3.30.28 \(\int \frac {e^{-7 x/3} ((192+e^{7 x/3} (-36-12 x+12 x^2)) \log ^3(x)+(-112 x+e^{7 x/3} (-3 x+6 x^2)) \log ^4(x))}{3 x} \, dx\)

Optimal. Leaf size=22 \[ \left (-3+16 e^{-7 x/3}-x+x^2\right ) \log ^4(x) \]

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Rubi [A]  time = 0.69, antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 26, number of rules used = 11, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 6742, 2202, 2357, 2296, 2295, 2302, 30, 2305, 2304, 2330} \begin {gather*} x^2 \log ^4(x)+16 e^{-7 x/3} \log ^4(x)-x \log ^4(x)-3 \log ^4(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((192 + E^((7*x)/3)*(-36 - 12*x + 12*x^2))*Log[x]^3 + (-112*x + E^((7*x)/3)*(-3*x + 6*x^2))*Log[x]^4)/(3*E
^((7*x)/3)*x),x]

[Out]

-3*Log[x]^4 + (16*Log[x]^4)/E^((7*x)/3) - x*Log[x]^4 + x^2*Log[x]^4

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2202

Int[Log[(d_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(x_)^(m_.)*((e_) + Log[(d_.)*(x_)]*(h_.)*((f_.) +
(g_.)*(x_))), x_Symbol] :> Simp[(e*x^(m + 1)*F^(c*(a + b*x))*Log[d*x]^(n + 1))/(n + 1), x] /; FreeQ[{F, a, b,
c, d, e, f, g, h, m, n}, x] && EqQ[e*(m + 1) - f*h*(n + 1), 0] && EqQ[g*h*(n + 1) - b*c*e*Log[F], 0] && NeQ[n,
-1]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

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}{3} \int \frac {e^{-7 x/3} \left (\left (192+e^{7 x/3} \left (-36-12 x+12 x^2\right )\right ) \log ^3(x)+\left (-112 x+e^{7 x/3} \left (-3 x+6 x^2\right )\right ) \log ^4(x)\right )}{x} \, dx\\ &=\frac {1}{3} \int \left (-\frac {16 e^{-7 x/3} \log ^3(x) (-12+7 x \log (x))}{x}+\frac {3 \log ^3(x) \left (-12-4 x+4 x^2-x \log (x)+2 x^2 \log (x)\right )}{x}\right ) \, dx\\ &=-\left (\frac {16}{3} \int \frac {e^{-7 x/3} \log ^3(x) (-12+7 x \log (x))}{x} \, dx\right )+\int \frac {\log ^3(x) \left (-12-4 x+4 x^2-x \log (x)+2 x^2 \log (x)\right )}{x} \, dx\\ &=16 e^{-7 x/3} \log ^4(x)+\int \left (\frac {4 \left (-3-x+x^2\right ) \log ^3(x)}{x}+(-1+2 x) \log ^4(x)\right ) \, dx\\ &=16 e^{-7 x/3} \log ^4(x)+4 \int \frac {\left (-3-x+x^2\right ) \log ^3(x)}{x} \, dx+\int (-1+2 x) \log ^4(x) \, dx\\ &=16 e^{-7 x/3} \log ^4(x)+4 \int \left (-\log ^3(x)-\frac {3 \log ^3(x)}{x}+x \log ^3(x)\right ) \, dx+\int \left (-\log ^4(x)+2 x \log ^4(x)\right ) \, dx\\ &=16 e^{-7 x/3} \log ^4(x)+2 \int x \log ^4(x) \, dx-4 \int \log ^3(x) \, dx+4 \int x \log ^3(x) \, dx-12 \int \frac {\log ^3(x)}{x} \, dx-\int \log ^4(x) \, dx\\ &=-4 x \log ^3(x)+2 x^2 \log ^3(x)+16 e^{-7 x/3} \log ^4(x)-x \log ^4(x)+x^2 \log ^4(x)+4 \int \log ^3(x) \, dx-4 \int x \log ^3(x) \, dx-6 \int x \log ^2(x) \, dx+12 \int \log ^2(x) \, dx-12 \operatorname {Subst}\left (\int x^3 \, dx,x,\log (x)\right )\\ &=12 x \log ^2(x)-3 x^2 \log ^2(x)-3 \log ^4(x)+16 e^{-7 x/3} \log ^4(x)-x \log ^4(x)+x^2 \log ^4(x)+6 \int x \log (x) \, dx+6 \int x \log ^2(x) \, dx-12 \int \log ^2(x) \, dx-24 \int \log (x) \, dx\\ &=24 x-\frac {3 x^2}{2}-24 x \log (x)+3 x^2 \log (x)-3 \log ^4(x)+16 e^{-7 x/3} \log ^4(x)-x \log ^4(x)+x^2 \log ^4(x)-6 \int x \log (x) \, dx+24 \int \log (x) \, dx\\ &=-3 \log ^4(x)+16 e^{-7 x/3} \log ^4(x)-x \log ^4(x)+x^2 \log ^4(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 22, normalized size = 1.00 \begin {gather*} \left (-3+16 e^{-7 x/3}-x+x^2\right ) \log ^4(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((192 + E^((7*x)/3)*(-36 - 12*x + 12*x^2))*Log[x]^3 + (-112*x + E^((7*x)/3)*(-3*x + 6*x^2))*Log[x]^4
)/(3*E^((7*x)/3)*x),x]

[Out]

(-3 + 16/E^((7*x)/3) - x + x^2)*Log[x]^4

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fricas [A]  time = 0.72, size = 24, normalized size = 1.09 \begin {gather*} {\left ({\left (x^{2} - x - 3\right )} e^{\left (\frac {7}{3} \, x\right )} + 16\right )} e^{\left (-\frac {7}{3} \, x\right )} \log \relax (x)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((6*x^2-3*x)*exp(7/3*x)-112*x)*log(x)^4+((12*x^2-12*x-36)*exp(7/3*x)+192)*log(x)^3)/x/exp(7/3*x
),x, algorithm="fricas")

[Out]

((x^2 - x - 3)*e^(7/3*x) + 16)*e^(-7/3*x)*log(x)^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (3 \, {\left (2 \, x^{2} - x\right )} e^{\left (\frac {7}{3} \, x\right )} - 112 \, x\right )} \log \relax (x)^{4} + 12 \, {\left ({\left (x^{2} - x - 3\right )} e^{\left (\frac {7}{3} \, x\right )} + 16\right )} \log \relax (x)^{3}\right )} e^{\left (-\frac {7}{3} \, x\right )}}{3 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((6*x^2-3*x)*exp(7/3*x)-112*x)*log(x)^4+((12*x^2-12*x-36)*exp(7/3*x)+192)*log(x)^3)/x/exp(7/3*x
),x, algorithm="giac")

[Out]

integrate(1/3*((3*(2*x^2 - x)*e^(7/3*x) - 112*x)*log(x)^4 + 12*((x^2 - x - 3)*e^(7/3*x) + 16)*log(x)^3)*e^(-7/
3*x)/x, x)

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maple [A]  time = 0.05, size = 33, normalized size = 1.50

method	result	size
risch	\(\left ({\mathrm e}^{\frac {7 x}{3}} x^{2}-{\mathrm e}^{\frac {7 x}{3}} x -3 \,{\mathrm e}^{\frac {7 x}{3}}+16\right ) {\mathrm e}^{-\frac {7 x}{3}} \ln \relax (x )^{4}\)	\(33\)
default	\(x^{2} \ln \relax (x )^{4}-x \ln \relax (x )^{4}+16 \ln \relax (x )^{4} {\mathrm e}^{-\frac {7 x}{3}}-3 \ln \relax (x )^{4}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(((6*x^2-3*x)*exp(7/3*x)-112*x)*ln(x)^4+((12*x^2-12*x-36)*exp(7/3*x)+192)*ln(x)^3)/x/exp(7/3*x),x,meth
od=_RETURNVERBOSE)

[Out]

(exp(7/3*x)*x^2-exp(7/3*x)*x-3*exp(7/3*x)+16)*exp(-7/3*x)*ln(x)^4

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maxima [B]  time = 0.56, size = 116, normalized size = 5.27 \begin {gather*} {\left (x^{2} - x\right )} \log \relax (x)^{4} + 16 \, e^{\left (-\frac {7}{3} \, x\right )} \log \relax (x)^{4} - 2 \, {\left (x^{2} - 2 \, x\right )} \log \relax (x)^{3} - 3 \, \log \relax (x)^{4} + \frac {1}{2} \, {\left (4 \, \log \relax (x)^{3} - 6 \, \log \relax (x)^{2} + 6 \, \log \relax (x) - 3\right )} x^{2} + 3 \, {\left (x^{2} - 4 \, x\right )} \log \relax (x)^{2} - 4 \, {\left (\log \relax (x)^{3} - 3 \, \log \relax (x)^{2} + 6 \, \log \relax (x) - 6\right )} x + \frac {3}{2} \, x^{2} - 3 \, {\left (x^{2} - 8 \, x\right )} \log \relax (x) - 24 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((6*x^2-3*x)*exp(7/3*x)-112*x)*log(x)^4+((12*x^2-12*x-36)*exp(7/3*x)+192)*log(x)^3)/x/exp(7/3*x
),x, algorithm="maxima")

[Out]

(x^2 - x)*log(x)^4 + 16*e^(-7/3*x)*log(x)^4 - 2*(x^2 - 2*x)*log(x)^3 - 3*log(x)^4 + 1/2*(4*log(x)^3 - 6*log(x)
^2 + 6*log(x) - 3)*x^2 + 3*(x^2 - 4*x)*log(x)^2 - 4*(log(x)^3 - 3*log(x)^2 + 6*log(x) - 6)*x + 3/2*x^2 - 3*(x^
2 - 8*x)*log(x) - 24*x

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mupad [B]  time = 1.82, size = 20, normalized size = 0.91 \begin {gather*} -{\ln \relax (x)}^4\,\left (x-16\,{\mathrm {e}}^{-\frac {7\,x}{3}}-x^2+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(7*x)/3)*((log(x)^3*(exp((7*x)/3)*(12*x - 12*x^2 + 36) - 192))/3 + (log(x)^4*(112*x + exp((7*x)/3)*
(3*x - 6*x^2)))/3))/x,x)

[Out]

-log(x)^4*(x - 16*exp(-(7*x)/3) - x^2 + 3)

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sympy [A]  time = 0.41, size = 24, normalized size = 1.09 \begin {gather*} \left (x^{2} - x - 3\right ) \log {\relax (x )}^{4} + 16 e^{- \frac {7 x}{3}} \log {\relax (x )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(((6*x**2-3*x)*exp(7/3*x)-112*x)*ln(x)**4+((12*x**2-12*x-36)*exp(7/3*x)+192)*ln(x)**3)/x/exp(7/3
*x),x)

[Out]

(x**2 - x - 3)*log(x)**4 + 16*exp(-7*x/3)*log(x)**4

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