3.33.1 $\int \frac{-16-4e^2+e(-16-8x)-16x-4x^2+e^2(10x^2+2e^2{x}^2+8x^3+2x^4+e(9x^2+4x^3))+e^{2+x^2}(8x^3+2e^2{x}^3+8x^4+2x^5+e(8x^3+4x^4))}{e^2(4x^2+e^2{x}^2+4x^3+x^4+e(4x^2+2x^3))}dx$

Optimal. Leaf size=25 $$e^{x^2}+\frac{4}{e^{2}x}+2x+\frac{x}{2+e+x}$$

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Rubi [B]  time = 1.07, antiderivative size = 213, normalized size of antiderivative = 8.52, number of steps used = 15, number of rules used = 9, integrand size = 148, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.061, Rules used = {6, 12, 6741, 27, 6742, 2209, 44, 77, 683} $$\begin{array}{c}{e}^{x^2}+2x+\frac{4(4+e^2)}{e^2(2+e{)}^2(x+e+2)}-\frac{2+e}{x+e+2}-\frac{16}{e^2(2+e)(x+e+2)}-\frac{8}{(2+e{)}^2(x+e+2)}+\frac{4}{e^2(x+e+2)}+\frac{4(4+e^2)}{e^2(2+e{)}^{2}x}+\frac{16}{e(2+e{)}^{2}x}+\frac{8(4+e^2)\log (x)}{e^2(2+e{)}^3}-\frac{16\log (x)}{e^2(2+e{)}^2}+\frac{8(2-e)\log (x)}{e(2+e{)}^3}-\frac{8(4+e^2)\log (x+e+2)}{e^2(2+e{)}^3}+\frac{16\log (x+e+2)}{e^2(2+e{)}^2}-\frac{8(2-e)\log (x+e+2)}{e(2+e{)}^3}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-16 - 4*E^2 + E*(-16 - 8*x) - 16*x - 4*x^2 + E^2*(10*x^2 + 2*E^2*x^2 + 8*x^3 + 2*x^4 + E*(9*x^2 + 4*x^3))
+ E^(2 + x^2)*(8*x^3 + 2*E^2*x^3 + 8*x^4 + 2*x^5 + E*(8*x^3 + 4*x^4)))/(E^2*(4*x^2 + E^2*x^2 + 4*x^3 + x^4 +
E*(4*x^2 + 2*x^3))),x]

[Out]

E^x^2 + 16/(E*(2 + E)^2*x) + (4*(4 + E^2))/(E^2*(2 + E)^2*x) + 2*x + 4/(E^2*(2 + E + x)) - 8/((2 + E)^2*(2 + E
+ x)) - 16/(E^2*(2 + E)*(2 + E + x)) - (2 + E)/(2 + E + x) + (4*(4 + E^2))/(E^2*(2 + E)^2*(2 + E + x)) + (8*(
2 - E)*Log[x])/(E*(2 + E)^3) - (16*Log[x])/(E^2*(2 + E)^2) + (8*(4 + E^2)*Log[x])/(E^2*(2 + E)^3) - (8*(2 - E)
*Log[2 + E + x])/(E*(2 + E)^3) + (16*Log[2 + E + x])/(E^2*(2 + E)^2) - (8*(4 + E^2)*Log[2 + E + x])/(E^2*(2 +
E)^3)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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{-16-4e^2+e(-16-8x)-16x-4x^2+e^2(10x^2+2e^2{x}^2+8x^3+2x^4+e(9x^2+4x^3))+e^{2+x^2}(8x^3+2e^2{x}^3+8x^4+2x^5+e(8x^3+4x^4))}{e^2((4+e^2)x^2+4x^3+x^4+e(4x^2+2x^3))}dx\\ & =\frac{\int \frac{-16-4e^2+e(-16-8x)-16x-4x^2+e^2(10x^2+2e^2{x}^2+8x^3+2x^4+e(9x^2+4x^3))+e^{2+x^2}(8x^3+2e^2{x}^3+8x^4+2x^5+e(8x^3+4x^4))}{(4+e^2)x^2+4x^3+x^4+e(4x^2+2x^3)}dx}{e^2}\\ & =\frac{\int \frac{-16(1+\frac{e^2}{4})+e(-16-8x)-16x-4x^2+e^2(10x^2+2e^2{x}^2+8x^3+2x^4+e(9x^2+4x^3))+e^{2+x^2}(8x^3+2e^2{x}^3+8x^4+2x^5+e(8x^3+4x^4))}{x^2((2+e{)}^2+2(2+e)x+x^2)}dx}{e^2}\\ & =\frac{\int \frac{-16(1+\frac{e^2}{4})+e(-16-8x)-16x-4x^2+e^2(10x^2+2e^2{x}^2+8x^3+2x^4+e(9x^2+4x^3))+e^{2+x^2}(8x^3+2e^2{x}^3+8x^4+2x^5+e(8x^3+4x^4))}{x^2(2+e+x{)}^2}dx}{e^2}\\ & =\frac{\int (2e^{2+x^2}x-\frac{4}{(2+e+x{)}^2}-\frac{4(4+e^2)}{x^2(2+e+x{)}^2}-\frac{16}{x(2+e+x{)}^2}-\frac{8e(2+x)}{x^2(2+e+x{)}^2}+\frac{e^2((2+e)(5+2e)+4(2+e)x+2x^2)}{(2+e+x{)}^2})dx}{e^2}\\ & =\frac{4}{e^2(2+e+x)}+\frac{2\int e^{2+x^2}xdx}{e^2}-\frac{16\int \frac{1}{x(2+e+x{)}^2}dx}{e^2}-\frac{8\int \frac{2+x}{x^2(2+e+x{)}^2}dx}{e}-\frac{\left(4(4+e^2)\right)\int \frac{1}{x^2(2+e+x{)}^2}dx}{e^2}+\int \frac{(2+e)(5+2e)+4(2+e)x+2x^2}{(2+e+x{)}^2}dx\\ & =e^{x^2}+\frac{4}{e^2(2+e+x)}-\frac{16\int (\frac{1}{(2+e{)}^{2}x}-\frac{1}{(2+e)(2+e+x{)}^2}-\frac{1}{(2+e{)}^2(2+e+x)})dx}{e^2}-\frac{8\int (\frac{2}{(2+e{)}^2{x}^2}+\frac{-2+e}{(2+e{)}^{3}x}-\frac{e}{(2+e{)}^2(2+e+x{)}^2}+\frac{2-e}{(2+e{)}^3(2+e+x)})dx}{e}-\frac{\left(4(4+e^2)\right)\int (\frac{1}{(2+e{)}^2{x}^2}-\frac{2}{(2+e{)}^{3}x}+\frac{1}{(2+e{)}^2(2+e+x{)}^2}+\frac{2}{(2+e{)}^3(2+e+x)})dx}{e^2}+\int (2+\frac{2+e}{(2+e+x{)}^2})dx\\ & =e^{x^2}+\frac{16}{e(2+e{)}^{2}x}+\frac{4(4+e^2)}{e^2(2+e{)}^{2}x}+2x+\frac{4}{e^2(2+e+x)}-\frac{8}{(2+e{)}^2(2+e+x)}-\frac{16}{e^2(2+e)(2+e+x)}-\frac{2+e}{2+e+x}+\frac{4(4+e^2)}{e^2(2+e{)}^2(2+e+x)}+\frac{8(2-e)\log (x)}{e(2+e{)}^3}-\frac{16\log (x)}{e^2(2+e{)}^2}+\frac{8(4+e^2)\log (x)}{e^2(2+e{)}^3}-\frac{8(2-e)\log (2+e+x)}{e(2+e{)}^3}+\frac{16\log (2+e+x)}{e^2(2+e{)}^2}-\frac{8(4+e^2)\log (2+e+x)}{e^2(2+e{)}^3}\end{array}\end{array}$$

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Mathematica [A]  time = 0.07, size = 37, normalized size = 1.48 $$\begin{array}{c}\frac{e^{2+x^2}+\frac{4}{x}+2e^{2}x-\frac{e^2(2+e)}{2+e+x}}{e^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 4*E^2 + E*(-16 - 8*x) - 16*x - 4*x^2 + E^2*(10*x^2 + 2*E^2*x^2 + 8*x^3 + 2*x^4 + E*(9*x^2 + 4
*x^3)) + E^(2 + x^2)*(8*x^3 + 2*E^2*x^3 + 8*x^4 + 2*x^5 + E*(8*x^3 + 4*x^4)))/(E^2*(4*x^2 + E^2*x^2 + 4*x^3 +
x^4 + E*(4*x^2 + 2*x^3))),x]

[Out]

(E^(2 + x^2) + 4/x + 2*E^2*x - (E^2*(2 + E))/(2 + E + x))/E^2

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fricas [B]  time = 0.63, size = 73, normalized size = 2.92 $$\begin{array}{c}\frac{(2x^2-x)e^3+2(x^3+2x^2-x)e^2+(x^2+xe+2x)e^{(x^2+2)}+4x+4e+8}{x{e}^3+(x^2+2x)e^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(1)^2+(4*x^4+8*x^3)*exp(1)+2*x^5+8*x^4+8*x^3)*exp(2)*exp(x^2)+(2*x^2*exp(1)^2+(4*x^3+9*x^
2)*exp(1)+2*x^4+8*x^3+10*x^2)*exp(2)-4*exp(1)^2+(-8*x-16)*exp(1)-4*x^2-16*x-16)/(x^2*exp(1)^2+(2*x^3+4*x^2)*ex
p(1)+x^4+4*x^3+4*x^2)/exp(2),x, algorithm="fricas")

[Out]

((2*x^2 - x)*e^3 + 2*(x^3 + 2*x^2 - x)*e^2 + (x^2 + x*e + 2*x)*e^(x^2 + 2) + 4*x + 4*e + 8)/(x*e^3 + (x^2 + 2*
x)*e^2)

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giac [B]  time = 0.32, size = 83, normalized size = 3.32 $$\begin{array}{c}\frac{(2x^3{e}^2+2x^2{e}^3+4x^2{e}^2+x^2{e}^{(x^2+2)}-x{e}^3-2x{e}^2+x{e}^{(x^2+3)}+2x{e}^{(x^2+2)}+4x+4e+8)e^{(-2)}}{x^2+xe+2x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(1)^2+(4*x^4+8*x^3)*exp(1)+2*x^5+8*x^4+8*x^3)*exp(2)*exp(x^2)+(2*x^2*exp(1)^2+(4*x^3+9*x^
2)*exp(1)+2*x^4+8*x^3+10*x^2)*exp(2)-4*exp(1)^2+(-8*x-16)*exp(1)-4*x^2-16*x-16)/(x^2*exp(1)^2+(2*x^3+4*x^2)*ex
p(1)+x^4+4*x^3+4*x^2)/exp(2),x, algorithm="giac")

[Out]

(2*x^3*e^2 + 2*x^2*e^3 + 4*x^2*e^2 + x^2*e^(x^2 + 2) - x*e^3 - 2*x*e^2 + x*e^(x^2 + 3) + 2*x*e^(x^2 + 2) + 4*x
+ 4*e + 8)*e^(-2)/(x^2 + x*e + 2*x)

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maple [A]  time = 0.62, size = 40, normalized size = 1.60

method	result	size
risch	$2x+\frac{{\mathrm{e}}^{-2}((-{\mathrm{e}}^3-2{\mathrm{e}}^2+4)x+4\mathrm{e}+8)}{(2+x+\mathrm{e})x}+{\mathrm{e}}^{x^2}$	$40$
norman	$\frac{x^2{\mathrm{e}}^{x^2}-(2{\left({\mathrm{e}}^2\right)}^2+9\mathrm{e}{\mathrm{e}}^2+10{\mathrm{e}}^2-4){\mathrm{e}}^{-2}x+x(\mathrm{e}+2){\mathrm{e}}^{x^2}+2x^3+4(\mathrm{e}+2){\mathrm{e}}^{-2}}{x(2+x+\mathrm{e})}$	$73$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3*exp(1)^2+(4*x^4+8*x^3)*exp(1)+2*x^5+8*x^4+8*x^3)*exp(2)*exp(x^2)+(2*x^2*exp(1)^2+(4*x^3+9*x^2)*exp
(1)+2*x^4+8*x^3+10*x^2)*exp(2)-4*exp(1)^2+(-8*x-16)*exp(1)-4*x^2-16*x-16)/(x^2*exp(1)^2+(2*x^3+4*x^2)*exp(1)+x
^4+4*x^3+4*x^2)/exp(2),x,method=_RETURNVERBOSE)

[Out]

2*x+exp(-2)*((-exp(3)-2*exp(2)+4)*x+4*exp(1)+8)/(2+x+exp(1))/x+exp(x^2)

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maxima [B]  time = 0.93, size = 480, normalized size = 19.20 $$\begin{array}{c}(4(\frac{e+2}{x+e+2}+\log (x+e+2))e^3-2(2(e+2)\log (x+e+2)-x+\frac{e^2+4e+4}{x+e+2})e^2+4(\frac{2x+e+2}{x^2(e^2+4e+4)+x(e^3+6e^2+12e+8)}-\frac{2\log (x+e+2)}{e^3+6e^2+12e+8}+\frac{2\log (x)}{e^3+6e^2+12e+8})e^2+8(\frac{e+2}{x+e+2}+\log (x+e+2))e^2+16(\frac{2x+e+2}{x^2(e^2+4e+4)+x(e^3+6e^2+12e+8)}-\frac{2\log (x+e+2)}{e^3+6e^2+12e+8}+\frac{2\log (x)}{e^3+6e^2+12e+8})e+8(\frac{\log (x+e+2)}{e^2+4e+4}-\frac{\log (x)}{e^2+4e+4}-\frac{1}{x(e+2)+e^2+4e+4})e+\frac{16(2x+e+2)}{x^2(e^2+4e+4)+x(e^3+6e^2+12e+8)}-\frac{2e^4}{x+e+2}-\frac{9e^3}{x+e+2}-\frac{10e^2}{x+e+2}-\frac{32\log (x+e+2)}{e^3+6e^2+12e+8}+\frac{16\log (x+e+2)}{e^2+4e+4}+\frac{32\log (x)}{e^3+6e^2+12e+8}-\frac{16\log (x)}{e^2+4e+4}-\frac{16}{x(e+2)+e^2+4e+4}+\frac{4}{x+e+2}+e^{(x^2+2)})e^{(-2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(1)^2+(4*x^4+8*x^3)*exp(1)+2*x^5+8*x^4+8*x^3)*exp(2)*exp(x^2)+(2*x^2*exp(1)^2+(4*x^3+9*x^
2)*exp(1)+2*x^4+8*x^3+10*x^2)*exp(2)-4*exp(1)^2+(-8*x-16)*exp(1)-4*x^2-16*x-16)/(x^2*exp(1)^2+(2*x^3+4*x^2)*ex
p(1)+x^4+4*x^3+4*x^2)/exp(2),x, algorithm="maxima")

[Out]

(4*((e + 2)/(x + e + 2) + log(x + e + 2))*e^3 - 2*(2*(e + 2)*log(x + e + 2) - x + (e^2 + 4*e + 4)/(x + e + 2))
*e^2 + 4*((2*x + e + 2)/(x^2*(e^2 + 4*e + 4) + x*(e^3 + 6*e^2 + 12*e + 8)) - 2*log(x + e + 2)/(e^3 + 6*e^2 + 1
2*e + 8) + 2*log(x)/(e^3 + 6*e^2 + 12*e + 8))*e^2 + 8*((e + 2)/(x + e + 2) + log(x + e + 2))*e^2 + 16*((2*x +
e + 2)/(x^2*(e^2 + 4*e + 4) + x*(e^3 + 6*e^2 + 12*e + 8)) - 2*log(x + e + 2)/(e^3 + 6*e^2 + 12*e + 8) + 2*log(
x)/(e^3 + 6*e^2 + 12*e + 8))*e + 8*(log(x + e + 2)/(e^2 + 4*e + 4) - log(x)/(e^2 + 4*e + 4) - 1/(x*(e + 2) + e
^2 + 4*e + 4))*e + 16*(2*x + e + 2)/(x^2*(e^2 + 4*e + 4) + x*(e^3 + 6*e^2 + 12*e + 8)) - 2*e^4/(x + e + 2) - 9
*e^3/(x + e + 2) - 10*e^2/(x + e + 2) - 32*log(x + e + 2)/(e^3 + 6*e^2 + 12*e + 8) + 16*log(x + e + 2)/(e^2 +
4*e + 4) + 32*log(x)/(e^3 + 6*e^2 + 12*e + 8) - 16*log(x)/(e^2 + 4*e + 4) - 16/(x*(e + 2) + e^2 + 4*e + 4) + 4
/(x + e + 2) + e^(x^2 + 2))*e^(-2)

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mupad [B]  time = 2.51, size = 44, normalized size = 1.76 $$\begin{array}{c}2x+{\mathrm{e}}^{x^2}+\frac{4\mathrm{e}-x(2{\mathrm{e}}^2+{\mathrm{e}}^3-4)+8}{{\mathrm{e}}^2{x}^2+(2{\mathrm{e}}^2+{\mathrm{e}}^3)x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*(16*x + 4*exp(2) + 4*x^2 - exp(2)*(exp(1)*(9*x^2 + 4*x^3) + 2*x^2*exp(2) + 10*x^2 + 8*x^3 + 2*x^
4) + exp(1)*(8*x + 16) - exp(x^2)*exp(2)*(exp(1)*(8*x^3 + 4*x^4) + 2*x^3*exp(2) + 8*x^3 + 8*x^4 + 2*x^5) + 16)
)/(exp(1)*(4*x^2 + 2*x^3) + x^2*exp(2) + 4*x^2 + 4*x^3 + x^4),x)

[Out]

2*x + exp(x^2) + (4*exp(1) - x*(2*exp(2) + exp(3) - 4) + 8)/(x*(2*exp(2) + exp(3)) + x^2*exp(2))

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sympy [A]  time = 1.02, size = 42, normalized size = 1.68 $$\begin{array}{c}2x+e^{x^2}+\frac{x(-e^3-2e^2+4)+8+4e}{x^2{e}^2+x(2e^2+e^3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3*exp(1)**2+(4*x**4+8*x**3)*exp(1)+2*x**5+8*x**4+8*x**3)*exp(2)*exp(x**2)+(2*x**2*exp(1)**2+(
4*x**3+9*x**2)*exp(1)+2*x**4+8*x**3+10*x**2)*exp(2)-4*exp(1)**2+(-8*x-16)*exp(1)-4*x**2-16*x-16)/(x**2*exp(1)*
*2+(2*x**3+4*x**2)*exp(1)+x**4+4*x**3+4*x**2)/exp(2),x)

[Out]

2*x + exp(x**2) + (x*(-exp(3) - 2*exp(2) + 4) + 8 + 4*E)/(x**2*exp(2) + x*(2*exp(2) + exp(3)))

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