3.41.52 $\int \frac{e^{\frac{x}{e^{12}(2+2x{)}^2}}(1-x)+e^{12}(1+x)(2+2x{)}^2}{e^{12}(1+x)(2+2x{)}^2}dx$

Optimal. Leaf size=17 $$4+e^{\frac{x}{e^{12}(2+2x{)}^2}}+x$$

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Rubi [A]  time = 0.19, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 51, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.078, Rules used = {12, 21, 6688, 6706} $$\begin{array}{c}x+e^{\frac{x}{4e^{12}(x+1{)}^2}}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(x/(E^12*(2 + 2*x)^2))*(1 - x) + E^12*(1 + x)*(2 + 2*x)^2)/(E^12*(1 + x)*(2 + 2*x)^2),x]

[Out]

E^(x/(4*E^12*(1 + x)^2)) + x

Rule 12

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int \frac{e^{\frac{x}{e^{12}(2+2x{)}^2}}(1-x)+e^{12}(1+x)(2+2x{)}^2}{(1+x)(2+2x{)}^2}dx}{e^{12}}\\ & =\frac{\int \frac{e^{\frac{x}{e^{12}(2+2x{)}^2}}(1-x)+e^{12}(1+x)(2+2x{)}^2}{(1+x{)}^3}dx}{4e^{12}}\\ & =\frac{\int (4e^{12}-\frac{e^{\frac{x}{e^{12}(2+2x{)}^2}}(-1+x)}{(1+x{)}^3})dx}{4e^{12}}\\ & =x-\frac{\int \frac{e^{\frac{x}{e^{12}(2+2x{)}^2}}(-1+x)}{(1+x{)}^3}dx}{4e^{12}}\\ & =e^{\frac{x}{4e^{12}(1+x{)}^2}}+x\end{array}\end{array}$$

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Mathematica [A]  time = 0.22, size = 17, normalized size = 1.00 $$\begin{array}{c}{e}^{\frac{x}{4e^{12}(1+x{)}^2}}+x\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/(E^12*(2 + 2*x)^2))*(1 - x) + E^12*(1 + x)*(2 + 2*x)^2)/(E^12*(1 + x)*(2 + 2*x)^2),x]

[Out]

E^(x/(4*E^12*(1 + x)^2)) + x

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fricas [A]  time = 0.62, size = 18, normalized size = 1.06 $$\begin{array}{c}x+e^{\left(\frac{x{e}^{(-12)}}{4(x^2+2x+1)}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + e^(1/4*x*e^(-12)/(x^2 + 2*x + 1))

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giac [B]  time = 0.15, size = 88, normalized size = 5.18 $$\begin{array}{c}x+e^{(-\frac{12x^2{e}^{12}}{x^2{e}^{12}+2x{e}^{12}+e^{12}}-\frac{24x{e}^{12}}{x^2{e}^{12}+2x{e}^{12}+e^{12}}+\frac{x}{4(x^2{e}^{12}+2x{e}^{12}+e^{12})}-\frac{12e^{12}}{x^2{e}^{12}+2x{e}^{12}+e^{12}}+12)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + e^(-12*x^2*e^12/(x^2*e^12 + 2*x*e^12 + e^12) - 24*x*e^12/(x^2*e^12 + 2*x*e^12 + e^12) + 1/4*x/(x^2*e^12 +
2*x*e^12 + e^12) - 12*e^12/(x^2*e^12 + 2*x*e^12 + e^12) + 12)

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maple [A]  time = 0.10, size = 14, normalized size = 0.82

method	result	size
risch	$x+{\mathrm{e}}^{\frac{x{\mathrm{e}}^{-12}}{4{(x+1)}^2}}$	$14$
default	${\mathrm{e}}^{-2}({\mathrm{e}}^{12}{\mathrm{e}}^{-10-\frac{{\mathrm{e}}^{-12}}{4{(x+1)}^2}+\frac{{\mathrm{e}}^{-12}}{4x+4}}+{\mathrm{e}}^{2}x)$	$37$
norman	$\frac{(x^3{\mathrm{e}}^5-3x{\mathrm{e}}^5+{\mathrm{e}}^5{\mathrm{e}}^{\frac{x{\mathrm{e}}^{-2}{\mathrm{e}}^{-10}}{{(2x+2)}^2}}+x^2{\mathrm{e}}^5{\mathrm{e}}^{\frac{x{\mathrm{e}}^{-2}{\mathrm{e}}^{-10}}{{(2x+2)}^2}}+2x{\mathrm{e}}^5{\mathrm{e}}^{\frac{x{\mathrm{e}}^{-2}{\mathrm{e}}^{-10}}{{(2x+2)}^2}}-2{\mathrm{e}}^5){\mathrm{e}}^{-5}}{{(x+1)}^2}$	$95$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+exp(1/4*x*exp(-12)/(x+1)^2)

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maxima [B]  time = 0.44, size = 129, normalized size = 7.59 $$\begin{array}{c}\frac{1}{2}((2x-\frac{6x+5}{x^2+2x+1}-6\log (x+1))e^{12}+3(\frac{4x+3}{x^2+2x+1}+2\log (x+1))e^{12}-\frac{3(2x+1)e^{12}}{x^2+2x+1}-\frac{e^{12}}{x^2+2x+1}+2e^{(-\frac{1}{4(x^2{e}^{12}+2x{e}^{12}+e^{12})}+\frac{1}{4(x{e}^{12}+e^{12})}+12)})e^{(-12)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*x - (6*x + 5)/(x^2 + 2*x + 1) - 6*log(x + 1))*e^12 + 3*((4*x + 3)/(x^2 + 2*x + 1) + 2*log(x + 1))*e^12
- 3*(2*x + 1)*e^12/(x^2 + 2*x + 1) - e^12/(x^2 + 2*x + 1) + 2*e^(-1/4/(x^2*e^12 + 2*x*e^12 + e^12) + 1/4/(x*e
^12 + e^12) + 12))*e^(-12)

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mupad [B]  time = 0.68, size = 19, normalized size = 1.12 $$\begin{array}{c}x+{\mathrm{e}}^{\frac{x{\mathrm{e}}^{-12}}{4x^2+8x+4}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*exp(- 2*log(2*x + 2) - 10)*(exp(x*exp(-2)*exp(- 2*log(2*x + 2) - 10))*(x - 1) - exp(2)*exp(2*log
(2*x + 2) + 10)*(x + 1)))/(x + 1),x)

[Out]

x + exp((x*exp(-12))/(8*x + 4*x^2 + 4))

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sympy [A]  time = 0.28, size = 14, normalized size = 0.82 $$\begin{array}{c}x+e^{\frac{x}{{(2x+2)}^2{e}^{12}}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(x*exp(-12)/(2*x + 2)**2)

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