∫ e3∕x(−12+3e3+3x) log(5)+(−4x2+e3x2+x3) log(5)+(−3e3∕x log(5)−x2 log(5)) log(x)+e4+x+log(5) log(4−e3−x+log(x)) ___________________________________________ log (5) (4x2−e3x2−x3+(x−x2) log(5)+x2 log(x)) _________________________________________________________________________________________________________________________________________________________________________________________________________________(4x2 −e3 x2 −x3 ) log (5)+x2 log (5) log (x) dx

3.86.100 $\int \frac {e^{3/x} (-12+3 e^3+3 x) \log (5)+(-4 x^2+e^3 x^2+x^3) \log (5)+(-3 e^{3/x} \log (5)-x^2 \log (5)) \log (x)+e^{\frac {4+x+\log (5) \log (4-e^3-x+\log (x))}{\log (5)}} (4 x^2-e^3 x^2-x^3+(x-x^2) \log (5)+x^2 \log (x))}{(4 x^2-e^3 x^2-x^3) \log (5)+x^2 \log (5) \log (x)} \, dx$

Optimal. Leaf size=34 \[ e^{3/x}-x+e^{\frac {4+x}{\log (5)}} \left (4-e^3-x+\log (x)\right ) \]

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Rubi [B] time = 0.73, antiderivative size = 95, normalized size of antiderivative = 2.79, number of steps used = 12, number of rules used = 8, integrand size = 158, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.051, Rules used = {6688, 2209, 2199, 2176, 2194, 2178, 2554, 12} \begin {gather*} -x+e^{3/x}+x \left (-e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}}\right )+e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}} \log (x)+\log (5) e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}}+\left (4-125^{\frac {1}{\log (5)}}-\log (5)\right ) e^{\frac {x}{\log (5)}+\frac {4}{\log (5)}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(E^(3/x)*(-12 + 3*E^3 + 3*x)*Log[5] + (-4*x^2 + E^3*x^2 + x^3)*Log[5] + (-3*E^(3/x)*Log[5] - x^2*Log[5])*L

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

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

[Out]

E^(3/x) - x - E^(4/Log[5] + x/Log[5])*x + E^(4/Log[5] + x/Log[5])*(4 - 125^Log[5]^(-1) - Log[5]) + E^(4/Log[5]

+ x/Log[5])*Log[5] + E^(4/Log[5] + x/Log[5])*Log[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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m

*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && 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 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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo

werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]

&& IntegerQ[m] && !$UseGamma === True

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*

F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &

& EqQ[d*e - c*f, 0]

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1-\frac {3 e^{3/x}}{x^2}-\frac {e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \left (x^2-x \left (4-125^{\frac {1}{\log (5)}}-\log (5)\right )-\log (5)\right )}{x \log (5)}+\frac {e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (x)}{\log (5)}\right ) \, dx\\ &=-x-3 \int \frac {e^{3/x}}{x^2} \, dx-\frac {\int \frac {e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \left (x^2-x \left (4-125^{\frac {1}{\log (5)}}-\log (5)\right )-\log (5)\right )}{x} \, dx}{\log (5)}+\frac {\int e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (x) \, dx}{\log (5)}\\ &=e^{3/x}-x+e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (x)-\frac {\int \frac {e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (5)}{x} \, dx}{\log (5)}-\frac {\int \left (e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} x-\frac {e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (5)}{x}+e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \left (-4+125^{\frac {1}{\log (5)}}+\log (5)\right )\right ) \, dx}{\log (5)}\\ &=e^{3/x}-x+e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (x)-\frac {\int e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} x \, dx}{\log (5)}+\frac {\left (4-125^{\frac {1}{\log (5)}}-\log (5)\right ) \int e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \, dx}{\log (5)}\\ &=e^{3/x}-x-e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} x+e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \left (4-125^{\frac {1}{\log (5)}}-\log (5)\right )+e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (x)+\int e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \, dx\\ &=e^{3/x}-x-e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} x+e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \left (4-125^{\frac {1}{\log (5)}}-\log (5)\right )+e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (5)+e^{\frac {4}{\log (5)}+\frac {x}{\log (5)}} \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.29, size = 45, normalized size = 1.32 \begin {gather*} e^{3/x}-x-e^{\frac {4+x}{\log (5)}} \left (-4+125^{\frac {1}{\log (5)}}+x\right )+e^{\frac {4+x}{\log (5)}} \log (x) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^(3/x)*(-12 + 3*E^3 + 3*x)*Log[5] + (-4*x^2 + E^3*x^2 + x^3)*Log[5] + (-3*E^(3/x)*Log[5] - x^2*Log

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

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

[Out]

E^(3/x) - x - E^((4 + x)/Log[5])*(-4 + 125^Log[5]^(-1) + x) + E^((4 + x)/Log[5])*Log[x]

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fricas [A] time = 0.87, size = 34, normalized size = 1.00 \begin {gather*} -x + e^{\left (\frac {\log \relax (5) \log \left (-x - e^{3} + \log \relax (x) + 4\right ) + x + 4}{\log \relax (5)}\right )} + e^{\frac {3}{x}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

-3*log(5)*exp(3/x)-x^2*log(5))*log(x)+(3*exp(3)+3*x-12)*log(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*log(5))/(x^2*lo

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

[Out]

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

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giac [B] time = 0.23, size = 77, normalized size = 2.26 \begin {gather*} -x e^{\left (\frac {x}{\log \relax (5)} + \frac {4}{\log \relax (5)}\right )} + e^{\left (\frac {x}{\log \relax (5)} + \frac {4}{\log \relax (5)}\right )} \log \relax (x) - x - e^{\left (\frac {x}{\log \relax (5)} + \frac {4}{\log \relax (5)} + 3\right )} + 4 \, e^{\left (\frac {x}{\log \relax (5)} + \frac {4}{\log \relax (5)}\right )} + e^{\frac {3}{x}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

-3*log(5)*exp(3/x)-x^2*log(5))*log(x)+(3*exp(3)+3*x-12)*log(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*log(5))/(x^2*lo

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

[Out]

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

+ 4/log(5)) + e^(3/x)

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maple [A] time = 0.32, size = 32, normalized size = 0.94


method	result	size

risch	${\mathrm e}^{\frac {3}{x}}-x +\left (\ln \relax (x )+4-{\mathrm e}^{3}-x \right ) {\mathrm e}^{\frac {4+x}{\ln \relax (5)}}$	$32$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

(3/x)-x^2*ln(5))*ln(x)+(3*exp(3)+3*x-12)*ln(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*ln(5))/(x^2*ln(5)*ln(x)+(-x^2*e

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

[Out]

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

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maxima [A] time = 0.48, size = 58, normalized size = 1.71 \begin {gather*} -{\left (x e^{\frac {4}{\log \relax (5)}} - e^{\frac {4}{\log \relax (5)}} \log \relax (x) - 4 \, e^{\frac {4}{\log \relax (5)}} + e^{\left (\frac {4}{\log \relax (5)} + 3\right )}\right )} e^{\frac {x}{\log \relax (5)}} - x + e^{\frac {3}{x}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

-3*log(5)*exp(3/x)-x^2*log(5))*log(x)+(3*exp(3)+3*x-12)*log(5)*exp(3/x)+(x^2*exp(3)+x^3-4*x^2)*log(5))/(x^2*lo

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

[Out]

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

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mupad [B] time = 6.60, size = 77, normalized size = 2.26 \begin {gather*} 4\,{\mathrm {e}}^{\frac {x}{\ln \relax (5)}+\frac {4}{\ln \relax (5)}}-x-{\mathrm {e}}^{\frac {x}{\ln \relax (5)}+\frac {4}{\ln \relax (5)}+3}+{\mathrm {e}}^{3/x}-x\,{\mathrm {e}}^{\frac {x}{\ln \relax (5)}+\frac {4}{\ln \relax (5)}}+{\mathrm {e}}^{\frac {x}{\ln \relax (5)}+\frac {4}{\ln \relax (5)}}\,\ln \relax (x) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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

3*x + 3*exp(3) - 12))/(log(5)*(x^2*exp(3) - 4*x^2 + x^3) - x^2*log(5)*log(x)),x)

[Out]

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

log(5) + 4/log(5))*log(x)

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sympy [A] time = 2.18, size = 29, normalized size = 0.85 \begin {gather*} - x + e^{\frac {3}{x}} + e^{\frac {x + \log {\relax (5 )} \log {\left (- x + \log {\relax (x )} - e^{3} + 4 \right )} + 4}{\log {\relax (5 )}}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

3*ln(5)*exp(3/x)-x**2*ln(5))*ln(x)+(3*exp(3)+3*x-12)*ln(5)*exp(3/x)+(x**2*exp(3)+x**3-4*x**2)*ln(5))/(x**2*ln(

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

[Out]

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

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3.86.100 \(\int \frac {e^{3/x} (-12+3 e^3+3 x) \log (5)+(-4 x^2+e^3 x^2+x^3) \log (5)+(-3 e^{3/x} \log (5)-x^2 \log (5)) \log (x)+e^{\frac {4+x+\log (5) \log (4-e^3-x+\log (x))}{\log (5)}} (4 x^2-e^3 x^2-x^3+(x-x^2) \log (5)+x^2 \log (x))}{(4 x^2-e^3 x^2-x^3) \log (5)+x^2 \log (5) \log (x)} \, dx\)