3.58.100 \(\int \frac {3 x^2+8 x^3+x^4+e^2 (5+2 x-x^2)+e^6 (e^2 x^2+2 x^3)}{1+8 x+18 x^2+e^{12} x^2+8 x^3+x^4+e^6 (2 x+8 x^2+2 x^3)} \, dx\)

Optimal. Leaf size=23 \[ \frac {x^2+e^2 (5+x)}{4+e^6+\frac {1}{x}+x} \]

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Rubi [B]  time = 0.23, antiderivative size = 81, normalized size of antiderivative = 3.52, number of steps used = 5, number of rules used = 5, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {6, 1680, 1814, 21, 8} \begin {gather*} \frac {\left (2+e^6\right ) \left (6+e^6\right ) \left (15+e^2+8 e^6-e^8+e^{12}\right ) x+\left (2+e^6\right ) \left (6+e^6\right ) \left (4-e^2+e^6\right )}{\left (12+8 e^6+e^{12}\right ) \left (x^2+\left (4+e^6\right ) x+1\right )}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*x^2 + 8*x^3 + x^4 + E^2*(5 + 2*x - x^2) + E^6*(E^2*x^2 + 2*x^3))/(1 + 8*x + 18*x^2 + E^12*x^2 + 8*x^3 +
x^4 + E^6*(2*x + 8*x^2 + 2*x^3)),x]

[Out]

x + ((2 + E^6)*(6 + E^6)*(4 - E^2 + E^6) + (2 + E^6)*(6 + E^6)*(15 + E^2 + 8*E^6 - E^8 + E^12)*x)/((12 + 8*E^6
+ E^12)*(1 + (4 + E^6)*x + x^2))

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 x^2+8 x^3+x^4+e^2 \left (5+2 x-x^2\right )+e^6 \left (e^2 x^2+2 x^3\right )}{1+8 x+\left (18+e^{12}\right ) x^2+8 x^3+x^4+e^6 \left (2 x+8 x^2+2 x^3\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {-\left (\left (2+e^6\right ) \left (6+e^6\right ) \left (48+4 e^2+24 e^6-4 e^8+3 e^{12}\right )\right )+16 \left (52+6 e^2+45 e^6-3 e^8+12 e^{12}-e^{14}+e^{18}\right ) x-8 \left (42+2 e^2+24 e^6-2 e^8+3 e^{12}\right ) x^2+16 x^4}{\left (12+8 e^6+e^{12}-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (8+2 e^6\right )+x\right )\\ &=\frac {\left (2+e^6\right ) \left (6+e^6\right ) \left (4-e^2+e^6\right )+\left (2+e^6\right ) \left (6+e^6\right ) \left (15+e^2+8 e^6-e^8+e^{12}\right ) x}{\left (12+8 e^6+e^{12}\right ) \left (1+\left (4+e^6\right ) x+x^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {-2 \left (12+8 e^6+e^{12}\right )^2+8 \left (12+8 e^6+e^{12}\right ) x^2}{12+8 e^6+e^{12}-4 x^2} \, dx,x,\frac {1}{4} \left (8+2 e^6\right )+x\right )}{2 \left (12+8 e^6+e^{12}\right )}\\ &=\frac {\left (2+e^6\right ) \left (6+e^6\right ) \left (4-e^2+e^6\right )+\left (2+e^6\right ) \left (6+e^6\right ) \left (15+e^2+8 e^6-e^8+e^{12}\right ) x}{\left (12+8 e^6+e^{12}\right ) \left (1+\left (4+e^6\right ) x+x^2\right )}+\operatorname {Subst}\left (\int 1 \, dx,x,\frac {1}{4} \left (8+2 e^6\right )+x\right )\\ &=x+\frac {\left (2+e^6\right ) \left (6+e^6\right ) \left (4-e^2+e^6\right )+\left (2+e^6\right ) \left (6+e^6\right ) \left (15+e^2+8 e^6-e^8+e^{12}\right ) x}{\left (12+8 e^6+e^{12}\right ) \left (1+\left (4+e^6\right ) x+x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 49, normalized size = 2.13 \begin {gather*} x+\frac {4+e^2 (-1+x)+15 x-e^8 x+e^{12} x+e^6 (1+8 x)}{1+\left (4+e^6\right ) x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*x^2 + 8*x^3 + x^4 + E^2*(5 + 2*x - x^2) + E^6*(E^2*x^2 + 2*x^3))/(1 + 8*x + 18*x^2 + E^12*x^2 + 8
*x^3 + x^4 + E^6*(2*x + 8*x^2 + 2*x^3)),x]

[Out]

x + (4 + E^2*(-1 + x) + 15*x - E^8*x + E^12*x + E^6*(1 + 8*x))/(1 + (4 + E^6)*x + x^2)

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fricas [B]  time = 0.64, size = 54, normalized size = 2.35 \begin {gather*} \frac {x^{3} + 4 \, x^{2} + x e^{12} - x e^{8} + {\left (x^{2} + 8 \, x + 1\right )} e^{6} + {\left (x - 1\right )} e^{2} + 16 \, x + 4}{x^{2} + x e^{6} + 4 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(2)+2*x^3)*exp(3)^2+(-x^2+2*x+5)*exp(2)+x^4+8*x^3+3*x^2)/(x^2*exp(3)^4+(2*x^3+8*x^2+2*x)*ex
p(3)^2+x^4+8*x^3+18*x^2+8*x+1),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(2)+2*x^3)*exp(3)^2+(-x^2+2*x+5)*exp(2)+x^4+8*x^3+3*x^2)/(x^2*exp(3)^4+(2*x^3+8*x^2+2*x)*ex
p(3)^2+x^4+8*x^3+18*x^2+8*x+1),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.12, size = 39, normalized size = 1.70

method	result	size
norman	\(\frac {x^{3}+\left (-{\mathrm e}^{6} {\mathrm e}^{2}+{\mathrm e}^{2}\right ) x -{\mathrm e}^{2}}{1+x \,{\mathrm e}^{6}+x^{2}+4 x}\)	\(39\)
gosper	\(-\frac {{\mathrm e}^{2} {\mathrm e}^{6} x -x^{3}-{\mathrm e}^{2} x +{\mathrm e}^{2}}{1+x \,{\mathrm e}^{6}+x^{2}+4 x}\)	\(40\)
risch	\(x +\frac {\left (8 \,{\mathrm e}^{6}+{\mathrm e}^{12}+15+{\mathrm e}^{2}-{\mathrm e}^{8}\right ) x +{\mathrm e}^{6}-{\mathrm e}^{2}+4}{1+x \,{\mathrm e}^{6}+x^{2}+4 x}\)	\(42\)
default	\(x -\frac {-\frac {\left ({\mathrm e}^{2} {\mathrm e}^{12}+8 \,{\mathrm e}^{18}+\left ({\mathrm e}^{12}\right )^{2}-{\mathrm e}^{12} {\mathrm e}^{8}+10 \,{\mathrm e}^{6} {\mathrm e}^{2}+91 \,{\mathrm e}^{12}+8 \,{\mathrm e}^{6} {\mathrm e}^{12}-8 \,{\mathrm e}^{6} {\mathrm e}^{8}+12 \,{\mathrm e}^{2}+216 \,{\mathrm e}^{6}-14 \,{\mathrm e}^{8}+180\right ) x}{{\mathrm e}^{12}+8 \,{\mathrm e}^{6}+12}+\frac {4 \,{\mathrm e}^{6} {\mathrm e}^{2}-12 \,{\mathrm e}^{12}-{\mathrm e}^{6} {\mathrm e}^{12}+{\mathrm e}^{6} {\mathrm e}^{8}+12 \,{\mathrm e}^{2}-44 \,{\mathrm e}^{6}+4 \,{\mathrm e}^{8}-48}{{\mathrm e}^{12}+8 \,{\mathrm e}^{6}+12}}{1+x \,{\mathrm e}^{6}+x^{2}+4 x}-\frac {4 \left ({\mathrm e}^{6} {\mathrm e}^{2}-{\mathrm e}^{8}\right ) \arctanh \left (\frac {{\mathrm e}^{6}+2 x +4}{\sqrt {{\mathrm e}^{12}+8 \,{\mathrm e}^{6}+12}}\right )}{\left ({\mathrm e}^{12}+8 \,{\mathrm e}^{6}+12\right )^{\frac {3}{2}}}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*exp(2)+2*x^3)*exp(3)^2+(-x^2+2*x+5)*exp(2)+x^4+8*x^3+3*x^2)/(x^2*exp(3)^4+(2*x^3+8*x^2+2*x)*exp(3)^2
+x^4+8*x^3+18*x^2+8*x+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 40, normalized size = 1.74 \begin {gather*} x + \frac {x {\left (e^{12} - e^{8} + 8 \, e^{6} + e^{2} + 15\right )} + e^{6} - e^{2} + 4}{x^{2} + x {\left (e^{6} + 4\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2*exp(2)+2*x^3)*exp(3)^2+(-x^2+2*x+5)*exp(2)+x^4+8*x^3+3*x^2)/(x^2*exp(3)^4+(2*x^3+8*x^2+2*x)*ex
p(3)^2+x^4+8*x^3+18*x^2+8*x+1),x, algorithm="maxima")

[Out]

x + (x*(e^12 - e^8 + 8*e^6 + e^2 + 15) + e^6 - e^2 + 4)/(x^2 + x*(e^6 + 4) + 1)

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mupad [B]  time = 0.18, size = 40, normalized size = 1.74 \begin {gather*} x+\frac {{\mathrm {e}}^6-{\mathrm {e}}^2+x\,\left ({\mathrm {e}}^2+8\,{\mathrm {e}}^6-{\mathrm {e}}^8+{\mathrm {e}}^{12}+15\right )+4}{x^2+\left ({\mathrm {e}}^6+4\right )\,x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)*(2*x - x^2 + 5) + 3*x^2 + 8*x^3 + x^4 + exp(6)*(x^2*exp(2) + 2*x^3))/(8*x + exp(6)*(2*x + 8*x^2 +
2*x^3) + x^2*exp(12) + 18*x^2 + 8*x^3 + x^4 + 1),x)

[Out]

x + (exp(6) - exp(2) + x*(exp(2) + 8*exp(6) - exp(8) + exp(12) + 15) + 4)/(x*(exp(6) + 4) + x^2 + 1)

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sympy [B]  time = 1.50, size = 39, normalized size = 1.70 \begin {gather*} x + \frac {x \left (- e^{8} + e^{2} + 15 + 8 e^{6} + e^{12}\right ) - e^{2} + 4 + e^{6}}{x^{2} + x \left (4 + e^{6}\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2*exp(2)+2*x**3)*exp(3)**2+(-x**2+2*x+5)*exp(2)+x**4+8*x**3+3*x**2)/(x**2*exp(3)**4+(2*x**3+8*x
**2+2*x)*exp(3)**2+x**4+8*x**3+18*x**2+8*x+1),x)

[Out]

x + (x*(-exp(8) + exp(2) + 15 + 8*exp(6) + exp(12)) - exp(2) + 4 + exp(6))/(x**2 + x*(4 + exp(6)) + 1)

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