3.103.73 \(\int \frac {2 e x+e^x (2 x-x^2)}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx\)

Optimal. Leaf size=15 \[ \frac {x^2}{\left (e+e^x\right ) \log (25)} \]

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Rubi [A]  time = 0.63, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 12, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6688, 12, 6742, 2184, 2190, 2279, 2391, 2531, 2282, 6589, 2185, 2191} \begin {gather*} \frac {x^2}{\left (e^x+e\right ) \log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E*x + E^x*(2*x - x^2))/(E^2*Log[25] + E^(2*x)*Log[25] + 2*E^(1 + x)*Log[25]),x]

[Out]

x^2/((E + E^x)*Log[25])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {2 e x-e^x (-2+x) x}{\left (e+e^x\right )^2 \log (25)} \, dx\\ &=\frac {\int \frac {2 e x-e^x (-2+x) x}{\left (e+e^x\right )^2} \, dx}{\log (25)}\\ &=\frac {\int \left (-\frac {(-2+x) x}{e+e^x}+\frac {e x^2}{\left (e+e^x\right )^2}\right ) \, dx}{\log (25)}\\ &=-\frac {\int \frac {(-2+x) x}{e+e^x} \, dx}{\log (25)}+\frac {e \int \frac {x^2}{\left (e+e^x\right )^2} \, dx}{\log (25)}\\ &=-\frac {\int \frac {e^x x^2}{\left (e+e^x\right )^2} \, dx}{\log (25)}+\frac {\int \frac {x^2}{e+e^x} \, dx}{\log (25)}-\frac {\int \left (-\frac {2 x}{e+e^x}+\frac {x^2}{e+e^x}\right ) \, dx}{\log (25)}\\ &=\frac {x^2}{\left (e+e^x\right ) \log (25)}+\frac {x^3}{3 e \log (25)}-\frac {\int \frac {x^2}{e+e^x} \, dx}{\log (25)}-\frac {\int \frac {e^x x^2}{e+e^x} \, dx}{e \log (25)}\\ &=\frac {x^2}{\left (e+e^x\right ) \log (25)}-\frac {x^2 \log \left (1+e^{-1+x}\right )}{e \log (25)}+\frac {\int \frac {e^x x^2}{e+e^x} \, dx}{e \log (25)}+\frac {2 \int x \log \left (1+e^{-1+x}\right ) \, dx}{e \log (25)}\\ &=\frac {x^2}{\left (e+e^x\right ) \log (25)}-\frac {2 x \text {Li}_2\left (-e^{-1+x}\right )}{e \log (25)}-\frac {2 \int x \log \left (1+e^{-1+x}\right ) \, dx}{e \log (25)}+\frac {2 \int \text {Li}_2\left (-e^{-1+x}\right ) \, dx}{e \log (25)}\\ &=\frac {x^2}{\left (e+e^x\right ) \log (25)}-\frac {2 \int \text {Li}_2\left (-e^{-1+x}\right ) \, dx}{e \log (25)}+\frac {2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{-1+x}\right )}{e \log (25)}\\ &=\frac {x^2}{\left (e+e^x\right ) \log (25)}+\frac {2 \text {Li}_3\left (-e^{-1+x}\right )}{e \log (25)}-\frac {2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{-1+x}\right )}{e \log (25)}\\ &=\frac {x^2}{\left (e+e^x\right ) \log (25)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 15, normalized size = 1.00 \begin {gather*} \frac {x^2}{\left (e+e^x\right ) \log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E*x + E^x*(2*x - x^2))/(E^2*Log[25] + E^(2*x)*Log[25] + 2*E^(1 + x)*Log[25]),x]

[Out]

x^2/((E + E^x)*Log[25])

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fricas [A]  time = 0.58, size = 22, normalized size = 1.47 \begin {gather*} \frac {x^{2} e}{2 \, {\left (e^{2} \log \relax (5) + e^{\left (x + 1\right )} \log \relax (5)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+2*x)*exp(x)+2*x*exp(1))/(2*log(5)*exp(x)^2+4*exp(1)*log(5)*exp(x)+2*exp(1)^2*log(5)),x, algor
ithm="fricas")

[Out]

1/2*x^2*e/(e^2*log(5) + e^(x + 1)*log(5))

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giac [A]  time = 0.13, size = 18, normalized size = 1.20 \begin {gather*} \frac {x^{2}}{2 \, {\left (e \log \relax (5) + e^{x} \log \relax (5)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+2*x)*exp(x)+2*x*exp(1))/(2*log(5)*exp(x)^2+4*exp(1)*log(5)*exp(x)+2*exp(1)^2*log(5)),x, algor
ithm="giac")

[Out]

1/2*x^2/(e*log(5) + e^x*log(5))

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maple [A]  time = 0.14, size = 17, normalized size = 1.13

method	result	size
norman	\(\frac {x^{2}}{2 \ln \relax (5) \left ({\mathrm e}+{\mathrm e}^{x}\right )}\)	\(17\)
risch	\(\frac {x^{2}}{2 \ln \relax (5) \left ({\mathrm e}+{\mathrm e}^{x}\right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+2*x)*exp(x)+2*x*exp(1))/(2*ln(5)*exp(x)^2+4*exp(1)*ln(5)*exp(x)+2*exp(1)^2*ln(5)),x,method=_RETURNV
ERBOSE)

[Out]

1/2*x^2/ln(5)/(exp(1)+exp(x))

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maxima [A]  time = 0.48, size = 18, normalized size = 1.20 \begin {gather*} \frac {x^{2}}{2 \, {\left (e \log \relax (5) + e^{x} \log \relax (5)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+2*x)*exp(x)+2*x*exp(1))/(2*log(5)*exp(x)^2+4*exp(1)*log(5)*exp(x)+2*exp(1)^2*log(5)),x, algor
ithm="maxima")

[Out]

1/2*x^2/(e*log(5) + e^x*log(5))

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mupad [B]  time = 6.80, size = 16, normalized size = 1.07 \begin {gather*} \frac {x^2}{2\,\ln \relax (5)\,\left (\mathrm {e}+{\mathrm {e}}^x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(1) + exp(x)*(2*x - x^2))/(2*exp(2*x)*log(5) + 2*exp(2)*log(5) + 4*exp(1)*exp(x)*log(5)),x)

[Out]

x^2/(2*log(5)*(exp(1) + exp(x)))

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sympy [A]  time = 0.13, size = 19, normalized size = 1.27 \begin {gather*} \frac {x^{2}}{2 e^{x} \log {\relax (5 )} + 2 e \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+2*x)*exp(x)+2*x*exp(1))/(2*ln(5)*exp(x)**2+4*exp(1)*ln(5)*exp(x)+2*exp(1)**2*ln(5)),x)

[Out]

x**2/(2*exp(x)*log(5) + 2*E*log(5))

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