3.3.9 \(\int \frac {-2 e^3 x^2+e^{3+\frac {1}{4} (4 e+x^2)} (-882-84 x+439 x^2+42 x^3+x^4)}{882 x^2+84 x^3+2 x^4} \, dx\)

Optimal. Leaf size=28 \[ 1+\frac {e^3 \left (e^{e+\frac {x^2}{4}}+\frac {x}{21+x}\right )}{x} \]

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Rubi [A]  time = 0.38, antiderivative size = 26, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {1594, 27, 12, 6688, 2288} \begin {gather*} \frac {e^{\frac {x^2}{4}+e+3}}{x}+\frac {e^3}{x+21} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^3*x^2 + E^(3 + (4*E + x^2)/4)*(-882 - 84*x + 439*x^2 + 42*x^3 + x^4))/(882*x^2 + 84*x^3 + 2*x^4),x]

[Out]

E^(3 + E + x^2/4)/x + E^3/(21 + 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 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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 e^3 x^2+e^{3+\frac {1}{4} \left (4 e+x^2\right )} \left (-882-84 x+439 x^2+42 x^3+x^4\right )}{x^2 \left (882+84 x+2 x^2\right )} \, dx\\ &=\int \frac {-2 e^3 x^2+e^{3+\frac {1}{4} \left (4 e+x^2\right )} \left (-882-84 x+439 x^2+42 x^3+x^4\right )}{2 x^2 (21+x)^2} \, dx\\ &=\frac {1}{2} \int \frac {-2 e^3 x^2+e^{3+\frac {1}{4} \left (4 e+x^2\right )} \left (-882-84 x+439 x^2+42 x^3+x^4\right )}{x^2 (21+x)^2} \, dx\\ &=\frac {1}{2} \int \left (e^{3+e+\frac {x^2}{4}} \left (1-\frac {2}{x^2}\right )-\frac {2 e^3}{(21+x)^2}\right ) \, dx\\ &=\frac {e^3}{21+x}+\frac {1}{2} \int e^{3+e+\frac {x^2}{4}} \left (1-\frac {2}{x^2}\right ) \, dx\\ &=\frac {e^{3+e+\frac {x^2}{4}}}{x}+\frac {e^3}{21+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 26, normalized size = 0.93 \begin {gather*} \frac {e^{3+e+\frac {x^2}{4}}}{x}+\frac {e^3}{21+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^3*x^2 + E^(3 + (4*E + x^2)/4)*(-882 - 84*x + 439*x^2 + 42*x^3 + x^4))/(882*x^2 + 84*x^3 + 2*x^
4),x]

[Out]

E^(3 + E + x^2/4)/x + E^3/(21 + x)

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fricas [A]  time = 0.75, size = 29, normalized size = 1.04 \begin {gather*} \frac {x e^{3} + {\left (x + 21\right )} e^{\left (\frac {1}{4} \, x^{2} + e + 3\right )}}{x^{2} + 21 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+42*x^3+439*x^2-84*x-882)*exp(3)*exp(exp(1)+1/4*x^2)-2*x^2*exp(3))/(2*x^4+84*x^3+882*x^2),x, al
gorithm="fricas")

[Out]

(x*e^3 + (x + 21)*e^(1/4*x^2 + e + 3))/(x^2 + 21*x)

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giac [A]  time = 0.45, size = 39, normalized size = 1.39 \begin {gather*} \frac {x e^{3} + x e^{\left (\frac {1}{4} \, x^{2} + e + 3\right )} + 21 \, e^{\left (\frac {1}{4} \, x^{2} + e + 3\right )}}{x^{2} + 21 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+42*x^3+439*x^2-84*x-882)*exp(3)*exp(exp(1)+1/4*x^2)-2*x^2*exp(3))/(2*x^4+84*x^3+882*x^2),x, al
gorithm="giac")

[Out]

(x*e^3 + x*e^(1/4*x^2 + e + 3) + 21*e^(1/4*x^2 + e + 3))/(x^2 + 21*x)

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maple [A]  time = 0.10, size = 24, normalized size = 0.86

method	result	size
risch	\(\frac {{\mathrm e}^{3}}{x +21}+\frac {{\mathrm e}^{3+{\mathrm e}+\frac {x^{2}}{4}}}{x}\)	\(24\)
default	\(\frac {{\mathrm e}^{3+{\mathrm e}} {\mathrm e}^{\frac {x^{2}}{4}}}{x}+\frac {{\mathrm e}^{3}}{x +21}\)	\(25\)
norman	\(\frac {x \,{\mathrm e}^{3}+x \,{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}+\frac {x^{2}}{4}}+21 \,{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}+\frac {x^{2}}{4}}}{x \left (x +21\right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4+42*x^3+439*x^2-84*x-882)*exp(3)*exp(exp(1)+1/4*x^2)-2*x^2*exp(3))/(2*x^4+84*x^3+882*x^2),x,method=_R
ETURNVERBOSE)

[Out]

exp(3)/(x+21)+1/x*exp(3+exp(1)+1/4*x^2)

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maxima [A]  time = 0.77, size = 23, normalized size = 0.82 \begin {gather*} \frac {e^{3}}{x + 21} + \frac {e^{\left (\frac {1}{4} \, x^{2} + e + 3\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4+42*x^3+439*x^2-84*x-882)*exp(3)*exp(exp(1)+1/4*x^2)-2*x^2*exp(3))/(2*x^4+84*x^3+882*x^2),x, al
gorithm="maxima")

[Out]

e^3/(x + 21) + e^(1/4*x^2 + e + 3)/x

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mupad [B]  time = 0.30, size = 37, normalized size = 1.32 \begin {gather*} \frac {21\,{\mathrm {e}}^{\frac {x^2}{4}+\mathrm {e}+3}+x\,\left ({\mathrm {e}}^{\frac {x^2}{4}+\mathrm {e}+3}+{\mathrm {e}}^3\right )}{x\,\left (x+21\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^2*exp(3) - exp(exp(1) + x^2/4)*exp(3)*(439*x^2 - 84*x + 42*x^3 + x^4 - 882))/(882*x^2 + 84*x^3 + 2*x
^4),x)

[Out]

(21*exp(exp(1) + x^2/4 + 3) + x*(exp(exp(1) + x^2/4 + 3) + exp(3)))/(x*(x + 21))

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sympy [A]  time = 0.17, size = 20, normalized size = 0.71 \begin {gather*} \frac {e^{3}}{x + 21} + \frac {e^{3} e^{\frac {x^{2}}{4} + e}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**4+42*x**3+439*x**2-84*x-882)*exp(3)*exp(exp(1)+1/4*x**2)-2*x**2*exp(3))/(2*x**4+84*x**3+882*x**
2),x)

[Out]

exp(3)/(x + 21) + exp(3)*exp(x**2/4 + E)/x

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