3.30.62 \(\int \frac {1223+16 x+(1223+8 x) \log (x)+405 \log ^2(x)+45 \log ^3(x)}{1215+1215 \log (x)+405 \log ^2(x)+45 \log ^3(x)} \, dx\)

Optimal. Leaf size=18 \[ x+\frac {4 x (2+x)}{45 (-3-\log (x))^2} \]

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Rubi [B]  time = 0.33, antiderivative size = 55, normalized size of antiderivative = 3.06, number of steps used = 26, number of rules used = 9, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.196, Rules used = {6688, 12, 6742, 2320, 2330, 2299, 2178, 2309, 2297} \begin {gather*} x-\frac {8 (x+1) x}{45 (\log (x)+3)}+\frac {8 (x+2) x}{45 (\log (x)+3)}-\frac {8 x}{45 (\log (x)+3)}+\frac {4 (x+2) x}{45 (\log (x)+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1223 + 16*x + (1223 + 8*x)*Log[x] + 405*Log[x]^2 + 45*Log[x]^3)/(1215 + 1215*Log[x] + 405*Log[x]^2 + 45*L
og[x]^3),x]

[Out]

x + (4*x*(2 + x))/(45*(3 + Log[x])^2) - (8*x)/(45*(3 + Log[x])) - (8*x*(1 + x))/(45*(3 + Log[x])) + (8*x*(2 +
x))/(45*(3 + Log[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 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 2299

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

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 2320

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(x*(d + e*x)^q*(a
+ b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), Int[(d + e*x)^q*(a + b*Log[c*x^n])^
(p + 1), x], x] + Dist[(d*q)/(b*n*(p + 1)), Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^(p + 1), x], x]) /; FreeQ
[{a, b, c, d, e, n}, x] && LtQ[p, -1] && GtQ[q, 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 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 {1223+16 x+(1223+8 x) \log (x)+405 \log ^2(x)+45 \log ^3(x)}{45 (3+\log (x))^3} \, dx\\ &=\frac {1}{45} \int \frac {1223+16 x+(1223+8 x) \log (x)+405 \log ^2(x)+45 \log ^3(x)}{(3+\log (x))^3} \, dx\\ &=\frac {1}{45} \int \left (45-\frac {8 (2+x)}{(3+\log (x))^3}+\frac {8 (1+x)}{(3+\log (x))^2}\right ) \, dx\\ &=x-\frac {8}{45} \int \frac {2+x}{(3+\log (x))^3} \, dx+\frac {8}{45} \int \frac {1+x}{(3+\log (x))^2} \, dx\\ &=x+\frac {4 x (2+x)}{45 (3+\log (x))^2}-\frac {8 x (1+x)}{45 (3+\log (x))}+\frac {8}{45} \int \frac {1}{(3+\log (x))^2} \, dx-\frac {8}{45} \int \frac {2+x}{(3+\log (x))^2} \, dx-\frac {8}{45} \int \frac {1}{3+\log (x)} \, dx+\frac {16}{45} \int \frac {1+x}{3+\log (x)} \, dx\\ &=x+\frac {4 x (2+x)}{45 (3+\log (x))^2}-\frac {8 x}{45 (3+\log (x))}-\frac {8 x (1+x)}{45 (3+\log (x))}+\frac {8 x (2+x)}{45 (3+\log (x))}+\frac {8}{45} \int \frac {1}{3+\log (x)} \, dx-\frac {8}{45} \operatorname {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (x)\right )+\frac {16}{45} \int \frac {1}{3+\log (x)} \, dx-\frac {16}{45} \int \frac {2+x}{3+\log (x)} \, dx+\frac {16}{45} \int \left (\frac {1}{3+\log (x)}+\frac {x}{3+\log (x)}\right ) \, dx\\ &=x-\frac {8 \text {Ei}(3+\log (x))}{45 e^3}+\frac {4 x (2+x)}{45 (3+\log (x))^2}-\frac {8 x}{45 (3+\log (x))}-\frac {8 x (1+x)}{45 (3+\log (x))}+\frac {8 x (2+x)}{45 (3+\log (x))}+\frac {8}{45} \operatorname {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (x)\right )+\frac {16}{45} \int \frac {1}{3+\log (x)} \, dx+\frac {16}{45} \int \frac {x}{3+\log (x)} \, dx-\frac {16}{45} \int \left (\frac {2}{3+\log (x)}+\frac {x}{3+\log (x)}\right ) \, dx+\frac {16}{45} \operatorname {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (x)\right )\\ &=x+\frac {16 \text {Ei}(3+\log (x))}{45 e^3}+\frac {4 x (2+x)}{45 (3+\log (x))^2}-\frac {8 x}{45 (3+\log (x))}-\frac {8 x (1+x)}{45 (3+\log (x))}+\frac {8 x (2+x)}{45 (3+\log (x))}-\frac {16}{45} \int \frac {x}{3+\log (x)} \, dx+\frac {16}{45} \operatorname {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (x)\right )+\frac {16}{45} \operatorname {Subst}\left (\int \frac {e^{2 x}}{3+x} \, dx,x,\log (x)\right )-\frac {32}{45} \int \frac {1}{3+\log (x)} \, dx\\ &=x+\frac {32 \text {Ei}(3+\log (x))}{45 e^3}+\frac {16 \text {Ei}(2 (3+\log (x)))}{45 e^6}+\frac {4 x (2+x)}{45 (3+\log (x))^2}-\frac {8 x}{45 (3+\log (x))}-\frac {8 x (1+x)}{45 (3+\log (x))}+\frac {8 x (2+x)}{45 (3+\log (x))}-\frac {16}{45} \operatorname {Subst}\left (\int \frac {e^{2 x}}{3+x} \, dx,x,\log (x)\right )-\frac {32}{45} \operatorname {Subst}\left (\int \frac {e^x}{3+x} \, dx,x,\log (x)\right )\\ &=x+\frac {4 x (2+x)}{45 (3+\log (x))^2}-\frac {8 x}{45 (3+\log (x))}-\frac {8 x (1+x)}{45 (3+\log (x))}+\frac {8 x (2+x)}{45 (3+\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 20, normalized size = 1.11 \begin {gather*} \frac {1}{45} \left (45 x+\frac {4 x (2+x)}{(3+\log (x))^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1223 + 16*x + (1223 + 8*x)*Log[x] + 405*Log[x]^2 + 45*Log[x]^3)/(1215 + 1215*Log[x] + 405*Log[x]^2
+ 45*Log[x]^3),x]

[Out]

(45*x + (4*x*(2 + x))/(3 + Log[x])^2)/45

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fricas [B]  time = 0.95, size = 35, normalized size = 1.94 \begin {gather*} \frac {45 \, x \log \relax (x)^{2} + 4 \, x^{2} + 270 \, x \log \relax (x) + 413 \, x}{45 \, {\left (\log \relax (x)^{2} + 6 \, \log \relax (x) + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(x)^3+405*log(x)^2+(8*x+1223)*log(x)+16*x+1223)/(45*log(x)^3+405*log(x)^2+1215*log(x)+1215),x
, algorithm="fricas")

[Out]

1/45*(45*x*log(x)^2 + 4*x^2 + 270*x*log(x) + 413*x)/(log(x)^2 + 6*log(x) + 9)

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giac [B]  time = 0.25, size = 68, normalized size = 3.78 \begin {gather*} \frac {x \log \relax (x)^{2}}{\log \relax (x)^{2} + 6 \, \log \relax (x) + 9} + \frac {4 \, x^{2}}{45 \, {\left (\log \relax (x)^{2} + 6 \, \log \relax (x) + 9\right )}} + \frac {6 \, x \log \relax (x)}{\log \relax (x)^{2} + 6 \, \log \relax (x) + 9} + \frac {413 \, x}{45 \, {\left (\log \relax (x)^{2} + 6 \, \log \relax (x) + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(x)^3+405*log(x)^2+(8*x+1223)*log(x)+16*x+1223)/(45*log(x)^3+405*log(x)^2+1215*log(x)+1215),x
, algorithm="giac")

[Out]

x*log(x)^2/(log(x)^2 + 6*log(x) + 9) + 4/45*x^2/(log(x)^2 + 6*log(x) + 9) + 6*x*log(x)/(log(x)^2 + 6*log(x) +
9) + 413/45*x/(log(x)^2 + 6*log(x) + 9)

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maple [A]  time = 0.03, size = 15, normalized size = 0.83

method	result	size
risch	\(x +\frac {4 x \left (2+x \right )}{45 \left (3+\ln \relax (x )\right )^{2}}\)	\(15\)
norman	\(\frac {x \ln \relax (x )^{2}+\frac {413 x}{45}+\frac {4 x^{2}}{45}+6 x \ln \relax (x )}{\left (3+\ln \relax (x )\right )^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*ln(x)^3+405*ln(x)^2+(8*x+1223)*ln(x)+16*x+1223)/(45*ln(x)^3+405*ln(x)^2+1215*ln(x)+1215),x,method=_RET
URNVERBOSE)

[Out]

x+4/45*x*(2+x)/(3+ln(x))^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {90 \, x \log \relax (x)^{2} + 120 \, x^{2} + {\left (32 \, x^{2} + 1763 \, x\right )} \log \relax (x) + 5718 \, x}{90 \, {\left (\log \relax (x)^{2} + 6 \, \log \relax (x) + 9\right )}} - \frac {1223 \, e^{\left (-3\right )} E_{3}\left (-\log \relax (x) - 3\right )}{45 \, {\left (\log \relax (x) + 3\right )}^{2}} - \frac {16 \, e^{\left (-6\right )} E_{3}\left (-2 \, \log \relax (x) - 6\right )}{45 \, {\left (\log \relax (x) + 3\right )}^{2}} - \frac {1}{45} \, \int \frac {64 \, x + 1223}{2 \, {\left (\log \relax (x) + 3\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(x)^3+405*log(x)^2+(8*x+1223)*log(x)+16*x+1223)/(45*log(x)^3+405*log(x)^2+1215*log(x)+1215),x
, algorithm="maxima")

[Out]

1/90*(90*x*log(x)^2 + 120*x^2 + (32*x^2 + 1763*x)*log(x) + 5718*x)/(log(x)^2 + 6*log(x) + 9) - 1223/45*e^(-3)*
exp_integral_e(3, -log(x) - 3)/(log(x) + 3)^2 - 16/45*e^(-6)*exp_integral_e(3, -2*log(x) - 6)/(log(x) + 3)^2 -
1/45*integrate(1/2*(64*x + 1223)/(log(x) + 3), x)

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mupad [B]  time = 1.89, size = 14, normalized size = 0.78 \begin {gather*} x+\frac {4\,x\,\left (x+2\right )}{45\,{\left (\ln \relax (x)+3\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x + 405*log(x)^2 + 45*log(x)^3 + log(x)*(8*x + 1223) + 1223)/(1215*log(x) + 405*log(x)^2 + 45*log(x)^3
+ 1215),x)

[Out]

x + (4*x*(x + 2))/(45*(log(x) + 3)^2)

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sympy [A]  time = 0.12, size = 22, normalized size = 1.22 \begin {gather*} x + \frac {4 x^{2} + 8 x}{45 \log {\relax (x )}^{2} + 270 \log {\relax (x )} + 405} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*ln(x)**3+405*ln(x)**2+(8*x+1223)*ln(x)+16*x+1223)/(45*ln(x)**3+405*ln(x)**2+1215*ln(x)+1215),x)

[Out]

x + (4*x**2 + 8*x)/(45*log(x)**2 + 270*log(x) + 405)

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