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

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

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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} $$\begin{array}{c}-\frac{1}{25}(7-5x{)}^2+\frac{2}{5}(x+1{)}^2-\frac{2}{5}(x+2{)}^2+\frac{4x}{5}-\frac{4(x+2)}{5(1-5e^x)}-\frac{4}{5}(x+1)\log (1-5e^x)+\frac{4}{5}(x+2)\log (1-5e^x)-\frac{4}{5}\log (1-5e^x)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2+e^{2x}(70-50x)-2x+e^x(-32+16x)}{{(1-5e^x)}^2}dx\\ & =\int (-\frac{4(1+x)}{5(-1+5e^x)}-\frac{4(2+x)}{5{(-1+5e^x)}^2}-\frac{2}{5}(-7+5x))dx\\ & =-\frac{1}{25}(7-5x{)}^2-\frac{4}{5}\int \frac{1+x}{-1+5e^x}dx-\frac{4}{5}\int \frac{2+x}{{(-1+5e^x)}^2}dx\\ & =-\frac{1}{25}(7-5x{)}^2+\frac{2}{5}(1+x{)}^2+\frac{4}{5}\int \frac{2+x}{-1+5e^x}dx-4\int \frac{e^x(1+x)}{-1+5e^x}dx-4\int \frac{e^x(2+x)}{{(-1+5e^x)}^2}dx\\ & =-\frac{1}{25}(7-5x{)}^2+\frac{2}{5}(1+x{)}^2-\frac{4(2+x)}{5(1-5e^x)}-\frac{2}{5}(2+x{)}^2-\frac{4}{5}(1+x)\log (1-5e^x)-\frac{4}{5}\int \frac{1}{-1+5e^x}dx+\frac{4}{5}\int \log (1-5e^x)dx+4\int \frac{e^x(2+x)}{-1+5e^x}dx\\ & =-\frac{1}{25}(7-5x{)}^2+\frac{2}{5}(1+x{)}^2-\frac{4(2+x)}{5(1-5e^x)}-\frac{2}{5}(2+x{)}^2-\frac{4}{5}(1+x)\log (1-5e^x)+\frac{4}{5}(2+x)\log (1-5e^x)-\frac{4}{5}\int \log (1-5e^x)dx-\frac{4}{5}\mathrm{Subst}(\int \frac{1}{x(-1+5x)}dx,x,e^x)+\frac{4}{5}\mathrm{Subst}(\int \frac{\log (1-5x)}{x}dx,x,e^x)\\ & =-\frac{1}{25}(7-5x{)}^2+\frac{2}{5}(1+x{)}^2-\frac{4(2+x)}{5(1-5e^x)}-\frac{2}{5}(2+x{)}^2-\frac{4}{5}(1+x)\log (1-5e^x)+\frac{4}{5}(2+x)\log (1-5e^x)-\frac{4{\text{Li}}_2\left(5e^x\right)}{5}+\frac{4}{5}\mathrm{Subst}(\int \frac{1}{x}dx,x,e^x)-\frac{4}{5}\mathrm{Subst}(\int \frac{\log (1-5x)}{x}dx,x,e^x)-4\mathrm{Subst}(\int \frac{1}{-1+5x}dx,x,e^x)\\ & =-\frac{1}{25}(7-5x{)}^2+\frac{4x}{5}+\frac{2}{5}(1+x{)}^2-\frac{4(2+x)}{5(1-5e^x)}-\frac{2}{5}(2+x{)}^2-\frac{4}{5}\log (1-5e^x)-\frac{4}{5}(1+x)\log (1-5e^x)+\frac{4}{5}(2+x)\log (1-5e^x)\end{array}\end{array}$$

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

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)))

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

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)

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

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)

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maple [A]  time = 0.07, size = 23, normalized size = 0.70

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

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)

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maxima [B]  time = 0.84, size = 60, normalized size = 1.82 $$\begin{array}{c}2x+\frac{5x^2-5(5x^2-4x)e^x+18}{5(5e^x-1)}-\frac{2}{5e^x-1}-2\log (5e^x-1)+2\log (e^x-\frac{1}{5})\end{array}$$

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)

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

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

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

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)

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