3.71.18 \(\int \frac {30+e^{2 x}+e^x (11-x)}{25+10 e^x+e^{2 x}} \, dx\)

Optimal. Leaf size=14 \[ \frac {35}{8}+x+\frac {x}{5+e^x} \]

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Rubi [B]  time = 0.30, antiderivative size = 77, normalized size of antiderivative = 5.50, number of steps used = 17, number of rules used = 12, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {6688, 6742, 2184, 2190, 2279, 2391, 2185, 2191, 2282, 36, 29, 31} \begin {gather*} \frac {x^2}{10}-\frac {1}{10} (1-x)^2+\frac {x}{e^x+5}+\frac {4 x}{5}-\frac {1}{5} (1-x) \log \left (\frac {e^x}{5}+1\right )-\frac {1}{5} x \log \left (\frac {e^x}{5}+1\right )+\frac {1}{5} \log \left (e^x+5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(30 + E^(2*x) + E^x*(11 - x))/(25 + 10*E^x + E^(2*x)),x]

[Out]

-1/10*(1 - x)^2 + (4*x)/5 + x/(5 + E^x) + x^2/10 - ((1 - x)*Log[1 + E^x/5])/5 - (x*Log[1 + E^x/5])/5 + Log[5 +
E^x]/5

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 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 {30+e^{2 x}-e^x (-11+x)}{\left (5+e^x\right )^2} \, dx\\ &=\int \left (1-\frac {-1+x}{5+e^x}+\frac {5 x}{\left (5+e^x\right )^2}\right ) \, dx\\ &=x+5 \int \frac {x}{\left (5+e^x\right )^2} \, dx-\int \frac {-1+x}{5+e^x} \, dx\\ &=-\frac {1}{10} (1-x)^2+x+\frac {1}{5} \int \frac {e^x (-1+x)}{5+e^x} \, dx-\int \frac {e^x x}{\left (5+e^x\right )^2} \, dx+\int \frac {x}{5+e^x} \, dx\\ &=-\frac {1}{10} (1-x)^2+x+\frac {x}{5+e^x}+\frac {x^2}{10}-\frac {1}{5} (1-x) \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} \int \frac {e^x x}{5+e^x} \, dx-\frac {1}{5} \int \log \left (1+\frac {e^x}{5}\right ) \, dx-\int \frac {1}{5+e^x} \, dx\\ &=-\frac {1}{10} (1-x)^2+x+\frac {x}{5+e^x}+\frac {x^2}{10}-\frac {1}{5} (1-x) \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} x \log \left (1+\frac {e^x}{5}\right )+\frac {1}{5} \int \log \left (1+\frac {e^x}{5}\right ) \, dx-\frac {1}{5} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )-\operatorname {Subst}\left (\int \frac {1}{x (5+x)} \, dx,x,e^x\right )\\ &=-\frac {1}{10} (1-x)^2+x+\frac {x}{5+e^x}+\frac {x^2}{10}-\frac {1}{5} (1-x) \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} x \log \left (1+\frac {e^x}{5}\right )+\frac {1}{5} \text {Li}_2\left (-\frac {e^x}{5}\right )-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{5+x} \, dx,x,e^x\right )+\frac {1}{5} \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx,x,e^x\right )\\ &=-\frac {1}{10} (1-x)^2+\frac {4 x}{5}+\frac {x}{5+e^x}+\frac {x^2}{10}-\frac {1}{5} (1-x) \log \left (1+\frac {e^x}{5}\right )-\frac {1}{5} x \log \left (1+\frac {e^x}{5}\right )+\frac {1}{5} \log \left (5+e^x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 11, normalized size = 0.79 \begin {gather*} x+\frac {x}{5+e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30 + E^(2*x) + E^x*(11 - x))/(25 + 10*E^x + E^(2*x)),x]

[Out]

x + x/(5 + E^x)

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fricas [A]  time = 0.64, size = 15, normalized size = 1.07 \begin {gather*} \frac {x e^{x} + 6 \, x}{e^{x} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2+(11-x)*exp(x)+30)/(exp(x)^2+10*exp(x)+25),x, algorithm="fricas")

[Out]

(x*e^x + 6*x)/(e^x + 5)

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giac [A]  time = 0.31, size = 15, normalized size = 1.07 \begin {gather*} \frac {x e^{x} + 6 \, x}{e^{x} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2+(11-x)*exp(x)+30)/(exp(x)^2+10*exp(x)+25),x, algorithm="giac")

[Out]

(x*e^x + 6*x)/(e^x + 5)

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maple [A]  time = 0.04, size = 11, normalized size = 0.79

method	result	size
risch	\(x +\frac {x}{{\mathrm e}^{x}+5}\)	\(11\)
norman	\(\frac {{\mathrm e}^{x} x +6 x}{{\mathrm e}^{x}+5}\)	\(16\)
default	\(\frac {6 \ln \left ({\mathrm e}^{x}\right )}{5}-\frac {x \,{\mathrm e}^{x}}{5 \left ({\mathrm e}^{x}+5\right )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)^2+(11-x)*exp(x)+30)/(exp(x)^2+10*exp(x)+25),x,method=_RETURNVERBOSE)

[Out]

x+x/(exp(x)+5)

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maxima [A]  time = 0.39, size = 15, normalized size = 1.07 \begin {gather*} \frac {6}{5} \, x - \frac {x e^{x}}{5 \, {\left (e^{x} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2+(11-x)*exp(x)+30)/(exp(x)^2+10*exp(x)+25),x, algorithm="maxima")

[Out]

6/5*x - 1/5*x*e^x/(e^x + 5)

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mupad [B]  time = 0.07, size = 10, normalized size = 0.71 \begin {gather*} x+\frac {x}{{\mathrm {e}}^x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x) - exp(x)*(x - 11) + 30)/(exp(2*x) + 10*exp(x) + 25),x)

[Out]

x + x/(exp(x) + 5)

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sympy [A]  time = 0.10, size = 7, normalized size = 0.50 \begin {gather*} x + \frac {x}{e^{x} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)**2+(11-x)*exp(x)+30)/(exp(x)**2+10*exp(x)+25),x)

[Out]

x + x/(exp(x) + 5)

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