3.66.24 \(\int \frac {-4+16 x-24 x^2+16 x^3-4 x^4+(24-80 x+96 x^2-48 x^3+8 x^4) \log (-3+x)}{(-3+x) \log ^3(-3+x)} \, dx\)

Optimal. Leaf size=15 \[ \frac {2 (1-x)^4}{\log ^2(-3+x)} \]

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Rubi [B]  time = 0.90, antiderivative size = 145, normalized size of antiderivative = 9.67, number of steps used = 48, number of rules used = 16, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {6688, 12, 6742, 2411, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178, 2400, 2399, 2389, 2390} \begin {gather*} \frac {2 (3-x)^4}{\log ^2(x-3)}-\frac {16 (3-x)^3}{\log ^2(x-3)}+\frac {48 (3-x)^2}{\log ^2(x-3)}-\frac {64 (3-x)}{\log ^2(x-3)}+\frac {32}{\log ^2(x-3)}+\frac {8 (3-x)^4}{\log (x-3)}-\frac {48 (3-x)^3}{\log (x-3)}+\frac {96 (3-x)^2}{\log (x-3)}-\frac {8 (1-x)^3 (3-x)}{\log (x-3)}-\frac {64 (3-x)}{\log (x-3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 16*x - 24*x^2 + 16*x^3 - 4*x^4 + (24 - 80*x + 96*x^2 - 48*x^3 + 8*x^4)*Log[-3 + x])/((-3 + x)*Log[-3
+ x]^3),x]

[Out]

32/Log[-3 + x]^2 - (64*(3 - x))/Log[-3 + x]^2 + (48*(3 - x)^2)/Log[-3 + x]^2 - (16*(3 - x)^3)/Log[-3 + x]^2 +
(2*(3 - x)^4)/Log[-3 + x]^2 - (64*(3 - x))/Log[-3 + x] - (8*(1 - x)^3*(3 - x))/Log[-3 + x] + (96*(3 - x)^2)/Lo
g[-3 + x] - (48*(3 - x)^3)/Log[-3 + x] + (8*(3 - x)^4)/Log[-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 30

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

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 2297

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

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

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 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2353

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

Rule 2389

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

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2399

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)), x_Symbol] :> Int[ExpandIn
tegrand[(f + g*x)^q/(a + b*Log[c*(d + e*x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && IGtQ[q, 0]

Rule 2400

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
d + e*x)*(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1))/(b*e*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), I
nt[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Dist[(q*(e*f - d*g))/(b*e*n*(p + 1)), Int[(f + g*x
)^(q - 1)*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x]) /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && LtQ[p, -1] && GtQ[q, 0]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 {4 (1-x)^3 (1-x+2 (-3+x) \log (-3+x))}{(3-x) \log ^3(-3+x)} \, dx\\ &=4 \int \frac {(1-x)^3 (1-x+2 (-3+x) \log (-3+x))}{(3-x) \log ^3(-3+x)} \, dx\\ &=4 \int \left (-\frac {(-1+x)^4}{(-3+x) \log ^3(-3+x)}+\frac {2 (-1+x)^3}{\log ^2(-3+x)}\right ) \, dx\\ &=-\left (4 \int \frac {(-1+x)^4}{(-3+x) \log ^3(-3+x)} \, dx\right )+8 \int \frac {(-1+x)^3}{\log ^2(-3+x)} \, dx\\ &=-\frac {8 (1-x)^3 (3-x)}{\log (-3+x)}-4 \operatorname {Subst}\left (\int \frac {(2+x)^4}{x \log ^3(x)} \, dx,x,-3+x\right )+32 \int \frac {(-1+x)^3}{\log (-3+x)} \, dx-48 \int \frac {(-1+x)^2}{\log (-3+x)} \, dx\\ &=-\frac {8 (1-x)^3 (3-x)}{\log (-3+x)}-4 \operatorname {Subst}\left (\int \left (\frac {32}{\log ^3(x)}+\frac {16}{x \log ^3(x)}+\frac {24 x}{\log ^3(x)}+\frac {8 x^2}{\log ^3(x)}+\frac {x^3}{\log ^3(x)}\right ) \, dx,x,-3+x\right )+32 \int \left (\frac {8}{\log (-3+x)}+\frac {12 (-3+x)}{\log (-3+x)}+\frac {6 (-3+x)^2}{\log (-3+x)}+\frac {(-3+x)^3}{\log (-3+x)}\right ) \, dx-48 \int \left (\frac {4}{\log (-3+x)}+\frac {4 (-3+x)}{\log (-3+x)}+\frac {(-3+x)^2}{\log (-3+x)}\right ) \, dx\\ &=-\frac {8 (1-x)^3 (3-x)}{\log (-3+x)}-4 \operatorname {Subst}\left (\int \frac {x^3}{\log ^3(x)} \, dx,x,-3+x\right )+32 \int \frac {(-3+x)^3}{\log (-3+x)} \, dx-32 \operatorname {Subst}\left (\int \frac {x^2}{\log ^3(x)} \, dx,x,-3+x\right )-48 \int \frac {(-3+x)^2}{\log (-3+x)} \, dx-64 \operatorname {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,-3+x\right )-96 \operatorname {Subst}\left (\int \frac {x}{\log ^3(x)} \, dx,x,-3+x\right )-128 \operatorname {Subst}\left (\int \frac {1}{\log ^3(x)} \, dx,x,-3+x\right )-192 \int \frac {1}{\log (-3+x)} \, dx-192 \int \frac {-3+x}{\log (-3+x)} \, dx+192 \int \frac {(-3+x)^2}{\log (-3+x)} \, dx+256 \int \frac {1}{\log (-3+x)} \, dx+384 \int \frac {-3+x}{\log (-3+x)} \, dx\\ &=-\frac {64 (3-x)}{\log ^2(-3+x)}+\frac {48 (3-x)^2}{\log ^2(-3+x)}-\frac {16 (3-x)^3}{\log ^2(-3+x)}+\frac {2 (3-x)^4}{\log ^2(-3+x)}-\frac {8 (1-x)^3 (3-x)}{\log (-3+x)}-8 \operatorname {Subst}\left (\int \frac {x^3}{\log ^2(x)} \, dx,x,-3+x\right )+32 \operatorname {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,-3+x\right )-48 \operatorname {Subst}\left (\int \frac {x^2}{\log ^2(x)} \, dx,x,-3+x\right )-48 \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-3+x\right )-64 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (-3+x)\right )-64 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-3+x\right )-96 \operatorname {Subst}\left (\int \frac {x}{\log ^2(x)} \, dx,x,-3+x\right )-192 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+x\right )-192 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-3+x\right )+192 \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-3+x\right )+256 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+x\right )+384 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-3+x\right )\\ &=\frac {32}{\log ^2(-3+x)}-\frac {64 (3-x)}{\log ^2(-3+x)}+\frac {48 (3-x)^2}{\log ^2(-3+x)}-\frac {16 (3-x)^3}{\log ^2(-3+x)}+\frac {2 (3-x)^4}{\log ^2(-3+x)}-\frac {64 (3-x)}{\log (-3+x)}-\frac {8 (1-x)^3 (3-x)}{\log (-3+x)}+\frac {96 (3-x)^2}{\log (-3+x)}-\frac {48 (3-x)^3}{\log (-3+x)}+\frac {8 (3-x)^4}{\log (-3+x)}+64 \text {li}(-3+x)+32 \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (-3+x)\right )-32 \operatorname {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,-3+x\right )-48 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-3+x)\right )-64 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+x\right )-144 \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-3+x\right )-192 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-3+x)\right )+192 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-3+x)\right )-192 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-3+x\right )+384 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-3+x)\right )\\ &=192 \text {Ei}(2 \log (-3+x))+144 \text {Ei}(3 \log (-3+x))+32 \text {Ei}(4 \log (-3+x))+\frac {32}{\log ^2(-3+x)}-\frac {64 (3-x)}{\log ^2(-3+x)}+\frac {48 (3-x)^2}{\log ^2(-3+x)}-\frac {16 (3-x)^3}{\log ^2(-3+x)}+\frac {2 (3-x)^4}{\log ^2(-3+x)}-\frac {64 (3-x)}{\log (-3+x)}-\frac {8 (1-x)^3 (3-x)}{\log (-3+x)}+\frac {96 (3-x)^2}{\log (-3+x)}-\frac {48 (3-x)^3}{\log (-3+x)}+\frac {8 (3-x)^4}{\log (-3+x)}-32 \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (-3+x)\right )-144 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-3+x)\right )-192 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-3+x)\right )\\ &=\frac {32}{\log ^2(-3+x)}-\frac {64 (3-x)}{\log ^2(-3+x)}+\frac {48 (3-x)^2}{\log ^2(-3+x)}-\frac {16 (3-x)^3}{\log ^2(-3+x)}+\frac {2 (3-x)^4}{\log ^2(-3+x)}-\frac {64 (3-x)}{\log (-3+x)}-\frac {8 (1-x)^3 (3-x)}{\log (-3+x)}+\frac {96 (3-x)^2}{\log (-3+x)}-\frac {48 (3-x)^3}{\log (-3+x)}+\frac {8 (3-x)^4}{\log (-3+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 13, normalized size = 0.87 \begin {gather*} \frac {2 (-1+x)^4}{\log ^2(-3+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 16*x - 24*x^2 + 16*x^3 - 4*x^4 + (24 - 80*x + 96*x^2 - 48*x^3 + 8*x^4)*Log[-3 + x])/((-3 + x)*
Log[-3 + x]^3),x]

[Out]

(2*(-1 + x)^4)/Log[-3 + x]^2

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fricas [A]  time = 0.60, size = 26, normalized size = 1.73 \begin {gather*} \frac {2 \, {\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1\right )}}{\log \left (x - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-48*x^3+96*x^2-80*x+24)*log(x-3)-4*x^4+16*x^3-24*x^2+16*x-4)/(x-3)/log(x-3)^3,x, algorithm="f
ricas")

[Out]

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

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giac [A]  time = 0.22, size = 26, normalized size = 1.73 \begin {gather*} \frac {2 \, {\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1\right )}}{\log \left (x - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-48*x^3+96*x^2-80*x+24)*log(x-3)-4*x^4+16*x^3-24*x^2+16*x-4)/(x-3)/log(x-3)^3,x, algorithm="g
iac")

[Out]

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

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maple [A]  time = 0.41, size = 27, normalized size = 1.80

method	result	size
risch	\(\frac {2 x^{4}-8 x^{3}+12 x^{2}-8 x +2}{\ln \left (x -3\right )^{2}}\)	\(27\)
norman	\(\frac {2 x^{4}-8 x^{3}+12 x^{2}-8 x +2}{\ln \left (x -3\right )^{2}}\)	\(28\)
derivativedivides	\(\frac {2 \left (x -3\right )^{4}}{\ln \left (x -3\right )^{2}}+\frac {16 \left (x -3\right )^{3}}{\ln \left (x -3\right )^{2}}+\frac {48 \left (x -3\right )^{2}}{\ln \left (x -3\right )^{2}}+\frac {64 x -192}{\ln \left (x -3\right )^{2}}+\frac {32}{\ln \left (x -3\right )^{2}}\)	\(60\)
default	\(\frac {2 \left (x -3\right )^{4}}{\ln \left (x -3\right )^{2}}+\frac {16 \left (x -3\right )^{3}}{\ln \left (x -3\right )^{2}}+\frac {48 \left (x -3\right )^{2}}{\ln \left (x -3\right )^{2}}+\frac {64 x -192}{\ln \left (x -3\right )^{2}}+\frac {32}{\ln \left (x -3\right )^{2}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^4-48*x^3+96*x^2-80*x+24)*ln(x-3)-4*x^4+16*x^3-24*x^2+16*x-4)/(x-3)/ln(x-3)^3,x,method=_RETURNVERBOSE
)

[Out]

2*(x^4-4*x^3+6*x^2-4*x+1)/ln(x-3)^2

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maxima [B]  time = 0.40, size = 48, normalized size = 3.20 \begin {gather*} \frac {2 \, {\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 12 \, \log \left (x - 3\right )\right )}}{\log \left (x - 3\right )^{2}} - \frac {24}{\log \left (x - 3\right )} + \frac {2}{\log \left (x - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-48*x^3+96*x^2-80*x+24)*log(x-3)-4*x^4+16*x^3-24*x^2+16*x-4)/(x-3)/log(x-3)^3,x, algorithm="m
axima")

[Out]

2*(x^4 - 4*x^3 + 6*x^2 - 4*x + 12*log(x - 3))/log(x - 3)^2 - 24/log(x - 3) + 2/log(x - 3)^2

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mupad [B]  time = 4.22, size = 13, normalized size = 0.87 \begin {gather*} \frac {2\,{\left (x-1\right )}^4}{{\ln \left (x-3\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + log(x - 3)*(96*x^2 - 80*x - 48*x^3 + 8*x^4 + 24) - 24*x^2 + 16*x^3 - 4*x^4 - 4)/(log(x - 3)^3*(x -
3)),x)

[Out]

(2*(x - 1)^4)/log(x - 3)^2

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sympy [B]  time = 0.11, size = 26, normalized size = 1.73 \begin {gather*} \frac {2 x^{4} - 8 x^{3} + 12 x^{2} - 8 x + 2}{\log {\left (x - 3 \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**4-48*x**3+96*x**2-80*x+24)*ln(x-3)-4*x**4+16*x**3-24*x**2+16*x-4)/(x-3)/ln(x-3)**3,x)

[Out]

(2*x**4 - 8*x**3 + 12*x**2 - 8*x + 2)/log(x - 3)**2

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