\(\int \frac {e^{-x} (4 \log ^2(5)+((-x-x^2) \log (3)+(4+4 x) \log ^2(5)) \log (\frac {-x \log (3)+4 \log ^2(5)}{x}))}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx\) [9833]

Optimal result

Integrand size = 73, antiderivative size = 32 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=e^{-x} \left (e^{1+x}-\frac {\log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x}\right ) \]

[Out]

(exp(1+x)-ln(4/x*ln(5)^2-ln(3))/x)/exp(x)

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78, number of steps used = 18, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.110, Rules used = {1607, 6820, 6874, 2209, 2208, 2228, 2634, 12} \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{-x} \log \left (\frac {4 \log ^2(5)}{x}-\log (3)\right )}{x} \]

[In]

Int[(4*Log[5]^2 + ((-x - x^2)*Log[3] + (4 + 4*x)*Log[5]^2)*Log[(-(x*Log[3]) + 4*Log[5]^2)/x])/(E^x*(-(x^3*Log[
3]) + 4*x^2*Log[5]^2)),x]

[Out]

-(Log[-Log[3] + (4*Log[5]^2)/x]/(E^x*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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{x^2 \left (-x \log (3)+4 \log ^2(5)\right )} \, dx \\ & = \int \frac {e^{-x} \left (-\frac {4 \log ^2(5)}{x \log (3)-4 \log ^2(5)}+(1+x) \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )\right )}{x^2} \, dx \\ & = \int \left (-\frac {4 e^{-x} \log ^2(5)}{x^2 \left (x \log (3)-4 \log ^2(5)\right )}+\frac {e^{-x} (1+x) \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x^2}\right ) \, dx \\ & = -\left (\left (4 \log ^2(5)\right ) \int \frac {e^{-x}}{x^2 \left (x \log (3)-4 \log ^2(5)\right )} \, dx\right )+\int \frac {e^{-x} (1+x) \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x^2} \, dx \\ & = -\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x}-\left (4 \log ^2(5)\right ) \int \left (-\frac {e^{-x} \log (3)}{16 x \log ^4(5)}-\frac {e^{-x}}{4 x^2 \log ^2(5)}-\frac {e^{-x} \log ^2(3)}{16 \log ^4(5) \left (-x \log (3)+4 \log ^2(5)\right )}\right ) \, dx-\int \frac {4 e^{-x} \log ^2(5)}{x^2 \left (-x \log (3)+4 \log ^2(5)\right )} \, dx \\ & = -\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x}+\frac {\log (3) \int \frac {e^{-x}}{x} \, dx}{4 \log ^2(5)}+\frac {\log ^2(3) \int \frac {e^{-x}}{-x \log (3)+4 \log ^2(5)} \, dx}{4 \log ^2(5)}-\left (4 \log ^2(5)\right ) \int \frac {e^{-x}}{x^2 \left (-x \log (3)+4 \log ^2(5)\right )} \, dx+\int \frac {e^{-x}}{x^2} \, dx \\ & = -\frac {e^{-x}}{x}+\frac {\operatorname {ExpIntegralEi}(-x) \log (3)}{4 \log ^2(5)}-\frac {e^{-\frac {4 \log ^2(5)}{\log (3)}} \operatorname {ExpIntegralEi}\left (-\frac {x \log (3)-4 \log ^2(5)}{\log (3)}\right ) \log (3)}{4 \log ^2(5)}-\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x}-\left (4 \log ^2(5)\right ) \int \left (\frac {e^{-x} \log (3)}{16 x \log ^4(5)}+\frac {e^{-x}}{4 x^2 \log ^2(5)}-\frac {e^{-x} \log ^2(3)}{16 \log ^4(5) \left (x \log (3)-4 \log ^2(5)\right )}\right ) \, dx-\int \frac {e^{-x}}{x} \, dx \\ & = -\frac {e^{-x}}{x}-\operatorname {ExpIntegralEi}(-x)+\frac {\operatorname {ExpIntegralEi}(-x) \log (3)}{4 \log ^2(5)}-\frac {e^{-\frac {4 \log ^2(5)}{\log (3)}} \operatorname {ExpIntegralEi}\left (-\frac {x \log (3)-4 \log ^2(5)}{\log (3)}\right ) \log (3)}{4 \log ^2(5)}-\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x}-\frac {\log (3) \int \frac {e^{-x}}{x} \, dx}{4 \log ^2(5)}+\frac {\log ^2(3) \int \frac {e^{-x}}{x \log (3)-4 \log ^2(5)} \, dx}{4 \log ^2(5)}-\int \frac {e^{-x}}{x^2} \, dx \\ & = -\operatorname {ExpIntegralEi}(-x)-\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x}+\int \frac {e^{-x}}{x} \, dx \\ & = -\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x} \]

[In]

Integrate[(4*Log[5]^2 + ((-x - x^2)*Log[3] + (4 + 4*x)*Log[5]^2)*Log[(-(x*Log[3]) + 4*Log[5]^2)/x])/(E^x*(-(x^
3*Log[3]) + 4*x^2*Log[5]^2)),x]

[Out]

-(Log[-Log[3] + (4*Log[5]^2)/x]/(E^x*x))

Maple [A] (verified)

Time = 1.61 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84

method	result	size
norman	\(-\frac {\ln \left (\frac {4 \ln \left (5\right )^{2}-x \ln \left (3\right )}{x}\right ) {\mathrm e}^{-x}}{x}\)	\(27\)
parallelrisch	\(-\frac {\ln \left (\frac {4 \ln \left (5\right )^{2}-x \ln \left (3\right )}{x}\right ) {\mathrm e}^{-x}}{x}\)	\(27\)
risch	\(-\frac {{\mathrm e}^{-x} \ln \left (\ln \left (5\right )^{2}-\frac {x \ln \left (3\right )}{4}\right )}{x}-\frac {\left (-i \pi \,\operatorname {csgn}\left (i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )}{x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+i \pi {\operatorname {csgn}\left (\frac {i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+i \pi {\operatorname {csgn}\left (\frac {i \left (-\ln \left (5\right )^{2}+\frac {x \ln \left (3\right )}{4}\right )}{x}\right )}^{3}+4 \ln \left (2\right )-2 \ln \left (x \right )\right ) {\mathrm e}^{-x}}{2 x}\)	\(184\)

[In]

int((((4+4*x)*ln(5)^2+(-x^2-x)*ln(3))*ln((4*ln(5)^2-x*ln(3))/x)+4*ln(5)^2)/(4*x^2*ln(5)^2-x^3*ln(3))/exp(x),x,
method=_RETURNVERBOSE)

[Out]

-ln((4*ln(5)^2-x*ln(3))/x)/exp(x)/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (-x\right )} \log \left (\frac {4 \, \log \left (5\right )^{2} - x \log \left (3\right )}{x}\right )}{x} \]

[In]

integrate((((4+4*x)*log(5)^2+(-x^2-x)*log(3))*log((4*log(5)^2-x*log(3))/x)+4*log(5)^2)/(4*x^2*log(5)^2-x^3*log
(3))/exp(x),x, algorithm="fricas")

[Out]

-e^(-x)*log((4*log(5)^2 - x*log(3))/x)/x

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=- \frac {e^{- x} \log {\left (\frac {- x \log {\left (3 \right )} + 4 \log {\left (5 \right )}^{2}}{x} \right )}}{x} \]

[In]

integrate((((4+4*x)*ln(5)**2+(-x**2-x)*ln(3))*ln((4*ln(5)**2-x*ln(3))/x)+4*ln(5)**2)/(4*x**2*ln(5)**2-x**3*ln(
3))/exp(x),x)

[Out]

-exp(-x)*log((-x*log(3) + 4*log(5)**2)/x)/x

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (-x\right )} \log \left (4 \, \log \left (5\right )^{2} - x \log \left (3\right )\right ) - e^{\left (-x\right )} \log \left (x\right )}{x} \]

[In]

integrate((((4+4*x)*log(5)^2+(-x^2-x)*log(3))*log((4*log(5)^2-x*log(3))/x)+4*log(5)^2)/(4*x^2*log(5)^2-x^3*log
(3))/exp(x),x, algorithm="maxima")

[Out]

-(e^(-x)*log(4*log(5)^2 - x*log(3)) - e^(-x)*log(x))/x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (-x\right )} \log \left (4 \, \log \left (5\right )^{2} - x \log \left (3\right )\right ) - e^{\left (-x\right )} \log \left (x\right )}{x} \]

[In]

integrate((((4+4*x)*log(5)^2+(-x^2-x)*log(3))*log((4*log(5)^2-x*log(3))/x)+4*log(5)^2)/(4*x^2*log(5)^2-x^3*log
(3))/exp(x),x, algorithm="giac")

[Out]

-(e^(-x)*log(4*log(5)^2 - x*log(3)) - e^(-x)*log(x))/x

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-x} \left (4 \log ^2(5)+\left (\left (-x-x^2\right ) \log (3)+(4+4 x) \log ^2(5)\right ) \log \left (\frac {-x \log (3)+4 \log ^2(5)}{x}\right )\right )}{-x^3 \log (3)+4 x^2 \log ^2(5)} \, dx=\int \frac {{\mathrm {e}}^{-x}\,\left (\ln \left (-\frac {x\,\ln \left (3\right )-4\,{\ln \left (5\right )}^2}{x}\right )\,\left ({\ln \left (5\right )}^2\,\left (4\,x+4\right )-\ln \left (3\right )\,\left (x^2+x\right )\right )+4\,{\ln \left (5\right )}^2\right )}{4\,x^2\,{\ln \left (5\right )}^2-x^3\,\ln \left (3\right )} \,d x \]

[In]

int((exp(-x)*(log(-(x*log(3) - 4*log(5)^2)/x)*(log(5)^2*(4*x + 4) - log(3)*(x + x^2)) + 4*log(5)^2))/(4*x^2*lo
g(5)^2 - x^3*log(3)),x)

[Out]

int((exp(-x)*(log(-(x*log(3) - 4*log(5)^2)/x)*(log(5)^2*(4*x + 4) - log(3)*(x + x^2)) + 4*log(5)^2))/(4*x^2*lo
g(5)^2 - x^3*log(3)), x)