3.15.20 \(\int \frac {5 x^2+(10 x-10 x^2) \log (1-x)+(-5+5 x) \log ^2(1-x)}{(-1+x) \log ^2(1-x)} \, dx\)

Optimal. Leaf size=17 \[ 5 \left (x-\frac {x^2}{\log (1-x)}\right ) \]

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Rubi [B]  time = 0.37, antiderivative size = 46, normalized size of antiderivative = 2.71, number of steps used = 19, number of rules used = 13, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.260, Rules used = {6688, 2411, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178, 2399, 2389, 2390} \begin {gather*} 5 x-\frac {5 (1-x)^2}{\log (1-x)}+\frac {10 (1-x)}{\log (1-x)}-\frac {5}{\log (1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*x^2 + (10*x - 10*x^2)*Log[1 - x] + (-5 + 5*x)*Log[1 - x]^2)/((-1 + x)*Log[1 - x]^2),x]

[Out]

5*x - 5/Log[1 - x] + (10*(1 - x))/Log[1 - x] - (5*(1 - x)^2)/Log[1 - 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 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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (5+\frac {5 x^2}{(-1+x) \log ^2(1-x)}-\frac {10 x}{\log (1-x)}\right ) \, dx\\ &=5 x+5 \int \frac {x^2}{(-1+x) \log ^2(1-x)} \, dx-10 \int \frac {x}{\log (1-x)} \, dx\\ &=5 x+5 \operatorname {Subst}\left (\int \frac {(1-x)^2}{x \log ^2(x)} \, dx,x,1-x\right )-10 \int \left (\frac {1}{\log (1-x)}-\frac {1-x}{\log (1-x)}\right ) \, dx\\ &=5 x+5 \operatorname {Subst}\left (\int \left (-\frac {2}{\log ^2(x)}+\frac {1}{x \log ^2(x)}+\frac {x}{\log ^2(x)}\right ) \, dx,x,1-x\right )-10 \int \frac {1}{\log (1-x)} \, dx+10 \int \frac {1-x}{\log (1-x)} \, dx\\ &=5 x+5 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,1-x\right )+5 \operatorname {Subst}\left (\int \frac {x}{\log ^2(x)} \, dx,x,1-x\right )-10 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,1-x\right )+10 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1-x\right )-10 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1-x\right )\\ &=5 x+\frac {10 (1-x)}{\log (1-x)}-\frac {5 (1-x)^2}{\log (1-x)}+10 \text {li}(1-x)+5 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (1-x)\right )-10 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1-x)\right )-10 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,1-x\right )+10 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,1-x\right )\\ &=5 x-10 \text {Ei}(2 \log (1-x))-\frac {5}{\log (1-x)}+\frac {10 (1-x)}{\log (1-x)}-\frac {5 (1-x)^2}{\log (1-x)}+10 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (1-x)\right )\\ &=5 x-\frac {5}{\log (1-x)}+\frac {10 (1-x)}{\log (1-x)}-\frac {5 (1-x)^2}{\log (1-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 16, normalized size = 0.94 \begin {gather*} 5 x \left (1-\frac {x}{\log (1-x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x^2 + (10*x - 10*x^2)*Log[1 - x] + (-5 + 5*x)*Log[1 - x]^2)/((-1 + x)*Log[1 - x]^2),x]

[Out]

5*x*(1 - x/Log[1 - x])

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fricas [A]  time = 0.53, size = 23, normalized size = 1.35 \begin {gather*} -\frac {5 \, {\left (x^{2} - x \log \left (-x + 1\right )\right )}}{\log \left (-x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*log(-x+1)^2+(-10*x^2+10*x)*log(-x+1)+5*x^2)/(x-1)/log(-x+1)^2,x, algorithm="fricas")

[Out]

-5*(x^2 - x*log(-x + 1))/log(-x + 1)

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giac [A]  time = 0.30, size = 17, normalized size = 1.00 \begin {gather*} 5 \, x - \frac {5 \, x^{2}}{\log \left (-x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*log(-x+1)^2+(-10*x^2+10*x)*log(-x+1)+5*x^2)/(x-1)/log(-x+1)^2,x, algorithm="giac")

[Out]

5*x - 5*x^2/log(-x + 1)

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maple [A]  time = 0.25, size = 18, normalized size = 1.06

method	result	size
risch	\(5 x -\frac {5 x^{2}}{\ln \left (1-x \right )}\)	\(18\)
norman	\(\frac {-5 x^{2}+5 x \ln \left (1-x \right )}{\ln \left (1-x \right )}\)	\(25\)
derivativedivides	\(-5+5 x -\frac {5 \left (1-x \right )^{2}}{\ln \left (1-x \right )}+\frac {-10 x +10}{\ln \left (1-x \right )}-\frac {5}{\ln \left (1-x \right )}\)	\(48\)
default	\(-5+5 x -\frac {5 \left (1-x \right )^{2}}{\ln \left (1-x \right )}+\frac {-10 x +10}{\ln \left (1-x \right )}-\frac {5}{\ln \left (1-x \right )}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x-5)*ln(1-x)^2+(-10*x^2+10*x)*ln(1-x)+5*x^2)/(x-1)/ln(1-x)^2,x,method=_RETURNVERBOSE)

[Out]

5*x-5*x^2/ln(1-x)

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maxima [B]  time = 0.68, size = 38, normalized size = 2.24 \begin {gather*} -\frac {5 \, {\left (x^{2} - x \log \left (-x + 1\right )\right )}}{\log \left (-x + 1\right )} + 5 \, \log \left (x - 1\right ) - 5 \, \log \left (-x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*log(-x+1)^2+(-10*x^2+10*x)*log(-x+1)+5*x^2)/(x-1)/log(-x+1)^2,x, algorithm="maxima")

[Out]

-5*(x^2 - x*log(-x + 1))/log(-x + 1) + 5*log(x - 1) - 5*log(-x + 1)

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mupad [B]  time = 0.13, size = 17, normalized size = 1.00 \begin {gather*} 5\,x-\frac {5\,x^2}{\ln \left (1-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(1 - x)*(10*x - 10*x^2) + log(1 - x)^2*(5*x - 5) + 5*x^2)/(log(1 - x)^2*(x - 1)),x)

[Out]

5*x - (5*x^2)/log(1 - x)

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sympy [A]  time = 0.10, size = 12, normalized size = 0.71 \begin {gather*} - \frac {5 x^{2}}{\log {\left (1 - x \right )}} + 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-5)*ln(-x+1)**2+(-10*x**2+10*x)*ln(-x+1)+5*x**2)/(x-1)/ln(-x+1)**2,x)

[Out]

-5*x**2/log(1 - x) + 5*x

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