3.12.1 \(\int \frac {6+e^{2 x}-125 e^9 \log ^2(4)+625 e^{18} \log ^4(4)+e^x (-5-x+50 e^9 \log ^2(4))}{9+e^{2 x}-150 e^9 \log ^2(4)+625 e^{18} \log ^4(4)+e^x (-6+50 e^9 \log ^2(4))} \, dx\)

Optimal. Leaf size=23 \[ x-\frac {x}{3-e^x-25 e^9 \log ^2(4)} \]

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Rubi [B]  time = 0.90, antiderivative size = 190, normalized size of antiderivative = 8.26, number of steps used = 17, number of rules used = 12, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6741, 6742, 2185, 2184, 2190, 2279, 2391, 2191, 2282, 36, 29, 31} \begin {gather*} -\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}+x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {\log \left (-e^x+3-25 e^9 \log ^2(4)\right )}{3-25 e^9 \log ^2(4)}+\frac {x \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {x}{-e^x+3-25 e^9 \log ^2(4)}+\frac {x}{3-25 e^9 \log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + E^(2*x) - 125*E^9*Log[4]^2 + 625*E^18*Log[4]^4 + E^x*(-5 - x + 50*E^9*Log[4]^2))/(9 + E^(2*x) - 150*E
^9*Log[4]^2 + 625*E^18*Log[4]^4 + E^x*(-6 + 50*E^9*Log[4]^2)),x]

[Out]

x + (1 - x)^2/(2*(3 - 25*E^9*Log[4]^2)) + x/(3 - 25*E^9*Log[4]^2) - x^2/(2*(3 - 25*E^9*Log[4]^2)) - x/(3 - E^x
- 25*E^9*Log[4]^2) - Log[3 - E^x - 25*E^9*Log[4]^2]/(3 - 25*E^9*Log[4]^2) + ((1 - x)*Log[1 - E^x/(3 - 25*E^9*
Log[4]^2)])/(3 - 25*E^9*Log[4]^2) + (x*Log[1 - E^x/(3 - 25*E^9*Log[4]^2)])/(3 - 25*E^9*Log[4]^2)

Rule 29

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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^{2 x}+e^x \left (-5-x+50 e^9 \log ^2(4)\right )+6 \left (1+\frac {125}{6} e^9 \log ^2(4) \left (-1+5 e^9 \log ^2(4)\right )\right )}{\left (e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )\right )^2} \, dx\\ &=\int \left (1+\frac {x \left (-3+25 e^9 \log ^2(4)\right )}{\left (e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )\right )^2}+\frac {1-x}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right ) \, dx\\ &=x+\left (-3+25 e^9 \log ^2(4)\right ) \int \frac {x}{\left (e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )\right )^2} \, dx+\int \frac {1-x}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}+\frac {\int \frac {e^x (1-x)}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx}{3-25 e^9 \log ^2(4)}-\int \frac {e^x x}{\left (e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )\right )^2} \, dx+\int \frac {x}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x}{3-e^x-25 e^9 \log ^2(4)}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}+\frac {\int \frac {e^x x}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx}{3-25 e^9 \log ^2(4)}+\frac {\int \log \left (1-\frac {e^x}{3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right ) \, dx}{3-25 e^9 \log ^2(4)}-\int \frac {1}{e^x-3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )} \, dx\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x}{3-e^x-25 e^9 \log ^2(4)}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}+\frac {x \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {\int \log \left (1-\frac {e^x}{3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right ) \, dx}{3-25 e^9 \log ^2(4)}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right )}{x} \, dx,x,e^x\right )}{3-25 e^9 \log ^2(4)}-\operatorname {Subst}\left (\int \frac {1}{x \left (-3+x+25 e^9 \log ^2(4)\right )} \, dx,x,e^x\right )\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x}{3-e^x-25 e^9 \log ^2(4)}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}+\frac {x \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {\text {Li}_2\left (\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3 \left (1-\frac {25}{3} e^9 \log ^2(4)\right )}\right )}{x} \, dx,x,e^x\right )}{3-25 e^9 \log ^2(4)}-\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{-3+25 e^9 \log ^2(4)}+\frac {\operatorname {Subst}\left (\int \frac {1}{-3+x+25 e^9 \log ^2(4)} \, dx,x,e^x\right )}{-3+25 e^9 \log ^2(4)}\\ &=x+\frac {(1-x)^2}{2 \left (3-25 e^9 \log ^2(4)\right )}+\frac {x}{3-25 e^9 \log ^2(4)}-\frac {x^2}{2 \left (3-25 e^9 \log ^2(4)\right )}-\frac {x}{3-e^x-25 e^9 \log ^2(4)}-\frac {\log \left (3-e^x-25 e^9 \log ^2(4)\right )}{3-25 e^9 \log ^2(4)}+\frac {(1-x) \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}+\frac {x \log \left (1-\frac {e^x}{3-25 e^9 \log ^2(4)}\right )}{3-25 e^9 \log ^2(4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 20, normalized size = 0.87 \begin {gather*} x+\frac {x}{-3+e^x+25 e^9 \log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + E^(2*x) - 125*E^9*Log[4]^2 + 625*E^18*Log[4]^4 + E^x*(-5 - x + 50*E^9*Log[4]^2))/(9 + E^(2*x) -
150*E^9*Log[4]^2 + 625*E^18*Log[4]^4 + E^x*(-6 + 50*E^9*Log[4]^2)),x]

[Out]

x + x/(-3 + E^x + 25*E^9*Log[4]^2)

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fricas [A]  time = 0.84, size = 32, normalized size = 1.39 \begin {gather*} \frac {100 \, x e^{9} \log \relax (2)^{2} + x e^{x} - 2 \, x}{100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2+(200*exp(9)*log(2)^2-x-5)*exp(x)+10000*exp(9)^2*log(2)^4-500*exp(9)*log(2)^2+6)/(exp(x)^2+
(200*exp(9)*log(2)^2-6)*exp(x)+10000*exp(9)^2*log(2)^4-600*exp(9)*log(2)^2+9),x, algorithm="fricas")

[Out]

(100*x*e^9*log(2)^2 + x*e^x - 2*x)/(100*e^9*log(2)^2 + e^x - 3)

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giac [B]  time = 0.55, size = 359, normalized size = 15.61 \begin {gather*} \frac {1000000 \, x e^{27} \log \relax (2)^{6} + 1000000 \, e^{27} \log \relax (2)^{6} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) - 1000000 \, e^{27} \log \relax (2)^{6} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) - 80000 \, x e^{18} \log \relax (2)^{4} + 10000 \, x e^{\left (x + 18\right )} \log \relax (2)^{4} - 30000 \, e^{18} \log \relax (2)^{4} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) + 10000 \, e^{\left (x + 18\right )} \log \relax (2)^{4} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) + 30000 \, e^{18} \log \relax (2)^{4} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) - 10000 \, e^{\left (x + 18\right )} \log \relax (2)^{4} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) + 2100 \, x e^{9} \log \relax (2)^{2} - 600 \, x e^{\left (x + 9\right )} \log \relax (2)^{2} + 600 \, e^{9} \log \relax (2)^{2} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) - 600 \, e^{9} \log \relax (2)^{2} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) + 9 \, x e^{x} + 6 \, e^{x} \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) - 6 \, e^{x} \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right ) - 18 \, x - 18 \, \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) + 18 \, \log \left (-100 \, e^{9} \log \relax (2)^{2} - e^{x} + 3\right )}{1000000 \, e^{27} \log \relax (2)^{6} - 90000 \, e^{18} \log \relax (2)^{4} + 10000 \, e^{\left (x + 18\right )} \log \relax (2)^{4} + 2700 \, e^{9} \log \relax (2)^{2} - 600 \, e^{\left (x + 9\right )} \log \relax (2)^{2} + 9 \, e^{x} - 27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2+(200*exp(9)*log(2)^2-x-5)*exp(x)+10000*exp(9)^2*log(2)^4-500*exp(9)*log(2)^2+6)/(exp(x)^2+
(200*exp(9)*log(2)^2-6)*exp(x)+10000*exp(9)^2*log(2)^4-600*exp(9)*log(2)^2+9),x, algorithm="giac")

[Out]

(1000000*x*e^27*log(2)^6 + 1000000*e^27*log(2)^6*log(100*e^9*log(2)^2 + e^x - 3) - 1000000*e^27*log(2)^6*log(-
100*e^9*log(2)^2 - e^x + 3) - 80000*x*e^18*log(2)^4 + 10000*x*e^(x + 18)*log(2)^4 - 30000*e^18*log(2)^4*log(10
0*e^9*log(2)^2 + e^x - 3) + 10000*e^(x + 18)*log(2)^4*log(100*e^9*log(2)^2 + e^x - 3) + 30000*e^18*log(2)^4*lo
g(-100*e^9*log(2)^2 - e^x + 3) - 10000*e^(x + 18)*log(2)^4*log(-100*e^9*log(2)^2 - e^x + 3) + 2100*x*e^9*log(2
)^2 - 600*x*e^(x + 9)*log(2)^2 + 600*e^9*log(2)^2*log(100*e^9*log(2)^2 + e^x - 3) - 600*e^9*log(2)^2*log(-100*
e^9*log(2)^2 - e^x + 3) + 9*x*e^x + 6*e^x*log(100*e^9*log(2)^2 + e^x - 3) - 6*e^x*log(-100*e^9*log(2)^2 - e^x
+ 3) - 18*x - 18*log(100*e^9*log(2)^2 + e^x - 3) + 18*log(-100*e^9*log(2)^2 - e^x + 3))/(1000000*e^27*log(2)^6
- 90000*e^18*log(2)^4 + 10000*e^(x + 18)*log(2)^4 + 2700*e^9*log(2)^2 - 600*e^(x + 9)*log(2)^2 + 9*e^x - 27)

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maple [A]  time = 0.25, size = 19, normalized size = 0.83

method	result	size
risch	\(x +\frac {x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}\)	\(19\)
norman	\(\frac {\left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-2\right ) x +{\mathrm e}^{x} x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}\)	\(33\)
default	\(-\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}+\frac {2}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}+\ln \left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3\right )+\frac {\frac {6}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}+\frac {6 x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}+\frac {6 x \,{\mathrm e}^{x}}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}-\frac {6 \ln \left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3\right )}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}+\frac {-\frac {500 \,{\mathrm e}^{9} \ln \relax (2)^{2}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}-\frac {500 \,{\mathrm e}^{9} \ln \relax (2)^{2} x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}-\frac {500 \,{\mathrm e}^{9} \ln \relax (2)^{2} x \,{\mathrm e}^{x}}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}+\frac {500 \,{\mathrm e}^{9} \ln \relax (2)^{2} \ln \left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3\right )}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}+\frac {\frac {10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}+\frac {10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4} x}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-3}+\frac {10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4} x \,{\mathrm e}^{x}}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3}-\frac {10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4} \ln \left (100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+{\mathrm e}^{x}-3\right )}{10000 \left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-600 \,{\mathrm e}^{9} \ln \relax (2)^{2}+9}-\frac {x \left (\ln \left (\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}+{\mathrm e}^{x}-3}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}-3}\right )-\ln \left (\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}+{\mathrm e}^{x}-3}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}-3}\right )\right )}{200 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}}-\frac {\dilog \left (\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}+{\mathrm e}^{x}-3}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}-100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}-3}\right )}{200 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}}+\frac {\dilog \left (\frac {100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}+{\mathrm e}^{x}-3}{100 \,{\mathrm e}^{9} \ln \relax (2)^{2}+100 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}-3}\right )}{200 \sqrt {\left ({\mathrm e}^{9}\right )^{2} \ln \relax (2)^{4}-{\mathrm e}^{18} \ln \relax (2)^{4}}}\)	\(786\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)^2+(200*exp(9)*ln(2)^2-x-5)*exp(x)+10000*exp(9)^2*ln(2)^4-500*exp(9)*ln(2)^2+6)/(exp(x)^2+(200*exp(
9)*ln(2)^2-6)*exp(x)+10000*exp(9)^2*ln(2)^4-600*exp(9)*ln(2)^2+9),x,method=_RETURNVERBOSE)

[Out]

x+x/(100*exp(9)*ln(2)^2+exp(x)-3)

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maxima [B]  time = 0.81, size = 437, normalized size = 19.00 \begin {gather*} 10000 \, {\left (\frac {x}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} - \frac {\log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right )}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} + \frac {1}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + {\left (100 \, e^{9} \log \relax (2)^{2} - 3\right )} e^{x} + 9}\right )} e^{18} \log \relax (2)^{4} - 500 \, {\left (\frac {x}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} - \frac {\log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right )}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} + \frac {1}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + {\left (100 \, e^{9} \log \relax (2)^{2} - 3\right )} e^{x} + 9}\right )} e^{9} \log \relax (2)^{2} - \frac {200 \, e^{18} \log \relax (2)^{2}}{100 \, e^{18} \log \relax (2)^{2} - 3 \, e^{9} + e^{\left (x + 9\right )}} - \frac {x e^{x}}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + {\left (100 \, e^{9} \log \relax (2)^{2} - 3\right )} e^{x} + 9} + \frac {100 \, e^{9} \log \relax (2)^{2} - 3}{100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3} + \frac {6 \, x}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} - \frac {6 \, \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right )}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + 9} + \frac {\log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right )}{100 \, e^{9} \log \relax (2)^{2} - 3} + \frac {6}{10000 \, e^{18} \log \relax (2)^{4} - 600 \, e^{9} \log \relax (2)^{2} + {\left (100 \, e^{9} \log \relax (2)^{2} - 3\right )} e^{x} + 9} + \frac {5}{100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3} + \log \left (100 \, e^{9} \log \relax (2)^{2} + e^{x} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2+(200*exp(9)*log(2)^2-x-5)*exp(x)+10000*exp(9)^2*log(2)^4-500*exp(9)*log(2)^2+6)/(exp(x)^2+
(200*exp(9)*log(2)^2-6)*exp(x)+10000*exp(9)^2*log(2)^4-600*exp(9)*log(2)^2+9),x, algorithm="maxima")

[Out]

10000*(x/(10000*e^18*log(2)^4 - 600*e^9*log(2)^2 + 9) - log(100*e^9*log(2)^2 + e^x - 3)/(10000*e^18*log(2)^4 -
600*e^9*log(2)^2 + 9) + 1/(10000*e^18*log(2)^4 - 600*e^9*log(2)^2 + (100*e^9*log(2)^2 - 3)*e^x + 9))*e^18*log
(2)^4 - 500*(x/(10000*e^18*log(2)^4 - 600*e^9*log(2)^2 + 9) - log(100*e^9*log(2)^2 + e^x - 3)/(10000*e^18*log(
2)^4 - 600*e^9*log(2)^2 + 9) + 1/(10000*e^18*log(2)^4 - 600*e^9*log(2)^2 + (100*e^9*log(2)^2 - 3)*e^x + 9))*e^
9*log(2)^2 - 200*e^18*log(2)^2/(100*e^18*log(2)^2 - 3*e^9 + e^(x + 9)) - x*e^x/(10000*e^18*log(2)^4 - 600*e^9*
log(2)^2 + (100*e^9*log(2)^2 - 3)*e^x + 9) + (100*e^9*log(2)^2 - 3)/(100*e^9*log(2)^2 + e^x - 3) + 6*x/(10000*
e^18*log(2)^4 - 600*e^9*log(2)^2 + 9) - 6*log(100*e^9*log(2)^2 + e^x - 3)/(10000*e^18*log(2)^4 - 600*e^9*log(2
)^2 + 9) + log(100*e^9*log(2)^2 + e^x - 3)/(100*e^9*log(2)^2 - 3) + 6/(10000*e^18*log(2)^4 - 600*e^9*log(2)^2
+ (100*e^9*log(2)^2 - 3)*e^x + 9) + 5/(100*e^9*log(2)^2 + e^x - 3) + log(100*e^9*log(2)^2 + e^x - 3)

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mupad [B]  time = 0.23, size = 18, normalized size = 0.78 \begin {gather*} x+\frac {x}{{\mathrm {e}}^x+100\,{\mathrm {e}}^9\,{\ln \relax (2)}^2-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x) - 500*exp(9)*log(2)^2 + 10000*exp(18)*log(2)^4 - exp(x)*(x - 200*exp(9)*log(2)^2 + 5) + 6)/(exp(
2*x) + exp(x)*(200*exp(9)*log(2)^2 - 6) - 600*exp(9)*log(2)^2 + 10000*exp(18)*log(2)^4 + 9),x)

[Out]

x + x/(exp(x) + 100*exp(9)*log(2)^2 - 3)

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sympy [A]  time = 0.12, size = 17, normalized size = 0.74 \begin {gather*} x + \frac {x}{e^{x} - 3 + 100 e^{9} \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)**2+(200*exp(9)*ln(2)**2-x-5)*exp(x)+10000*exp(9)**2*ln(2)**4-500*exp(9)*ln(2)**2+6)/(exp(x)*
*2+(200*exp(9)*ln(2)**2-6)*exp(x)+10000*exp(9)**2*ln(2)**4-600*exp(9)*ln(2)**2+9),x)

[Out]

x + x/(exp(x) - 3 + 100*exp(9)*log(2)**2)

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