\(\int \frac {15 e^5-30 e^5 \log (\frac {4}{x})-e^{e^x+x} x^3 \log ^2(\frac {4}{x})}{x^3 \log ^2(\frac {4}{x})} \, dx\) [8236]

Optimal result

Integrand size = 49, antiderivative size = 24 \[ \int \frac {15 e^5-30 e^5 \log \left (\frac {4}{x}\right )-e^{e^x+x} x^3 \log ^2\left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx=-e^{e^x}+\frac {15 e^5}{x^2 \log \left (\frac {4}{x}\right )} \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 4.42, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {6874, 2320, 2225, 2343, 2347, 2209, 2413, 6617} \[ \int \frac {15 e^5-30 e^5 \log \left (\frac {4}{x}\right )-e^{e^x+x} x^3 \log ^2\left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx=-\frac {15}{8} e^5 \left (1-2 \log \left (\frac {4}{x}\right )\right ) \operatorname {ExpIntegralEi}\left (2 \log \left (\frac {4}{x}\right )\right )-\frac {15}{4} e^5 \log \left (\frac {4}{x}\right ) \operatorname {ExpIntegralEi}\left (2 \log \left (\frac {4}{x}\right )\right )+\frac {15}{8} e^5 \operatorname {ExpIntegralEi}\left (2 \log \left (\frac {4}{x}\right )\right )+\frac {30 e^5}{x^2}+\frac {15 e^5 \left (1-2 \log \left (\frac {4}{x}\right )\right )}{x^2 \log \left (\frac {4}{x}\right )}-e^{e^x} \]

[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 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 2225

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 2320

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 2343

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 2347

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)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2413

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 6617

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 6874

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

Rubi steps \begin{align*} \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 \operatorname {ExpIntegralEi}\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 {\operatorname {ExpIntegralEi}\left (2 \log \left (\frac {4}{x}\right )\right )}{8 x}+\frac {1}{x^3 \log \left (\frac {4}{x}\right )}\right ) \, dx-\text {Subst}\left (\int e^x \, dx,x,e^x\right ) \\ & = -e^{e^x}-\frac {15}{8} e^5 \operatorname {ExpIntegralEi}\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 {\operatorname {ExpIntegralEi}\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 \operatorname {ExpIntegralEi}\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 ) \text {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 ) \text {Subst}\left (\int \operatorname {ExpIntegralEi}(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 \operatorname {ExpIntegralEi}\left (2 \log \left (\frac {4}{x}\right )\right )-\frac {15}{8} e^5 \operatorname {ExpIntegralEi}\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 \operatorname {ExpIntegralEi}\left (2 \log \left (\frac {4}{x}\right )\right ) \log \left (\frac {4}{x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {15 e^5-30 e^5 \log \left (\frac {4}{x}\right )-e^{e^x+x} x^3 \log ^2\left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx=-e^{e^x}+\frac {15 e^5}{x^2 \log \left (\frac {4}{x}\right )} \]

[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])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 6.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17

method	result	size
risch	\(-\frac {30 i {\mathrm e}^{5}}{x^{2} \left (-4 i \ln \left (2\right )+2 i \ln \left (x \right )\right )}-{\mathrm e}^{{\mathrm e}^{x}}\)	\(28\)
parallelrisch	\(\frac {-{\mathrm e}^{{\mathrm e}^{x}} \ln \left (\frac {4}{x}\right ) x^{2}+15 \,{\mathrm e}^{5}}{x^{2} \ln \left (\frac {4}{x}\right )}\)	\(32\)
default	\(-{\mathrm e}^{{\mathrm e}^{x}}+15 \,{\mathrm e}^{5} \left (\frac {1}{x^{2} \ln \left (\frac {4}{x}\right )}+\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (\frac {4}{x}\right )\right )}{8}\right )-\frac {15 \,{\mathrm e}^{5} \operatorname {Ei}_{1}\left (-2 \ln \left (\frac {4}{x}\right )\right )}{8}\)	\(50\)
parts	\(-{\mathrm e}^{{\mathrm e}^{x}}+15 \,{\mathrm e}^{5} \left (\frac {1}{x^{2} \ln \left (\frac {4}{x}\right )}+\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (\frac {4}{x}\right )\right )}{8}\right )-\frac {15 \,{\mathrm e}^{5} \operatorname {Ei}_{1}\left (-2 \ln \left (\frac {4}{x}\right )\right )}{8}\)	\(50\)

[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]

-30*I*exp(5)/x^2/(-4*I*ln(2)+2*I*ln(x))-exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {15 e^5-30 e^5 \log \left (\frac {4}{x}\right )-e^{e^x+x} x^3 \log ^2\left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx=-\frac {{\left (x^{2} e^{\left (x + e^{x}\right )} \log \left (\frac {4}{x}\right ) - 15 \, e^{\left (x + 5\right )}\right )} e^{\left (-x\right )}}{x^{2} \log \left (\frac {4}{x}\right )} \]

[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))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {15 e^5-30 e^5 \log \left (\frac {4}{x}\right )-e^{e^x+x} x^3 \log ^2\left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx=- e^{e^{x}} + \frac {15 e^{5}}{x^{2} \log {\left (\frac {4}{x} \right )}} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (21) = 42\).

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {15 e^5-30 e^5 \log \left (\frac {4}{x}\right )-e^{e^x+x} x^3 \log ^2\left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx=-\frac {{\left (2 \, x^{2} \log \left (2\right ) - x^{2} \log \left (x\right )\right )} e^{\left (e^{x}\right )} - 15 \, e^{5}}{2 \, x^{2} \log \left (2\right ) - x^{2} \log \left (x\right )} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {15 e^5-30 e^5 \log \left (\frac {4}{x}\right )-e^{e^x+x} x^3 \log ^2\left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx=-\frac {{\left (x^{2} e^{\left (x + e^{x}\right )} \log \left (\frac {4}{x}\right ) - 15 \, e^{\left (x + 5\right )}\right )} e^{\left (-x\right )}}{x^{2} \log \left (\frac {4}{x}\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 13.76 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {15 e^5-30 e^5 \log \left (\frac {4}{x}\right )-e^{e^x+x} x^3 \log ^2\left (\frac {4}{x}\right )}{x^3 \log ^2\left (\frac {4}{x}\right )} \, dx=\frac {15\,{\mathrm {e}}^5}{x^2\,\ln \left (\frac {4}{x}\right )}-{\mathrm {e}}^{{\mathrm {e}}^x} \]

[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))