3.65.8 \(\int \frac {e^{-6 x} (-6 x+e^{3 x} (-4-36 x-18 x^2)+e^{6 x} (-28+30 x+18 x^2)+(e^{6 x} (4-6 x)+6 e^{3 x} x) \log (\frac {x^2}{4}))}{x} \, dx\)

Optimal. Leaf size=22 \[ \left (7+e^{-3 x}+3 x-\log \left (\frac {x^2}{4}\right )\right )^2 \]

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Rubi [B]  time = 1.19, antiderivative size = 58, normalized size of antiderivative = 2.64, number of steps used = 18, number of rules used = 9, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {6688, 12, 6742, 2194, 2199, 2178, 2176, 2554, 6686} \begin {gather*} \left (-\log \left (x^2\right )+3 x+7+\log (4)\right )^2-2 e^{-3 x} \log \left (x^2\right )+e^{-6 x}+14 e^{-3 x}+6 e^{-3 x} x+\frac {2}{3} e^{-3 x} \log (64) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*x + E^(3*x)*(-4 - 36*x - 18*x^2) + E^(6*x)*(-28 + 30*x + 18*x^2) + (E^(6*x)*(4 - 6*x) + 6*E^(3*x)*x)*L
og[x^2/4])/(E^(6*x)*x),x]

[Out]

E^(-6*x) + 14/E^(3*x) + (6*x)/E^(3*x) + (2*Log[64])/(3*E^(3*x)) + (7 + 3*x + Log[4] - Log[x^2])^2 - (2*Log[x^2
])/E^(3*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 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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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} &=\int \frac {2 e^{-6 x} \left (3 x-e^{3 x} (-2+3 x)\right ) \left (-1-e^{3 x} (7+3 x)+e^{3 x} \log \left (\frac {x^2}{4}\right )\right )}{x} \, dx\\ &=2 \int \frac {e^{-6 x} \left (3 x-e^{3 x} (-2+3 x)\right ) \left (-1-e^{3 x} (7+3 x)+e^{3 x} \log \left (\frac {x^2}{4}\right )\right )}{x} \, dx\\ &=2 \int \left (-3 e^{-6 x}-\frac {e^{-3 x} \left (2+18 x+9 x^2-3 x \log \left (\frac {x^2}{4}\right )\right )}{x}+\frac {(2-3 x) \left (-3 x-7 \left (1+\frac {2 \log (2)}{7}\right )+\log \left (x^2\right )\right )}{x}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-3 x} \left (2+18 x+9 x^2-3 x \log \left (\frac {x^2}{4}\right )\right )}{x} \, dx\right )+2 \int \frac {(2-3 x) \left (-3 x-7 \left (1+\frac {2 \log (2)}{7}\right )+\log \left (x^2\right )\right )}{x} \, dx-6 \int e^{-6 x} \, dx\\ &=e^{-6 x}+\left (7+3 x+\log (4)-\log \left (x^2\right )\right )^2-2 \int \left (\frac {e^{-3 x} \left (2+18 x+9 x^2\right )}{x}+e^{-3 x} \log (64)-3 e^{-3 x} \log \left (x^2\right )\right ) \, dx\\ &=e^{-6 x}+\left (7+3 x+\log (4)-\log \left (x^2\right )\right )^2-2 \int \frac {e^{-3 x} \left (2+18 x+9 x^2\right )}{x} \, dx+6 \int e^{-3 x} \log \left (x^2\right ) \, dx-(2 \log (64)) \int e^{-3 x} \, dx\\ &=e^{-6 x}+\frac {2}{3} e^{-3 x} \log (64)+\left (7+3 x+\log (4)-\log \left (x^2\right )\right )^2-2 e^{-3 x} \log \left (x^2\right )-2 \int \left (18 e^{-3 x}+\frac {2 e^{-3 x}}{x}+9 e^{-3 x} x\right ) \, dx-6 \int -\frac {2 e^{-3 x}}{3 x} \, dx\\ &=e^{-6 x}+\frac {2}{3} e^{-3 x} \log (64)+\left (7+3 x+\log (4)-\log \left (x^2\right )\right )^2-2 e^{-3 x} \log \left (x^2\right )-18 \int e^{-3 x} x \, dx-36 \int e^{-3 x} \, dx\\ &=e^{-6 x}+12 e^{-3 x}+6 e^{-3 x} x+\frac {2}{3} e^{-3 x} \log (64)+\left (7+3 x+\log (4)-\log \left (x^2\right )\right )^2-2 e^{-3 x} \log \left (x^2\right )-6 \int e^{-3 x} \, dx\\ &=e^{-6 x}+14 e^{-3 x}+6 e^{-3 x} x+\frac {2}{3} e^{-3 x} \log (64)+\left (7+3 x+\log (4)-\log \left (x^2\right )\right )^2-2 e^{-3 x} \log \left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.36, size = 61, normalized size = 2.77 \begin {gather*} e^{-6 x}+42 x+9 x^2+2 e^{-3 x} (7+3 x)-28 \log (x)+2 \left (-e^{-3 x}-3 x\right ) \log \left (\frac {x^2}{4}\right )+\log ^2\left (\frac {x^2}{4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x + E^(3*x)*(-4 - 36*x - 18*x^2) + E^(6*x)*(-28 + 30*x + 18*x^2) + (E^(6*x)*(4 - 6*x) + 6*E^(3*x
)*x)*Log[x^2/4])/(E^(6*x)*x),x]

[Out]

E^(-6*x) + 42*x + 9*x^2 + (2*(7 + 3*x))/E^(3*x) - 28*Log[x] + 2*(-E^(-3*x) - 3*x)*Log[x^2/4] + Log[x^2/4]^2

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fricas [B]  time = 0.64, size = 69, normalized size = 3.14 \begin {gather*} {\left (e^{\left (6 \, x\right )} \log \left (\frac {1}{4} \, x^{2}\right )^{2} + 3 \, {\left (3 \, x^{2} + 14 \, x\right )} e^{\left (6 \, x\right )} + 2 \, {\left (3 \, x + 7\right )} e^{\left (3 \, x\right )} - 2 \, {\left ({\left (3 \, x + 7\right )} e^{\left (6 \, x\right )} + e^{\left (3 \, x\right )}\right )} \log \left (\frac {1}{4} \, x^{2}\right ) + 1\right )} e^{\left (-6 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x+4)*exp(x)^6+6*x*exp(x)^3)*log(1/4*x^2)+(18*x^2+30*x-28)*exp(x)^6+(-18*x^2-36*x-4)*exp(x)^3-6
*x)/x/exp(x)^6,x, algorithm="fricas")

[Out]

(e^(6*x)*log(1/4*x^2)^2 + 3*(3*x^2 + 14*x)*e^(6*x) + 2*(3*x + 7)*e^(3*x) - 2*((3*x + 7)*e^(6*x) + e^(3*x))*log
(1/4*x^2) + 1)*e^(-6*x)

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giac [B]  time = 0.31, size = 77, normalized size = 3.50 \begin {gather*} 9 \, x^{2} + 6 \, x e^{\left (-3 \, x\right )} + 12 \, x \log \relax (2) + 4 \, e^{\left (-3 \, x\right )} \log \relax (2) - 12 \, x \log \left (x \mathrm {sgn}\relax (x)\right ) - 4 \, e^{\left (-3 \, x\right )} \log \left (x \mathrm {sgn}\relax (x)\right ) + 4 \, \log \left (x \mathrm {sgn}\relax (x)\right )^{2} - 8 \, \log \relax (2) \log \relax (x) + 42 \, x + 14 \, e^{\left (-3 \, x\right )} + e^{\left (-6 \, x\right )} - 28 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x+4)*exp(x)^6+6*x*exp(x)^3)*log(1/4*x^2)+(18*x^2+30*x-28)*exp(x)^6+(-18*x^2-36*x-4)*exp(x)^3-6
*x)/x/exp(x)^6,x, algorithm="giac")

[Out]

9*x^2 + 6*x*e^(-3*x) + 12*x*log(2) + 4*e^(-3*x)*log(2) - 12*x*log(x*sgn(x)) - 4*e^(-3*x)*log(x*sgn(x)) + 4*log
(x*sgn(x))^2 - 8*log(2)*log(x) + 42*x + 14*e^(-3*x) + e^(-6*x) - 28*log(x)

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maple [B]  time = 0.25, size = 94, normalized size = 4.27

method	result	size
default	\(4 \ln \relax (2) {\mathrm e}^{-3 x}-2 \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right ) {\mathrm e}^{-3 x}+14 \,{\mathrm e}^{-3 x}+6 x \,{\mathrm e}^{-3 x}-4 \ln \relax (x ) {\mathrm e}^{-3 x}+{\mathrm e}^{-6 x}+9 x^{2}+42 x -28 \ln \relax (x )-6 x \ln \left (x^{2}\right )+4 \ln \relax (x ) \ln \left (x^{2}\right )-4 \ln \relax (x )^{2}-8 \ln \relax (2) \ln \relax (x )+12 x \ln \relax (2)\)	\(94\)
risch	\(4 \ln \relax (x )^{2}-4 \left (3 x \,{\mathrm e}^{3 x}+1\right ) {\mathrm e}^{-3 x} \ln \relax (x )+\left (1-8 \ln \relax (x ) \ln \relax (2) {\mathrm e}^{6 x}+12 \ln \relax (2) x \,{\mathrm e}^{6 x}+14 \,{\mathrm e}^{3 x}-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{3 x}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{6 x}+4 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{6 x}+3 i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{6 x}+42 x \,{\mathrm e}^{6 x}+9 x^{2} {\mathrm e}^{6 x}+6 x \,{\mathrm e}^{3 x}+i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{3 x}+4 \ln \relax (2) {\mathrm e}^{3 x}-28 \ln \relax (x ) {\mathrm e}^{6 x}-6 i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{6 x}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{6 x}+3 i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{6 x}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{3 x}\right ) {\mathrm e}^{-6 x}\)	\(288\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-6*x+4)*exp(x)^6+6*x*exp(x)^3)*ln(1/4*x^2)+(18*x^2+30*x-28)*exp(x)^6+(-18*x^2-36*x-4)*exp(x)^3-6*x)/x/e
xp(x)^6,x,method=_RETURNVERBOSE)

[Out]

4*ln(2)*exp(-3*x)-2*(ln(x^2)-2*ln(x))*exp(-3*x)+14*exp(-3*x)+6*x*exp(-3*x)-4*ln(x)*exp(-3*x)+1/exp(x)^6+9*x^2+
42*x-28*ln(x)-6*x*ln(x^2)+4*ln(x)*ln(x^2)-4*ln(x)^2-8*ln(2)*ln(x)+12*x*ln(2)

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maxima [B]  time = 0.41, size = 63, normalized size = 2.86 \begin {gather*} 9 \, x^{2} + 2 \, {\left (3 \, x + 1\right )} e^{\left (-3 \, x\right )} - 6 \, x \log \left (\frac {1}{4} \, x^{2}\right ) - 2 \, e^{\left (-3 \, x\right )} \log \left (\frac {1}{4} \, x^{2}\right ) + \log \left (\frac {1}{4} \, x^{2}\right )^{2} + 42 \, x + 12 \, e^{\left (-3 \, x\right )} + e^{\left (-6 \, x\right )} - 28 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x+4)*exp(x)^6+6*x*exp(x)^3)*log(1/4*x^2)+(18*x^2+30*x-28)*exp(x)^6+(-18*x^2-36*x-4)*exp(x)^3-6
*x)/x/exp(x)^6,x, algorithm="maxima")

[Out]

9*x^2 + 2*(3*x + 1)*e^(-3*x) - 6*x*log(1/4*x^2) - 2*e^(-3*x)*log(1/4*x^2) + log(1/4*x^2)^2 + 42*x + 12*e^(-3*x
) + e^(-6*x) - 28*log(x)

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mupad [B]  time = 4.79, size = 63, normalized size = 2.86 \begin {gather*} 30\,x+14\,{\mathrm {e}}^{-3\,x}+{\mathrm {e}}^{-6\,x}-28\,\ln \relax (x)+{\ln \left (\frac {x^2}{4}\right )}^2+6\,x\,{\mathrm {e}}^{-3\,x}-x\,\left (6\,\ln \left (\frac {x^2}{4}\right )-12\right )-2\,{\mathrm {e}}^{-3\,x}\,\ln \left (\frac {x^2}{4}\right )+9\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-6*x)*(6*x + exp(3*x)*(36*x + 18*x^2 + 4) - exp(6*x)*(30*x + 18*x^2 - 28) - log(x^2/4)*(6*x*exp(3*x)
- exp(6*x)*(6*x - 4))))/x,x)

[Out]

30*x + 14*exp(-3*x) + exp(-6*x) - 28*log(x) + log(x^2/4)^2 + 6*x*exp(-3*x) - x*(6*log(x^2/4) - 12) - 2*exp(-3*
x)*log(x^2/4) + 9*x^2

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sympy [B]  time = 0.43, size = 56, normalized size = 2.55 \begin {gather*} 9 x^{2} - 6 x \log {\left (\frac {x^{2}}{4} \right )} + 42 x + \left (6 x - 2 \log {\left (\frac {x^{2}}{4} \right )} + 14\right ) e^{- 3 x} - 28 \log {\relax (x )} + \log {\left (\frac {x^{2}}{4} \right )}^{2} + e^{- 6 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x+4)*exp(x)**6+6*x*exp(x)**3)*ln(1/4*x**2)+(18*x**2+30*x-28)*exp(x)**6+(-18*x**2-36*x-4)*exp(x
)**3-6*x)/x/exp(x)**6,x)

[Out]

9*x**2 - 6*x*log(x**2/4) + 42*x + (6*x - 2*log(x**2/4) + 14)*exp(-3*x) - 28*log(x) + log(x**2/4)**2 + exp(-6*x
)

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