3.70.96 \(\int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} (729+486 \log (5)+81 \log ^2(5))+e^x (108 x+90 x \log (5)+18 x \log ^2(5))+e^x (e (-27 x+27 x^2)+e (-9 x+9 x^2) \log (5)) \log (x)}{(4 x^3+4 x^3 \log (5)+x^3 \log ^2(5)+e^{2 x} (729 x+486 x \log (5)+81 x \log ^2(5))+e^x (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5))) \log (x)} \, dx\)

Optimal. Leaf size=25 \[ -\frac {e}{2+\log (5)+\frac {9 e^x (3+\log (5))}{x}}+\log (\log (x)) \]

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Rubi [A]  time = 0.88, antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 10, number of rules used = 6, integrand size = 166, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6, 6688, 6711, 32, 2302, 29} \begin {gather*} \log (\log (x))+\frac {9 e (3+\log (5))}{(2+\log (5)) \left (e^{-x} x (2+\log (5))+9 (3+\log (5))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x^2 + 4*x^2*Log[5] + x^2*Log[5]^2 + E^(2*x)*(729 + 486*Log[5] + 81*Log[5]^2) + E^x*(108*x + 90*x*Log[5]
+ 18*x*Log[5]^2) + E^x*(E*(-27*x + 27*x^2) + E*(-9*x + 9*x^2)*Log[5])*Log[x])/((4*x^3 + 4*x^3*Log[5] + x^3*Lo
g[5]^2 + E^(2*x)*(729*x + 486*x*Log[5] + 81*x*Log[5]^2) + E^x*(108*x^2 + 90*x^2*Log[5] + 18*x^2*Log[5]^2))*Log
[x]),x]

[Out]

(9*E*(3 + Log[5]))/((2 + Log[5])*((x*(2 + Log[5]))/E^x + 9*(3 + Log[5]))) + Log[Log[x]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2302

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

Rule 6688

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

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (x^3 \log ^2(5)+x^3 (4+4 \log (5))+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx\\ &=\int \frac {4 x^2+4 x^2 \log (5)+x^2 \log ^2(5)+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (x^3 \left (4+4 \log (5)+\log ^2(5)\right )+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx\\ &=\int \frac {x^2 \log ^2(5)+x^2 (4+4 \log (5))+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (x^3 \left (4+4 \log (5)+\log ^2(5)\right )+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx\\ &=\int \frac {x^2 \left (4+4 \log (5)+\log ^2(5)\right )+e^{2 x} \left (729+486 \log (5)+81 \log ^2(5)\right )+e^x \left (108 x+90 x \log (5)+18 x \log ^2(5)\right )+e^x \left (e \left (-27 x+27 x^2\right )+e \left (-9 x+9 x^2\right ) \log (5)\right ) \log (x)}{\left (x^3 \left (4+4 \log (5)+\log ^2(5)\right )+e^{2 x} \left (729 x+486 x \log (5)+81 x \log ^2(5)\right )+e^x \left (108 x^2+90 x^2 \log (5)+18 x^2 \log ^2(5)\right )\right ) \log (x)} \, dx\\ &=\int \left (\frac {9 e^{1+x} (-1+x) (3+\log (5))}{\left (x (2+\log (5))+9 e^x (3+\log (5))\right )^2}+\frac {1}{x \log (x)}\right ) \, dx\\ &=(9 (3+\log (5))) \int \frac {e^{1+x} (-1+x)}{\left (x (2+\log (5))+9 e^x (3+\log (5))\right )^2} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=-\left ((9 e (3+\log (5))) \operatorname {Subst}\left (\int \frac {1}{(x (2+\log (5))+9 (3+\log (5)))^2} \, dx,x,e^{-x} x\right )\right )+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=\frac {9 e (3+\log (5))}{(2+\log (5)) \left (e^{-x} x (2+\log (5))+9 (3+\log (5))\right )}+\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 30, normalized size = 1.20 \begin {gather*} -\frac {e x}{27 e^x+2 x+9 e^x \log (5)+x \log (5)}+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + 4*x^2*Log[5] + x^2*Log[5]^2 + E^(2*x)*(729 + 486*Log[5] + 81*Log[5]^2) + E^x*(108*x + 90*x*
Log[5] + 18*x*Log[5]^2) + E^x*(E*(-27*x + 27*x^2) + E*(-9*x + 9*x^2)*Log[5])*Log[x])/((4*x^3 + 4*x^3*Log[5] +
x^3*Log[5]^2 + E^(2*x)*(729*x + 486*x*Log[5] + 81*x*Log[5]^2) + E^x*(108*x^2 + 90*x^2*Log[5] + 18*x^2*Log[5]^2
))*Log[x]),x]

[Out]

-((E*x)/(27*E^x + 2*x + 9*E^x*Log[5] + x*Log[5])) + Log[Log[x]]

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fricas [A]  time = 1.14, size = 46, normalized size = 1.84 \begin {gather*} -\frac {x e - {\left (9 \, {\left (\log \relax (5) + 3\right )} e^{x} + x \log \relax (5) + 2 \, x\right )} \log \left (\log \relax (x)\right )}{9 \, {\left (\log \relax (5) + 3\right )} e^{x} + x \log \relax (5) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2-9*x)*exp(1)*log(5)+(27*x^2-27*x)*exp(1))*exp(x)*log(x)+(81*log(5)^2+486*log(5)+729)*exp(x)^
2+(18*x*log(5)^2+90*x*log(5)+108*x)*exp(x)+x^2*log(5)^2+4*x^2*log(5)+4*x^2)/((81*x*log(5)^2+486*x*log(5)+729*x
)*exp(x)^2+(18*x^2*log(5)^2+90*x^2*log(5)+108*x^2)*exp(x)+x^3*log(5)^2+4*x^3*log(5)+4*x^3)/log(x),x, algorithm
="fricas")

[Out]

-(x*e - (9*(log(5) + 3)*e^x + x*log(5) + 2*x)*log(log(x)))/(9*(log(5) + 3)*e^x + x*log(5) + 2*x)

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giac [B]  time = 0.27, size = 56, normalized size = 2.24 \begin {gather*} \frac {x \log \relax (5) \log \left (\log \relax (x)\right ) + 9 \, e^{x} \log \relax (5) \log \left (\log \relax (x)\right ) - x e + 2 \, x \log \left (\log \relax (x)\right ) + 27 \, e^{x} \log \left (\log \relax (x)\right )}{x \log \relax (5) + 9 \, e^{x} \log \relax (5) + 2 \, x + 27 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2-9*x)*exp(1)*log(5)+(27*x^2-27*x)*exp(1))*exp(x)*log(x)+(81*log(5)^2+486*log(5)+729)*exp(x)^
2+(18*x*log(5)^2+90*x*log(5)+108*x)*exp(x)+x^2*log(5)^2+4*x^2*log(5)+4*x^2)/((81*x*log(5)^2+486*x*log(5)+729*x
)*exp(x)^2+(18*x^2*log(5)^2+90*x^2*log(5)+108*x^2)*exp(x)+x^3*log(5)^2+4*x^3*log(5)+4*x^3)/log(x),x, algorithm
="giac")

[Out]

(x*log(5)*log(log(x)) + 9*e^x*log(5)*log(log(x)) - x*e + 2*x*log(log(x)) + 27*e^x*log(log(x)))/(x*log(5) + 9*e
^x*log(5) + 2*x + 27*e^x)

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maple [A]  time = 0.06, size = 30, normalized size = 1.20

method	result	size
risch	\(-\frac {x \,{\mathrm e}}{9 \,{\mathrm e}^{x} \ln \relax (5)+x \ln \relax (5)+27 \,{\mathrm e}^{x}+2 x}+\ln \left (\ln \relax (x )\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((9*x^2-9*x)*exp(1)*ln(5)+(27*x^2-27*x)*exp(1))*exp(x)*ln(x)+(81*ln(5)^2+486*ln(5)+729)*exp(x)^2+(18*x*ln
(5)^2+90*x*ln(5)+108*x)*exp(x)+x^2*ln(5)^2+4*x^2*ln(5)+4*x^2)/((81*x*ln(5)^2+486*x*ln(5)+729*x)*exp(x)^2+(18*x
^2*ln(5)^2+90*x^2*ln(5)+108*x^2)*exp(x)+x^3*ln(5)^2+4*x^3*ln(5)+4*x^3)/ln(x),x,method=_RETURNVERBOSE)

[Out]

-x*exp(1)/(9*exp(x)*ln(5)+x*ln(5)+27*exp(x)+2*x)+ln(ln(x))

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maxima [A]  time = 0.55, size = 26, normalized size = 1.04 \begin {gather*} -\frac {x e}{x {\left (\log \relax (5) + 2\right )} + 9 \, {\left (\log \relax (5) + 3\right )} e^{x}} + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2-9*x)*exp(1)*log(5)+(27*x^2-27*x)*exp(1))*exp(x)*log(x)+(81*log(5)^2+486*log(5)+729)*exp(x)^
2+(18*x*log(5)^2+90*x*log(5)+108*x)*exp(x)+x^2*log(5)^2+4*x^2*log(5)+4*x^2)/((81*x*log(5)^2+486*x*log(5)+729*x
)*exp(x)^2+(18*x^2*log(5)^2+90*x^2*log(5)+108*x^2)*exp(x)+x^3*log(5)^2+4*x^3*log(5)+4*x^3)/log(x),x, algorithm
="maxima")

[Out]

-x*e/(x*(log(5) + 2) + 9*(log(5) + 3)*e^x) + log(log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x^2\,{\ln \relax (5)}^2+{\mathrm {e}}^x\,\left (108\,x+90\,x\,\ln \relax (5)+18\,x\,{\ln \relax (5)}^2\right )+4\,x^2\,\ln \relax (5)+4\,x^2+{\mathrm {e}}^{2\,x}\,\left (486\,\ln \relax (5)+81\,{\ln \relax (5)}^2+729\right )-{\mathrm {e}}^x\,\ln \relax (x)\,\left (\mathrm {e}\,\left (27\,x-27\,x^2\right )+\mathrm {e}\,\ln \relax (5)\,\left (9\,x-9\,x^2\right )\right )}{\ln \relax (x)\,\left (x^3\,{\ln \relax (5)}^2+{\mathrm {e}}^x\,\left (18\,x^2\,{\ln \relax (5)}^2+90\,x^2\,\ln \relax (5)+108\,x^2\right )+4\,x^3\,\ln \relax (5)+{\mathrm {e}}^{2\,x}\,\left (729\,x+486\,x\,\ln \relax (5)+81\,x\,{\ln \relax (5)}^2\right )+4\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*log(5)^2 + exp(x)*(108*x + 90*x*log(5) + 18*x*log(5)^2) + 4*x^2*log(5) + 4*x^2 + exp(2*x)*(486*log(5)
+ 81*log(5)^2 + 729) - exp(x)*log(x)*(exp(1)*(27*x - 27*x^2) + exp(1)*log(5)*(9*x - 9*x^2)))/(log(x)*(x^3*log
(5)^2 + exp(x)*(18*x^2*log(5)^2 + 90*x^2*log(5) + 108*x^2) + 4*x^3*log(5) + exp(2*x)*(729*x + 486*x*log(5) + 8
1*x*log(5)^2) + 4*x^3)),x)

[Out]

int((x^2*log(5)^2 + exp(x)*(108*x + 90*x*log(5) + 18*x*log(5)^2) + 4*x^2*log(5) + 4*x^2 + exp(2*x)*(486*log(5)
+ 81*log(5)^2 + 729) - exp(x)*log(x)*(exp(1)*(27*x - 27*x^2) + exp(1)*log(5)*(9*x - 9*x^2)))/(log(x)*(x^3*log
(5)^2 + exp(x)*(18*x^2*log(5)^2 + 90*x^2*log(5) + 108*x^2) + 4*x^3*log(5) + exp(2*x)*(729*x + 486*x*log(5) + 8
1*x*log(5)^2) + 4*x^3)), x)

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sympy [A]  time = 0.47, size = 27, normalized size = 1.08 \begin {gather*} - \frac {e x}{x \log {\relax (5 )} + 2 x + \left (9 \log {\relax (5 )} + 27\right ) e^{x}} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x**2-9*x)*exp(1)*ln(5)+(27*x**2-27*x)*exp(1))*exp(x)*ln(x)+(81*ln(5)**2+486*ln(5)+729)*exp(x)**
2+(18*x*ln(5)**2+90*x*ln(5)+108*x)*exp(x)+x**2*ln(5)**2+4*x**2*ln(5)+4*x**2)/((81*x*ln(5)**2+486*x*ln(5)+729*x
)*exp(x)**2+(18*x**2*ln(5)**2+90*x**2*ln(5)+108*x**2)*exp(x)+x**3*ln(5)**2+4*x**3*ln(5)+4*x**3)/ln(x),x)

[Out]

-E*x/(x*log(5) + 2*x + (9*log(5) + 27)*exp(x)) + log(log(x))

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