\(\int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx\) [2970]

Optimal result

Integrand size = 42, antiderivative size = 33 \[ \int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx=5+2 x-x^2+\frac {4 (2+x)}{4-e^{-x} \left (1-e^x\right )} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(93\) vs. \(2(33)=66\).

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.82, number of steps used = 17, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6873, 6874, 2215, 2221, 2317, 2438, 2216, 2222, 2320, 36, 29, 31} \[ \int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx=-\frac {1}{25} (7-5 x)^2+\frac {2}{5} (x+1)^2-\frac {2}{5} (x+2)^2+\frac {4 x}{5}-\frac {4 (x+2)}{5 \left (1-5 e^x\right )}-\frac {4}{5} (x+1) \log \left (1-5 e^x\right )+\frac {4}{5} (x+2) \log \left (1-5 e^x\right )-\frac {4}{5} \log \left (1-5 e^x\right ) \]

[In]

Int[(2 + E^(2*x)*(70 - 50*x) - 2*x + E^x*(-32 + 16*x))/(1 - 10*E^x + 25*E^(2*x)),x]

[Out]

-1/25*(7 - 5*x)^2 + (4*x)/5 + (2*(1 + x)^2)/5 - (4*(2 + x))/(5*(1 - 5*E^x)) - (2*(2 + x)^2)/5 - (4*Log[1 - 5*E
^x])/5 - (4*(1 + x)*Log[1 - 5*E^x])/5 + (4*(2 + x)*Log[1 - 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 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 6873

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

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^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{\left (1-5 e^x\right )^2} \, dx \\ & = \int \left (-\frac {4 (1+x)}{5 \left (-1+5 e^x\right )}-\frac {4 (2+x)}{5 \left (-1+5 e^x\right )^2}-\frac {2}{5} (-7+5 x)\right ) \, dx \\ & = -\frac {1}{25} (7-5 x)^2-\frac {4}{5} \int \frac {1+x}{-1+5 e^x} \, dx-\frac {4}{5} \int \frac {2+x}{\left (-1+5 e^x\right )^2} \, dx \\ & = -\frac {1}{25} (7-5 x)^2+\frac {2}{5} (1+x)^2+\frac {4}{5} \int \frac {2+x}{-1+5 e^x} \, dx-4 \int \frac {e^x (1+x)}{-1+5 e^x} \, dx-4 \int \frac {e^x (2+x)}{\left (-1+5 e^x\right )^2} \, dx \\ & = -\frac {1}{25} (7-5 x)^2+\frac {2}{5} (1+x)^2-\frac {4 (2+x)}{5 \left (1-5 e^x\right )}-\frac {2}{5} (2+x)^2-\frac {4}{5} (1+x) \log \left (1-5 e^x\right )-\frac {4}{5} \int \frac {1}{-1+5 e^x} \, dx+\frac {4}{5} \int \log \left (1-5 e^x\right ) \, dx+4 \int \frac {e^x (2+x)}{-1+5 e^x} \, dx \\ & = -\frac {1}{25} (7-5 x)^2+\frac {2}{5} (1+x)^2-\frac {4 (2+x)}{5 \left (1-5 e^x\right )}-\frac {2}{5} (2+x)^2-\frac {4}{5} (1+x) \log \left (1-5 e^x\right )+\frac {4}{5} (2+x) \log \left (1-5 e^x\right )-\frac {4}{5} \int \log \left (1-5 e^x\right ) \, dx-\frac {4}{5} \text {Subst}\left (\int \frac {1}{x (-1+5 x)} \, dx,x,e^x\right )+\frac {4}{5} \text {Subst}\left (\int \frac {\log (1-5 x)}{x} \, dx,x,e^x\right ) \\ & = -\frac {1}{25} (7-5 x)^2+\frac {2}{5} (1+x)^2-\frac {4 (2+x)}{5 \left (1-5 e^x\right )}-\frac {2}{5} (2+x)^2-\frac {4}{5} (1+x) \log \left (1-5 e^x\right )+\frac {4}{5} (2+x) \log \left (1-5 e^x\right )-\frac {4 \text {Li}_2\left (5 e^x\right )}{5}+\frac {4}{5} \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\frac {4}{5} \text {Subst}\left (\int \frac {\log (1-5 x)}{x} \, dx,x,e^x\right )-4 \text {Subst}\left (\int \frac {1}{-1+5 x} \, dx,x,e^x\right ) \\ & = -\frac {1}{25} (7-5 x)^2+\frac {4 x}{5}+\frac {2}{5} (1+x)^2-\frac {4 (2+x)}{5 \left (1-5 e^x\right )}-\frac {2}{5} (2+x)^2-\frac {4}{5} \log \left (1-5 e^x\right )-\frac {4}{5} (1+x) \log \left (1-5 e^x\right )+\frac {4}{5} (2+x) \log \left (1-5 e^x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx=-2 \left (-\frac {7 x}{5}+\frac {x^2}{2}-\frac {2 (2+x)}{5 \left (-1+5 e^x\right )}\right ) \]

[In]

Integrate[(2 + E^(2*x)*(70 - 50*x) - 2*x + E^x*(-32 + 16*x))/(1 - 10*E^x + 25*E^(2*x)),x]

[Out]

-2*((-7*x)/5 + x^2/2 - (2*(2 + x))/(5*(-1 + 5*E^x)))

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70

method	result	size
risch	\(-x^{2}+\frac {14 x}{5}+\frac {\frac {4 x}{5}+\frac {8}{5}}{5 \,{\mathrm e}^{x}-1}\)	\(23\)
norman	\(\frac {x^{2}-2 x +14 \,{\mathrm e}^{x} x -5 \,{\mathrm e}^{x} x^{2}+\frac {8}{5}}{5 \,{\mathrm e}^{x}-1}\)	\(30\)
parallelrisch	\(-\frac {25 \,{\mathrm e}^{x} x^{2}-70 \,{\mathrm e}^{x} x -5 x^{2}+10 x -8}{5 \left (5 \,{\mathrm e}^{x}-1\right )}\)	\(33\)
default	\(2 \ln \left ({\mathrm e}^{x}\right )+\frac {8}{5 \left (5 \,{\mathrm e}^{x}-1\right )}-x^{2}+\frac {4 x \,{\mathrm e}^{x}}{5 \,{\mathrm e}^{x}-1}\)	\(35\)

[In]

int(((-50*x+70)*exp(x)^2+(16*x-32)*exp(x)-2*x+2)/(25*exp(x)^2-10*exp(x)+1),x,method=_RETURNVERBOSE)

[Out]

-x^2+14/5*x+4/5*(2+x)/(5*exp(x)-1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx=\frac {5 \, x^{2} - 5 \, {\left (5 \, x^{2} - 14 \, x\right )} e^{x} - 10 \, x + 8}{5 \, {\left (5 \, e^{x} - 1\right )}} \]

[In]

integrate(((-50*x+70)*exp(x)^2+(16*x-32)*exp(x)-2*x+2)/(25*exp(x)^2-10*exp(x)+1),x, algorithm="fricas")

[Out]

1/5*(5*x^2 - 5*(5*x^2 - 14*x)*e^x - 10*x + 8)/(5*e^x - 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.58 \[ \int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx=- x^{2} + \frac {14 x}{5} + \frac {4 x + 8}{25 e^{x} - 5} \]

[In]

integrate(((-50*x+70)*exp(x)**2+(16*x-32)*exp(x)-2*x+2)/(25*exp(x)**2-10*exp(x)+1),x)

[Out]

-x**2 + 14*x/5 + (4*x + 8)/(25*exp(x) - 5)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (28) = 56\).

Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.82 \[ \int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx=2 \, x + \frac {5 \, x^{2} - 5 \, {\left (5 \, x^{2} - 4 \, x\right )} e^{x} + 18}{5 \, {\left (5 \, e^{x} - 1\right )}} - \frac {2}{5 \, e^{x} - 1} - 2 \, \log \left (5 \, e^{x} - 1\right ) + 2 \, \log \left (e^{x} - \frac {1}{5}\right ) \]

[In]

integrate(((-50*x+70)*exp(x)^2+(16*x-32)*exp(x)-2*x+2)/(25*exp(x)^2-10*exp(x)+1),x, algorithm="maxima")

[Out]

2*x + 1/5*(5*x^2 - 5*(5*x^2 - 4*x)*e^x + 18)/(5*e^x - 1) - 2/(5*e^x - 1) - 2*log(5*e^x - 1) + 2*log(e^x - 1/5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx=-\frac {25 \, x^{2} e^{x} - 5 \, x^{2} - 70 \, x e^{x} + 10 \, x - 8}{5 \, {\left (5 \, e^{x} - 1\right )}} \]

[In]

integrate(((-50*x+70)*exp(x)^2+(16*x-32)*exp(x)-2*x+2)/(25*exp(x)^2-10*exp(x)+1),x, algorithm="giac")

[Out]

-1/5*(25*x^2*e^x - 5*x^2 - 70*x*e^x + 10*x - 8)/(5*e^x - 1)

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.70 \[ \int \frac {2+e^{2 x} (70-50 x)-2 x+e^x (-32+16 x)}{1-10 e^x+25 e^{2 x}} \, dx=\frac {14\,x}{5}+\frac {\frac {4\,x}{5}+\frac {8}{5}}{5\,{\mathrm {e}}^x-1}-x^2 \]

[In]

int(-(2*x - exp(x)*(16*x - 32) + exp(2*x)*(50*x - 70) - 2)/(25*exp(2*x) - 10*exp(x) + 1),x)

[Out]

(14*x)/5 + ((4*x)/5 + 8/5)/(5*exp(x) - 1) - x^2