3.33.35 $\int \frac{10+20x+75x^2-40x^3+e^2(5+10x+75x^2-20x^3)+e^x(2+6x+17x^2-3x^3-2x^4+e^2(1+3x+16x^2+x^3-x^4))}{2+e^2}dx$

Optimal. Leaf size=29 $$(5+e^x)x(1+x(1+(5-\frac{5}{2+e^2}-x)x))$$

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Rubi [B]  time = 0.40, antiderivative size = 227, normalized size of antiderivative = 7.83, number of steps used = 37, number of rules used = 4, integrand size = 88, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.045, Rules used = {12, 2196, 2194, 2176} $$\begin{array}{c}-\frac{2e^x{x}^4}{2+e^2}-\frac{e^{x+2}{x}^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{10x^4}{2+e^2}+\frac{5e^x{x}^3}{2+e^2}+\frac{5e^{x+2}{x}^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{2e^x{x}^2}{2+e^2}+\frac{e^{x+2}{x}^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{2e^{x}x}{2+e^2}+\frac{e^{x+2}x}{2+e^2}+\frac{5e^{2}x}{2+e^2}+\frac{10x}{2+e^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(10 + 20*x + 75*x^2 - 40*x^3 + E^2*(5 + 10*x + 75*x^2 - 20*x^3) + E^x*(2 + 6*x + 17*x^2 - 3*x^3 - 2*x^4 +
E^2*(1 + 3*x + 16*x^2 + x^3 - x^4)))/(2 + E^2),x]

[Out]

(10*x)/(2 + E^2) + (5*E^2*x)/(2 + E^2) + (2*E^x*x)/(2 + E^2) + (E^(2 + x)*x)/(2 + E^2) + (10*x^2)/(2 + E^2) +
(5*E^2*x^2)/(2 + E^2) + (2*E^x*x^2)/(2 + E^2) + (E^(2 + x)*x^2)/(2 + E^2) + (25*x^3)/(2 + E^2) + (25*E^2*x^3)/
(2 + E^2) + (5*E^x*x^3)/(2 + E^2) + (5*E^(2 + x)*x^3)/(2 + E^2) - (10*x^4)/(2 + E^2) - (5*E^2*x^4)/(2 + E^2) -
(2*E^x*x^4)/(2 + E^2) - (E^(2 + x)*x^4)/(2 + E^2)

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int (10+20x+75x^2-40x^3+e^2(5+10x+75x^2-20x^3)+e^x(2+6x+17x^2-3x^3-2x^4+e^2(1+3x+16x^2+x^3-x^4)))dx}{2+e^2}\\ & =\frac{10x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{25x^3}{2+e^2}-\frac{10x^4}{2+e^2}+\frac{\int e^x(2+6x+17x^2-3x^3-2x^4+e^2(1+3x+16x^2+x^3-x^4))dx}{2+e^2}+\frac{e^2\int (5+10x+75x^2-20x^3)dx}{2+e^2}\\ & =\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}+\frac{\int (2e^x+6e^{x}x+17e^x{x}^2-3e^x{x}^3-2e^x{x}^4-e^{2+x}(-1-3x-16x^2-x^3+x^4))dx}{2+e^2}\\ & =\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{\int e^{2+x}(-1-3x-16x^2-x^3+x^4)dx}{2+e^2}+\frac{2\int e^{x}dx}{2+e^2}-\frac{2\int e^x{x}^{4}dx}{2+e^2}-\frac{3\int e^x{x}^{3}dx}{2+e^2}+\frac{6\int e^{x}xdx}{2+e^2}+\frac{17\int e^x{x}^{2}dx}{2+e^2}\\ & =\frac{2e^x}{2+e^2}+\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}+\frac{6e^{x}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{17e^x{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}-\frac{3e^x{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{2e^x{x}^4}{2+e^2}-\frac{\int (-e^{2+x}-3e^{2+x}x-16e^{2+x}{x}^2-e^{2+x}{x}^3+e^{2+x}{x}^4)dx}{2+e^2}-\frac{6\int e^{x}dx}{2+e^2}+\frac{8\int e^x{x}^{3}dx}{2+e^2}+\frac{9\int e^x{x}^{2}dx}{2+e^2}-\frac{34\int e^{x}xdx}{2+e^2}\\ & =-\frac{4e^x}{2+e^2}+\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}-\frac{28e^{x}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{26e^x{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}+\frac{5e^x{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{2e^x{x}^4}{2+e^2}+\frac{\int e^{2+x}dx}{2+e^2}+\frac{\int e^{2+x}{x}^{3}dx}{2+e^2}-\frac{\int e^{2+x}{x}^{4}dx}{2+e^2}+\frac{3\int e^{2+x}xdx}{2+e^2}+\frac{16\int e^{2+x}{x}^{2}dx}{2+e^2}-\frac{18\int e^{x}xdx}{2+e^2}-\frac{24\int e^x{x}^{2}dx}{2+e^2}+\frac{34\int e^{x}dx}{2+e^2}\\ & =\frac{30e^x}{2+e^2}+\frac{e^{2+x}}{2+e^2}+\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}-\frac{46e^{x}x}{2+e^2}+\frac{3e^{2+x}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{2e^x{x}^2}{2+e^2}+\frac{16e^{2+x}{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}+\frac{5e^x{x}^3}{2+e^2}+\frac{e^{2+x}{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{2e^x{x}^4}{2+e^2}-\frac{e^{2+x}{x}^4}{2+e^2}-\frac{3\int e^{2+x}dx}{2+e^2}-\frac{3\int e^{2+x}{x}^{2}dx}{2+e^2}+\frac{4\int e^{2+x}{x}^{3}dx}{2+e^2}+\frac{18\int e^{x}dx}{2+e^2}-\frac{32\int e^{2+x}xdx}{2+e^2}+\frac{48\int e^{x}xdx}{2+e^2}\\ & =\frac{48e^x}{2+e^2}-\frac{2e^{2+x}}{2+e^2}+\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}+\frac{2e^{x}x}{2+e^2}-\frac{29e^{2+x}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{2e^x{x}^2}{2+e^2}+\frac{13e^{2+x}{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}+\frac{5e^x{x}^3}{2+e^2}+\frac{5e^{2+x}{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{2e^x{x}^4}{2+e^2}-\frac{e^{2+x}{x}^4}{2+e^2}+\frac{6\int e^{2+x}xdx}{2+e^2}-\frac{12\int e^{2+x}{x}^{2}dx}{2+e^2}+\frac{32\int e^{2+x}dx}{2+e^2}-\frac{48\int e^{x}dx}{2+e^2}\\ & =\frac{30e^{2+x}}{2+e^2}+\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}+\frac{2e^{x}x}{2+e^2}-\frac{23e^{2+x}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{2e^x{x}^2}{2+e^2}+\frac{e^{2+x}{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}+\frac{5e^x{x}^3}{2+e^2}+\frac{5e^{2+x}{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{2e^x{x}^4}{2+e^2}-\frac{e^{2+x}{x}^4}{2+e^2}-\frac{6\int e^{2+x}dx}{2+e^2}+\frac{24\int e^{2+x}xdx}{2+e^2}\\ & =\frac{24e^{2+x}}{2+e^2}+\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}+\frac{2e^{x}x}{2+e^2}+\frac{e^{2+x}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{2e^x{x}^2}{2+e^2}+\frac{e^{2+x}{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}+\frac{5e^x{x}^3}{2+e^2}+\frac{5e^{2+x}{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{2e^x{x}^4}{2+e^2}-\frac{e^{2+x}{x}^4}{2+e^2}-\frac{24\int e^{2+x}dx}{2+e^2}\\ & =\frac{10x}{2+e^2}+\frac{5e^{2}x}{2+e^2}+\frac{2e^{x}x}{2+e^2}+\frac{e^{2+x}x}{2+e^2}+\frac{10x^2}{2+e^2}+\frac{5e^2{x}^2}{2+e^2}+\frac{2e^x{x}^2}{2+e^2}+\frac{e^{2+x}{x}^2}{2+e^2}+\frac{25x^3}{2+e^2}+\frac{25e^2{x}^3}{2+e^2}+\frac{5e^x{x}^3}{2+e^2}+\frac{5e^{2+x}{x}^3}{2+e^2}-\frac{10x^4}{2+e^2}-\frac{5e^2{x}^4}{2+e^2}-\frac{2e^x{x}^4}{2+e^2}-\frac{e^{2+x}{x}^4}{2+e^2}\end{array}\end{array}$$

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Mathematica [A]  time = 0.15, size = 47, normalized size = 1.62 $$\begin{array}{c}-\frac{(5+e^x)x(-2-2x-5x^2+2x^3+e^2(-1-x-5x^2+x^3))}{2+e^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(10 + 20*x + 75*x^2 - 40*x^3 + E^2*(5 + 10*x + 75*x^2 - 20*x^3) + E^x*(2 + 6*x + 17*x^2 - 3*x^3 - 2*
x^4 + E^2*(1 + 3*x + 16*x^2 + x^3 - x^4)))/(2 + E^2),x]

[Out]

-(((5 + E^x)*x*(-2 - 2*x - 5*x^2 + 2*x^3 + E^2*(-1 - x - 5*x^2 + x^3)))/(2 + E^2))

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fricas [B]  time = 0.66, size = 90, normalized size = 3.10 $$\begin{array}{c}-\frac{10x^4-25x^3-10x^2+5(x^4-5x^3-x^2-x)e^2+(2x^4-5x^3-2x^2+(x^4-5x^3-x^2-x)e^2-2x)e^x-10x}{e^2+2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+x^3+16*x^2+3*x+1)*exp(2)-2*x^4-3*x^3+17*x^2+6*x+2)*exp(x)+(-20*x^3+75*x^2+10*x+5)*exp(2)-40*
x^3+75*x^2+20*x+10)/(exp(2)+2),x, algorithm="fricas")

[Out]

-(10*x^4 - 25*x^3 - 10*x^2 + 5*(x^4 - 5*x^3 - x^2 - x)*e^2 + (2*x^4 - 5*x^3 - 2*x^2 + (x^4 - 5*x^3 - x^2 - x)*
e^2 - 2*x)*e^x - 10*x)/(e^2 + 2)

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giac [B]  time = 0.14, size = 92, normalized size = 3.17 $$\begin{array}{c}-\frac{10x^4-25x^3-10x^2+5(x^4-5x^3-x^2-x)e^2+(x^4-5x^3-x^2-x)e^{(x+2)}+(2x^4-5x^3-2x^2-2x)e^x-10x}{e^2+2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+x^3+16*x^2+3*x+1)*exp(2)-2*x^4-3*x^3+17*x^2+6*x+2)*exp(x)+(-20*x^3+75*x^2+10*x+5)*exp(2)-40*
x^3+75*x^2+20*x+10)/(exp(2)+2),x, algorithm="giac")

[Out]

-(10*x^4 - 25*x^3 - 10*x^2 + 5*(x^4 - 5*x^3 - x^2 - x)*e^2 + (x^4 - 5*x^3 - x^2 - x)*e^(x + 2) + (2*x^4 - 5*x^
3 - 2*x^2 - 2*x)*e^x - 10*x)/(e^2 + 2)

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maple [B]  time = 0.04, size = 64, normalized size = 2.21

method	result	size
norman	${\mathrm{e}}^{x}x+{\mathrm{e}}^x{x}^2+5x+5x^2-5x^4-{\mathrm{e}}^x{x}^4+\frac{25({\mathrm{e}}^2+1)x^3}{{\mathrm{e}}^2+2}+\frac{5({\mathrm{e}}^2+1)x^3{\mathrm{e}}^x}{{\mathrm{e}}^2+2}$	$64$
risch	$-\frac{5x^4{\mathrm{e}}^2}{{\mathrm{e}}^2+2}+\frac{25x^3{\mathrm{e}}^2}{{\mathrm{e}}^2+2}-\frac{10x^4}{{\mathrm{e}}^2+2}+\frac{5x^2{\mathrm{e}}^2}{{\mathrm{e}}^2+2}+\frac{25x^3}{{\mathrm{e}}^2+2}+\frac{5{\mathrm{e}}^{2}x}{{\mathrm{e}}^2+2}+\frac{10x^2}{{\mathrm{e}}^2+2}+\frac{10x}{{\mathrm{e}}^2+2}+\frac{(-x^4{\mathrm{e}}^2+5x^3{\mathrm{e}}^2-2x^4+x^2{\mathrm{e}}^2+5x^3+{\mathrm{e}}^{2}x+2x^2+2x){\mathrm{e}}^x}{{\mathrm{e}}^2+2}$	$146$
default	$\frac{10x+{\mathrm{e}}^2{\mathrm{e}}^x+{\mathrm{e}}^2({\mathrm{e}}^x{x}^3-3{\mathrm{e}}^x{x}^2+6{\mathrm{e}}^{x}x-6{\mathrm{e}}^x)+2{\mathrm{e}}^{x}x+2{\mathrm{e}}^x{x}^2+5{\mathrm{e}}^x{x}^3-2{\mathrm{e}}^x{x}^4+3{\mathrm{e}}^2({\mathrm{e}}^{x}x-{\mathrm{e}}^x)+16{\mathrm{e}}^2({\mathrm{e}}^x{x}^2-2{\mathrm{e}}^{x}x+2{\mathrm{e}}^x)-{\mathrm{e}}^2({\mathrm{e}}^x{x}^4-4{\mathrm{e}}^x{x}^3+12{\mathrm{e}}^x{x}^2-24{\mathrm{e}}^{x}x+24{\mathrm{e}}^x)+{\mathrm{e}}^2(-5x^4+25x^3+5x^2+5x)+10x^2+25x^3-10x^4}{{\mathrm{e}}^2+2}$	$173$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^4+x^3+16*x^2+3*x+1)*exp(2)-2*x^4-3*x^3+17*x^2+6*x+2)*exp(x)+(-20*x^3+75*x^2+10*x+5)*exp(2)-40*x^3+75
*x^2+20*x+10)/(exp(2)+2),x,method=_RETURNVERBOSE)

[Out]

exp(x)*x+exp(x)*x^2+5*x+5*x^2-5*x^4-exp(x)*x^4+25*(exp(2)+1)/(exp(2)+2)*x^3+5*(exp(2)+1)/(exp(2)+2)*x^3*exp(x)

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maxima [B]  time = 0.38, size = 85, normalized size = 2.93 $$\begin{array}{c}-\frac{10x^4-25x^3-10x^2+5(x^4-5x^3-x^2-x)e^2+(x^4(e^2+2)-5x^3(e^2+1)-x^2(e^2+2)-x(e^2+2))e^x-10x}{e^2+2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+x^3+16*x^2+3*x+1)*exp(2)-2*x^4-3*x^3+17*x^2+6*x+2)*exp(x)+(-20*x^3+75*x^2+10*x+5)*exp(2)-40*
x^3+75*x^2+20*x+10)/(exp(2)+2),x, algorithm="maxima")

[Out]

-(10*x^4 - 25*x^3 - 10*x^2 + 5*(x^4 - 5*x^3 - x^2 - x)*e^2 + (x^4*(e^2 + 2) - 5*x^3*(e^2 + 1) - x^2*(e^2 + 2)
- x*(e^2 + 2))*e^x - 10*x)/(e^2 + 2)

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mupad [B]  time = 2.04, size = 47, normalized size = 1.62 $$\begin{array}{c}\frac{x({\mathrm{e}}^x+5)(2x+{\mathrm{e}}^2+x{\mathrm{e}}^2+5x^2{\mathrm{e}}^2-x^3{\mathrm{e}}^2+5x^2-2x^3+2)}{{\mathrm{e}}^2+2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x + exp(x)*(6*x + exp(2)*(3*x + 16*x^2 + x^3 - x^4 + 1) + 17*x^2 - 3*x^3 - 2*x^4 + 2) + exp(2)*(10*x +
75*x^2 - 20*x^3 + 5) + 75*x^2 - 40*x^3 + 10)/(exp(2) + 2),x)

[Out]

(x*(exp(x) + 5)*(2*x + exp(2) + x*exp(2) + 5*x^2*exp(2) - x^3*exp(2) + 5*x^2 - 2*x^3 + 2))/(exp(2) + 2)

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sympy [B]  time = 0.23, size = 82, normalized size = 2.83 $$\begin{array}{c}-5x^4+\frac{x^3(25+25e^2)}{2+e^2}+5x^2+5x+\frac{(-x^4{e}^2-2x^4+5x^3+5x^3{e}^2+2x^2+x^2{e}^2+2x+x{e}^2)e^x}{2+e^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**4+x**3+16*x**2+3*x+1)*exp(2)-2*x**4-3*x**3+17*x**2+6*x+2)*exp(x)+(-20*x**3+75*x**2+10*x+5)*ex
p(2)-40*x**3+75*x**2+20*x+10)/(exp(2)+2),x)

[Out]

-5*x**4 + x**3*(25 + 25*exp(2))/(2 + exp(2)) + 5*x**2 + 5*x + (-x**4*exp(2) - 2*x**4 + 5*x**3 + 5*x**3*exp(2)
+ 2*x**2 + x**2*exp(2) + 2*x + x*exp(2))*exp(x)/(2 + exp(2))

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