3.19.30 \(\int \frac {e^3 x^2-5 e^8 x^3+e^5 (-e^6+10 e^{11} x-25 e^{16} x^2)+(-e^3 x^2+e^5 (-e^6+10 e^{11} x-25 e^{16} x^2)) \log (x)}{(-e^3 x^3+5 e^8 x^4+e^5 (e^6 x-10 e^{11} x^2+25 e^{16} x^3)) \log (x)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (\frac {\frac {e^5}{x}+\frac {x}{-e^3+5 e^8 x}}{\log (x)}\right ) \]

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Rubi [A]  time = 2.49, antiderivative size = 37, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 7, integrand size = 126, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6741, 12, 6742, 6728, 628, 2302, 29} \begin {gather*} \log \left (-x^2-5 e^{13} x+e^8\right )-\log (x)-\log \left (1-5 e^5 x\right )-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*x^2 - 5*E^8*x^3 + E^5*(-E^6 + 10*E^11*x - 25*E^16*x^2) + (-(E^3*x^2) + E^5*(-E^6 + 10*E^11*x - 25*E^1
6*x^2))*Log[x])/((-(E^3*x^3) + 5*E^8*x^4 + E^5*(E^6*x - 10*E^11*x^2 + 25*E^16*x^3))*Log[x]),x]

[Out]

-Log[x] - Log[1 - 5*E^5*x] + Log[E^8 - 5*E^13*x - x^2] - Log[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 29

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 6741

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

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 \frac {e^3 \left (-e^8+10 e^{13} x+\left (1-25 e^{18}\right ) x^2-5 e^5 x^3-e^8 \log (x)+10 e^{13} x \log (x)-\left (1+25 e^{18}\right ) x^2 \log (x)\right )}{x \left (e^{11}-10 e^{16} x+e^3 \left (-1+25 e^{18}\right ) x^2+5 e^8 x^3\right ) \log (x)} \, dx\\ &=e^3 \int \frac {-e^8+10 e^{13} x+\left (1-25 e^{18}\right ) x^2-5 e^5 x^3-e^8 \log (x)+10 e^{13} x \log (x)-\left (1+25 e^{18}\right ) x^2 \log (x)}{x \left (e^{11}-10 e^{16} x+e^3 \left (-1+25 e^{18}\right ) x^2+5 e^8 x^3\right ) \log (x)} \, dx\\ &=e^3 \int \frac {-e^8+10 e^{13} x+\left (1-25 e^{18}\right ) x^2-5 e^5 x^3-e^8 \log (x)+10 e^{13} x \log (x)-\left (1+25 e^{18}\right ) x^2 \log (x)}{x \left (e^{11}-10 e^{16} x-e^3 \left (1-25 e^{18}\right ) x^2+5 e^8 x^3\right ) \log (x)} \, dx\\ &=e^3 \int \left (\frac {-e^8+10 e^{13} x-\left (1+25 e^{18}\right ) x^2}{e^3 x \left (1-5 e^5 x\right ) \left (e^8-5 e^{13} x-x^2\right )}-\frac {1}{e^3 x \log (x)}\right ) \, dx\\ &=\int \frac {-e^8+10 e^{13} x-\left (1+25 e^{18}\right ) x^2}{x \left (1-5 e^5 x\right ) \left (e^8-5 e^{13} x-x^2\right )} \, dx-\int \frac {1}{x \log (x)} \, dx\\ &=\int \left (-\frac {1}{x}-\frac {5 e^5}{-1+5 e^5 x}+\frac {5 e^{13}+2 x}{-e^8+5 e^{13} x+x^2}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-\log (x)-\log \left (1-5 e^5 x\right )-\log (\log (x))+\int \frac {5 e^{13}+2 x}{-e^8+5 e^{13} x+x^2} \, dx\\ &=-\log (x)-\log \left (1-5 e^5 x\right )+\log \left (e^8-5 e^{13} x-x^2\right )-\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 37, normalized size = 1.23 \begin {gather*} -\log (x)-\log \left (1-5 e^5 x\right )+\log \left (-e^8+5 e^{13} x+x^2\right )-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*x^2 - 5*E^8*x^3 + E^5*(-E^6 + 10*E^11*x - 25*E^16*x^2) + (-(E^3*x^2) + E^5*(-E^6 + 10*E^11*x -
25*E^16*x^2))*Log[x])/((-(E^3*x^3) + 5*E^8*x^4 + E^5*(E^6*x - 10*E^11*x^2 + 25*E^16*x^3))*Log[x]),x]

[Out]

-Log[x] - Log[1 - 5*E^5*x] + Log[-E^8 + 5*E^13*x + x^2] - Log[Log[x]]

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fricas [A]  time = 0.66, size = 34, normalized size = 1.13 \begin {gather*} -\log \left (5 \, x^{2} e^{5} - x\right ) + \log \left (x^{2} + 5 \, x e^{13} - e^{8}\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*exp(3))*log(x)+(-25*x^2*exp(4)^4+10*x*
exp(3)*exp(4)^2-exp(3)^2)*exp(5)-5*x^3*exp(4)^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^
2)*exp(5)+5*x^4*exp(4)^2-x^3*exp(3))/log(x),x, algorithm="fricas")

[Out]

-log(5*x^2*e^5 - x) + log(x^2 + 5*x*e^13 - e^8) - log(log(x))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*exp(3))*log(x)+(-25*x^2*exp(4)^4+10*x*
exp(3)*exp(4)^2-exp(3)^2)*exp(5)-5*x^3*exp(4)^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^
2)*exp(5)+5*x^4*exp(4)^2-x^3*exp(3))/log(x),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.16, size = 33, normalized size = 1.10

method	result	size
risch	\(-\ln \left (-5 x^{2} {\mathrm e}^{5}+x \right )+\ln \left (-5 \,{\mathrm e}^{13} x +{\mathrm e}^{8}-x^{2}\right )-\ln \left (\ln \relax (x )\right )\)	\(33\)
norman	\(-\ln \relax (x )-\ln \left (\ln \relax (x )\right )-\ln \left (5 x \,{\mathrm e}^{8}-{\mathrm e}^{3}\right )+\ln \left (5 \,{\mathrm e}^{5} {\mathrm e}^{8} x -{\mathrm e}^{3} {\mathrm e}^{5}+x^{2}\right )\)	\(46\)
default	\(-\ln \left (\ln \relax (x )\right )-\left (\munderset {\textit {\_R} =\RootOf \left (5 \textit {\_Z}^{3} {\mathrm e}^{8}+\left (25 \,{\mathrm e}^{21}-{\mathrm e}^{3}\right ) \textit {\_Z}^{2}-10 \textit {\_Z} \,{\mathrm e}^{16}+{\mathrm e}^{11}\right )}{\sum }\frac {\textit {\_R} \left (-5 \textit {\_R} \,{\mathrm e}^{8}+2 \,{\mathrm e}^{3}\right ) \ln \left (x -\textit {\_R} \right )}{50 \textit {\_R} \,{\mathrm e}^{21}-10 \,{\mathrm e}^{16}+15 \textit {\_R}^{2} {\mathrm e}^{8}-2 \textit {\_R} \,{\mathrm e}^{3}}\right )-\ln \relax (x )\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*exp(3))*ln(x)+(-25*x^2*exp(4)^4+10*x*exp(3)*
exp(4)^2-exp(3)^2)*exp(5)-5*x^3*exp(4)^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^2)*exp(
5)+5*x^4*exp(4)^2-x^3*exp(3))/ln(x),x,method=_RETURNVERBOSE)

[Out]

-ln(-5*x^2*exp(5)+x)+ln(-5*exp(13)*x+exp(8)-x^2)-ln(ln(x))

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maxima [A]  time = 0.80, size = 34, normalized size = 1.13 \begin {gather*} \log \left (x^{2} + 5 \, x e^{13} - e^{8}\right ) - \log \left (5 \, x e^{5} - 1\right ) - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-25*x^2*exp(4)^4+10*x*exp(3)*exp(4)^2-exp(3)^2)*exp(5)-x^2*exp(3))*log(x)+(-25*x^2*exp(4)^4+10*x*
exp(3)*exp(4)^2-exp(3)^2)*exp(5)-5*x^3*exp(4)^2+x^2*exp(3))/((25*x^3*exp(4)^4-10*x^2*exp(3)*exp(4)^2+x*exp(3)^
2)*exp(5)+5*x^4*exp(4)^2-x^3*exp(3))/log(x),x, algorithm="maxima")

[Out]

log(x^2 + 5*x*e^13 - e^8) - log(5*x*e^5 - 1) - log(x) - log(log(x))

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mupad [B]  time = 2.38, size = 85, normalized size = 2.83 \begin {gather*} -\ln \left (\ln \relax (x)\right )-\mathrm {atan}\left (\frac {{\mathrm {e}}^8\,1{}\mathrm {i}-x\,{\mathrm {e}}^{13}\,25{}\mathrm {i}-x\,{\mathrm {e}}^{31}\,125{}\mathrm {i}+x^2\,{\mathrm {e}}^{18}\,100{}\mathrm {i}+x^2\,{\mathrm {e}}^{36}\,625{}\mathrm {i}-x^2\,1{}\mathrm {i}}{{\mathrm {e}}^8+15\,x\,{\mathrm {e}}^{13}+125\,x\,{\mathrm {e}}^{31}-100\,x^2\,{\mathrm {e}}^{18}-625\,x^2\,{\mathrm {e}}^{36}-x^2}\right )\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5)*(exp(6) - 10*x*exp(11) + 25*x^2*exp(16)) - x^2*exp(3) + 5*x^3*exp(8) + log(x)*(exp(5)*(exp(6) - 1
0*x*exp(11) + 25*x^2*exp(16)) + x^2*exp(3)))/(log(x)*(exp(5)*(x*exp(6) - 10*x^2*exp(11) + 25*x^3*exp(16)) - x^
3*exp(3) + 5*x^4*exp(8))),x)

[Out]

- log(log(x)) - atan((exp(8)*1i - x*exp(13)*25i - x*exp(31)*125i + x^2*exp(18)*100i + x^2*exp(36)*625i - x^2*1
i)/(exp(8) + 15*x*exp(13) + 125*x*exp(31) - 100*x^2*exp(18) - 625*x^2*exp(36) - x^2))*2i

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sympy [A]  time = 1.50, size = 31, normalized size = 1.03 \begin {gather*} - \log {\left (x^{2} - \frac {x}{5 e^{5}} \right )} + \log {\left (x^{2} + 5 x e^{13} - e^{8} \right )} - \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-25*x**2*exp(4)**4+10*x*exp(3)*exp(4)**2-exp(3)**2)*exp(5)-x**2*exp(3))*ln(x)+(-25*x**2*exp(4)**4
+10*x*exp(3)*exp(4)**2-exp(3)**2)*exp(5)-5*x**3*exp(4)**2+x**2*exp(3))/((25*x**3*exp(4)**4-10*x**2*exp(3)*exp(
4)**2+x*exp(3)**2)*exp(5)+5*x**4*exp(4)**2-x**3*exp(3))/ln(x),x)

[Out]

-log(x**2 - x*exp(-5)/5) + log(x**2 + 5*x*exp(13) - exp(8)) - log(log(x))

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