∫ e−2x(−18x−12x2+12x3−2x4+e2x(−18x4+6x5)+(−27−6x+11x2−2x3) log(3e))________________________________________________________

3.12.49 $\int \frac {e^{-2 x} (-18 x-12 x^2+12 x^3-2 x^4+e^{2 x} (-18 x^4+6 x^5)+(-27-6 x+11 x^2-2 x^3) \log (3 e))}{x^4} \, dx$

Optimal. Leaf size=23 \[ (-3+x)^2 \left (3+\frac {e^{-2 x} (x+\log (3 e))}{x^3}\right ) \]

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Rubi [C] time = 1.86, antiderivative size = 145, normalized size of antiderivative = 6.30, number of steps used = 16, number of rules used = 6, integrand size = 65, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.092, Rules used = {6741, 6742, 2199, 2194, 2177, 2178} \begin {gather*} 2 (1-\log (177147)) \text {Ei}(-2 x)-2 (24+\log (729)) \text {Ei}(-2 x)+(10-\log (9)) \text {Ei}(-2 x)+36 (1+\log (3)) \text {Ei}(-2 x)+\frac {9 e^{-2 x} (1+\log (3))}{x^3}+\frac {e^{-2 x} (24+\log (729))}{2 x^2}-\frac {9 e^{-2 x} (1+\log (3))}{x^2}+3 (3-x)^2+e^{-2 x}+\frac {e^{-2 x} (1-\log (177147))}{x}-\frac {e^{-2 x} (24+\log (729))}{x}+\frac {18 e^{-2 x} (1+\log (3))}{x} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(-18*x - 12*x^2 + 12*x^3 - 2*x^4 + E^(2*x)*(-18*x^4 + 6*x^5) + (-27 - 6*x + 11*x^2 - 2*x^3)*Log[3*E])/(E^(

2*x)*x^4),x]

[Out]

E^(-2*x) + 3*(3 - x)^2 + (9*(1 + Log[3]))/(E^(2*x)*x^3) - (9*(1 + Log[3]))/(E^(2*x)*x^2) + (18*(1 + Log[3]))/(

E^(2*x)*x) + 36*ExpIntegralEi[-2*x]*(1 + Log[3]) + ExpIntegralEi[-2*x]*(10 - Log[9]) + (24 + Log[729])/(2*E^(2

*x)*x^2) - (24 + Log[729])/(E^(2*x)*x) - 2*ExpIntegralEi[-2*x]*(24 + Log[729]) + (1 - Log[177147])/(E^(2*x)*x)

+ 2*ExpIntegralEi[-2*x]*(1 - Log[177147])

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 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 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-2 x} (3-x) \left (2 x^3-6 e^{2 x} x^4-4 x^2 \left (1-\frac {\log (3)}{2}\right )-11 x \left (1+\frac {5 \log (3)}{11}\right )-9 (1+\log (3))\right )}{x^4} \, dx\\ &=\int \left (6 (-3+x)+\frac {e^{-2 x} (3-x) \left (2 x^3-9 (1+\log (3))-x^2 (4-\log (9))-x (11+\log (243))\right )}{x^4}\right ) \, dx\\ &=3 (3-x)^2+\int \frac {e^{-2 x} (3-x) \left (2 x^3-9 (1+\log (3))-x^2 (4-\log (9))-x (11+\log (243))\right )}{x^4} \, dx\\ &=3 (3-x)^2+\int \left (-2 e^{-2 x}-\frac {27 e^{-2 x} (1+\log (3))}{x^4}+\frac {e^{-2 x} (10-\log (9))}{x}+\frac {e^{-2 x} (-24-\log (729))}{x^3}+\frac {e^{-2 x} (-1+\log (177147))}{x^2}\right ) \, dx\\ &=3 (3-x)^2-2 \int e^{-2 x} \, dx-(27 (1+\log (3))) \int \frac {e^{-2 x}}{x^4} \, dx+(10-\log (9)) \int \frac {e^{-2 x}}{x} \, dx+(-24-\log (729)) \int \frac {e^{-2 x}}{x^3} \, dx+(-1+\log (177147)) \int \frac {e^{-2 x}}{x^2} \, dx\\ &=e^{-2 x}+3 (3-x)^2+\frac {9 e^{-2 x} (1+\log (3))}{x^3}+\text {Ei}(-2 x) (10-\log (9))+\frac {e^{-2 x} (24+\log (729))}{2 x^2}+\frac {e^{-2 x} (1-\log (177147))}{x}+(18 (1+\log (3))) \int \frac {e^{-2 x}}{x^3} \, dx+(24+\log (729)) \int \frac {e^{-2 x}}{x^2} \, dx+(2 (1-\log (177147))) \int \frac {e^{-2 x}}{x} \, dx\\ &=e^{-2 x}+3 (3-x)^2+\frac {9 e^{-2 x} (1+\log (3))}{x^3}-\frac {9 e^{-2 x} (1+\log (3))}{x^2}+\text {Ei}(-2 x) (10-\log (9))+\frac {e^{-2 x} (24+\log (729))}{2 x^2}-\frac {e^{-2 x} (24+\log (729))}{x}+\frac {e^{-2 x} (1-\log (177147))}{x}+2 \text {Ei}(-2 x) (1-\log (177147))-(18 (1+\log (3))) \int \frac {e^{-2 x}}{x^2} \, dx-(2 (24+\log (729))) \int \frac {e^{-2 x}}{x} \, dx\\ &=e^{-2 x}+3 (3-x)^2+\frac {9 e^{-2 x} (1+\log (3))}{x^3}-\frac {9 e^{-2 x} (1+\log (3))}{x^2}+\frac {18 e^{-2 x} (1+\log (3))}{x}+\text {Ei}(-2 x) (10-\log (9))+\frac {e^{-2 x} (24+\log (729))}{2 x^2}-\frac {e^{-2 x} (24+\log (729))}{x}-2 \text {Ei}(-2 x) (24+\log (729))+\frac {e^{-2 x} (1-\log (177147))}{x}+2 \text {Ei}(-2 x) (1-\log (177147))+(36 (1+\log (3))) \int \frac {e^{-2 x}}{x} \, dx\\ &=e^{-2 x}+3 (3-x)^2+\frac {9 e^{-2 x} (1+\log (3))}{x^3}-\frac {9 e^{-2 x} (1+\log (3))}{x^2}+\frac {18 e^{-2 x} (1+\log (3))}{x}+36 \text {Ei}(-2 x) (1+\log (3))+\text {Ei}(-2 x) (10-\log (9))+\frac {e^{-2 x} (24+\log (729))}{2 x^2}-\frac {e^{-2 x} (24+\log (729))}{x}-2 \text {Ei}(-2 x) (24+\log (729))+\frac {e^{-2 x} (1-\log (177147))}{x}+2 \text {Ei}(-2 x) (1-\log (177147))\\ \end {aligned} \end {gather*}

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Mathematica [B] time = 0.92, size = 60, normalized size = 2.61 \begin {gather*} \frac {e^{-2 x} \left (2 x^3-36 e^{2 x} x^4+6 e^{2 x} x^5+2 x^2 (-5+\log (3))-3 x (-2+\log (81))+2 (9+\log (19683))\right )}{2 x^3} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-18*x - 12*x^2 + 12*x^3 - 2*x^4 + E^(2*x)*(-18*x^4 + 6*x^5) + (-27 - 6*x + 11*x^2 - 2*x^3)*Log[3*E]

)/(E^(2*x)*x^4),x]

[Out]

(2*x^3 - 36*E^(2*x)*x^4 + 6*E^(2*x)*x^5 + 2*x^2*(-5 + Log[3]) - 3*x*(-2 + Log[81]) + 2*(9 + Log[19683]))/(2*E^

(2*x)*x^3)

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fricas [B] time = 0.66, size = 47, normalized size = 2.04 \begin {gather*} \frac {{\left (x^{3} - 5 \, x^{2} + 3 \, {\left (x^{5} - 6 \, x^{4}\right )} e^{\left (2 \, x\right )} + {\left (x^{2} - 6 \, x + 9\right )} \log \relax (3) + 3 \, x + 9\right )} e^{\left (-2 \, x\right )}}{x^{3}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+11*x^2-6*x-27)*log(3*exp(1))+(6*x^5-18*x^4)*exp(x)^2-2*x^4+12*x^3-12*x^2-18*x)/exp(x)^2/x^4

,x, algorithm="fricas")

[Out]

(x^3 - 5*x^2 + 3*(x^5 - 6*x^4)*e^(2*x) + (x^2 - 6*x + 9)*log(3) + 3*x + 9)*e^(-2*x)/x^3

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giac [B] time = 0.34, size = 72, normalized size = 3.13 \begin {gather*} \frac {3 \, x^{5} - 18 \, x^{4} + x^{3} e^{\left (-2 \, x\right )} + x^{2} e^{\left (-2 \, x\right )} \log \relax (3) - 5 \, x^{2} e^{\left (-2 \, x\right )} - 6 \, x e^{\left (-2 \, x\right )} \log \relax (3) + 3 \, x e^{\left (-2 \, x\right )} + 9 \, e^{\left (-2 \, x\right )} \log \relax (3) + 9 \, e^{\left (-2 \, x\right )}}{x^{3}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+11*x^2-6*x-27)*log(3*exp(1))+(6*x^5-18*x^4)*exp(x)^2-2*x^4+12*x^3-12*x^2-18*x)/exp(x)^2/x^4

,x, algorithm="giac")

[Out]

(3*x^5 - 18*x^4 + x^3*e^(-2*x) + x^2*e^(-2*x)*log(3) - 5*x^2*e^(-2*x) - 6*x*e^(-2*x)*log(3) + 3*x*e^(-2*x) + 9

*e^(-2*x)*log(3) + 9*e^(-2*x))/x^3

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maple [A] time = 0.08, size = 46, normalized size = 2.00


method	result	size

risch	$3 x^{2}-18 x +\frac {\left (x^{2} \ln \relax (3)+x^{3}-6 x \ln \relax (3)-5 x^{2}+9 \ln \relax (3)+3 x +9\right ) {\mathrm e}^{-2 x}}{x^{3}}$	$46$
norman	$\frac {\left (x^{3}+\left (3-6 \ln \relax (3)\right ) x +\left (\ln \relax (3)-5\right ) x^{2}+3 x^{5} {\mathrm e}^{2 x}-18 \,{\mathrm e}^{2 x} x^{4}+9 \ln \relax (3)+9\right ) {\mathrm e}^{-2 x}}{x^{3}}$	$52$
default	$3 x^{2}-18 x +{\mathrm e}^{-2 x}+\frac {9 \,{\mathrm e}^{-2 x}}{x^{3}}+\frac {3 \,{\mathrm e}^{-2 x}}{x^{2}}-\frac {5 \,{\mathrm e}^{-2 x}}{x}+\frac {9 \,{\mathrm e}^{-2 x} \ln \relax (3)}{x^{3}}-\frac {6 \,{\mathrm e}^{-2 x} \ln \relax (3)}{x^{2}}+\frac {{\mathrm e}^{-2 x} \ln \relax (3)}{x}$	$73$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+11*x^2-6*x-27)*ln(3*exp(1))+(6*x^5-18*x^4)*exp(x)^2-2*x^4+12*x^3-12*x^2-18*x)/exp(x)^2/x^4,x,meth

od=_RETURNVERBOSE)

[Out]

3*x^2-18*x+(x^2*ln(3)+x^3-6*x*ln(3)-5*x^2+9*ln(3)+3*x+9)/x^3*exp(-2*x)

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maxima [C] time = 0.50, size = 80, normalized size = 3.48 \begin {gather*} 3 \, x^{2} - 2 \, {\rm Ei}\left (-2 \, x\right ) \log \left (3 \, e\right ) - 22 \, \Gamma \left (-1, 2 \, x\right ) \log \left (3 \, e\right ) + 24 \, \Gamma \left (-2, 2 \, x\right ) \log \left (3 \, e\right ) + 216 \, \Gamma \left (-3, 2 \, x\right ) \log \left (3 \, e\right ) - 18 \, x + 12 \, {\rm Ei}\left (-2 \, x\right ) + e^{\left (-2 \, x\right )} + 24 \, \Gamma \left (-1, 2 \, x\right ) + 72 \, \Gamma \left (-2, 2 \, x\right ) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+11*x^2-6*x-27)*log(3*exp(1))+(6*x^5-18*x^4)*exp(x)^2-2*x^4+12*x^3-12*x^2-18*x)/exp(x)^2/x^4

,x, algorithm="maxima")

[Out]

3*x^2 - 2*Ei(-2*x)*log(3*e) - 22*gamma(-1, 2*x)*log(3*e) + 24*gamma(-2, 2*x)*log(3*e) + 216*gamma(-3, 2*x)*log

(3*e) - 18*x + 12*Ei(-2*x) + e^(-2*x) + 24*gamma(-1, 2*x) + 72*gamma(-2, 2*x)

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mupad [B] time = 0.11, size = 52, normalized size = 2.26 \begin {gather*} \frac {{\mathrm {e}}^{-2\,x}\,\left (9\,\ln \relax (3)-x\,\left (6\,\ln \relax (3)-3\right )+x^2\,\left (\ln \relax (3)-5\right )-18\,x^4\,{\mathrm {e}}^{2\,x}+3\,x^5\,{\mathrm {e}}^{2\,x}+x^3+9\right )}{x^3} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x)*(18*x + log(3*exp(1))*(6*x - 11*x^2 + 2*x^3 + 27) + exp(2*x)*(18*x^4 - 6*x^5) + 12*x^2 - 12*x^

3 + 2*x^4))/x^4,x)

[Out]

(exp(-2*x)*(9*log(3) - x*(6*log(3) - 3) + x^2*(log(3) - 5) - 18*x^4*exp(2*x) + 3*x^5*exp(2*x) + x^3 + 9))/x^3

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sympy [B] time = 0.18, size = 48, normalized size = 2.09 \begin {gather*} 3 x^{2} - 18 x + \frac {\left (x^{3} - 5 x^{2} + x^{2} \log {\relax (3 )} - 6 x \log {\relax (3 )} + 3 x + 9 + 9 \log {\relax (3 )}\right ) e^{- 2 x}}{x^{3}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+11*x**2-6*x-27)*ln(3*exp(1))+(6*x**5-18*x**4)*exp(x)**2-2*x**4+12*x**3-12*x**2-18*x)/exp(x

)**2/x**4,x)

[Out]

3*x**2 - 18*x + (x**3 - 5*x**2 + x**2*log(3) - 6*x*log(3) + 3*x + 9 + 9*log(3))*exp(-2*x)/x**3

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3.12.49 \(\int \frac {e^{-2 x} (-18 x-12 x^2+12 x^3-2 x^4+e^{2 x} (-18 x^4+6 x^5)+(-27-6 x+11 x^2-2 x^3) \log (3 e))}{x^4} \, dx\)