3.4.87 \(\int \frac {e^{-2 x} (8+23 x-24 x^2-10 x^3+(2+4 x-5 x^2-2 x^3) \log (3 x))}{5 x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{5} e^{-2 x} \left (3-\frac {2}{x}+x\right ) (5+\log (3 x)) \]

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Rubi [B]  time = 0.66, antiderivative size = 68, normalized size of antiderivative = 2.96, number of steps used = 18, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {12, 6742, 2199, 2194, 2177, 2178, 2176, 2554} \begin {gather*} e^{-2 x} x+3 e^{-2 x}-\frac {2 e^{-2 x}}{x}+\frac {1}{5} e^{-2 x} x \log (3 x)+\frac {3}{5} e^{-2 x} \log (3 x)-\frac {2 e^{-2 x} \log (3 x)}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 + 23*x - 24*x^2 - 10*x^3 + (2 + 4*x - 5*x^2 - 2*x^3)*Log[3*x])/(5*E^(2*x)*x^2),x]

[Out]

3/E^(2*x) - 2/(E^(2*x)*x) + x/E^(2*x) + (3*Log[3*x])/(5*E^(2*x)) - (2*Log[3*x])/(5*E^(2*x)*x) + (x*Log[3*x])/(
5*E^(2*x))

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

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}{5} \int \frac {e^{-2 x} \left (8+23 x-24 x^2-10 x^3+\left (2+4 x-5 x^2-2 x^3\right ) \log (3 x)\right )}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{-2 x} \left (8+23 x-24 x^2-10 x^3\right )}{x^2}-\frac {e^{-2 x} \left (-2-4 x+5 x^2+2 x^3\right ) \log (3 x)}{x^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{-2 x} \left (8+23 x-24 x^2-10 x^3\right )}{x^2} \, dx-\frac {1}{5} \int \frac {e^{-2 x} \left (-2-4 x+5 x^2+2 x^3\right ) \log (3 x)}{x^2} \, dx\\ &=\frac {3}{5} e^{-2 x} \log (3 x)-\frac {2 e^{-2 x} \log (3 x)}{5 x}+\frac {1}{5} e^{-2 x} x \log (3 x)+\frac {1}{5} \int \left (-24 e^{-2 x}+\frac {8 e^{-2 x}}{x^2}+\frac {23 e^{-2 x}}{x}-10 e^{-2 x} x\right ) \, dx+\frac {1}{5} \int \frac {e^{-2 x} \left (2-3 x-x^2\right )}{x^2} \, dx\\ &=\frac {3}{5} e^{-2 x} \log (3 x)-\frac {2 e^{-2 x} \log (3 x)}{5 x}+\frac {1}{5} e^{-2 x} x \log (3 x)+\frac {1}{5} \int \left (-e^{-2 x}+\frac {2 e^{-2 x}}{x^2}-\frac {3 e^{-2 x}}{x}\right ) \, dx+\frac {8}{5} \int \frac {e^{-2 x}}{x^2} \, dx-2 \int e^{-2 x} x \, dx+\frac {23}{5} \int \frac {e^{-2 x}}{x} \, dx-\frac {24}{5} \int e^{-2 x} \, dx\\ &=\frac {12 e^{-2 x}}{5}-\frac {8 e^{-2 x}}{5 x}+e^{-2 x} x+\frac {23 \text {Ei}(-2 x)}{5}+\frac {3}{5} e^{-2 x} \log (3 x)-\frac {2 e^{-2 x} \log (3 x)}{5 x}+\frac {1}{5} e^{-2 x} x \log (3 x)-\frac {1}{5} \int e^{-2 x} \, dx+\frac {2}{5} \int \frac {e^{-2 x}}{x^2} \, dx-\frac {3}{5} \int \frac {e^{-2 x}}{x} \, dx-\frac {16}{5} \int \frac {e^{-2 x}}{x} \, dx-\int e^{-2 x} \, dx\\ &=3 e^{-2 x}-\frac {2 e^{-2 x}}{x}+e^{-2 x} x+\frac {4 \text {Ei}(-2 x)}{5}+\frac {3}{5} e^{-2 x} \log (3 x)-\frac {2 e^{-2 x} \log (3 x)}{5 x}+\frac {1}{5} e^{-2 x} x \log (3 x)-\frac {4}{5} \int \frac {e^{-2 x}}{x} \, dx\\ &=3 e^{-2 x}-\frac {2 e^{-2 x}}{x}+e^{-2 x} x+\frac {3}{5} e^{-2 x} \log (3 x)-\frac {2 e^{-2 x} \log (3 x)}{5 x}+\frac {1}{5} e^{-2 x} x \log (3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 26, normalized size = 1.13 \begin {gather*} \frac {e^{-2 x} \left (-2+3 x+x^2\right ) (5+\log (3 x))}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + 23*x - 24*x^2 - 10*x^3 + (2 + 4*x - 5*x^2 - 2*x^3)*Log[3*x])/(5*E^(2*x)*x^2),x]

[Out]

((-2 + 3*x + x^2)*(5 + Log[3*x]))/(5*E^(2*x)*x)

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fricas [A]  time = 0.69, size = 37, normalized size = 1.61 \begin {gather*} \frac {{\left (x^{2} + 3 \, x - 2\right )} e^{\left (-2 \, x\right )} \log \left (3 \, x\right ) + 5 \, {\left (x^{2} + 3 \, x - 2\right )} e^{\left (-2 \, x\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*x^3-5*x^2+4*x+2)*log(3*x)-10*x^3-24*x^2+23*x+8)/exp(2*x)/x^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.31, size = 61, normalized size = 2.65 \begin {gather*} \frac {x^{2} e^{\left (-2 \, x\right )} \log \left (3 \, x\right ) + 5 \, x^{2} e^{\left (-2 \, x\right )} + 3 \, x e^{\left (-2 \, x\right )} \log \left (3 \, x\right ) + 15 \, x e^{\left (-2 \, x\right )} - 2 \, e^{\left (-2 \, x\right )} \log \left (3 \, x\right ) - 10 \, e^{\left (-2 \, x\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*x^3-5*x^2+4*x+2)*log(3*x)-10*x^3-24*x^2+23*x+8)/exp(2*x)/x^2,x, algorithm="giac")

[Out]

1/5*(x^2*e^(-2*x)*log(3*x) + 5*x^2*e^(-2*x) + 3*x*e^(-2*x)*log(3*x) + 15*x*e^(-2*x) - 2*e^(-2*x)*log(3*x) - 10
*e^(-2*x))/x

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maple [A]  time = 0.06, size = 39, normalized size = 1.70

method	result	size
risch	\(\frac {\left (x^{2}+3 x -2\right ) {\mathrm e}^{-2 x} \ln \left (3 x \right )}{5 x}+\frac {\left (x^{2}+3 x -2\right ) {\mathrm e}^{-2 x}}{x}\)	\(39\)
norman	\(\frac {\left (-2+x^{2}+3 x +\frac {3 x \ln \left (3 x \right )}{5}+\frac {x^{2} \ln \left (3 x \right )}{5}-\frac {2 \ln \left (3 x \right )}{5}\right ) {\mathrm e}^{-2 x}}{x}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-2*x^3-5*x^2+4*x+2)*ln(3*x)-10*x^3-24*x^2+23*x+8)/exp(2*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{2} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + \frac {1}{2} \, e^{\left (-2 \, x\right )} \log \left (3 \, x\right ) + \frac {{\left (2 \, x^{2} + x - 4\right )} e^{\left (-2 \, x\right )} \log \relax (x)}{10 \, x} + \frac {41}{10} \, {\rm Ei}\left (-2 \, x\right ) + \frac {12}{5} \, e^{\left (-2 \, x\right )} - \frac {16}{5} \, \Gamma \left (-1, 2 \, x\right ) - \frac {1}{10} \, \int \frac {{\left (4 \, x^{3} \log \relax (3) + 2 \, x^{2} - x {\left (8 \, \log \relax (3) - 1\right )} - 4 \, \log \relax (3) - 4\right )} e^{\left (-2 \, x\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*x^3-5*x^2+4*x+2)*log(3*x)-10*x^3-24*x^2+23*x+8)/exp(2*x)/x^2,x, algorithm="maxima")

[Out]

1/2*(2*x + 1)*e^(-2*x) + 1/2*e^(-2*x)*log(3*x) + 1/10*(2*x^2 + x - 4)*e^(-2*x)*log(x)/x + 41/10*Ei(-2*x) + 12/
5*e^(-2*x) - 16/5*gamma(-1, 2*x) - 1/10*integrate((4*x^3*log(3) + 2*x^2 - x*(8*log(3) - 1) - 4*log(3) - 4)*e^(
-2*x)/x^2, x)

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mupad [B]  time = 0.55, size = 23, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^{-2\,x}\,\left (\ln \left (3\,x\right )+5\right )\,\left (x^2+3\,x-2\right )}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x)*((23*x)/5 + (log(3*x)*(4*x - 5*x^2 - 2*x^3 + 2))/5 - (24*x^2)/5 - 2*x^3 + 8/5))/x^2,x)

[Out]

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

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sympy [A]  time = 0.32, size = 41, normalized size = 1.78 \begin {gather*} \frac {\left (x^{2} \log {\left (3 x \right )} + 5 x^{2} + 3 x \log {\left (3 x \right )} + 15 x - 2 \log {\left (3 x \right )} - 10\right ) e^{- 2 x}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*x**3-5*x**2+4*x+2)*ln(3*x)-10*x**3-24*x**2+23*x+8)/exp(2*x)/x**2,x)

[Out]

(x**2*log(3*x) + 5*x**2 + 3*x*log(3*x) + 15*x - 2*log(3*x) - 10)*exp(-2*x)/(5*x)

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