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

Optimal. Leaf size=32 \[ e^{-x} \left (e^{1+x}-\frac {\log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x}\right ) \]

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Rubi [A]  time = 1.25, antiderivative size = 25, normalized size of antiderivative = 0.78, number of steps used = 18, number of rules used = 8, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.110, Rules used = {1593, 6688, 6742, 2178, 2177, 2197, 2554, 12} \begin {gather*} -\frac {e^{-x} \log \left (\frac {4 \log ^2(5)}{x}-\log (3)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[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 1593

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 2177

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] &&  !$UseGamma ===
True

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 2197

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 2554

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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {\text {Ei}(-x) \log (3)}{4 \log ^2(5)}-\frac {e^{-\frac {4 \log ^2(5)}{\log (3)}} \text {Ei}\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}-\text {Ei}(-x)+\frac {\text {Ei}(-x) \log (3)}{4 \log ^2(5)}-\frac {e^{-\frac {4 \log ^2(5)}{\log (3)}} \text {Ei}\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\\ &=-\text {Ei}(-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 {aligned} \end {gather*}

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Mathematica [A]  time = 5.04, size = 25, normalized size = 0.78 \begin {gather*} -\frac {e^{-x} \log \left (-\log (3)+\frac {4 \log ^2(5)}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.68, size = 26, normalized size = 0.81 \begin {gather*} -\frac {e^{\left (-x\right )} \log \left (\frac {4 \, \log \relax (5)^{2} - x \log \relax (3)}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*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

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giac [A]  time = 0.26, size = 32, normalized size = 1.00 \begin {gather*} -\frac {e^{\left (-x\right )} \log \left (4 \, \log \relax (5)^{2} - x \log \relax (3)\right ) - e^{\left (-x\right )} \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*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

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maple [A]  time = 0.23, size = 27, normalized size = 0.84

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x+4)*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

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maxima [A]  time = 0.49, size = 32, normalized size = 1.00 \begin {gather*} -\frac {e^{\left (-x\right )} \log \left (4 \, \log \relax (5)^{2} - x \log \relax (3)\right ) - e^{\left (-x\right )} \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{-x}\,\left (\ln \left (-\frac {x\,\ln \relax (3)-4\,{\ln \relax (5)}^2}{x}\right )\,\left ({\ln \relax (5)}^2\,\left (4\,x+4\right )-\ln \relax (3)\,\left (x^2+x\right )\right )+4\,{\ln \relax (5)}^2\right )}{4\,x^2\,{\ln \relax (5)}^2-x^3\,\ln \relax (3)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.40, size = 20, normalized size = 0.62 \begin {gather*} - \frac {e^{- x} \log {\left (\frac {- x \log {\relax (3 )} + 4 \log {\relax (5 )}^{2}}{x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*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

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