3.83.36 \(\int \frac {15 e^5-30 e^5 \log (\frac {4}{x})-e^{e^x+x} x^3 \log ^2(\frac {4}{x})}{x^3 \log ^2(\frac {4}{x})} \, dx\)

Optimal. Leaf size=24 \[ -e^{e^x}+\frac {15 e^5}{x^2 \log \left (\frac {4}{x}\right )} \]

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Rubi [C]  time = 0.57, antiderivative size = 106, normalized size of antiderivative = 4.42, number of steps used = 10, number of rules used = 8, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6742, 2282, 2194, 2306, 2310, 2178, 2366, 6482} \begin {gather*} -\frac {15}{8} e^5 \left (1-2 \log \left (\frac {4}{x}\right )\right ) \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right )-\frac {15}{4} e^5 \log \left (\frac {4}{x}\right ) \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right )+\frac {15}{8} e^5 \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right )+\frac {30 e^5}{x^2}+\frac {15 e^5 \left (1-2 \log \left (\frac {4}{x}\right )\right )}{x^2 \log \left (\frac {4}{x}\right )}-e^{e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15*E^5 - 30*E^5*Log[4/x] - E^(E^x + x)*x^3*Log[4/x]^2)/(x^3*Log[4/x]^2),x]

[Out]

-E^E^x + (30*E^5)/x^2 + (15*E^5*ExpIntegralEi[2*Log[4/x]])/8 - (15*E^5*ExpIntegralEi[2*Log[4/x]]*(1 - 2*Log[4/
x]))/8 + (15*E^5*(1 - 2*Log[4/x]))/(x^2*Log[4/x]) - (15*E^5*ExpIntegralEi[2*Log[4/x]]*Log[4/x])/4

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[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 2310

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

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} &=\int \left (-e^{e^x+x}-\frac {15 e^5 \left (-1+2 \log \left (\frac {4}{x}\right )\right )}{x^3 \log ^2\left (\frac {4}{x}\right )}\right ) \, dx\\ &=-\left (\left (15 e^5\right ) \int \frac {-1+2 \log \left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx\right )-\int e^{e^x+x} \, dx\\ &=-\frac {15}{8} e^5 \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right ) \left (1-2 \log \left (\frac {4}{x}\right )\right )+\frac {15 e^5 \left (1-2 \log \left (\frac {4}{x}\right )\right )}{x^2 \log \left (\frac {4}{x}\right )}-\left (30 e^5\right ) \int \left (-\frac {\text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right )}{8 x}+\frac {1}{x^3 \log \left (\frac {4}{x}\right )}\right ) \, dx-\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=-e^{e^x}-\frac {15}{8} e^5 \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right ) \left (1-2 \log \left (\frac {4}{x}\right )\right )+\frac {15 e^5 \left (1-2 \log \left (\frac {4}{x}\right )\right )}{x^2 \log \left (\frac {4}{x}\right )}+\frac {1}{4} \left (15 e^5\right ) \int \frac {\text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right )}{x} \, dx-\left (30 e^5\right ) \int \frac {1}{x^3 \log \left (\frac {4}{x}\right )} \, dx\\ &=-e^{e^x}-\frac {15}{8} e^5 \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right ) \left (1-2 \log \left (\frac {4}{x}\right )\right )+\frac {15 e^5 \left (1-2 \log \left (\frac {4}{x}\right )\right )}{x^2 \log \left (\frac {4}{x}\right )}+\frac {1}{8} \left (15 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log \left (\frac {4}{x}\right )\right )-\frac {1}{4} \left (15 e^5\right ) \operatorname {Subst}\left (\int \text {Ei}(2 x) \, dx,x,\log \left (\frac {4}{x}\right )\right )\\ &=-e^{e^x}+\frac {30 e^5}{x^2}+\frac {15}{8} e^5 \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right )-\frac {15}{8} e^5 \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right ) \left (1-2 \log \left (\frac {4}{x}\right )\right )+\frac {15 e^5 \left (1-2 \log \left (\frac {4}{x}\right )\right )}{x^2 \log \left (\frac {4}{x}\right )}-\frac {15}{4} e^5 \text {Ei}\left (2 \log \left (\frac {4}{x}\right )\right ) \log \left (\frac {4}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 24, normalized size = 1.00 \begin {gather*} -e^{e^x}+\frac {15 e^5}{x^2 \log \left (\frac {4}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*E^5 - 30*E^5*Log[4/x] - E^(E^x + x)*x^3*Log[4/x]^2)/(x^3*Log[4/x]^2),x]

[Out]

-E^E^x + (15*E^5)/(x^2*Log[4/x])

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fricas [A]  time = 0.97, size = 39, normalized size = 1.62 \begin {gather*} -\frac {{\left (x^{2} e^{\left (x + e^{x}\right )} \log \left (\frac {4}{x}\right ) - 15 \, e^{\left (x + 5\right )}\right )} e^{\left (-x\right )}}{x^{2} \log \left (\frac {4}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(x)*log(4/x)^2*exp(exp(x))-30*exp(5)*log(4/x)+15*exp(5))/x^3/log(4/x)^2,x, algorithm="frica
s")

[Out]

-(x^2*e^(x + e^x)*log(4/x) - 15*e^(x + 5))*e^(-x)/(x^2*log(4/x))

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giac [A]  time = 0.17, size = 39, normalized size = 1.62 \begin {gather*} -\frac {{\left (x^{2} e^{\left (x + e^{x}\right )} \log \left (\frac {4}{x}\right ) - 15 \, e^{\left (x + 5\right )}\right )} e^{\left (-x\right )}}{x^{2} \log \left (\frac {4}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(x)*log(4/x)^2*exp(exp(x))-30*exp(5)*log(4/x)+15*exp(5))/x^3/log(4/x)^2,x, algorithm="giac"
)

[Out]

-(x^2*e^(x + e^x)*log(4/x) - 15*e^(x + 5))*e^(-x)/(x^2*log(4/x))

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maple [A]  time = 0.08, size = 22, normalized size = 0.92

method	result	size
default	\(\frac {15 \,{\mathrm e}^{5}}{x^{2} \ln \left (\frac {4}{x}\right )}-{\mathrm e}^{{\mathrm e}^{x}}\)	\(22\)
risch	\(\frac {30 i {\mathrm e}^{5}}{x^{2} \left (4 i \ln \relax (2)-2 i \ln \relax (x )\right )}-{\mathrm e}^{{\mathrm e}^{x}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^3*exp(x)*ln(4/x)^2*exp(exp(x))-30*exp(5)*ln(4/x)+15*exp(5))/x^3/ln(4/x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.48, size = 43, normalized size = 1.79 \begin {gather*} -\frac {{\left (2 \, x^{2} \log \relax (2) - x^{2} \log \relax (x)\right )} e^{\left (e^{x}\right )} - 15 \, e^{5}}{2 \, x^{2} \log \relax (2) - x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3*exp(x)*log(4/x)^2*exp(exp(x))-30*exp(5)*log(4/x)+15*exp(5))/x^3/log(4/x)^2,x, algorithm="maxim
a")

[Out]

-((2*x^2*log(2) - x^2*log(x))*e^(e^x) - 15*e^5)/(2*x^2*log(2) - x^2*log(x))

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mupad [B]  time = 4.91, size = 21, normalized size = 0.88 \begin {gather*} \frac {15\,{\mathrm {e}}^5}{x^2\,\ln \left (\frac {4}{x}\right )}-{\mathrm {e}}^{{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*exp(5)*log(4/x) - 15*exp(5) + x^3*exp(exp(x))*exp(x)*log(4/x)^2)/(x^3*log(4/x)^2),x)

[Out]

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

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sympy [A]  time = 0.35, size = 17, normalized size = 0.71 \begin {gather*} - e^{e^{x}} + \frac {15 e^{5}}{x^{2} \log {\left (\frac {4}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**3*exp(x)*ln(4/x)**2*exp(exp(x))-30*exp(5)*ln(4/x)+15*exp(5))/x**3/ln(4/x)**2,x)

[Out]

-exp(exp(x)) + 15*exp(5)/(x**2*log(4/x))

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