3.34.70 \(\int \frac {e^{8-x} (e^4 (-3-x)-x^2-x^3+(-2 x-x^2) \log (2))}{5 x^4} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^{8-x} \left (\frac {e^4}{x}+x+\log (2)\right )}{5 x^2} \]

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Rubi [C]  time = 0.27, antiderivative size = 155, normalized size of antiderivative = 6.20, number of steps used = 13, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {12, 2199, 2177, 2178} \begin {gather*} \frac {e^{12} \text {Ei}(-x)}{10}-\frac {e^8 \text {Ei}(-x)}{5}-\frac {1}{10} e^8 \left (e^4+\log (4)\right ) \text {Ei}(-x)+\frac {1}{5} e^8 (1+\log (2)) \text {Ei}(-x)+\frac {e^{12-x}}{5 x^3}-\frac {e^{12-x}}{10 x^2}+\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x^2}+\frac {e^{12-x}}{10 x}-\frac {e^{8-x} \left (e^4+\log (4)\right )}{10 x}+\frac {e^{8-x} (1+\log (2))}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(8 - x)*(E^4*(-3 - x) - x^2 - x^3 + (-2*x - x^2)*Log[2]))/(5*x^4),x]

[Out]

E^(12 - x)/(5*x^3) - E^(12 - x)/(10*x^2) + E^(12 - x)/(10*x) - (E^8*ExpIntegralEi[-x])/5 + (E^12*ExpIntegralEi
[-x])/10 + (E^(8 - x)*(1 + Log[2]))/(5*x) + (E^8*ExpIntegralEi[-x]*(1 + Log[2]))/5 + (E^(8 - x)*(E^4 + Log[4])
)/(10*x^2) - (E^(8 - x)*(E^4 + Log[4]))/(10*x) - (E^8*ExpIntegralEi[-x]*(E^4 + Log[4]))/10

Rule 12

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

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 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

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 28, normalized size = 1.12 \begin {gather*} \frac {e^{8-x} \left (2 e^4+x (2 x+\log (4))\right )}{10 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(8 - x)*(E^4*(-3 - x) - x^2 - x^3 + (-2*x - x^2)*Log[2]))/(5*x^4),x]

[Out]

(E^(8 - x)*(2*E^4 + x*(2*x + Log[4])))/(10*x^3)

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fricas [A]  time = 0.80, size = 21, normalized size = 0.84 \begin {gather*} \frac {{\left (x^{2} + x \log \relax (2) + e^{4}\right )} e^{\left (-x + 8\right )}}{5 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x^2-2*x)*log(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x, algorithm="fricas")

[Out]

1/5*(x^2 + x*log(2) + e^4)*e^(-x + 8)/x^3

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giac [A]  time = 0.14, size = 32, normalized size = 1.28 \begin {gather*} \frac {x^{2} e^{\left (-x + 8\right )} + x e^{\left (-x + 8\right )} \log \relax (2) + e^{\left (-x + 12\right )}}{5 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x^2-2*x)*log(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x, algorithm="giac")

[Out]

1/5*(x^2*e^(-x + 8) + x*e^(-x + 8)*log(2) + e^(-x + 12))/x^3

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maple [A]  time = 0.09, size = 22, normalized size = 0.88

method	result	size
gosper	\(\frac {{\mathrm e}^{8-x} \left (x \ln \relax (2)+x^{2}+{\mathrm e}^{4}\right )}{5 x^{3}}\)	\(22\)
risch	\(\frac {{\mathrm e}^{8-x} \left (x \ln \relax (2)+x^{2}+{\mathrm e}^{4}\right )}{5 x^{3}}\)	\(22\)
norman	\(\frac {\frac {x^{2} {\mathrm e}^{8-x}}{5}+\frac {{\mathrm e}^{4} {\mathrm e}^{8-x}}{5}+\frac {x \ln \relax (2) {\mathrm e}^{8-x}}{5}}{x^{3}}\)	\(38\)
derivativedivides	\(\frac {{\mathrm e}^{8-x}}{5 x}-\frac {{\mathrm e}^{4} \left (-\frac {11 \,{\mathrm e}^{8-x}}{6 x^{2}}+\frac {11 \,{\mathrm e}^{8-x}}{6 x}-\frac {11 \,{\mathrm e}^{8} \expIntegralEi \left (1, x\right )}{6}+\frac {8 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}+\frac {11 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{8-x}}{3 x^{3}}-\frac {{\mathrm e}^{8-x}}{6 x^{2}}+\frac {{\mathrm e}^{8-x}}{6 x}-\frac {{\mathrm e}^{8} \expIntegralEi \left (1, x\right )}{6}\right )}{5}+\frac {\ln \relax (2) {\mathrm e}^{8-x}}{5 x^{2}}\)	\(116\)
default	\(\frac {{\mathrm e}^{8-x}}{5 x}-\frac {{\mathrm e}^{4} \left (-\frac {11 \,{\mathrm e}^{8-x}}{6 x^{2}}+\frac {11 \,{\mathrm e}^{8-x}}{6 x}-\frac {11 \,{\mathrm e}^{8} \expIntegralEi \left (1, x\right )}{6}+\frac {8 \,{\mathrm e}^{8-x}}{3 x^{3}}\right )}{5}+\frac {11 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{8-x}}{3 x^{3}}-\frac {{\mathrm e}^{8-x}}{6 x^{2}}+\frac {{\mathrm e}^{8-x}}{6 x}-\frac {{\mathrm e}^{8} \expIntegralEi \left (1, x\right )}{6}\right )}{5}+\frac {\ln \relax (2) {\mathrm e}^{8-x}}{5 x^{2}}\)	\(116\)
meijerg	\(\frac {\left (-\ln \relax (2)-1\right ) {\mathrm e}^{8} \left (-\frac {1}{x}+1+\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )\right )}{5}+\frac {\left (-{\mathrm e}^{4}-2 \ln \relax (2)\right ) {\mathrm e}^{8} \left (-\frac {1}{2 x^{2}}+\frac {1}{x}-\frac {3}{4}+\frac {9 x^{2}-12 x +6}{12 x^{2}}-\frac {\left (-3 x +3\right ) {\mathrm e}^{-x}}{6 x^{2}}-\frac {\expIntegralEi \left (1, x\right )}{2}\right )}{5}+\frac {{\mathrm e}^{8} \expIntegralEi \left (1, x\right )}{5}-\frac {3 \,{\mathrm e}^{12} \left (-\frac {1}{3 x^{3}}+\frac {1}{2 x^{2}}-\frac {1}{2 x}+\frac {11}{36}+\frac {-22 x^{3}+36 x^{2}-36 x +24}{72 x^{3}}-\frac {\left (4 x^{2}-4 x +8\right ) {\mathrm e}^{-x}}{24 x^{3}}+\frac {\expIntegralEi \left (1, x\right )}{6}\right )}{5}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-x^2-2*x)*ln(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/5*exp(8-x)*(x*ln(2)+x^2+exp(4))/x^3

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maxima [C]  time = 0.47, size = 48, normalized size = 1.92 \begin {gather*} \frac {1}{5} \, e^{8} \Gamma \left (-1, x\right ) \log \relax (2) + \frac {2}{5} \, e^{8} \Gamma \left (-2, x\right ) \log \relax (2) - \frac {1}{5} \, {\rm Ei}\left (-x\right ) e^{8} + \frac {1}{5} \, e^{8} \Gamma \left (-1, x\right ) + \frac {1}{5} \, e^{12} \Gamma \left (-2, x\right ) + \frac {3}{5} \, e^{12} \Gamma \left (-3, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x^2-2*x)*log(2)+(-3-x)*exp(4)-x^3-x^2)*exp(8-x)/x^4,x, algorithm="maxima")

[Out]

1/5*e^8*gamma(-1, x)*log(2) + 2/5*e^8*gamma(-2, x)*log(2) - 1/5*Ei(-x)*e^8 + 1/5*e^8*gamma(-1, x) + 1/5*e^12*g
amma(-2, x) + 3/5*e^12*gamma(-3, x)

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mupad [B]  time = 2.05, size = 36, normalized size = 1.44 \begin {gather*} \frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^8}{5\,x}+\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{12}}{5\,x^3}+\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^8\,\ln \relax (2)}{5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8 - x)*(exp(4)*(x + 3) + x^2 + x^3 + log(2)*(2*x + x^2)))/(5*x^4),x)

[Out]

(exp(-x)*exp(8))/(5*x) + (exp(-x)*exp(12))/(5*x^3) + (exp(-x)*exp(8)*log(2))/(5*x^2)

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sympy [A]  time = 0.15, size = 20, normalized size = 0.80 \begin {gather*} \frac {\left (x^{2} + x \log {\relax (2 )} + e^{4}\right ) e^{8 - x}}{5 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x**2-2*x)*ln(2)+(-3-x)*exp(4)-x**3-x**2)*exp(8-x)/x**4,x)

[Out]

(x**2 + x*log(2) + exp(4))*exp(8 - x)/(5*x**3)

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