3.21.38 \(\int \frac {e^{-6-2 x} (-4-8 x+e^{3+x} (64+64 x-4 x^2)+e^{6+2 x} (-256-16 e^{25}+x^2))}{16 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^{25}+\left (4-\frac {e^{-3-x}}{2}-\frac {x}{4}\right )^2}{x} \]

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Rubi [A]  time = 0.50, antiderivative size = 52, normalized size of antiderivative = 1.86, number of steps used = 12, number of rules used = 8, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.151, Rules used = {12, 6742, 2197, 14, 2199, 2194, 2177, 2178} \begin {gather*} \frac {x}{16}+\frac {e^{-x-3}}{4}+\frac {e^{-2 x-6}}{4 x}-\frac {4 e^{-x-3}}{x}+\frac {16+e^{25}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-6 - 2*x)*(-4 - 8*x + E^(3 + x)*(64 + 64*x - 4*x^2) + E^(6 + 2*x)*(-256 - 16*E^25 + x^2)))/(16*x^2),x]

[Out]

E^(-3 - x)/4 + E^(-6 - 2*x)/(4*x) - (4*E^(-3 - x))/x + (16 + E^25)/x + x/16

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \frac {e^{-6-2 x} \left (-4-8 x+e^{3+x} \left (64+64 x-4 x^2\right )+e^{6+2 x} \left (-256-16 e^{25}+x^2\right )\right )}{x^2} \, dx\\ &=\frac {1}{16} \int \left (-\frac {4 e^{-6-2 x} (1+2 x)}{x^2}-\frac {256+16 e^{25}-x^2}{x^2}-\frac {4 e^{-3-x} \left (-16-16 x+x^2\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{16} \int \frac {256+16 e^{25}-x^2}{x^2} \, dx\right )-\frac {1}{4} \int \frac {e^{-6-2 x} (1+2 x)}{x^2} \, dx-\frac {1}{4} \int \frac {e^{-3-x} \left (-16-16 x+x^2\right )}{x^2} \, dx\\ &=\frac {e^{-6-2 x}}{4 x}-\frac {1}{16} \int \left (-1+\frac {16 \left (16+e^{25}\right )}{x^2}\right ) \, dx-\frac {1}{4} \int \left (e^{-3-x}-\frac {16 e^{-3-x}}{x^2}-\frac {16 e^{-3-x}}{x}\right ) \, dx\\ &=\frac {e^{-6-2 x}}{4 x}+\frac {16+e^{25}}{x}+\frac {x}{16}-\frac {1}{4} \int e^{-3-x} \, dx+4 \int \frac {e^{-3-x}}{x^2} \, dx+4 \int \frac {e^{-3-x}}{x} \, dx\\ &=\frac {e^{-3-x}}{4}+\frac {e^{-6-2 x}}{4 x}-\frac {4 e^{-3-x}}{x}+\frac {16+e^{25}}{x}+\frac {x}{16}+\frac {4 \text {Ei}(-x)}{e^3}-4 \int \frac {e^{-3-x}}{x} \, dx\\ &=\frac {e^{-3-x}}{4}+\frac {e^{-6-2 x}}{4 x}-\frac {4 e^{-3-x}}{x}+\frac {16+e^{25}}{x}+\frac {x}{16}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 49, normalized size = 1.75 \begin {gather*} \frac {1}{16} \left (\frac {4 e^{-6-2 x}}{x}+\frac {16 \left (16+e^{25}\right )}{x}+x-\frac {4 e^{-3-x} \left (16 x-x^2\right )}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-6 - 2*x)*(-4 - 8*x + E^(3 + x)*(64 + 64*x - 4*x^2) + E^(6 + 2*x)*(-256 - 16*E^25 + x^2)))/(16*x
^2),x]

[Out]

((4*E^(-6 - 2*x))/x + (16*(16 + E^25))/x + x - (4*E^(-3 - x)*(16*x - x^2))/x^2)/16

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fricas [A]  time = 0.63, size = 38, normalized size = 1.36 \begin {gather*} \frac {{\left ({\left (x^{2} + 16 \, e^{25} + 256\right )} e^{\left (2 \, x + 6\right )} + 4 \, {\left (x - 16\right )} e^{\left (x + 3\right )} + 4\right )} e^{\left (-2 \, x - 6\right )}}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((-16*exp(25)+x^2-256)*exp(3+x)^2+(-4*x^2+64*x+64)*exp(3+x)-8*x-4)/x^2/exp(3+x)^2,x, algorithm=
"fricas")

[Out]

1/16*((x^2 + 16*e^25 + 256)*e^(2*x + 6) + 4*(x - 16)*e^(x + 3) + 4)*e^(-2*x - 6)/x

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giac [B]  time = 0.25, size = 47, normalized size = 1.68 \begin {gather*} \frac {{\left (x^{2} e^{9} + 4 \, x e^{\left (-x + 6\right )} + 16 \, e^{34} + 256 \, e^{9} - 64 \, e^{\left (-x + 6\right )} + 4 \, e^{\left (-2 \, x + 3\right )}\right )} e^{\left (-9\right )}}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((-16*exp(25)+x^2-256)*exp(3+x)^2+(-4*x^2+64*x+64)*exp(3+x)-8*x-4)/x^2/exp(3+x)^2,x, algorithm=
"giac")

[Out]

1/16*(x^2*e^9 + 4*x*e^(-x + 6) + 16*e^34 + 256*e^9 - 64*e^(-x + 6) + 4*e^(-2*x + 3))*e^(-9)/x

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maple [A]  time = 0.22, size = 41, normalized size = 1.46

method	result	size
risch	\(\frac {x}{16}+\frac {{\mathrm e}^{25}}{x}+\frac {16}{x}+\frac {\left (x -16\right ) {\mathrm e}^{-3-x}}{4 x}+\frac {{\mathrm e}^{-2 x -6}}{4 x}\)	\(41\)
derivativedivides	\(\frac {3}{16}+\frac {x}{16}+\frac {16}{x}+\frac {{\mathrm e}^{25}}{x}+\frac {{\mathrm e}^{-2 x -6}}{4 x}-\frac {4 \,{\mathrm e}^{-3-x}}{x}+\frac {{\mathrm e}^{-3-x}}{4}\)	\(47\)
default	\(\frac {3}{16}+\frac {x}{16}+\frac {16}{x}+\frac {{\mathrm e}^{25}}{x}+\frac {{\mathrm e}^{-2 x -6}}{4 x}-\frac {4 \,{\mathrm e}^{-3-x}}{x}+\frac {{\mathrm e}^{-3-x}}{4}\)	\(47\)
norman	\(\frac {\left (\frac {1}{4}+\left ({\mathrm e}^{25}+16\right ) {\mathrm e}^{2 x +6}+\frac {x^{2} {\mathrm e}^{2 x +6}}{16}+\frac {{\mathrm e}^{3+x} x}{4}-4 \,{\mathrm e}^{3+x}\right ) {\mathrm e}^{-2 x -6}}{x}\)	\(48\)
meijerg	\(-\frac {{\mathrm e}^{-2 x -12+2 x \,{\mathrm e}^{-6}} \left (\frac {{\mathrm e}^{6} \left (2-4 x \,{\mathrm e}^{-6}\right )}{4 x}-\frac {{\mathrm e}^{6-2 x \,{\mathrm e}^{-6}}}{2 x}+\ln \left (2 x \,{\mathrm e}^{-6}\right )+\expIntegralEi \left (1, 2 x \,{\mathrm e}^{-6}\right )+7-\ln \relax (x )-\ln \relax (2)-\frac {{\mathrm e}^{6}}{2 x}\right )}{2}-\frac {{\mathrm e}^{-2 x \,{\mathrm e}^{6}+2 x \,{\mathrm e}^{-6}-6} \left (1-{\mathrm e}^{2 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )}\right )}{32 \left (1-{\mathrm e}^{-12}\right )}-\frac {{\mathrm e}^{3-2 x +2 x \,{\mathrm e}^{-6}} \left (1-{\mathrm e}^{-x \,{\mathrm e}^{-6} \left (-{\mathrm e}^{6}+2\right )}\right )}{4 \left (-{\mathrm e}^{6}+2\right )}+4 \,{\mathrm e}^{-2 x -3+2 x \,{\mathrm e}^{-6}} \left (-\ln \left (x \,{\mathrm e}^{-6} \left (-{\mathrm e}^{6}+2\right )\right )-\expIntegralEi \left (1, x \,{\mathrm e}^{-6} \left (-{\mathrm e}^{6}+2\right )\right )+\ln \relax (x )-6+\ln \left (-{\mathrm e}^{6}+2\right )\right )-\frac {{\mathrm e}^{-2 x -6+2 x \,{\mathrm e}^{-6}} \left (-\ln \left (2 x \,{\mathrm e}^{-6}\right )-\expIntegralEi \left (1, 2 x \,{\mathrm e}^{-6}\right )+\ln \relax (x )+\ln \relax (2)-6\right )}{2}+2 \,{\mathrm e}^{31-2 x \,{\mathrm e}^{6}+2 x \,{\mathrm e}^{-6}} \left (1-{\mathrm e}^{-12}\right ) \left (-\frac {{\mathrm e}^{-6} \left (2+4 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )\right )}{4 x \left (1-{\mathrm e}^{-12}\right )}+\frac {{\mathrm e}^{-6+2 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )}}{2 x \left (1-{\mathrm e}^{-12}\right )}+\ln \left (-2 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )\right )+\expIntegralEi \left (1, -2 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )\right )-5-\ln \relax (x )-\ln \relax (2)-i \pi -\ln \left (1-{\mathrm e}^{-12}\right )+\frac {{\mathrm e}^{-6}}{2 x \left (1-{\mathrm e}^{-12}\right )}\right )+32 \,{\mathrm e}^{-2 x \,{\mathrm e}^{6}+6+2 x \,{\mathrm e}^{-6}} \left (1-{\mathrm e}^{-12}\right ) \left (-\frac {{\mathrm e}^{-6} \left (2+4 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )\right )}{4 x \left (1-{\mathrm e}^{-12}\right )}+\frac {{\mathrm e}^{-6+2 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )}}{2 x \left (1-{\mathrm e}^{-12}\right )}+\ln \left (-2 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )\right )+\expIntegralEi \left (1, -2 x \,{\mathrm e}^{6} \left (1-{\mathrm e}^{-12}\right )\right )-5-\ln \relax (x )-\ln \relax (2)-i \pi -\ln \left (1-{\mathrm e}^{-12}\right )+\frac {{\mathrm e}^{-6}}{2 x \left (1-{\mathrm e}^{-12}\right )}\right )+4 \,{\mathrm e}^{-2 x -9+2 x \,{\mathrm e}^{-6}} \left (-{\mathrm e}^{6}+2\right ) \left (\frac {{\mathrm e}^{6} \left (2-2 x \,{\mathrm e}^{-6} \left (-{\mathrm e}^{6}+2\right )\right )}{2 x \left (-{\mathrm e}^{6}+2\right )}-\frac {{\mathrm e}^{6-x \,{\mathrm e}^{-6} \left (-{\mathrm e}^{6}+2\right )}}{x \left (-{\mathrm e}^{6}+2\right )}+\ln \left (x \,{\mathrm e}^{-6} \left (-{\mathrm e}^{6}+2\right )\right )+\expIntegralEi \left (1, x \,{\mathrm e}^{-6} \left (-{\mathrm e}^{6}+2\right )\right )+7-\ln \relax (x )-\ln \left (-{\mathrm e}^{6}+2\right )-\frac {{\mathrm e}^{6}}{x \left (-{\mathrm e}^{6}+2\right )}\right )\)	\(640\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*((-16*exp(25)+x^2-256)*exp(3+x)^2+(-4*x^2+64*x+64)*exp(3+x)-8*x-4)/x^2/exp(3+x)^2,x,method=_RETURNVER
BOSE)

[Out]

1/16*x+exp(25)/x+16/x+1/4*(x-16)/x*exp(-3-x)+1/4*exp(-2*x-6)/x

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maxima [C]  time = 0.52, size = 55, normalized size = 1.96 \begin {gather*} 4 \, {\rm Ei}\left (-x\right ) e^{\left (-3\right )} - \frac {1}{2} \, {\rm Ei}\left (-2 \, x\right ) e^{\left (-6\right )} + \frac {1}{2} \, e^{\left (-6\right )} \Gamma \left (-1, 2 \, x\right ) - 4 \, e^{\left (-3\right )} \Gamma \left (-1, x\right ) + \frac {1}{16} \, x + \frac {e^{25}}{x} + \frac {16}{x} + \frac {1}{4} \, e^{\left (-x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((-16*exp(25)+x^2-256)*exp(3+x)^2+(-4*x^2+64*x+64)*exp(3+x)-8*x-4)/x^2/exp(3+x)^2,x, algorithm=
"maxima")

[Out]

4*Ei(-x)*e^(-3) - 1/2*Ei(-2*x)*e^(-6) + 1/2*e^(-6)*gamma(-1, 2*x) - 4*e^(-3)*gamma(-1, x) + 1/16*x + e^25/x +
16/x + 1/4*e^(-x - 3)

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mupad [B]  time = 1.26, size = 38, normalized size = 1.36 \begin {gather*} \frac {x}{16}+\frac {{\mathrm {e}}^{-2\,x-6}}{4\,x}+\frac {{\mathrm {e}}^{25}+16}{x}+\frac {{\mathrm {e}}^{-x-3}\,\left (\frac {x}{4}-4\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- 2*x - 6)*(x/2 - (exp(x + 3)*(64*x - 4*x^2 + 64))/16 + (exp(2*x + 6)*(16*exp(25) - x^2 + 256))/16 +
1/4))/x^2,x)

[Out]

x/16 + exp(- 2*x - 6)/(4*x) + (exp(25) + 16)/x + (exp(- x - 3)*(x/4 - 4))/x

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sympy [B]  time = 0.18, size = 44, normalized size = 1.57 \begin {gather*} \frac {x}{16} + \frac {256 + 16 e^{25}}{16 x} + \frac {4 x e^{- 2 x - 6} + \left (4 x^{2} - 64 x\right ) e^{- x - 3}}{16 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*((-16*exp(25)+x**2-256)*exp(3+x)**2+(-4*x**2+64*x+64)*exp(3+x)-8*x-4)/x**2/exp(3+x)**2,x)

[Out]

x/16 + (256 + 16*exp(25))/(16*x) + (4*x*exp(-2*x - 6) + (4*x**2 - 64*x)*exp(-x - 3))/(16*x**2)

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