\(\int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx\) [1766]

Optimal result

Integrand size = 44, antiderivative size = 25 \[ \int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx=\frac {-e^5-\frac {1}{7} e^{e^3} (-1+x)}{\log (-3+x)} \]

[Out]

(-1/7*(-1+x)*exp(exp(3))-exp(5))/ln(-3+x)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6874, 2458, 2395, 2334, 2335, 2339, 30, 2436} \[ \int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx=\frac {e^{e^3} (3-x)}{7 \log (x-3)}-\frac {7 e^5+2 e^{e^3}}{7 \log (x-3)} \]

[In]

Int[(7*E^5 + E^E^3*(-1 + x) + E^E^3*(3 - x)*Log[-3 + x])/((-21 + 7*x)*Log[-3 + x]^2),x]

[Out]

-1/7*(7*E^5 + 2*E^E^3)/Log[-3 + x] + (E^E^3*(3 - x))/(7*Log[-3 + 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 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {7 e^5-e^{e^3}+e^{e^3} x}{7 (-3+x) \log ^2(-3+x)}-\frac {e^{e^3}}{7 \log (-3+x)}\right ) \, dx \\ & = \frac {1}{7} \int \frac {7 e^5-e^{e^3}+e^{e^3} x}{(-3+x) \log ^2(-3+x)} \, dx-\frac {1}{7} e^{e^3} \int \frac {1}{\log (-3+x)} \, dx \\ & = \frac {1}{7} \text {Subst}\left (\int \frac {7 e^5+2 e^{e^3}+e^{e^3} x}{x \log ^2(x)} \, dx,x,-3+x\right )-\frac {1}{7} e^{e^3} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+x\right ) \\ & = -\frac {1}{7} e^{e^3} \operatorname {LogIntegral}(-3+x)+\frac {1}{7} \text {Subst}\left (\int \left (\frac {e^{e^3}}{\log ^2(x)}+\frac {7 e^5+2 e^{e^3}}{x \log ^2(x)}\right ) \, dx,x,-3+x\right ) \\ & = -\frac {1}{7} e^{e^3} \operatorname {LogIntegral}(-3+x)+\frac {1}{7} e^{e^3} \text {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-3+x\right )+\frac {1}{7} \left (7 e^5+2 e^{e^3}\right ) \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-3+x\right ) \\ & = \frac {e^{e^3} (3-x)}{7 \log (-3+x)}-\frac {1}{7} e^{e^3} \operatorname {LogIntegral}(-3+x)+\frac {1}{7} e^{e^3} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+x\right )+\frac {1}{7} \left (7 e^5+2 e^{e^3}\right ) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-3+x)\right ) \\ & = -\frac {7 e^5+2 e^{e^3}}{7 \log (-3+x)}+\frac {e^{e^3} (3-x)}{7 \log (-3+x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx=-\frac {7 e^5+e^{e^3} (-1+x)}{7 \log (-3+x)} \]

[In]

Integrate[(7*E^5 + E^E^3*(-1 + x) + E^E^3*(3 - x)*Log[-3 + x])/((-21 + 7*x)*Log[-3 + x]^2),x]

[Out]

-1/7*(7*E^5 + E^E^3*(-1 + x))/Log[-3 + x]

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
norman	\(\frac {-\frac {x \,{\mathrm e}^{{\mathrm e}^{3}}}{7}+\frac {{\mathrm e}^{{\mathrm e}^{3}}}{7}-{\mathrm e}^{5}}{\ln \left (-3+x \right )}\)	\(24\)
risch	\(-\frac {7 \,{\mathrm e}^{5}+x \,{\mathrm e}^{{\mathrm e}^{3}}-{\mathrm e}^{{\mathrm e}^{3}}}{7 \ln \left (-3+x \right )}\)	\(24\)
parallelrisch	\(-\frac {7 \,{\mathrm e}^{5}+x \,{\mathrm e}^{{\mathrm e}^{3}}-{\mathrm e}^{{\mathrm e}^{3}}}{7 \ln \left (-3+x \right )}\)	\(24\)
derivativedivides	\(\frac {{\mathrm e}^{{\mathrm e}^{3}} \operatorname {Ei}_{1}\left (-\ln \left (-3+x \right )\right )}{7}+\frac {{\mathrm e}^{{\mathrm e}^{3}} \left (-\frac {-3+x}{\ln \left (-3+x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (-3+x \right )\right )\right )}{7}-\frac {{\mathrm e}^{5}}{\ln \left (-3+x \right )}-\frac {2 \,{\mathrm e}^{{\mathrm e}^{3}}}{7 \ln \left (-3+x \right )}\)	\(63\)
default	\(\frac {{\mathrm e}^{{\mathrm e}^{3}} \operatorname {Ei}_{1}\left (-\ln \left (-3+x \right )\right )}{7}+\frac {{\mathrm e}^{{\mathrm e}^{3}} \left (-\frac {-3+x}{\ln \left (-3+x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (-3+x \right )\right )\right )}{7}-\frac {{\mathrm e}^{5}}{\ln \left (-3+x \right )}-\frac {2 \,{\mathrm e}^{{\mathrm e}^{3}}}{7 \ln \left (-3+x \right )}\)	\(63\)
parts	\(\frac {{\mathrm e}^{{\mathrm e}^{3}} \operatorname {Ei}_{1}\left (-\ln \left (-3+x \right )\right )}{7}+\frac {{\mathrm e}^{{\mathrm e}^{3}} \left (-\frac {-3+x}{\ln \left (-3+x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (-3+x \right )\right )\right )}{7}-\frac {{\mathrm e}^{5}}{\ln \left (-3+x \right )}-\frac {2 \,{\mathrm e}^{{\mathrm e}^{3}}}{7 \ln \left (-3+x \right )}\)	\(63\)

[In]

int(((-x+3)*exp(exp(3))*ln(-3+x)+(-1+x)*exp(exp(3))+7*exp(5))/(7*x-21)/ln(-3+x)^2,x,method=_RETURNVERBOSE)

[Out]

(-1/7*x*exp(exp(3))+1/7*exp(exp(3))-exp(5))/ln(-3+x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx=-\frac {{\left (x - 1\right )} e^{\left (e^{3}\right )} + 7 \, e^{5}}{7 \, \log \left (x - 3\right )} \]

[In]

integrate(((-x+3)*exp(exp(3))*log(-3+x)+(-1+x)*exp(exp(3))+7*exp(5))/(7*x-21)/log(-3+x)^2,x, algorithm="fricas
")

[Out]

-1/7*((x - 1)*e^(e^3) + 7*e^5)/log(x - 3)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx=\frac {- x e^{e^{3}} - 7 e^{5} + e^{e^{3}}}{7 \log {\left (x - 3 \right )}} \]

[In]

integrate(((-x+3)*exp(exp(3))*ln(-3+x)+(-1+x)*exp(exp(3))+7*exp(5))/(7*x-21)/ln(-3+x)**2,x)

[Out]

(-x*exp(exp(3)) - 7*exp(5) + exp(exp(3)))/(7*log(x - 3))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx=-\frac {x e^{\left (e^{3}\right )}}{7 \, \log \left (x - 3\right )} - \frac {e^{5}}{\log \left (x - 3\right )} + \frac {e^{\left (e^{3}\right )}}{7 \, \log \left (x - 3\right )} - \frac {3}{7} \, e^{\left (e^{3}\right )} \]

[In]

integrate(((-x+3)*exp(exp(3))*log(-3+x)+(-1+x)*exp(exp(3))+7*exp(5))/(7*x-21)/log(-3+x)^2,x, algorithm="maxima
")

[Out]

-1/7*x*e^(e^3)/log(x - 3) - e^5/log(x - 3) + 1/7*e^(e^3)/log(x - 3) - 3/7*e^(e^3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx=-\frac {x e^{\left (e^{3}\right )} + 7 \, e^{5} - e^{\left (e^{3}\right )}}{7 \, \log \left (x - 3\right )} \]

[In]

integrate(((-x+3)*exp(exp(3))*log(-3+x)+(-1+x)*exp(exp(3))+7*exp(5))/(7*x-21)/log(-3+x)^2,x, algorithm="giac")

[Out]

-1/7*(x*e^(e^3) + 7*e^5 - e^(e^3))/log(x - 3)

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {7 e^5+e^{e^3} (-1+x)+e^{e^3} (3-x) \log (-3+x)}{(-21+7 x) \log ^2(-3+x)} \, dx=-\frac {7\,{\mathrm {e}}^5-{\mathrm {e}}^{{\mathrm {e}}^3}+x\,{\mathrm {e}}^{{\mathrm {e}}^3}}{7\,\ln \left (x-3\right )} \]

[In]

int((7*exp(5) + exp(exp(3))*(x - 1) - log(x - 3)*exp(exp(3))*(x - 3))/(log(x - 3)^2*(7*x - 21)),x)

[Out]

-(7*exp(5) - exp(exp(3)) + x*exp(exp(3)))/(7*log(x - 3))