3.31.47 \(\int \frac {792+e^{3 x} (-4+x)+177 x-150 x^2+e^{2 x} (-36+9 x)+e^x (-408-323 x+50 x^3)}{-81+e^{3 x} (-3+x)+27 x+e^{2 x} (-27+9 x)+e^x (-81+27 x)} \, dx\)

Optimal. Leaf size=27 \[ 1+x-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\log (3 (-4+x)+x) \]

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Rubi [C]  time = 1.28, antiderivative size = 246, normalized size of antiderivative = 9.11, number of steps used = 54, number of rules used = 16, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.198, Rules used = {6688, 6742, 43, 2185, 2184, 2190, 2531, 2282, 6589, 2191, 2279, 2391, 36, 29, 31, 44} \begin {gather*} -\frac {100}{9} x \text {Li}_2\left (-\frac {e^x}{3}\right )+\frac {100}{9} (x+2) \text {Li}_2\left (-\frac {e^x}{3}\right )-\frac {200}{9} \text {Li}_2\left (-\frac {e^x}{3}\right )+\frac {50 x^3}{27}+\frac {50 x^2}{3 \left (e^x+3\right )}+\frac {25 x^2}{9}-\frac {50}{9} x^2 \log \left (\frac {e^x}{3}+1\right )+\frac {50 x}{e^x+3}-\frac {91 x}{9}-\frac {50}{27} (x+2)^3-\frac {50 (x+2)^2}{3 \left (e^x+3\right )}-\frac {25 (x+2)^2}{\left (e^x+3\right )^2}+\frac {25}{3} (x+2)^2+\frac {50 (x+2)}{3 \left (e^x+3\right )}+\frac {100}{3 \left (e^x+3\right )}-\frac {50}{9} x \log \left (\frac {e^x}{3}+1\right )+\frac {50}{9} (x+2)^2 \log \left (\frac {e^x}{3}+1\right )-\frac {50}{3} (x+2) \log \left (\frac {e^x}{3}+1\right )+\frac {100}{9} \log \left (e^x+3\right )-\log (3-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(792 + E^(3*x)*(-4 + x) + 177*x - 150*x^2 + E^(2*x)*(-36 + 9*x) + E^x*(-408 - 323*x + 50*x^3))/(-81 + E^(3
*x)*(-3 + x) + 27*x + E^(2*x)*(-27 + 9*x) + E^x*(-81 + 27*x)),x]

[Out]

100/(3*(3 + E^x)) - (91*x)/9 + (50*x)/(3 + E^x) + (25*x^2)/9 + (50*x^2)/(3*(3 + E^x)) + (50*x^3)/27 + (50*(2 +
x))/(3*(3 + E^x)) + (25*(2 + x)^2)/3 - (25*(2 + x)^2)/(3 + E^x)^2 - (50*(2 + x)^2)/(3*(3 + E^x)) - (50*(2 + x
)^3)/27 - (50*x*Log[1 + E^x/3])/9 - (50*x^2*Log[1 + E^x/3])/9 - (50*(2 + x)*Log[1 + E^x/3])/3 + (50*(2 + x)^2*
Log[1 + E^x/3])/9 + (100*Log[3 + E^x])/9 - Log[3 - x] - (200*PolyLog[2, -1/3*E^x])/9 - (100*x*PolyLog[2, -1/3*
E^x])/9 + (100*(2 + x)*PolyLog[2, -1/3*E^x])/9

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6688

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

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 {-792-9 e^{2 x} (-4+x)-e^{3 x} (-4+x)-177 x+150 x^2-e^x \left (-408-323 x+50 x^3\right )}{\left (3+e^x\right )^3 (3-x)} \, dx\\ &=\int \left (\frac {-4+x}{-3+x}-\frac {150 (2+x)^2}{\left (3+e^x\right )^3}+\frac {50 \left (2+3 x+x^2\right )}{\left (3+e^x\right )^2}\right ) \, dx\\ &=50 \int \frac {2+3 x+x^2}{\left (3+e^x\right )^2} \, dx-150 \int \frac {(2+x)^2}{\left (3+e^x\right )^3} \, dx+\int \frac {-4+x}{-3+x} \, dx\\ &=50 \int \frac {e^x (2+x)^2}{\left (3+e^x\right )^3} \, dx-50 \int \frac {(2+x)^2}{\left (3+e^x\right )^2} \, dx+50 \int \left (\frac {2}{\left (3+e^x\right )^2}+\frac {3 x}{\left (3+e^x\right )^2}+\frac {x^2}{\left (3+e^x\right )^2}\right ) \, dx+\int \left (1+\frac {1}{3-x}\right ) \, dx\\ &=x-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\log (3-x)+\frac {50}{3} \int \frac {e^x (2+x)^2}{\left (3+e^x\right )^2} \, dx-\frac {50}{3} \int \frac {(2+x)^2}{3+e^x} \, dx+50 \int \frac {x^2}{\left (3+e^x\right )^2} \, dx+50 \int \frac {2+x}{\left (3+e^x\right )^2} \, dx+100 \int \frac {1}{\left (3+e^x\right )^2} \, dx+150 \int \frac {x}{\left (3+e^x\right )^2} \, dx\\ &=x-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\frac {50 (2+x)^2}{3 \left (3+e^x\right )}-\frac {50}{27} (2+x)^3-\log (3-x)+\frac {50}{9} \int \frac {e^x (2+x)^2}{3+e^x} \, dx-\frac {50}{3} \int \frac {e^x x^2}{\left (3+e^x\right )^2} \, dx+\frac {50}{3} \int \frac {x^2}{3+e^x} \, dx-\frac {50}{3} \int \frac {e^x (2+x)}{\left (3+e^x\right )^2} \, dx+\frac {50}{3} \int \frac {2+x}{3+e^x} \, dx+\frac {100}{3} \int \frac {2+x}{3+e^x} \, dx-50 \int \frac {e^x x}{\left (3+e^x\right )^2} \, dx+50 \int \frac {x}{3+e^x} \, dx+100 \operatorname {Subst}\left (\int \frac {1}{x (3+x)^2} \, dx,x,e^x\right )\\ &=x+\frac {50 x}{3+e^x}+\frac {25 x^2}{3}+\frac {50 x^2}{3 \left (3+e^x\right )}+\frac {50 x^3}{27}+\frac {50 (2+x)}{3 \left (3+e^x\right )}+\frac {25}{3} (2+x)^2-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\frac {50 (2+x)^2}{3 \left (3+e^x\right )}-\frac {50}{27} (2+x)^3+\frac {50}{9} (2+x)^2 \log \left (1+\frac {e^x}{3}\right )-\log (3-x)-\frac {50}{9} \int \frac {e^x x^2}{3+e^x} \, dx-\frac {50}{9} \int \frac {e^x (2+x)}{3+e^x} \, dx-\frac {100}{9} \int \frac {e^x (2+x)}{3+e^x} \, dx-\frac {100}{9} \int (2+x) \log \left (1+\frac {e^x}{3}\right ) \, dx-\frac {50}{3} \int \frac {1}{3+e^x} \, dx-\frac {50}{3} \int \frac {e^x x}{3+e^x} \, dx-\frac {100}{3} \int \frac {x}{3+e^x} \, dx-50 \int \frac {1}{3+e^x} \, dx+100 \operatorname {Subst}\left (\int \left (\frac {1}{9 x}-\frac {1}{3 (3+x)^2}-\frac {1}{9 (3+x)}\right ) \, dx,x,e^x\right )\\ &=\frac {100}{3 \left (3+e^x\right )}+\frac {109 x}{9}+\frac {50 x}{3+e^x}+\frac {25 x^2}{9}+\frac {50 x^2}{3 \left (3+e^x\right )}+\frac {50 x^3}{27}+\frac {50 (2+x)}{3 \left (3+e^x\right )}+\frac {25}{3} (2+x)^2-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\frac {50 (2+x)^2}{3 \left (3+e^x\right )}-\frac {50}{27} (2+x)^3-\frac {50}{3} x \log \left (1+\frac {e^x}{3}\right )-\frac {50}{9} x^2 \log \left (1+\frac {e^x}{3}\right )-\frac {50}{3} (2+x) \log \left (1+\frac {e^x}{3}\right )+\frac {50}{9} (2+x)^2 \log \left (1+\frac {e^x}{3}\right )-\frac {100}{9} \log \left (3+e^x\right )-\log (3-x)+\frac {100}{9} (2+x) \text {Li}_2\left (-\frac {e^x}{3}\right )+\frac {50}{9} \int \log \left (1+\frac {e^x}{3}\right ) \, dx+\frac {100}{9} \int \frac {e^x x}{3+e^x} \, dx+\frac {100}{9} \int \log \left (1+\frac {e^x}{3}\right ) \, dx+\frac {100}{9} \int x \log \left (1+\frac {e^x}{3}\right ) \, dx-\frac {100}{9} \int \text {Li}_2\left (-\frac {e^x}{3}\right ) \, dx+\frac {50}{3} \int \log \left (1+\frac {e^x}{3}\right ) \, dx-\frac {50}{3} \operatorname {Subst}\left (\int \frac {1}{x (3+x)} \, dx,x,e^x\right )-50 \operatorname {Subst}\left (\int \frac {1}{x (3+x)} \, dx,x,e^x\right )\\ &=\frac {100}{3 \left (3+e^x\right )}+\frac {109 x}{9}+\frac {50 x}{3+e^x}+\frac {25 x^2}{9}+\frac {50 x^2}{3 \left (3+e^x\right )}+\frac {50 x^3}{27}+\frac {50 (2+x)}{3 \left (3+e^x\right )}+\frac {25}{3} (2+x)^2-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\frac {50 (2+x)^2}{3 \left (3+e^x\right )}-\frac {50}{27} (2+x)^3-\frac {50}{9} x \log \left (1+\frac {e^x}{3}\right )-\frac {50}{9} x^2 \log \left (1+\frac {e^x}{3}\right )-\frac {50}{3} (2+x) \log \left (1+\frac {e^x}{3}\right )+\frac {50}{9} (2+x)^2 \log \left (1+\frac {e^x}{3}\right )-\frac {100}{9} \log \left (3+e^x\right )-\log (3-x)-\frac {100}{9} x \text {Li}_2\left (-\frac {e^x}{3}\right )+\frac {100}{9} (2+x) \text {Li}_2\left (-\frac {e^x}{3}\right )-\frac {50}{9} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {50}{9} \operatorname {Subst}\left (\int \frac {1}{3+x} \, dx,x,e^x\right )+\frac {50}{9} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{3}\right )}{x} \, dx,x,e^x\right )-\frac {100}{9} \int \log \left (1+\frac {e^x}{3}\right ) \, dx+\frac {100}{9} \int \text {Li}_2\left (-\frac {e^x}{3}\right ) \, dx+\frac {100}{9} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{3}\right )}{x} \, dx,x,e^x\right )-\frac {100}{9} \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{3}\right )}{x} \, dx,x,e^x\right )-\frac {50}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {50}{3} \operatorname {Subst}\left (\int \frac {1}{3+x} \, dx,x,e^x\right )+\frac {50}{3} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{3}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {100}{3 \left (3+e^x\right )}-\frac {91 x}{9}+\frac {50 x}{3+e^x}+\frac {25 x^2}{9}+\frac {50 x^2}{3 \left (3+e^x\right )}+\frac {50 x^3}{27}+\frac {50 (2+x)}{3 \left (3+e^x\right )}+\frac {25}{3} (2+x)^2-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\frac {50 (2+x)^2}{3 \left (3+e^x\right )}-\frac {50}{27} (2+x)^3-\frac {50}{9} x \log \left (1+\frac {e^x}{3}\right )-\frac {50}{9} x^2 \log \left (1+\frac {e^x}{3}\right )-\frac {50}{3} (2+x) \log \left (1+\frac {e^x}{3}\right )+\frac {50}{9} (2+x)^2 \log \left (1+\frac {e^x}{3}\right )+\frac {100}{9} \log \left (3+e^x\right )-\log (3-x)-\frac {100}{3} \text {Li}_2\left (-\frac {e^x}{3}\right )-\frac {100}{9} x \text {Li}_2\left (-\frac {e^x}{3}\right )+\frac {100}{9} (2+x) \text {Li}_2\left (-\frac {e^x}{3}\right )-\frac {100}{9} \text {Li}_3\left (-\frac {e^x}{3}\right )-\frac {100}{9} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{3}\right )}{x} \, dx,x,e^x\right )+\frac {100}{9} \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {x}{3}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {100}{3 \left (3+e^x\right )}-\frac {91 x}{9}+\frac {50 x}{3+e^x}+\frac {25 x^2}{9}+\frac {50 x^2}{3 \left (3+e^x\right )}+\frac {50 x^3}{27}+\frac {50 (2+x)}{3 \left (3+e^x\right )}+\frac {25}{3} (2+x)^2-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\frac {50 (2+x)^2}{3 \left (3+e^x\right )}-\frac {50}{27} (2+x)^3-\frac {50}{9} x \log \left (1+\frac {e^x}{3}\right )-\frac {50}{9} x^2 \log \left (1+\frac {e^x}{3}\right )-\frac {50}{3} (2+x) \log \left (1+\frac {e^x}{3}\right )+\frac {50}{9} (2+x)^2 \log \left (1+\frac {e^x}{3}\right )+\frac {100}{9} \log \left (3+e^x\right )-\log (3-x)-\frac {200}{9} \text {Li}_2\left (-\frac {e^x}{3}\right )-\frac {100}{9} x \text {Li}_2\left (-\frac {e^x}{3}\right )+\frac {100}{9} (2+x) \text {Li}_2\left (-\frac {e^x}{3}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 24, normalized size = 0.89 \begin {gather*} x-\frac {25 (2+x)^2}{\left (3+e^x\right )^2}-\log (3-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(792 + E^(3*x)*(-4 + x) + 177*x - 150*x^2 + E^(2*x)*(-36 + 9*x) + E^x*(-408 - 323*x + 50*x^3))/(-81
+ E^(3*x)*(-3 + x) + 27*x + E^(2*x)*(-27 + 9*x) + E^x*(-81 + 27*x)),x]

[Out]

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

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fricas [B]  time = 0.60, size = 51, normalized size = 1.89 \begin {gather*} -\frac {25 \, x^{2} - x e^{\left (2 \, x\right )} - 6 \, x e^{x} + {\left (e^{\left (2 \, x\right )} + 6 \, e^{x} + 9\right )} \log \left (x - 3\right ) + 91 \, x + 100}{e^{\left (2 \, x\right )} + 6 \, e^{x} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*exp(x)^3+(9*x-36)*exp(x)^2+(50*x^3-323*x-408)*exp(x)-150*x^2+177*x+792)/((x-3)*exp(x)^3+(9*x-
27)*exp(x)^2+(27*x-81)*exp(x)+27*x-81),x, algorithm="fricas")

[Out]

-(25*x^2 - x*e^(2*x) - 6*x*e^x + (e^(2*x) + 6*e^x + 9)*log(x - 3) + 91*x + 100)/(e^(2*x) + 6*e^x + 9)

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giac [B]  time = 0.20, size = 59, normalized size = 2.19 \begin {gather*} -\frac {25 \, x^{2} - x e^{\left (2 \, x\right )} - 6 \, x e^{x} + e^{\left (2 \, x\right )} \log \left (x - 3\right ) + 6 \, e^{x} \log \left (x - 3\right ) + 91 \, x + 9 \, \log \left (x - 3\right ) + 100}{e^{\left (2 \, x\right )} + 6 \, e^{x} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*exp(x)^3+(9*x-36)*exp(x)^2+(50*x^3-323*x-408)*exp(x)-150*x^2+177*x+792)/((x-3)*exp(x)^3+(9*x-
27)*exp(x)^2+(27*x-81)*exp(x)+27*x-81),x, algorithm="giac")

[Out]

-(25*x^2 - x*e^(2*x) - 6*x*e^x + e^(2*x)*log(x - 3) + 6*e^x*log(x - 3) + 91*x + 9*log(x - 3) + 100)/(e^(2*x) +
6*e^x + 9)

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maple [A]  time = 0.07, size = 25, normalized size = 0.93

method	result	size
risch	\(x -\ln \left (x -3\right )-\frac {25 \left (x^{2}+4 x +4\right )}{\left (3+{\mathrm e}^{x}\right )^{2}}\)	\(25\)
norman	\(\frac {x \,{\mathrm e}^{2 x}-91 x -25 x^{2}-100+6 \,{\mathrm e}^{x} x}{\left (3+{\mathrm e}^{x}\right )^{2}}-\ln \left (x -3\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-4)*exp(x)^3+(9*x-36)*exp(x)^2+(50*x^3-323*x-408)*exp(x)-150*x^2+177*x+792)/((x-3)*exp(x)^3+(9*x-27)*ex
p(x)^2+(27*x-81)*exp(x)+27*x-81),x,method=_RETURNVERBOSE)

[Out]

x-ln(x-3)-25*(x^2+4*x+4)/(3+exp(x))^2

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maxima [A]  time = 0.87, size = 43, normalized size = 1.59 \begin {gather*} -\frac {25 \, x^{2} - x e^{\left (2 \, x\right )} - 6 \, x e^{x} + 91 \, x + 100}{e^{\left (2 \, x\right )} + 6 \, e^{x} + 9} - \log \left (x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*exp(x)^3+(9*x-36)*exp(x)^2+(50*x^3-323*x-408)*exp(x)-150*x^2+177*x+792)/((x-3)*exp(x)^3+(9*x-
27)*exp(x)^2+(27*x-81)*exp(x)+27*x-81),x, algorithm="maxima")

[Out]

-(25*x^2 - x*e^(2*x) - 6*x*e^x + 91*x + 100)/(e^(2*x) + 6*e^x + 9) - log(x - 3)

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mupad [B]  time = 1.86, size = 50, normalized size = 1.85 \begin {gather*} \frac {\frac {100\,{\mathrm {e}}^{2\,x}}{9}-91\,x+\frac {200\,{\mathrm {e}}^x}{3}+x\,{\mathrm {e}}^{2\,x}+6\,x\,{\mathrm {e}}^x-25\,x^2}{{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^x+9}-\ln \left (x-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((177*x + exp(3*x)*(x - 4) - exp(x)*(323*x - 50*x^3 + 408) + exp(2*x)*(9*x - 36) - 150*x^2 + 792)/(27*x + e
xp(3*x)*(x - 3) + exp(x)*(27*x - 81) + exp(2*x)*(9*x - 27) - 81),x)

[Out]

((100*exp(2*x))/9 - 91*x + (200*exp(x))/3 + x*exp(2*x) + 6*x*exp(x) - 25*x^2)/(exp(2*x) + 6*exp(x) + 9) - log(
x - 3)

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sympy [A]  time = 0.16, size = 29, normalized size = 1.07 \begin {gather*} x + \frac {- 25 x^{2} - 100 x - 100}{e^{2 x} + 6 e^{x} + 9} - \log {\left (x - 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-4)*exp(x)**3+(9*x-36)*exp(x)**2+(50*x**3-323*x-408)*exp(x)-150*x**2+177*x+792)/((x-3)*exp(x)**3+
(9*x-27)*exp(x)**2+(27*x-81)*exp(x)+27*x-81),x)

[Out]

x + (-25*x**2 - 100*x - 100)/(exp(2*x) + 6*exp(x) + 9) - log(x - 3)

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