\(\int \frac {2 e x+e^x (2 x-x^2)}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx\) [10273]

Optimal result

Integrand size = 45, antiderivative size = 15 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^2}{\left (e+e^x\right ) \log (25)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6820, 12, 6874, 2215, 2221, 2317, 2438, 2611, 2320, 6724, 2216, 2222} \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^2}{\left (e^x+e\right ) \log (25)} \]

[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 2215

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 2216

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2222

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 2317

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 2320

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 2438

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 2611

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 6724

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 6820

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

Rule 6874

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

Rubi steps \begin{align*} \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 \operatorname {PolyLog}\left (2,-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 \operatorname {PolyLog}\left (2,-e^{-1+x}\right ) \, dx}{e \log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)}-\frac {2 \int \operatorname {PolyLog}\left (2,-e^{-1+x}\right ) \, dx}{e \log (25)}+\frac {2 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{-1+x}\right )}{e \log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)}+\frac {2 \operatorname {PolyLog}\left (3,-e^{-1+x}\right )}{e \log (25)}-\frac {2 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{-1+x}\right )}{e \log (25)} \\ & = \frac {x^2}{\left (e+e^x\right ) \log (25)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^2}{\left (e+e^x\right ) \log (25)} \]

[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])

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^{2} e}{2 \, {\left (e^{2} \log \left (5\right ) + e^{\left (x + 1\right )} \log \left (5\right )\right )}} \]

[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))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^{2}}{2 e^{x} \log {\left (5 \right )} + 2 e \log {\left (5 \right )}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^{2}}{2 \, {\left (e \log \left (5\right ) + e^{x} \log \left (5\right )\right )}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^{2}}{2 \, {\left (e \log \left (5\right ) + e^{x} \log \left (5\right )\right )}} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {2 e x+e^x \left (2 x-x^2\right )}{e^2 \log (25)+e^{2 x} \log (25)+2 e^{1+x} \log (25)} \, dx=\frac {x^2}{2\,\ln \left (5\right )\,\left (\mathrm {e}+{\mathrm {e}}^x\right )} \]

[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)))