3.1.55 \(\int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log (\frac {1}{5} (-1+5 x))+(-1+5 x) \log ^2(\frac {1}{5} (-1+5 x))}{(-1+5 x) \log ^2(\frac {1}{5} (-1+5 x))} \, dx\)

Optimal. Leaf size=27 \[ x+\frac {4+2 \left (-2-\left (-3+e^3\right )^2+x\right )}{\log \left (-\frac {1}{5}+x\right )} \]

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Rubi [A]  time = 0.28, antiderivative size = 43, normalized size of antiderivative = 1.59, number of steps used = 12, number of rules used = 9, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6688, 2411, 12, 2353, 2297, 2298, 2302, 30, 2389} \begin {gather*} x-\frac {2 (1-5 x)}{5 \log \left (x-\frac {1}{5}\right )}-\frac {2 \left (44-30 e^3+5 e^6\right )}{5 \log \left (x-\frac {1}{5}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(90 - 60*E^3 + 10*E^6 - 10*x + (-2 + 10*x)*Log[(-1 + 5*x)/5] + (-1 + 5*x)*Log[(-1 + 5*x)/5]^2)/((-1 + 5*x)
*Log[(-1 + 5*x)/5]^2),x]

[Out]

x - (2*(44 - 30*E^3 + 5*E^6))/(5*Log[-1/5 + x]) - (2*(1 - 5*x))/(5*Log[-1/5 + 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 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 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 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 (1+\frac {10 \left (9-6 e^3+e^6-x\right )}{(-1+5 x) \log ^2\left (-\frac {1}{5}+x\right )}+\frac {2}{\log \left (-\frac {1}{5}+x\right )}\right ) \, dx\\ &=x+2 \int \frac {1}{\log \left (-\frac {1}{5}+x\right )} \, dx+10 \int \frac {9-6 e^3+e^6-x}{(-1+5 x) \log ^2\left (-\frac {1}{5}+x\right )} \, dx\\ &=x+2 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-\frac {1}{5}+x\right )+10 \operatorname {Subst}\left (\int \frac {\frac {44}{5}-6 e^3+e^6-x}{5 x \log ^2(x)} \, dx,x,-\frac {1}{5}+x\right )\\ &=x+2 \text {li}\left (-\frac {1}{5}+x\right )+2 \operatorname {Subst}\left (\int \frac {\frac {44}{5}-6 e^3+e^6-x}{x \log ^2(x)} \, dx,x,-\frac {1}{5}+x\right )\\ &=x+2 \text {li}\left (-\frac {1}{5}+x\right )+2 \operatorname {Subst}\left (\int \left (-\frac {1}{\log ^2(x)}+\frac {44-30 e^3+5 e^6}{5 x \log ^2(x)}\right ) \, dx,x,-\frac {1}{5}+x\right )\\ &=x+2 \text {li}\left (-\frac {1}{5}+x\right )-2 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-\frac {1}{5}+x\right )+\frac {1}{5} \left (2 \left (44-30 e^3+5 e^6\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-\frac {1}{5}+x\right )\\ &=x-\frac {2 (1-5 x)}{5 \log \left (-\frac {1}{5}+x\right )}+2 \text {li}\left (-\frac {1}{5}+x\right )-2 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-\frac {1}{5}+x\right )+\frac {1}{5} \left (2 \left (44-30 e^3+5 e^6\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (-\frac {1}{5}+x\right )\right )\\ &=x-\frac {2 \left (44-30 e^3+5 e^6\right )}{5 \log \left (-\frac {1}{5}+x\right )}-\frac {2 (1-5 x)}{5 \log \left (-\frac {1}{5}+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.07, size = 42, normalized size = 1.56 \begin {gather*} x-2 \text {Ei}\left (\log \left (-\frac {1}{5}+x\right )\right )+\frac {2 \left (-9+6 e^3-e^6+x\right )}{\log \left (-\frac {1}{5}+x\right )}+2 \text {li}\left (-\frac {1}{5}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(90 - 60*E^3 + 10*E^6 - 10*x + (-2 + 10*x)*Log[(-1 + 5*x)/5] + (-1 + 5*x)*Log[(-1 + 5*x)/5]^2)/((-1
+ 5*x)*Log[(-1 + 5*x)/5]^2),x]

[Out]

x - 2*ExpIntegralEi[Log[-1/5 + x]] + (2*(-9 + 6*E^3 - E^6 + x))/Log[-1/5 + x] + 2*LogIntegral[-1/5 + x]

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fricas [A]  time = 1.02, size = 26, normalized size = 0.96 \begin {gather*} \frac {x \log \left (x - \frac {1}{5}\right ) + 2 \, x - 2 \, e^{6} + 12 \, e^{3} - 18}{\log \left (x - \frac {1}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(x-1/5)^2+(10*x-2)*log(x-1/5)+10*exp(3)^2-60*exp(3)-10*x+90)/(5*x-1)/log(x-1/5)^2,x, alg
orithm="fricas")

[Out]

(x*log(x - 1/5) + 2*x - 2*e^6 + 12*e^3 - 18)/log(x - 1/5)

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giac [A]  time = 0.25, size = 26, normalized size = 0.96 \begin {gather*} \frac {x \log \left (x - \frac {1}{5}\right ) + 2 \, x - 2 \, e^{6} + 12 \, e^{3} - 18}{\log \left (x - \frac {1}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(x-1/5)^2+(10*x-2)*log(x-1/5)+10*exp(3)^2-60*exp(3)-10*x+90)/(5*x-1)/log(x-1/5)^2,x, alg
orithm="giac")

[Out]

(x*log(x - 1/5) + 2*x - 2*e^6 + 12*e^3 - 18)/log(x - 1/5)

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maple [A]  time = 0.06, size = 22, normalized size = 0.81

method	result	size
risch	\(x -\frac {2 \left ({\mathrm e}^{6}-6 \,{\mathrm e}^{3}-x +9\right )}{\ln \left (x -\frac {1}{5}\right )}\)	\(22\)
norman	\(\frac {x \ln \left (x -\frac {1}{5}\right )+2 x -18-2 \,{\mathrm e}^{6}+12 \,{\mathrm e}^{3}}{\ln \left (x -\frac {1}{5}\right )}\)	\(29\)
derivativedivides	\(x -\frac {1}{5}-\frac {2 \,{\mathrm e}^{6}}{\ln \left (x -\frac {1}{5}\right )}+\frac {12 \,{\mathrm e}^{3}}{\ln \left (x -\frac {1}{5}\right )}-\frac {88}{5 \ln \left (x -\frac {1}{5}\right )}+\frac {2 x -\frac {2}{5}}{\ln \left (x -\frac {1}{5}\right )}\)	\(45\)
default	\(x -\frac {1}{5}-\frac {2 \,{\mathrm e}^{6}}{\ln \left (x -\frac {1}{5}\right )}+\frac {12 \,{\mathrm e}^{3}}{\ln \left (x -\frac {1}{5}\right )}-\frac {88}{5 \ln \left (x -\frac {1}{5}\right )}+\frac {2 x -\frac {2}{5}}{\ln \left (x -\frac {1}{5}\right )}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x-1)*ln(x-1/5)^2+(10*x-2)*ln(x-1/5)+10*exp(3)^2-60*exp(3)-10*x+90)/(5*x-1)/ln(x-1/5)^2,x,method=_RETUR
NVERBOSE)

[Out]

x-2*(exp(6)-6*exp(3)-x+9)/ln(x-1/5)

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maxima [B]  time = 0.59, size = 171, normalized size = 6.33 \begin {gather*} -\frac {2}{5} \, {\left (\log \relax (5) - \log \left (5 \, x - 1\right )\right )} \log \left (-\log \relax (5) + \log \left (5 \, x - 1\right )\right ) - \frac {2}{5} \, \log \left (x - \frac {1}{5}\right ) \log \left (-\log \relax (5) + \log \left (5 \, x - 1\right )\right ) - \frac {\log \left (x - \frac {1}{5}\right )^{2}}{5 \, {\left (\log \relax (5) - \log \left (5 \, x - 1\right )\right )}} + \frac {x {\left (\log \relax (5) - 2\right )} - x \log \left (5 \, x - 1\right )}{\log \relax (5) - \log \left (5 \, x - 1\right )} + \frac {2 \, e^{6}}{\log \relax (5) - \log \left (5 \, x - 1\right )} - \frac {12 \, e^{3}}{\log \relax (5) - \log \left (5 \, x - 1\right )} - \frac {2 \, \log \left (x - \frac {1}{5}\right )}{5 \, {\left (\log \relax (5) - \log \left (5 \, x - 1\right )\right )}} + \frac {18}{\log \relax (5) - \log \left (5 \, x - 1\right )} - \frac {1}{5} \, \log \left (5 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(x-1/5)^2+(10*x-2)*log(x-1/5)+10*exp(3)^2-60*exp(3)-10*x+90)/(5*x-1)/log(x-1/5)^2,x, alg
orithm="maxima")

[Out]

-2/5*(log(5) - log(5*x - 1))*log(-log(5) + log(5*x - 1)) - 2/5*log(x - 1/5)*log(-log(5) + log(5*x - 1)) - 1/5*
log(x - 1/5)^2/(log(5) - log(5*x - 1)) + (x*(log(5) - 2) - x*log(5*x - 1))/(log(5) - log(5*x - 1)) + 2*e^6/(lo
g(5) - log(5*x - 1)) - 12*e^3/(log(5) - log(5*x - 1)) - 2/5*log(x - 1/5)/(log(5) - log(5*x - 1)) + 18/(log(5)
- log(5*x - 1)) - 1/5*log(5*x - 1)

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mupad [B]  time = 0.35, size = 22, normalized size = 0.81 \begin {gather*} x+\frac {2\,x+12\,{\mathrm {e}}^3-2\,{\mathrm {e}}^6-18}{\ln \left (x-\frac {1}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(6) - 60*exp(3) - 10*x + log(x - 1/5)^2*(5*x - 1) + log(x - 1/5)*(10*x - 2) + 90)/(log(x - 1/5)^2*(
5*x - 1)),x)

[Out]

x + (2*x + 12*exp(3) - 2*exp(6) - 18)/log(x - 1/5)

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sympy [A]  time = 0.10, size = 22, normalized size = 0.81 \begin {gather*} x + \frac {2 x - 2 e^{6} - 18 + 12 e^{3}}{\log {\left (x - \frac {1}{5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*ln(x-1/5)**2+(10*x-2)*ln(x-1/5)+10*exp(3)**2-60*exp(3)-10*x+90)/(5*x-1)/ln(x-1/5)**2,x)

[Out]

x + (2*x - 2*exp(6) - 18 + 12*exp(3))/log(x - 1/5)

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