3.16.4 \(\int \frac {16-64 x-8 x^2+32 x^3+x^4-4 x^5+e^{\frac {-32+x+8 x^2+x^3}{-4+x^2}} (16-4 x-8 x^2-13 x^3+x^4+x^5)}{32-16 x^2+2 x^4} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{2} \left (1+e^{8+x-\frac {5 x}{4-x^2}}-2 x\right ) x \]

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Rubi [B]  time = 0.93, antiderivative size = 143, normalized size of antiderivative = 5.30, number of steps used = 18, number of rules used = 10, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.124, Rules used = {28, 6742, 199, 207, 261, 288, 266, 43, 321, 2288} \begin {gather*} -x^2-\frac {x}{4-x^2}+\frac {x^3}{4 \left (4-x^2\right )}+\frac {e^{\frac {-x^3-8 x^2-x+32}{4-x^2}} \left (-x^5+13 x^3+4 x\right )}{2 \left (4-x^2\right )^2 \left (\frac {3 x^2+16 x+1}{4-x^2}-\frac {2 x \left (-x^3-8 x^2-x+32\right )}{\left (4-x^2\right )^2}\right )}+\frac {3 x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 - 64*x - 8*x^2 + 32*x^3 + x^4 - 4*x^5 + E^((-32 + x + 8*x^2 + x^3)/(-4 + x^2))*(16 - 4*x - 8*x^2 - 13*
x^3 + x^4 + x^5))/(32 - 16*x^2 + 2*x^4),x]

[Out]

(3*x)/4 - x^2 - x/(4 - x^2) + x^3/(4*(4 - x^2)) + (E^((32 - x - 8*x^2 - x^3)/(4 - x^2))*(4*x + 13*x^3 - x^5))/
(2*(4 - x^2)^2*((1 + 16*x + 3*x^2)/(4 - x^2) - (2*x*(32 - x - 8*x^2 - x^3))/(4 - x^2)^2))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 \int \frac {16-64 x-8 x^2+32 x^3+x^4-4 x^5+e^{\frac {-32+x+8 x^2+x^3}{-4+x^2}} \left (16-4 x-8 x^2-13 x^3+x^4+x^5\right )}{\left (-8+2 x^2\right )^2} \, dx\\ &=2 \int \left (\frac {4}{\left (-4+x^2\right )^2}-\frac {16 x}{\left (-4+x^2\right )^2}-\frac {2 x^2}{\left (-4+x^2\right )^2}+\frac {8 x^3}{\left (-4+x^2\right )^2}+\frac {x^4}{4 \left (-4+x^2\right )^2}-\frac {x^5}{\left (-4+x^2\right )^2}+\frac {e^{\frac {-32+x+8 x^2+x^3}{-4+x^2}} \left (16-4 x-8 x^2-13 x^3+x^4+x^5\right )}{4 \left (4-x^2\right )^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {x^4}{\left (-4+x^2\right )^2} \, dx+\frac {1}{2} \int \frac {e^{\frac {-32+x+8 x^2+x^3}{-4+x^2}} \left (16-4 x-8 x^2-13 x^3+x^4+x^5\right )}{\left (4-x^2\right )^2} \, dx-2 \int \frac {x^5}{\left (-4+x^2\right )^2} \, dx-4 \int \frac {x^2}{\left (-4+x^2\right )^2} \, dx+8 \int \frac {1}{\left (-4+x^2\right )^2} \, dx+16 \int \frac {x^3}{\left (-4+x^2\right )^2} \, dx-32 \int \frac {x}{\left (-4+x^2\right )^2} \, dx\\ &=-\frac {16}{4-x^2}-\frac {x}{4-x^2}+\frac {x^3}{4 \left (4-x^2\right )}+\frac {e^{\frac {32-x-8 x^2-x^3}{4-x^2}} \left (4 x+13 x^3-x^5\right )}{2 \left (4-x^2\right )^2 \left (\frac {1+16 x+3 x^2}{4-x^2}-\frac {2 x \left (32-x-8 x^2-x^3\right )}{\left (4-x^2\right )^2}\right )}+\frac {3}{4} \int \frac {x^2}{-4+x^2} \, dx-2 \int \frac {1}{-4+x^2} \, dx+8 \operatorname {Subst}\left (\int \frac {x}{(-4+x)^2} \, dx,x,x^2\right )-\int \frac {1}{-4+x^2} \, dx-\operatorname {Subst}\left (\int \frac {x^2}{(-4+x)^2} \, dx,x,x^2\right )\\ &=\frac {3 x}{4}-\frac {16}{4-x^2}-\frac {x}{4-x^2}+\frac {x^3}{4 \left (4-x^2\right )}+\frac {e^{\frac {32-x-8 x^2-x^3}{4-x^2}} \left (4 x+13 x^3-x^5\right )}{2 \left (4-x^2\right )^2 \left (\frac {1+16 x+3 x^2}{4-x^2}-\frac {2 x \left (32-x-8 x^2-x^3\right )}{\left (4-x^2\right )^2}\right )}+\frac {3}{2} \tanh ^{-1}\left (\frac {x}{2}\right )+3 \int \frac {1}{-4+x^2} \, dx+8 \operatorname {Subst}\left (\int \left (\frac {4}{(-4+x)^2}+\frac {1}{-4+x}\right ) \, dx,x,x^2\right )-\operatorname {Subst}\left (\int \left (1+\frac {16}{(-4+x)^2}+\frac {8}{-4+x}\right ) \, dx,x,x^2\right )\\ &=\frac {3 x}{4}-x^2-\frac {x}{4-x^2}+\frac {x^3}{4 \left (4-x^2\right )}+\frac {e^{\frac {32-x-8 x^2-x^3}{4-x^2}} \left (4 x+13 x^3-x^5\right )}{2 \left (4-x^2\right )^2 \left (\frac {1+16 x+3 x^2}{4-x^2}-\frac {2 x \left (32-x-8 x^2-x^3\right )}{\left (4-x^2\right )^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 31, normalized size = 1.15 \begin {gather*} \frac {1}{2} \left (1+e^{\frac {-32+x+8 x^2+x^3}{-4+x^2}}-2 x\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 - 64*x - 8*x^2 + 32*x^3 + x^4 - 4*x^5 + E^((-32 + x + 8*x^2 + x^3)/(-4 + x^2))*(16 - 4*x - 8*x^2
- 13*x^3 + x^4 + x^5))/(32 - 16*x^2 + 2*x^4),x]

[Out]

((1 + E^((-32 + x + 8*x^2 + x^3)/(-4 + x^2)) - 2*x)*x)/2

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fricas [A]  time = 0.65, size = 32, normalized size = 1.19 \begin {gather*} -x^{2} + \frac {1}{2} \, x e^{\left (\frac {x^{3} + 8 \, x^{2} + x - 32}{x^{2} - 4}\right )} + \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5+x^4-13*x^3-8*x^2-4*x+16)*exp((x^3+8*x^2+x-32)/(x^2-4))-4*x^5+x^4+32*x^3-8*x^2-64*x+16)/(2*x^4-
16*x^2+32),x, algorithm="fricas")

[Out]

-x^2 + 1/2*x*e^((x^3 + 8*x^2 + x - 32)/(x^2 - 4)) + 1/2*x

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giac [A]  time = 0.34, size = 28, normalized size = 1.04 \begin {gather*} -x^{2} + \frac {1}{2} \, x e^{\left (\frac {x^{3} + x}{x^{2} - 4} + 8\right )} + \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5+x^4-13*x^3-8*x^2-4*x+16)*exp((x^3+8*x^2+x-32)/(x^2-4))-4*x^5+x^4+32*x^3-8*x^2-64*x+16)/(2*x^4-
16*x^2+32),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 36, normalized size = 1.33

method	result	size
risch	\(-x^{2}+\frac {x}{2}+\frac {{\mathrm e}^{\frac {x^{3}+8 x^{2}+x -32}{\left (x -2\right ) \left (2+x \right )}} x}{2}\)	\(36\)
norman	\(\frac {-2 x +\frac {x^{3}}{2}-x^{4}-2 \,{\mathrm e}^{\frac {x^{3}+8 x^{2}+x -32}{x^{2}-4}} x +\frac {{\mathrm e}^{\frac {x^{3}+8 x^{2}+x -32}{x^{2}-4}} x^{3}}{2}+16}{x^{2}-4}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^5+x^4-13*x^3-8*x^2-4*x+16)*exp((x^3+8*x^2+x-32)/(x^2-4))-4*x^5+x^4+32*x^3-8*x^2-64*x+16)/(2*x^4-16*x^2
+32),x,method=_RETURNVERBOSE)

[Out]

-x^2+1/2*x+1/2*exp((x^3+8*x^2+x-32)/(x-2)/(2+x))*x

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maxima [A]  time = 0.72, size = 30, normalized size = 1.11 \begin {gather*} -x^{2} + \frac {1}{2} \, x e^{\left (x + \frac {5}{2 \, {\left (x + 2\right )}} + \frac {5}{2 \, {\left (x - 2\right )}} + 8\right )} + \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^5+x^4-13*x^3-8*x^2-4*x+16)*exp((x^3+8*x^2+x-32)/(x^2-4))-4*x^5+x^4+32*x^3-8*x^2-64*x+16)/(2*x^4-
16*x^2+32),x, algorithm="maxima")

[Out]

-x^2 + 1/2*x*e^(x + 5/2/(x + 2) + 5/2/(x - 2) + 8) + 1/2*x

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mupad [B]  time = 1.16, size = 57, normalized size = 2.11 \begin {gather*} \frac {x}{2}-x^2+\frac {x\,{\mathrm {e}}^{\frac {x^3}{x^2-4}}\,{\mathrm {e}}^{\frac {8\,x^2}{x^2-4}}\,{\mathrm {e}}^{-\frac {32}{x^2-4}}\,{\mathrm {e}}^{\frac {x}{x^2-4}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(64*x + exp((x + 8*x^2 + x^3 - 32)/(x^2 - 4))*(4*x + 8*x^2 + 13*x^3 - x^4 - x^5 - 16) + 8*x^2 - 32*x^3 -
x^4 + 4*x^5 - 16)/(2*x^4 - 16*x^2 + 32),x)

[Out]

x/2 - x^2 + (x*exp(x^3/(x^2 - 4))*exp((8*x^2)/(x^2 - 4))*exp(-32/(x^2 - 4))*exp(x/(x^2 - 4)))/2

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sympy [A]  time = 0.24, size = 27, normalized size = 1.00 \begin {gather*} - x^{2} + \frac {x e^{\frac {x^{3} + 8 x^{2} + x - 32}{x^{2} - 4}}}{2} + \frac {x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**5+x**4-13*x**3-8*x**2-4*x+16)*exp((x**3+8*x**2+x-32)/(x**2-4))-4*x**5+x**4+32*x**3-8*x**2-64*x+
16)/(2*x**4-16*x**2+32),x)

[Out]

-x**2 + x*exp((x**3 + 8*x**2 + x - 32)/(x**2 - 4))/2 + x/2

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