3.103.18 \(\int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ 4 \left (5-e^{10-x}+\frac {x^2}{5 e^3 \log (x)}\right ) \]

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Rubi [C]  time = 0.20, antiderivative size = 84, normalized size of antiderivative = 3.11, number of steps used = 10, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {12, 6742, 2194, 2306, 2309, 2178, 2366, 6482} \begin {gather*} \frac {8 \text {Ei}(2 \log (x))}{5 e^3}-\frac {8 (1-2 \log (x)) \text {Ei}(2 \log (x))}{5 e^3}-\frac {16 \log (x) \text {Ei}(2 \log (x))}{5 e^3}+\frac {8 x^2}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}-4 e^{10-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*x + 8*x*Log[x] + 20*E^(13 - x)*Log[x]^2)/(5*E^3*Log[x]^2),x]

[Out]

-4*E^(10 - x) + (8*x^2)/(5*E^3) + (8*ExpIntegralEi[2*Log[x]])/(5*E^3) - (8*ExpIntegralEi[2*Log[x]]*(1 - 2*Log[
x]))/(5*E^3) + (4*x^2*(1 - 2*Log[x]))/(5*E^3*Log[x]) - (16*ExpIntegralEi[2*Log[x]]*Log[x])/(5*E^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 6482

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, 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} &=\frac {\int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{\log ^2(x)} \, dx}{5 e^3}\\ &=\frac {\int \left (20 e^{13-x}+\frac {4 x (-1+2 \log (x))}{\log ^2(x)}\right ) \, dx}{5 e^3}\\ &=\frac {4 \int \frac {x (-1+2 \log (x))}{\log ^2(x)} \, dx}{5 e^3}+\frac {4 \int e^{13-x} \, dx}{e^3}\\ &=-4 e^{10-x}-\frac {8 \text {Ei}(2 \log (x)) (1-2 \log (x))}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}-\frac {8 \int \left (\frac {2 \text {Ei}(2 \log (x))}{x}-\frac {x}{\log (x)}\right ) \, dx}{5 e^3}\\ &=-4 e^{10-x}-\frac {8 \text {Ei}(2 \log (x)) (1-2 \log (x))}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}+\frac {8 \int \frac {x}{\log (x)} \, dx}{5 e^3}-\frac {16 \int \frac {\text {Ei}(2 \log (x))}{x} \, dx}{5 e^3}\\ &=-4 e^{10-x}-\frac {8 \text {Ei}(2 \log (x)) (1-2 \log (x))}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}+\frac {8 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )}{5 e^3}-\frac {16 \operatorname {Subst}(\int \text {Ei}(2 x) \, dx,x,\log (x))}{5 e^3}\\ &=-4 e^{10-x}+\frac {8 x^2}{5 e^3}+\frac {8 \text {Ei}(2 \log (x))}{5 e^3}-\frac {8 \text {Ei}(2 \log (x)) (1-2 \log (x))}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}-\frac {16 \text {Ei}(2 \log (x)) \log (x)}{5 e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 26, normalized size = 0.96 \begin {gather*} \frac {-20 e^{13-x}+\frac {4 x^2}{\log (x)}}{5 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x + 8*x*Log[x] + 20*E^(13 - x)*Log[x]^2)/(5*E^3*Log[x]^2),x]

[Out]

(-20*E^(13 - x) + (4*x^2)/Log[x])/(5*E^3)

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fricas [A]  time = 0.84, size = 22, normalized size = 0.81 \begin {gather*} \frac {4 \, {\left (x^{2} - 5 \, e^{\left (-x + 13\right )} \log \relax (x)\right )} e^{\left (-3\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*exp(3)*exp(10-x)*log(x)^2+8*x*log(x)-4*x)/exp(3)/log(x)^2,x, algorithm="fricas")

[Out]

4/5*(x^2 - 5*e^(-x + 13)*log(x))*e^(-3)/log(x)

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giac [A]  time = 0.19, size = 22, normalized size = 0.81 \begin {gather*} \frac {4 \, {\left (x^{2} - 5 \, e^{\left (-x + 13\right )} \log \relax (x)\right )} e^{\left (-3\right )}}{5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*exp(3)*exp(10-x)*log(x)^2+8*x*log(x)-4*x)/exp(3)/log(x)^2,x, algorithm="giac")

[Out]

4/5*(x^2 - 5*e^(-x + 13)*log(x))*e^(-3)/log(x)

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maple [A]  time = 0.07, size = 21, normalized size = 0.78

method	result	size
risch	\(-4 \,{\mathrm e}^{10-x}+\frac {4 x^{2} {\mathrm e}^{-3}}{5 \ln \relax (x )}\)	\(21\)
default	\(\frac {4 \,{\mathrm e}^{-3} \left (\frac {x^{2}}{\ln \relax (x )}-5 \,{\mathrm e}^{3} {\mathrm e}^{10-x}\right )}{5}\)	\(26\)
norman	\(\frac {\frac {4 x^{2} {\mathrm e}^{-3}}{5}-4 \,{\mathrm e}^{10-x} \ln \relax (x )}{\ln \relax (x )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(20*exp(3)*exp(10-x)*ln(x)^2+8*x*ln(x)-4*x)/exp(3)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-4*exp(10-x)+4/5*x^2/ln(x)*exp(-3)

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maxima [C]  time = 0.37, size = 28, normalized size = 1.04 \begin {gather*} \frac {4}{5} \, {\left (2 \, {\rm Ei}\left (2 \, \log \relax (x)\right ) - 5 \, e^{\left (-x + 13\right )} - 2 \, \Gamma \left (-1, -2 \, \log \relax (x)\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*exp(3)*exp(10-x)*log(x)^2+8*x*log(x)-4*x)/exp(3)/log(x)^2,x, algorithm="maxima")

[Out]

4/5*(2*Ei(2*log(x)) - 5*e^(-x + 13) - 2*gamma(-1, -2*log(x)))*e^(-3)

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mupad [B]  time = 6.61, size = 20, normalized size = 0.74 \begin {gather*} \frac {4\,x^2\,{\mathrm {e}}^{-3}}{5\,\ln \relax (x)}-4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*((8*x*log(x))/5 - (4*x)/5 + 4*exp(3)*exp(10 - x)*log(x)^2))/log(x)^2,x)

[Out]

(4*x^2*exp(-3))/(5*log(x)) - 4*exp(-x)*exp(10)

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sympy [A]  time = 0.28, size = 19, normalized size = 0.70 \begin {gather*} \frac {4 x^{2}}{5 e^{3} \log {\relax (x )}} - 4 e^{10 - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(20*exp(3)*exp(10-x)*ln(x)**2+8*x*ln(x)-4*x)/exp(3)/ln(x)**2,x)

[Out]

4*x**2*exp(-3)/(5*log(x)) - 4*exp(10 - x)

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