\(\int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log (\frac {1}{5} (-1+5 x))+(-1+5 x) \log ^2(\frac {1}{5} (-1+5 x))}{(-1+5 x) \log ^2(\frac {1}{5} (-1+5 x))} \, dx\) [55]

Optimal result

Integrand size = 69, antiderivative size = 27 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x+\frac {4+2 \left (-2-\left (-3+e^3\right )^2+x\right )}{\log \left (-\frac {1}{5}+x\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6820, 2458, 12, 2395, 2334, 2335, 2339, 30, 2436} \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x-\frac {2 (1-5 x)}{5 \log \left (x-\frac {1}{5}\right )}-\frac {2 \left (44-30 e^3+5 e^6\right )}{5 \log \left (x-\frac {1}{5}\right )} \]

[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 2334

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 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2339

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 2395

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 2436

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 2458

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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \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 \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-\frac {1}{5}+x\right )+10 \text {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 \operatorname {LogIntegral}\left (-\frac {1}{5}+x\right )+2 \text {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 \operatorname {LogIntegral}\left (-\frac {1}{5}+x\right )+2 \text {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 \operatorname {LogIntegral}\left (-\frac {1}{5}+x\right )-2 \text {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 ) \text {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 \operatorname {LogIntegral}\left (-\frac {1}{5}+x\right )-2 \text {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 ) \text {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{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x-\frac {2 \left (9-6 e^3+e^6-x\right )}{\log \left (-\frac {1}{5}+x\right )} \]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 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	\(\frac {x \ln \left (x -\frac {1}{5}\right )+12 \,{\mathrm e}^{3}-2 \,{\mathrm e}^{6}+2 x -18}{\ln \left (x -\frac {1}{5}\right )}\)	\(29\)
parallelrisch	\(-\frac {2 \,{\mathrm e}^{6}-x \ln \left (x -\frac {1}{5}\right )+18-12 \,{\mathrm e}^{3}-2 x}{\ln \left (x -\frac {1}{5}\right )}\)	\(31\)
parts	\(x -\frac {2 \,{\mathrm e}^{6}}{\ln \left (x -\frac {1}{5}\right )}+\frac {12 \,{\mathrm e}^{3}}{\ln \left (x -\frac {1}{5}\right )}+\frac {2 x -\frac {2}{5}}{\ln \left (x -\frac {1}{5}\right )}-\frac {88}{5 \ln \left (x -\frac {1}{5}\right )}\)	\(44\)
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	\(x -\frac {1}{5}-\frac {2 \,{\mathrm e}^{6}}{\ln \left (x -\frac {1}{5}\right )}+\frac {12 \,{\mathrm e}^{3}}{\ln \left (x -\frac {1}{5}\right )}-\frac {88}{5 \ln \left (x -\frac {1}{5}\right )}+\frac {2 x -\frac {2}{5}}{\ln \left (x -\frac {1}{5}\right )}\)	\(45\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=\frac {x \log \left (x - \frac {1}{5}\right ) + 2 \, x - 2 \, e^{6} + 12 \, e^{3} - 18}{\log \left (x - \frac {1}{5}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x + \frac {2 x - 2 e^{6} - 18 + 12 e^{3}}{\log {\left (x - \frac {1}{5} \right )}} \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (20) = 40\).

Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 6.33 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=-\frac {2}{5} \, {\left (\log \left (5\right ) - \log \left (5 \, x - 1\right )\right )} \log \left (-\log \left (5\right ) + \log \left (5 \, x - 1\right )\right ) - \frac {2}{5} \, \log \left (x - \frac {1}{5}\right ) \log \left (-\log \left (5\right ) + \log \left (5 \, x - 1\right )\right ) - \frac {\log \left (x - \frac {1}{5}\right )^{2}}{5 \, {\left (\log \left (5\right ) - \log \left (5 \, x - 1\right )\right )}} + \frac {x {\left (\log \left (5\right ) - 2\right )} - x \log \left (5 \, x - 1\right )}{\log \left (5\right ) - \log \left (5 \, x - 1\right )} + \frac {2 \, e^{6}}{\log \left (5\right ) - \log \left (5 \, x - 1\right )} - \frac {12 \, e^{3}}{\log \left (5\right ) - \log \left (5 \, x - 1\right )} - \frac {2 \, \log \left (x - \frac {1}{5}\right )}{5 \, {\left (\log \left (5\right ) - \log \left (5 \, x - 1\right )\right )}} + \frac {18}{\log \left (5\right ) - \log \left (5 \, x - 1\right )} - \frac {1}{5} \, \log \left (5 \, x - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=\frac {x \log \left (x - \frac {1}{5}\right ) + 2 \, x - 2 \, e^{6} + 12 \, e^{3} - 18}{\log \left (x - \frac {1}{5}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 8.46 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {90-60 e^3+10 e^6-10 x+(-2+10 x) \log \left (\frac {1}{5} (-1+5 x)\right )+(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )}{(-1+5 x) \log ^2\left (\frac {1}{5} (-1+5 x)\right )} \, dx=x+\frac {2\,x+12\,{\mathrm {e}}^3-2\,{\mathrm {e}}^6-18}{\ln \left (x-\frac {1}{5}\right )} \]

[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)