3.75.74 \(\int \frac {-25 x^2-10 x^3-x^4+(-100 x-10 x^2+22 x^3+4 x^4) \log (-2+x)}{(-150+75 x) \log ^2(-2+x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {\left (x+\frac {x^2}{5}\right )^2}{3 \log (-2+x)} \]

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Rubi [B]  time = 1.26, antiderivative size = 78, normalized size of antiderivative = 3.71, number of steps used = 77, 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, 2390, 2302, 30, 2400, 2399, 2309, 2178, 2417} \begin {gather*} -\frac {(2-x) x^3}{75 \log (x-2)}-\frac {4 (2-x) x^2}{25 \log (x-2)}-\frac {49 (2-x) x}{75 \log (x-2)}-\frac {98 (2-x)}{75 \log (x-2)}+\frac {196}{75 \log (x-2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25*x^2 - 10*x^3 - x^4 + (-100*x - 10*x^2 + 22*x^3 + 4*x^4)*Log[-2 + x])/((-150 + 75*x)*Log[-2 + x]^2),x]

[Out]

196/(75*Log[-2 + x]) - (98*(2 - x))/(75*Log[-2 + x]) - (49*(2 - x)*x)/(75*Log[-2 + x]) - (4*(2 - x)*x^2)/(25*L
og[-2 + x]) - ((2 - x)*x^3)/(75*Log[-2 + 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 2417

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Poly
x*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, n, p}, x] && PolynomialQ[Polyx, x]

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 {x (5+x) \left (x (5+x)-2 \left (-10+x+2 x^2\right ) \log (-2+x)\right )}{75 (2-x) \log ^2(-2+x)} \, dx\\ &=\frac {1}{75} \int \frac {x (5+x) \left (x (5+x)-2 \left (-10+x+2 x^2\right ) \log (-2+x)\right )}{(2-x) \log ^2(-2+x)} \, dx\\ &=\frac {1}{75} \int \left (-\frac {x^2 (5+x)^2}{(-2+x) \log ^2(-2+x)}+\frac {2 x (5+x) (5+2 x)}{\log (-2+x)}\right ) \, dx\\ &=-\left (\frac {1}{75} \int \frac {x^2 (5+x)^2}{(-2+x) \log ^2(-2+x)} \, dx\right )+\frac {2}{75} \int \frac {x (5+x) (5+2 x)}{\log (-2+x)} \, dx\\ &=-\left (\frac {1}{75} \int \left (\frac {98}{\log ^2(-2+x)}+\frac {196}{(-2+x) \log ^2(-2+x)}+\frac {49 x}{\log ^2(-2+x)}+\frac {12 x^2}{\log ^2(-2+x)}+\frac {x^3}{\log ^2(-2+x)}\right ) \, dx\right )+\frac {2}{75} \int \left (\frac {126}{\log (-2+x)}+\frac {109 (-2+x)}{\log (-2+x)}+\frac {27 (-2+x)^2}{\log (-2+x)}+\frac {2 (-2+x)^3}{\log (-2+x)}\right ) \, dx\\ &=-\left (\frac {1}{75} \int \frac {x^3}{\log ^2(-2+x)} \, dx\right )+\frac {4}{75} \int \frac {(-2+x)^3}{\log (-2+x)} \, dx-\frac {4}{25} \int \frac {x^2}{\log ^2(-2+x)} \, dx-\frac {49}{75} \int \frac {x}{\log ^2(-2+x)} \, dx+\frac {18}{25} \int \frac {(-2+x)^2}{\log (-2+x)} \, dx-\frac {98}{75} \int \frac {1}{\log ^2(-2+x)} \, dx-\frac {196}{75} \int \frac {1}{(-2+x) \log ^2(-2+x)} \, dx+\frac {218}{75} \int \frac {-2+x}{\log (-2+x)} \, dx+\frac {84}{25} \int \frac {1}{\log (-2+x)} \, dx\\ &=-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}-\frac {4}{75} \int \frac {x^3}{\log (-2+x)} \, dx+\frac {4}{75} \operatorname {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,-2+x\right )+\frac {2}{25} \int \frac {x^2}{\log (-2+x)} \, dx-\frac {12}{25} \int \frac {x^2}{\log (-2+x)} \, dx+\frac {16}{25} \int \frac {x}{\log (-2+x)} \, dx+\frac {18}{25} \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-2+x\right )+\frac {98}{75} \int \frac {1}{\log (-2+x)} \, dx-\frac {98}{75} \int \frac {x}{\log (-2+x)} \, dx-\frac {98}{75} \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-2+x\right )-\frac {196}{75} \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-2+x\right )+\frac {218}{75} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-2+x\right )+\frac {84}{25} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )\\ &=-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}+\frac {84 \text {li}(-2+x)}{25}-\frac {4}{75} \int \left (\frac {8}{\log (-2+x)}+\frac {12 (-2+x)}{\log (-2+x)}+\frac {6 (-2+x)^2}{\log (-2+x)}+\frac {(-2+x)^3}{\log (-2+x)}\right ) \, dx+\frac {4}{75} \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (-2+x)\right )+\frac {2}{25} \int \left (\frac {4}{\log (-2+x)}+\frac {4 (-2+x)}{\log (-2+x)}+\frac {(-2+x)^2}{\log (-2+x)}\right ) \, dx-\frac {12}{25} \int \left (\frac {4}{\log (-2+x)}+\frac {4 (-2+x)}{\log (-2+x)}+\frac {(-2+x)^2}{\log (-2+x)}\right ) \, dx+\frac {16}{25} \int \left (\frac {2}{\log (-2+x)}+\frac {-2+x}{\log (-2+x)}\right ) \, dx+\frac {18}{25} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {98}{75} \int \left (\frac {2}{\log (-2+x)}+\frac {-2+x}{\log (-2+x)}\right ) \, dx-\frac {196}{75} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-2+x)\right )+\frac {218}{75} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-2+x)\right )\\ &=\frac {218}{75} \text {Ei}(2 \log (-2+x))+\frac {18}{25} \text {Ei}(3 \log (-2+x))+\frac {4}{75} \text {Ei}(4 \log (-2+x))+\frac {196}{75 \log (-2+x)}-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}+\frac {84 \text {li}(-2+x)}{25}-\frac {4}{75} \int \frac {(-2+x)^3}{\log (-2+x)} \, dx+\frac {2}{25} \int \frac {(-2+x)^2}{\log (-2+x)} \, dx+\frac {8}{25} \int \frac {1}{\log (-2+x)} \, dx+\frac {8}{25} \int \frac {-2+x}{\log (-2+x)} \, dx-\frac {8}{25} \int \frac {(-2+x)^2}{\log (-2+x)} \, dx-\frac {32}{75} \int \frac {1}{\log (-2+x)} \, dx-\frac {12}{25} \int \frac {(-2+x)^2}{\log (-2+x)} \, dx+\frac {32}{25} \int \frac {1}{\log (-2+x)} \, dx-\frac {98}{75} \int \frac {-2+x}{\log (-2+x)} \, dx-\frac {48}{25} \int \frac {1}{\log (-2+x)} \, dx-\frac {48}{25} \int \frac {-2+x}{\log (-2+x)} \, dx-\frac {196}{75} \int \frac {1}{\log (-2+x)} \, dx\\ &=\frac {218}{75} \text {Ei}(2 \log (-2+x))+\frac {18}{25} \text {Ei}(3 \log (-2+x))+\frac {4}{75} \text {Ei}(4 \log (-2+x))+\frac {196}{75 \log (-2+x)}-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}+\frac {84 \text {li}(-2+x)}{25}-\frac {4}{75} \operatorname {Subst}\left (\int \frac {x^3}{\log (x)} \, dx,x,-2+x\right )+\frac {2}{25} \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-2+x\right )+\frac {8}{25} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )+\frac {8}{25} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-2+x\right )-\frac {8}{25} \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-2+x\right )-\frac {32}{75} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )-\frac {12}{25} \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-2+x\right )+\frac {32}{25} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )-\frac {98}{75} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-2+x\right )-\frac {48}{25} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )-\frac {48}{25} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-2+x\right )-\frac {196}{75} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )\\ &=\frac {218}{75} \text {Ei}(2 \log (-2+x))+\frac {18}{25} \text {Ei}(3 \log (-2+x))+\frac {4}{75} \text {Ei}(4 \log (-2+x))+\frac {196}{75 \log (-2+x)}-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}-\frac {4}{75} \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (-2+x)\right )+\frac {2}{25} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-2+x)\right )+\frac {8}{25} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {8}{25} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {12}{25} \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {98}{75} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-2+x)\right )-\frac {48}{25} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-2+x)\right )\\ &=\frac {196}{75 \log (-2+x)}-\frac {98 (2-x)}{75 \log (-2+x)}-\frac {49 (2-x) x}{75 \log (-2+x)}-\frac {4 (2-x) x^2}{25 \log (-2+x)}-\frac {(2-x) x^3}{75 \log (-2+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 18, normalized size = 0.86 \begin {gather*} \frac {x^2 (5+x)^2}{75 \log (-2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*x^2 - 10*x^3 - x^4 + (-100*x - 10*x^2 + 22*x^3 + 4*x^4)*Log[-2 + x])/((-150 + 75*x)*Log[-2 + x]
^2),x]

[Out]

(x^2*(5 + x)^2)/(75*Log[-2 + x])

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fricas [A]  time = 0.91, size = 22, normalized size = 1.05 \begin {gather*} \frac {x^{4} + 10 \, x^{3} + 25 \, x^{2}}{75 \, \log \left (x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+22*x^3-10*x^2-100*x)*log(x-2)-x^4-10*x^3-25*x^2)/(75*x-150)/log(x-2)^2,x, algorithm="fricas"
)

[Out]

1/75*(x^4 + 10*x^3 + 25*x^2)/log(x - 2)

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giac [A]  time = 0.15, size = 22, normalized size = 1.05 \begin {gather*} \frac {x^{4} + 10 \, x^{3} + 25 \, x^{2}}{75 \, \log \left (x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+22*x^3-10*x^2-100*x)*log(x-2)-x^4-10*x^3-25*x^2)/(75*x-150)/log(x-2)^2,x, algorithm="giac")

[Out]

1/75*(x^4 + 10*x^3 + 25*x^2)/log(x - 2)

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maple [A]  time = 0.17, size = 20, normalized size = 0.95




method result size



risch \(\frac {x^{2} \left (x^{2}+10 x +25\right )}{75 \ln \left (x -2\right )}\) \(20\)
norman \(\frac {\frac {1}{3} x^{2}+\frac {2}{15} x^{3}+\frac {1}{75} x^{4}}{\ln \left (x -2\right )}\) \(24\)
derivativedivides \(\frac {\left (x -2\right )^{4}}{75 \ln \left (x -2\right )}+\frac {6 \left (x -2\right )^{3}}{25 \ln \left (x -2\right )}+\frac {109 \left (x -2\right )^{2}}{75 \ln \left (x -2\right )}+\frac {\frac {84 x}{25}-\frac {168}{25}}{\ln \left (x -2\right )}+\frac {196}{75 \ln \left (x -2\right )}\) \(60\)
default \(\frac {\left (x -2\right )^{4}}{75 \ln \left (x -2\right )}+\frac {6 \left (x -2\right )^{3}}{25 \ln \left (x -2\right )}+\frac {109 \left (x -2\right )^{2}}{75 \ln \left (x -2\right )}+\frac {\frac {84 x}{25}-\frac {168}{25}}{\ln \left (x -2\right )}+\frac {196}{75 \ln \left (x -2\right )}\) \(60\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^4+22*x^3-10*x^2-100*x)*ln(x-2)-x^4-10*x^3-25*x^2)/(75*x-150)/ln(x-2)^2,x,method=_RETURNVERBOSE)

[Out]

1/75*x^2*(x^2+10*x+25)/ln(x-2)

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maxima [A]  time = 0.41, size = 22, normalized size = 1.05 \begin {gather*} \frac {x^{4} + 10 \, x^{3} + 25 \, x^{2}}{75 \, \log \left (x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+22*x^3-10*x^2-100*x)*log(x-2)-x^4-10*x^3-25*x^2)/(75*x-150)/log(x-2)^2,x, algorithm="maxima"
)

[Out]

1/75*(x^4 + 10*x^3 + 25*x^2)/log(x - 2)

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mupad [B]  time = 4.53, size = 16, normalized size = 0.76 \begin {gather*} \frac {x^2\,{\left (x+5\right )}^2}{75\,\ln \left (x-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x - 2)*(100*x + 10*x^2 - 22*x^3 - 4*x^4) + 25*x^2 + 10*x^3 + x^4)/(log(x - 2)^2*(75*x - 150)),x)

[Out]

(x^2*(x + 5)^2)/(75*log(x - 2))

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sympy [A]  time = 0.12, size = 19, normalized size = 0.90 \begin {gather*} \frac {x^{4} + 10 x^{3} + 25 x^{2}}{75 \log {\left (x - 2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**4+22*x**3-10*x**2-100*x)*ln(x-2)-x**4-10*x**3-25*x**2)/(75*x-150)/ln(x-2)**2,x)

[Out]

(x**4 + 10*x**3 + 25*x**2)/(75*log(x - 2))

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