3.91.75 \(\int \frac {e^{-1-e^x} (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x (-16+16 x+15 x^3+15 e^2 x^3))}{16 x^2+16 e^2 x^2} \, dx\)

Optimal. Leaf size=29 \[ e^{-1-e^x} \left (-\frac {15 x}{16}+\frac {e^x}{x+e^2 x}\right ) \]

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Rubi [F]  time = 0.74, 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^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{16 x^2+16 e^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-1 - E^x)*(-16*E^(2*x)*x - 15*x^2 - 15*E^2*x^2 + E^x*(-16 + 16*x + 15*x^3 + 15*E^2*x^3)))/(16*x^2 + 16
*E^2*x^2),x]

[Out]

(-15*ExpIntegralEi[-E^x])/(16*E) - Defer[Int][E^(-1 - E^x + x)/x^2, x]/(1 + E^2) + Defer[Int][E^(-1 - E^x + x)
/x, x]/(1 + E^2) - Defer[Int][E^(-1 - E^x + 2*x)/x, x]/(1 + E^2) + (15*Defer[Int][E^(-1 - E^x + x)*x, x])/16

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-1-e^x} \left (-16 e^{2 x} x-15 x^2-15 e^2 x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{\left (16+16 e^2\right ) x^2} \, dx\\ &=\int \frac {e^{-1-e^x} \left (-16 e^{2 x} x+\left (-15-15 e^2\right ) x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{\left (16+16 e^2\right ) x^2} \, dx\\ &=\frac {\int \frac {e^{-1-e^x} \left (-16 e^{2 x} x+\left (-15-15 e^2\right ) x^2+e^x \left (-16+16 x+15 x^3+15 e^2 x^3\right )\right )}{x^2} \, dx}{16 \left (1+e^2\right )}\\ &=\frac {\int \left (-15 e^{-1-e^x} \left (1+e^2\right )-\frac {16 e^{-1-e^x+2 x}}{x}+\frac {e^{-1-e^x+x} \left (-16+16 x+15 \left (1+e^2\right ) x^3\right )}{x^2}\right ) \, dx}{16 \left (1+e^2\right )}\\ &=-\left (\frac {15}{16} \int e^{-1-e^x} \, dx\right )+\frac {\int \frac {e^{-1-e^x+x} \left (-16+16 x+15 \left (1+e^2\right ) x^3\right )}{x^2} \, dx}{16 \left (1+e^2\right )}-\frac {\int \frac {e^{-1-e^x+2 x}}{x} \, dx}{1+e^2}\\ &=-\left (\frac {15}{16} \operatorname {Subst}\left (\int \frac {e^{-1-x}}{x} \, dx,x,e^x\right )\right )+\frac {\int \left (-\frac {16 e^{-1-e^x+x}}{x^2}+\frac {16 e^{-1-e^x+x}}{x}+15 e^{-1-e^x+x} \left (1+e^2\right ) x\right ) \, dx}{16 \left (1+e^2\right )}-\frac {\int \frac {e^{-1-e^x+2 x}}{x} \, dx}{1+e^2}\\ &=-\frac {15 \text {Ei}\left (-e^x\right )}{16 e}+\frac {15}{16} \int e^{-1-e^x+x} x \, dx-\frac {\int \frac {e^{-1-e^x+x}}{x^2} \, dx}{1+e^2}+\frac {\int \frac {e^{-1-e^x+x}}{x} \, dx}{1+e^2}-\frac {\int \frac {e^{-1-e^x+2 x}}{x} \, dx}{1+e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 40, normalized size = 1.38 \begin {gather*} \frac {e^{-e^x} \left (\frac {16 e^{-1+x}}{x}-\frac {15 \left (1+e^2\right ) x}{e}\right )}{16 \left (1+e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 - E^x)*(-16*E^(2*x)*x - 15*x^2 - 15*E^2*x^2 + E^x*(-16 + 16*x + 15*x^3 + 15*E^2*x^3)))/(16*x^
2 + 16*E^2*x^2),x]

[Out]

((16*E^(-1 + x))/x - (15*(1 + E^2)*x)/E)/(16*E^E^x*(1 + E^2))

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fricas [A]  time = 0.54, size = 34, normalized size = 1.17 \begin {gather*} -\frac {{\left (15 \, x^{2} e^{2} + 15 \, x^{2} - 16 \, e^{x}\right )} e^{\left (-e^{x} - 1\right )}}{16 \, {\left (x e^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(x)^2+(15*x^3*exp(2)+15*x^3+16*x-16)*exp(x)-15*x^2*exp(2)-15*x^2)/(16*x^2*exp(2)+16*x^2)/e
xp(exp(x)+1),x, algorithm="fricas")

[Out]

-1/16*(15*x^2*e^2 + 15*x^2 - 16*e^x)*e^(-e^x - 1)/(x*e^2 + x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (15 \, x^{2} e^{2} + 15 \, x^{2} + 16 \, x e^{\left (2 \, x\right )} - {\left (15 \, x^{3} e^{2} + 15 \, x^{3} + 16 \, x - 16\right )} e^{x}\right )} e^{\left (-e^{x} - 1\right )}}{16 \, {\left (x^{2} e^{2} + x^{2}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(x)^2+(15*x^3*exp(2)+15*x^3+16*x-16)*exp(x)-15*x^2*exp(2)-15*x^2)/(16*x^2*exp(2)+16*x^2)/e
xp(exp(x)+1),x, algorithm="giac")

[Out]

integrate(-1/16*(15*x^2*e^2 + 15*x^2 + 16*x*e^(2*x) - (15*x^3*e^2 + 15*x^3 + 16*x - 16)*e^x)*e^(-e^x - 1)/(x^2
*e^2 + x^2), x)

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maple [A]  time = 0.51, size = 27, normalized size = 0.93

method	result	size
norman	\(\frac {\left (\frac {{\mathrm e}^{x}}{{\mathrm e}^{2}+1}-\frac {15 x^{2}}{16}\right ) {\mathrm e}^{-{\mathrm e}^{x}-1}}{x}\)	\(27\)
risch	\(-\frac {\left (15 x^{2} {\mathrm e}^{2}+15 x^{2}-16 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-{\mathrm e}^{x}-1}}{16 x \left ({\mathrm e}^{2}+1\right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*x*exp(x)^2+(15*x^3*exp(2)+15*x^3+16*x-16)*exp(x)-15*x^2*exp(2)-15*x^2)/(16*x^2*exp(2)+16*x^2)/exp(exp
(x)+1),x,method=_RETURNVERBOSE)

[Out]

(1/(exp(2)+1)*exp(x)-15/16*x^2)/x/exp(exp(x)+1)

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maxima [A]  time = 0.49, size = 31, normalized size = 1.07 \begin {gather*} -\frac {{\left (15 \, x^{2} {\left (e^{2} + 1\right )} - 16 \, e^{x}\right )} e^{\left (-e^{x}\right )}}{16 \, x {\left (e^{3} + e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(x)^2+(15*x^3*exp(2)+15*x^3+16*x-16)*exp(x)-15*x^2*exp(2)-15*x^2)/(16*x^2*exp(2)+16*x^2)/e
xp(exp(x)+1),x, algorithm="maxima")

[Out]

-1/16*(15*x^2*(e^2 + 1) - 16*e^x)*e^(-e^x)/(x*(e^3 + e))

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mupad [B]  time = 5.93, size = 37, normalized size = 1.28 \begin {gather*} \frac {{\mathrm {e}}^{x-{\mathrm {e}}^x-1}-\frac {x^2\,{\mathrm {e}}^{-{\mathrm {e}}^x-1}\,\left (15\,{\mathrm {e}}^2+15\right )}{16}}{x\,\left ({\mathrm {e}}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- exp(x) - 1)*(16*x*exp(2*x) - exp(x)*(16*x + 15*x^3*exp(2) + 15*x^3 - 16) + 15*x^2*exp(2) + 15*x^2)
)/(16*x^2*exp(2) + 16*x^2),x)

[Out]

(exp(x - exp(x) - 1) - (x^2*exp(- exp(x) - 1)*(15*exp(2) + 15))/16)/(x*(exp(2) + 1))

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sympy [A]  time = 0.19, size = 36, normalized size = 1.24 \begin {gather*} \frac {\left (- 15 x^{2} e^{2} - 15 x^{2} + 16 e^{x}\right ) e^{- e^{x} - 1}}{16 x + 16 x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(x)**2+(15*x**3*exp(2)+15*x**3+16*x-16)*exp(x)-15*x**2*exp(2)-15*x**2)/(16*x**2*exp(2)+16*
x**2)/exp(exp(x)+1),x)

[Out]

(-15*x**2*exp(2) - 15*x**2 + 16*exp(x))*exp(-exp(x) - 1)/(16*x + 16*x*exp(2))

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