3.11.27 \(\int \frac {72 x^3+48 x^4+8 x^5+e^{2 x} (27-36 x-33 x^2-6 x^3)+e^{\frac {x-x^2}{3+x}} (-6 x^2+12 x^3+2 x^4)}{36 x^2+24 x^3+4 x^4} \, dx\)

Optimal. Leaf size=36 \[ -2-\frac {1}{2} e^{\frac {x-x^2}{3+x}}-\frac {3 e^{2 x}}{4 x}+x^2 \]

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Rubi [A]  time = 1.11, antiderivative size = 34, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {1594, 27, 12, 6688, 2197, 6706} \begin {gather*} x^2-\frac {1}{2} e^{\frac {(1-x) x}{x+3}}-\frac {3 e^{2 x}}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(72*x^3 + 48*x^4 + 8*x^5 + E^(2*x)*(27 - 36*x - 33*x^2 - 6*x^3) + E^((x - x^2)/(3 + x))*(-6*x^2 + 12*x^3 +
2*x^4))/(36*x^2 + 24*x^3 + 4*x^4),x]

[Out]

-1/2*E^(((1 - x)*x)/(3 + x)) - (3*E^(2*x))/(4*x) + x^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 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, 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 {gather*} \begin {aligned} \text {integral} &=\int \frac {72 x^3+48 x^4+8 x^5+e^{2 x} \left (27-36 x-33 x^2-6 x^3\right )+e^{\frac {x-x^2}{3+x}} \left (-6 x^2+12 x^3+2 x^4\right )}{x^2 \left (36+24 x+4 x^2\right )} \, dx\\ &=\int \frac {72 x^3+48 x^4+8 x^5+e^{2 x} \left (27-36 x-33 x^2-6 x^3\right )+e^{\frac {x-x^2}{3+x}} \left (-6 x^2+12 x^3+2 x^4\right )}{4 x^2 (3+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {72 x^3+48 x^4+8 x^5+e^{2 x} \left (27-36 x-33 x^2-6 x^3\right )+e^{\frac {x-x^2}{3+x}} \left (-6 x^2+12 x^3+2 x^4\right )}{x^2 (3+x)^2} \, dx\\ &=\frac {1}{4} \int \left (8 x-\frac {3 e^{2 x} (-1+2 x)}{x^2}+\frac {2 e^{-\frac {(-1+x) x}{3+x}} \left (-3+6 x+x^2\right )}{(3+x)^2}\right ) \, dx\\ &=x^2+\frac {1}{2} \int \frac {e^{-\frac {(-1+x) x}{3+x}} \left (-3+6 x+x^2\right )}{(3+x)^2} \, dx-\frac {3}{4} \int \frac {e^{2 x} (-1+2 x)}{x^2} \, dx\\ &=-\frac {1}{2} e^{\frac {(1-x) x}{3+x}}-\frac {3 e^{2 x}}{4 x}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 34, normalized size = 0.94 \begin {gather*} -\frac {1}{2} e^{4-x-\frac {12}{3+x}}-\frac {3 e^{2 x}}{4 x}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(72*x^3 + 48*x^4 + 8*x^5 + E^(2*x)*(27 - 36*x - 33*x^2 - 6*x^3) + E^((x - x^2)/(3 + x))*(-6*x^2 + 12
*x^3 + 2*x^4))/(36*x^2 + 24*x^3 + 4*x^4),x]

[Out]

-1/2*E^(4 - x - 12/(3 + x)) - (3*E^(2*x))/(4*x) + x^2

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fricas [A]  time = 0.76, size = 35, normalized size = 0.97 \begin {gather*} \frac {4 \, x^{3} - 2 \, x e^{\left (-\frac {x^{2} - x}{x + 3}\right )} - 3 \, e^{\left (2 \, x\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3-33*x^2-36*x+27)*exp(2*x)+(2*x^4+12*x^3-6*x^2)*exp((-x^2+x)/(3+x))+8*x^5+48*x^4+72*x^3)/(4*x
^4+24*x^3+36*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.35, size = 35, normalized size = 0.97 \begin {gather*} \frac {4 \, x^{3} - 2 \, x e^{\left (-\frac {x^{2} - x}{x + 3}\right )} - 3 \, e^{\left (2 \, x\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3-33*x^2-36*x+27)*exp(2*x)+(2*x^4+12*x^3-6*x^2)*exp((-x^2+x)/(3+x))+8*x^5+48*x^4+72*x^3)/(4*x
^4+24*x^3+36*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 28, normalized size = 0.78

method	result	size
risch	\(x^{2}-\frac {3 \,{\mathrm e}^{2 x}}{4 x}-\frac {{\mathrm e}^{-\frac {x \left (x -1\right )}{3+x}}}{2}\)	\(28\)
norman	\(\frac {x^{4}+3 x^{3}-\frac {3 x \,{\mathrm e}^{\frac {-x^{2}+x}{3+x}}}{2}-\frac {3 x \,{\mathrm e}^{2 x}}{4}-\frac {x^{2} {\mathrm e}^{\frac {-x^{2}+x}{3+x}}}{2}-\frac {9 \,{\mathrm e}^{2 x}}{4}}{\left (3+x \right ) x}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x^3-33*x^2-36*x+27)*exp(2*x)+(2*x^4+12*x^3-6*x^2)*exp((-x^2+x)/(3+x))+8*x^5+48*x^4+72*x^3)/(4*x^4+24*
x^3+36*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.72, size = 33, normalized size = 0.92 \begin {gather*} x^{2} - \frac {{\left (2 \, x e^{\left (-\frac {12}{x + 3} + 4\right )} + 3 \, e^{\left (3 \, x\right )}\right )} e^{\left (-x\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3-33*x^2-36*x+27)*exp(2*x)+(2*x^4+12*x^3-6*x^2)*exp((-x^2+x)/(3+x))+8*x^5+48*x^4+72*x^3)/(4*x
^4+24*x^3+36*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.04, size = 34, normalized size = 0.94 \begin {gather*} x^2-\frac {{\mathrm {e}}^{\frac {x}{x+3}}\,{\mathrm {e}}^{-\frac {x^2}{x+3}}}{2}-\frac {3\,{\mathrm {e}}^{2\,x}}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x - x^2)/(x + 3))*(12*x^3 - 6*x^2 + 2*x^4) - exp(2*x)*(36*x + 33*x^2 + 6*x^3 - 27) + 72*x^3 + 48*x^4
+ 8*x^5)/(36*x^2 + 24*x^3 + 4*x^4),x)

[Out]

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

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sympy [A]  time = 0.31, size = 24, normalized size = 0.67 \begin {gather*} x^{2} - \frac {e^{\frac {- x^{2} + x}{x + 3}}}{2} - \frac {3 e^{2 x}}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x**3-33*x**2-36*x+27)*exp(2*x)+(2*x**4+12*x**3-6*x**2)*exp((-x**2+x)/(3+x))+8*x**5+48*x**4+72*x
**3)/(4*x**4+24*x**3+36*x**2),x)

[Out]

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

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