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

Optimal. Leaf size=33 \[ 5+2 x-x^2+\frac {4 (2+x)}{4-e^{-x} \left (1-e^x\right )} \]

________________________________________________________________________________________

Rubi [B]  time = 0.38, antiderivative size = 93, normalized size of antiderivative = 2.82, number of steps used = 17, number of rules used = 12, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6741, 6742, 2184, 2190, 2279, 2391, 2185, 2191, 2282, 36, 29, 31} \begin {gather*} -\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 ) \end {gather*}

Antiderivative was successfully verified.

[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 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 6741

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

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^{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} \operatorname {Subst}\left (\int \frac {1}{x (-1+5 x)} \, dx,x,e^x\right )+\frac {4}{5} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\frac {4}{5} \operatorname {Subst}\left (\int \frac {\log (1-5 x)}{x} \, dx,x,e^x\right )-4 \operatorname {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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 31, normalized size = 0.94 \begin {gather*} -2 \left (-\frac {7 x}{5}+\frac {x^2}{2}-\frac {2 (2+x)}{5 \left (-1+5 e^x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[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)))

________________________________________________________________________________________

fricas [A]  time = 0.92, size = 33, normalized size = 1.00 \begin {gather*} \frac {5 \, x^{2} - 5 \, {\left (5 \, x^{2} - 14 \, x\right )} e^{x} - 10 \, x + 8}{5 \, {\left (5 \, e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 32, normalized size = 0.97 \begin {gather*} -\frac {25 \, x^{2} e^{x} - 5 \, x^{2} - 70 \, x e^{x} + 10 \, x - 8}{5 \, {\left (5 \, e^{x} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 23, normalized size = 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	\(\frac {x^{2}+8 \,{\mathrm e}^{x}-2 x +14 \,{\mathrm e}^{x} x -5 \,{\mathrm e}^{x} x^{2}}{5 \,{\mathrm e}^{x}-1}\)	\(33\)
default	\(2 \ln \left ({\mathrm e}^{x}\right )+\frac {8}{5 \left (5 \,{\mathrm e}^{x}-1\right )}-x^{2}+\frac {4 x \,{\mathrm e}^{x}}{5 \,{\mathrm e}^{x}-1}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [B]  time = 0.84, size = 60, normalized size = 1.82 \begin {gather*} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 1.79, size = 23, normalized size = 0.70 \begin {gather*} \frac {14\,x}{5}+\frac {\frac {4\,x}{5}+\frac {8}{5}}{5\,{\mathrm {e}}^x-1}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 19, normalized size = 0.58 \begin {gather*} - x^{2} + \frac {14 x}{5} + \frac {4 x + 8}{25 e^{x} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________