3.36.14 \(\int \frac {-64-32 x^2-4 x^4+(-80 x+64 x^2-20 x^3+16 x^4) \log (5-4 x)}{(-5+4 x) \log ^2(5-4 x)} \, dx\)

Optimal. Leaf size=18 \[ 2+\frac {\left (4+x^2\right )^2}{\log (5-4 x)} \]

________________________________________________________________________________________

Rubi [B]  time = 1.11, antiderivative size = 88, normalized size of antiderivative = 4.89, number of steps used = 86, number of rules used = 15, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6688, 12, 6742, 2418, 2389, 2297, 2298, 2400, 2399, 2390, 2309, 2178, 2302, 30, 2416} \begin {gather*} -\frac {(5-4 x) x^3}{4 \log (5-4 x)}-\frac {5 (5-4 x) x^2}{16 \log (5-4 x)}-\frac {153 (5-4 x) x}{64 \log (5-4 x)}-\frac {765 (5-4 x)}{256 \log (5-4 x)}+\frac {7921}{256 \log (5-4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64 - 32*x^2 - 4*x^4 + (-80*x + 64*x^2 - 20*x^3 + 16*x^4)*Log[5 - 4*x])/((-5 + 4*x)*Log[5 - 4*x]^2),x]

[Out]

7921/(256*Log[5 - 4*x]) - (765*(5 - 4*x))/(256*Log[5 - 4*x]) - (153*(5 - 4*x)*x)/(64*Log[5 - 4*x]) - (5*(5 - 4
*x)*x^2)/(16*Log[5 - 4*x]) - ((5 - 4*x)*x^3)/(4*Log[5 - 4*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 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 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 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

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 \left (4+x^2\right ) \left (4+x^2-x (-5+4 x) \log (5-4 x)\right )}{(5-4 x) \log ^2(5-4 x)} \, dx\\ &=4 \int \frac {\left (4+x^2\right ) \left (4+x^2-x (-5+4 x) \log (5-4 x)\right )}{(5-4 x) \log ^2(5-4 x)} \, dx\\ &=4 \int \left (-\frac {\left (4+x^2\right )^2}{(-5+4 x) \log ^2(5-4 x)}+\frac {x \left (4+x^2\right )}{\log (5-4 x)}\right ) \, dx\\ &=-\left (4 \int \frac {\left (4+x^2\right )^2}{(-5+4 x) \log ^2(5-4 x)} \, dx\right )+4 \int \frac {x \left (4+x^2\right )}{\log (5-4 x)} \, dx\\ &=-\left (4 \int \left (\frac {765}{256 \log ^2(5-4 x)}+\frac {153 x}{64 \log ^2(5-4 x)}+\frac {5 x^2}{16 \log ^2(5-4 x)}+\frac {x^3}{4 \log ^2(5-4 x)}+\frac {7921}{256 (-5+4 x) \log ^2(5-4 x)}\right ) \, dx\right )+4 \int \left (\frac {4 x}{\log (5-4 x)}+\frac {x^3}{\log (5-4 x)}\right ) \, dx\\ &=-\left (\frac {5}{4} \int \frac {x^2}{\log ^2(5-4 x)} \, dx\right )+4 \int \frac {x^3}{\log (5-4 x)} \, dx-\frac {153}{16} \int \frac {x}{\log ^2(5-4 x)} \, dx-\frac {765}{64} \int \frac {1}{\log ^2(5-4 x)} \, dx+16 \int \frac {x}{\log (5-4 x)} \, dx-\frac {7921}{64} \int \frac {1}{(-5+4 x) \log ^2(5-4 x)} \, dx-\int \frac {x^3}{\log ^2(5-4 x)} \, dx\\ &=-\frac {153 (5-4 x) x}{64 \log (5-4 x)}-\frac {5 (5-4 x) x^2}{16 \log (5-4 x)}-\frac {(5-4 x) x^3}{4 \log (5-4 x)}+\frac {765}{256} \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,5-4 x\right )+\frac {25}{8} \int \frac {x}{\log (5-4 x)} \, dx+4 \int \left (\frac {125}{64 \log (5-4 x)}-\frac {75 (5-4 x)}{64 \log (5-4 x)}+\frac {15 (5-4 x)^2}{64 \log (5-4 x)}-\frac {(5-4 x)^3}{64 \log (5-4 x)}\right ) \, dx-4 \int \frac {x^3}{\log (5-4 x)} \, dx+\frac {765}{64} \int \frac {1}{\log (5-4 x)} \, dx+16 \int \left (\frac {5}{4 \log (5-4 x)}-\frac {5-4 x}{4 \log (5-4 x)}\right ) \, dx-\frac {153}{8} \int \frac {x}{\log (5-4 x)} \, dx-\frac {7921}{256} \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,5-4 x\right )\\ &=-\frac {765 (5-4 x)}{256 \log (5-4 x)}-\frac {153 (5-4 x) x}{64 \log (5-4 x)}-\frac {5 (5-4 x) x^2}{16 \log (5-4 x)}-\frac {(5-4 x) x^3}{4 \log (5-4 x)}-\frac {1}{16} \int \frac {(5-4 x)^3}{\log (5-4 x)} \, dx+\frac {15}{16} \int \frac {(5-4 x)^2}{\log (5-4 x)} \, dx+\frac {25}{8} \int \left (\frac {5}{4 \log (5-4 x)}-\frac {5-4 x}{4 \log (5-4 x)}\right ) \, dx-4 \int \left (\frac {125}{64 \log (5-4 x)}-\frac {75 (5-4 x)}{64 \log (5-4 x)}+\frac {15 (5-4 x)^2}{64 \log (5-4 x)}-\frac {(5-4 x)^3}{64 \log (5-4 x)}\right ) \, dx-4 \int \frac {5-4 x}{\log (5-4 x)} \, dx-\frac {75}{16} \int \frac {5-4 x}{\log (5-4 x)} \, dx+\frac {125}{16} \int \frac {1}{\log (5-4 x)} \, dx-\frac {153}{8} \int \left (\frac {5}{4 \log (5-4 x)}-\frac {5-4 x}{4 \log (5-4 x)}\right ) \, dx+20 \int \frac {1}{\log (5-4 x)} \, dx-\frac {7921}{256} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (5-4 x)\right )\\ &=\frac {7921}{256 \log (5-4 x)}-\frac {765 (5-4 x)}{256 \log (5-4 x)}-\frac {153 (5-4 x) x}{64 \log (5-4 x)}-\frac {5 (5-4 x) x^2}{16 \log (5-4 x)}-\frac {(5-4 x) x^3}{4 \log (5-4 x)}+\frac {1}{64} \operatorname {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,5-4 x\right )+\frac {1}{16} \int \frac {(5-4 x)^3}{\log (5-4 x)} \, dx-\frac {15}{64} \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,5-4 x\right )-\frac {25}{32} \int \frac {5-4 x}{\log (5-4 x)} \, dx-\frac {15}{16} \int \frac {(5-4 x)^2}{\log (5-4 x)} \, dx+\frac {75}{64} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-4 x\right )-\frac {125}{64} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-4 x\right )+\frac {125}{32} \int \frac {1}{\log (5-4 x)} \, dx+\frac {75}{16} \int \frac {5-4 x}{\log (5-4 x)} \, dx+\frac {153}{32} \int \frac {5-4 x}{\log (5-4 x)} \, dx-5 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-4 x\right )-\frac {125}{16} \int \frac {1}{\log (5-4 x)} \, dx-\frac {765}{32} \int \frac {1}{\log (5-4 x)} \, dx+\operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-4 x\right )\\ &=\frac {7921}{256 \log (5-4 x)}-\frac {765 (5-4 x)}{256 \log (5-4 x)}-\frac {153 (5-4 x) x}{64 \log (5-4 x)}-\frac {5 (5-4 x) x^2}{16 \log (5-4 x)}-\frac {(5-4 x) x^3}{4 \log (5-4 x)}-\frac {445 \text {li}(5-4 x)}{64}+\frac {1}{64} \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (5-4 x)\right )-\frac {1}{64} \operatorname {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,5-4 x\right )+\frac {25}{128} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-4 x\right )-\frac {15}{64} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (5-4 x)\right )+\frac {15}{64} \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,5-4 x\right )-\frac {125}{128} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-4 x\right )+\frac {75}{64} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-4 x)\right )-\frac {75}{64} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-4 x\right )-\frac {153}{128} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-4 x\right )+\frac {125}{64} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-4 x\right )+\frac {765}{128} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-4 x\right )+\operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-4 x)\right )\\ &=\frac {139}{64} \text {Ei}(2 \log (5-4 x))-\frac {15}{64} \text {Ei}(3 \log (5-4 x))+\frac {1}{64} \text {Ei}(4 \log (5-4 x))+\frac {7921}{256 \log (5-4 x)}-\frac {765 (5-4 x)}{256 \log (5-4 x)}-\frac {153 (5-4 x) x}{64 \log (5-4 x)}-\frac {5 (5-4 x) x^2}{16 \log (5-4 x)}-\frac {(5-4 x) x^3}{4 \log (5-4 x)}-\frac {1}{64} \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (5-4 x)\right )+\frac {25}{128} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-4 x)\right )+\frac {15}{64} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (5-4 x)\right )-\frac {75}{64} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-4 x)\right )-\frac {153}{128} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-4 x)\right )\\ &=\frac {7921}{256 \log (5-4 x)}-\frac {765 (5-4 x)}{256 \log (5-4 x)}-\frac {153 (5-4 x) x}{64 \log (5-4 x)}-\frac {5 (5-4 x) x^2}{16 \log (5-4 x)}-\frac {(5-4 x) x^3}{4 \log (5-4 x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 16, normalized size = 0.89 \begin {gather*} \frac {\left (4+x^2\right )^2}{\log (5-4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64 - 32*x^2 - 4*x^4 + (-80*x + 64*x^2 - 20*x^3 + 16*x^4)*Log[5 - 4*x])/((-5 + 4*x)*Log[5 - 4*x]^2)
,x]

[Out]

(4 + x^2)^2/Log[5 - 4*x]

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 19, normalized size = 1.06 \begin {gather*} \frac {x^{4} + 8 \, x^{2} + 16}{\log \left (-4 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^4-20*x^3+64*x^2-80*x)*log(-4*x+5)-4*x^4-32*x^2-64)/(4*x-5)/log(-4*x+5)^2,x, algorithm="fricas
")

[Out]

(x^4 + 8*x^2 + 16)/log(-4*x + 5)

________________________________________________________________________________________

giac [A]  time = 0.17, size = 19, normalized size = 1.06 \begin {gather*} \frac {x^{4} + 8 \, x^{2} + 16}{\log \left (-4 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^4-20*x^3+64*x^2-80*x)*log(-4*x+5)-4*x^4-32*x^2-64)/(4*x-5)/log(-4*x+5)^2,x, algorithm="giac")

[Out]

(x^4 + 8*x^2 + 16)/log(-4*x + 5)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 20, normalized size = 1.11

method	result	size
norman	\(\frac {x^{4}+8 x^{2}+16}{\ln \left (-4 x +5\right )}\)	\(20\)
risch	\(\frac {x^{4}+8 x^{2}+16}{\ln \left (-4 x +5\right )}\)	\(20\)
derivativedivides	\(\frac {\left (-4 x +5\right )^{4}}{256 \ln \left (-4 x +5\right )}-\frac {5 \left (-4 x +5\right )^{3}}{64 \ln \left (-4 x +5\right )}+\frac {139 \left (-4 x +5\right )^{2}}{128 \ln \left (-4 x +5\right )}-\frac {445 \left (-4 x +5\right )}{64 \ln \left (-4 x +5\right )}+\frac {7921}{256 \ln \left (-4 x +5\right )}\)	\(78\)
default	\(\frac {\left (-4 x +5\right )^{4}}{256 \ln \left (-4 x +5\right )}-\frac {5 \left (-4 x +5\right )^{3}}{64 \ln \left (-4 x +5\right )}+\frac {139 \left (-4 x +5\right )^{2}}{128 \ln \left (-4 x +5\right )}-\frac {445 \left (-4 x +5\right )}{64 \ln \left (-4 x +5\right )}+\frac {7921}{256 \ln \left (-4 x +5\right )}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x^4-20*x^3+64*x^2-80*x)*ln(-4*x+5)-4*x^4-32*x^2-64)/(4*x-5)/ln(-4*x+5)^2,x,method=_RETURNVERBOSE)

[Out]

(x^4+8*x^2+16)/ln(-4*x+5)

________________________________________________________________________________________

maxima [A]  time = 0.76, size = 29, normalized size = 1.61 \begin {gather*} \frac {x^{4} + 8 \, x^{2}}{\log \left (-4 \, x + 5\right )} + \frac {16}{\log \left (-4 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^4-20*x^3+64*x^2-80*x)*log(-4*x+5)-4*x^4-32*x^2-64)/(4*x-5)/log(-4*x+5)^2,x, algorithm="maxima
")

[Out]

(x^4 + 8*x^2)/log(-4*x + 5) + 16/log(-4*x + 5)

________________________________________________________________________________________

mupad [B]  time = 0.14, size = 16, normalized size = 0.89 \begin {gather*} \frac {{\left (x^2+4\right )}^2}{\ln \left (5-4\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x^2 + 4*x^4 + log(5 - 4*x)*(80*x - 64*x^2 + 20*x^3 - 16*x^4) + 64)/(log(5 - 4*x)^2*(4*x - 5)),x)

[Out]

(x^2 + 4)^2/log(5 - 4*x)

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 15, normalized size = 0.83 \begin {gather*} \frac {x^{4} + 8 x^{2} + 16}{\log {\left (5 - 4 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x**4-20*x**3+64*x**2-80*x)*ln(-4*x+5)-4*x**4-32*x**2-64)/(4*x-5)/ln(-4*x+5)**2,x)

[Out]

(x**4 + 8*x**2 + 16)/log(5 - 4*x)

________________________________________________________________________________________