - 5\right )}{x + 5} + 25 \, \int \frac {e^{x}}{x^{2} + 10 \, x + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x^2+125*x-1)/(25+5*x)))/(5*x^2+50*x+1
25),x, algorithm="maxima")

[Out]

-(x*e^(6*x) - e^(-1251/5/(x + 5) + 50))*e^(-5*x) + 25*e^(-5)*exp_integral_e(2, -x - 5)/(x + 5) + 25*integrate(
e^x/(x^2 + 10*x + 25), x)

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mupad [B]  time = 6.23, size = 41, normalized size = 1.52 \begin {gather*} {\mathrm {e}}^{-\frac {25\,x^2}{5\,x+25}}\,{\mathrm {e}}^{-\frac {1}{5\,x+25}}\,{\mathrm {e}}^{\frac {125\,x}{5\,x+25}}-x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(25*x^2 - 125*x + 1)/(5*x + 25))*(250*x + 25*x^2 - 626) + exp(x)*(175*x + 55*x^2 + 5*x^3 + 125))/(5
0*x + 5*x^2 + 125),x)

[Out]

exp(-(25*x^2)/(5*x + 25))*exp(-1/(5*x + 25))*exp((125*x)/(5*x + 25)) - x*exp(x)

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sympy [A]  time = 0.31, size = 20, normalized size = 0.74 \begin {gather*} - x e^{x} + e^{\frac {- 25 x^{2} + 125 x - 1}{5 x + 25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**3-55*x**2-175*x-125)*exp(x)+(-25*x**2-250*x+626)*exp((-25*x**2+125*x-1)/(25+5*x)))/(5*x**2+5
0*x+125),x)

[Out]

-x*exp(x) + exp((-25*x**2 + 125*x - 1)/(5*x + 25))

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