3.20.46 \(\int \frac {32-40 x-96 x^2+157 x^3-111 x^5-9 x^7+3 x^9+e^{2 x^2} (2-10 x^2+4 x^4)+e^{x^2} (16-10 x-64 x^2+60 x^3+16 x^4-70 x^5+20 x^7)}{-x^3+3 x^5-3 x^7+x^9} \, dx\)

Optimal. Leaf size=25 \[ 3 x+\left (5+\frac {4+e^{x^2}}{-x+x^3}\right )^2 \]

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Rubi [B]  time = 2.87, antiderivative size = 258, normalized size of antiderivative = 10.32, number of steps used = 55, number of rules used = 16, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6741, 6742, 199, 207, 266, 44, 290, 325, 288, 321, 2177, 2178, 2220, 2204, 2214, 2210} \begin {gather*} -\frac {609 x}{8 \left (1-x^2\right )}-\frac {23 x}{2 \left (1-x^2\right )^2}+\frac {e^{x^2}}{1-x}+\frac {11 e^{x^2}}{x+1}+\frac {e^{2 x^2}}{1-x^2}+\frac {16}{1-x^2}+\frac {2 e^{x^2}}{(1-x)^2}+\frac {2 e^{x^2}}{(x+1)^2}+\frac {e^{2 x^2}}{\left (1-x^2\right )^2}+\frac {16}{\left (1-x^2\right )^2}-\frac {10 e^{x^2}}{x}+\frac {25}{\left (1-x^2\right ) x}+\frac {10}{\left (1-x^2\right )^2 x}+\frac {8 e^{x^2}}{x^2}+\frac {e^{2 x^2}}{x^2}+\frac {16}{x^2}-\frac {3 x^5}{4 \left (1-x^2\right )^2}+\frac {15 x^3}{8 \left (1-x^2\right )}+\frac {9 x^3}{4 \left (1-x^2\right )^2}+\frac {45 x}{8}-\frac {75}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 - 40*x - 96*x^2 + 157*x^3 - 111*x^5 - 9*x^7 + 3*x^9 + E^(2*x^2)*(2 - 10*x^2 + 4*x^4) + E^x^2*(16 - 10*
x - 64*x^2 + 60*x^3 + 16*x^4 - 70*x^5 + 20*x^7))/(-x^3 + 3*x^5 - 3*x^7 + x^9),x]

[Out]

(2*E^x^2)/(1 - x)^2 + E^x^2/(1 - x) + 16/x^2 + (8*E^x^2)/x^2 + E^(2*x^2)/x^2 - 75/x - (10*E^x^2)/x + (45*x)/8
+ (2*E^x^2)/(1 + x)^2 + (11*E^x^2)/(1 + x) + 16/(1 - x^2)^2 + E^(2*x^2)/(1 - x^2)^2 + 10/(x*(1 - x^2)^2) - (23
*x)/(2*(1 - x^2)^2) + (9*x^3)/(4*(1 - x^2)^2) - (3*x^5)/(4*(1 - x^2)^2) + 16/(1 - x^2) + E^(2*x^2)/(1 - x^2) +
25/(x*(1 - x^2)) - (609*x)/(8*(1 - x^2)) + (15*x^3)/(8*(1 - x^2))

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 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 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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 325

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

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2210

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

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2220

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[(f*(e + f*x)^(m +
1)*F^(a + b*(c + d*x)^2))/((m + 1)*f^2), x] + (-Dist[(2*b*d^2*Log[F])/(f^2*(m + 1)), Int[(e + f*x)^(m + 2)*F^
(a + b*(c + d*x)^2), x], x] + Dist[(2*b*d*(d*e - c*f)*Log[F])/(f^2*(m + 1)), Int[(e + f*x)^(m + 1)*F^(a + b*(c
+ d*x)^2), x], x]) /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && LtQ[m, -1]

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-32+40 x+96 x^2-157 x^3+111 x^5+9 x^7-3 x^9-e^{2 x^2} \left (2-10 x^2+4 x^4\right )-e^{x^2} \left (16-10 x-64 x^2+60 x^3+16 x^4-70 x^5+20 x^7\right )}{x^3 \left (1-x^2\right )^3} \, dx\\ &=\int \left (\frac {157}{\left (-1+x^2\right )^3}+\frac {32}{x^3 \left (-1+x^2\right )^3}-\frac {40}{x^2 \left (-1+x^2\right )^3}-\frac {96}{x \left (-1+x^2\right )^3}-\frac {111 x^2}{\left (-1+x^2\right )^3}-\frac {9 x^4}{\left (-1+x^2\right )^3}+\frac {3 x^6}{\left (-1+x^2\right )^3}+\frac {2 e^{2 x^2} \left (1-5 x^2+2 x^4\right )}{x^3 \left (-1+x^2\right )^3}+\frac {2 e^{x^2} \left (8-5 x-32 x^2+30 x^3+8 x^4-35 x^5+10 x^7\right )}{x^3 \left (-1+x^2\right )^3}\right ) \, dx\\ &=2 \int \frac {e^{2 x^2} \left (1-5 x^2+2 x^4\right )}{x^3 \left (-1+x^2\right )^3} \, dx+2 \int \frac {e^{x^2} \left (8-5 x-32 x^2+30 x^3+8 x^4-35 x^5+10 x^7\right )}{x^3 \left (-1+x^2\right )^3} \, dx+3 \int \frac {x^6}{\left (-1+x^2\right )^3} \, dx-9 \int \frac {x^4}{\left (-1+x^2\right )^3} \, dx+32 \int \frac {1}{x^3 \left (-1+x^2\right )^3} \, dx-40 \int \frac {1}{x^2 \left (-1+x^2\right )^3} \, dx-96 \int \frac {1}{x \left (-1+x^2\right )^3} \, dx-111 \int \frac {x^2}{\left (-1+x^2\right )^3} \, dx+157 \int \frac {1}{\left (-1+x^2\right )^3} \, dx\\ &=\frac {10}{x \left (1-x^2\right )^2}-\frac {23 x}{2 \left (1-x^2\right )^2}+\frac {9 x^3}{4 \left (1-x^2\right )^2}-\frac {3 x^5}{4 \left (1-x^2\right )^2}+2 \int \left (-\frac {2 e^{x^2}}{(-1+x)^3}+\frac {5 e^{x^2}}{2 (-1+x)^2}+\frac {e^{x^2}}{-1+x}-\frac {8 e^{x^2}}{x^3}+\frac {5 e^{x^2}}{x^2}+\frac {8 e^{x^2}}{x}-\frac {2 e^{x^2}}{(1+x)^3}-\frac {15 e^{x^2}}{2 (1+x)^2}-\frac {9 e^{x^2}}{1+x}\right ) \, dx+\frac {15}{4} \int \frac {x^4}{\left (-1+x^2\right )^2} \, dx-\frac {27}{4} \int \frac {x^2}{\left (-1+x^2\right )^2} \, dx+16 \operatorname {Subst}\left (\int \frac {1}{(-1+x)^3 x^2} \, dx,x,x^2\right )-\frac {111}{4} \int \frac {1}{\left (-1+x^2\right )^2} \, dx-48 \operatorname {Subst}\left (\int \frac {1}{(-1+x)^3 x} \, dx,x,x^2\right )+50 \int \frac {1}{x^2 \left (-1+x^2\right )^2} \, dx-\frac {471}{4} \int \frac {1}{\left (-1+x^2\right )^2} \, dx+\operatorname {Subst}\left (\int \frac {e^{2 x} \left (1-5 x+2 x^2\right )}{(-1+x)^3 x^2} \, dx,x,x^2\right )\\ &=\frac {10}{x \left (1-x^2\right )^2}-\frac {23 x}{2 \left (1-x^2\right )^2}+\frac {9 x^3}{4 \left (1-x^2\right )^2}-\frac {3 x^5}{4 \left (1-x^2\right )^2}+\frac {25}{x \left (1-x^2\right )}-\frac {609 x}{8 \left (1-x^2\right )}+\frac {15 x^3}{8 \left (1-x^2\right )}+2 \int \frac {e^{x^2}}{-1+x} \, dx-\frac {27}{8} \int \frac {1}{-1+x^2} \, dx-4 \int \frac {e^{x^2}}{(-1+x)^3} \, dx-4 \int \frac {e^{x^2}}{(1+x)^3} \, dx+5 \int \frac {e^{x^2}}{(-1+x)^2} \, dx+\frac {45}{8} \int \frac {x^2}{-1+x^2} \, dx+10 \int \frac {e^{x^2}}{x^2} \, dx+\frac {111}{8} \int \frac {1}{-1+x^2} \, dx-15 \int \frac {e^{x^2}}{(1+x)^2} \, dx-16 \int \frac {e^{x^2}}{x^3} \, dx+16 \int \frac {e^{x^2}}{x} \, dx+16 \operatorname {Subst}\left (\int \left (\frac {1}{(-1+x)^3}-\frac {2}{(-1+x)^2}+\frac {3}{-1+x}-\frac {1}{x^2}-\frac {3}{x}\right ) \, dx,x,x^2\right )-18 \int \frac {e^{x^2}}{1+x} \, dx-48 \operatorname {Subst}\left (\int \left (\frac {1}{(-1+x)^3}-\frac {1}{(-1+x)^2}+\frac {1}{-1+x}-\frac {1}{x}\right ) \, dx,x,x^2\right )+\frac {471}{8} \int \frac {1}{-1+x^2} \, dx-75 \int \frac {1}{x^2 \left (-1+x^2\right )} \, dx+\operatorname {Subst}\left (\int \left (-\frac {2 e^{2 x}}{(-1+x)^3}+\frac {3 e^{2 x}}{(-1+x)^2}-\frac {2 e^{2 x}}{-1+x}-\frac {e^{2 x}}{x^2}+\frac {2 e^{2 x}}{x}\right ) \, dx,x,x^2\right )\\ &=\frac {2 e^{x^2}}{(1-x)^2}+\frac {5 e^{x^2}}{1-x}+\frac {16}{x^2}+\frac {8 e^{x^2}}{x^2}-\frac {75}{x}-\frac {10 e^{x^2}}{x}+\frac {45 x}{8}+\frac {2 e^{x^2}}{(1+x)^2}+\frac {15 e^{x^2}}{1+x}+\frac {16}{\left (1-x^2\right )^2}+\frac {10}{x \left (1-x^2\right )^2}-\frac {23 x}{2 \left (1-x^2\right )^2}+\frac {9 x^3}{4 \left (1-x^2\right )^2}-\frac {3 x^5}{4 \left (1-x^2\right )^2}+\frac {16}{1-x^2}+\frac {25}{x \left (1-x^2\right )}-\frac {609 x}{8 \left (1-x^2\right )}+\frac {15 x^3}{8 \left (1-x^2\right )}-\frac {555}{8} \tanh ^{-1}(x)+8 \text {Ei}\left (x^2\right )+2 \int \frac {e^{x^2}}{-1+x} \, dx-2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{(-1+x)^3} \, dx,x,x^2\right )-2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{-1+x} \, dx,x,x^2\right )+2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,x^2\right )+3 \operatorname {Subst}\left (\int \frac {e^{2 x}}{(-1+x)^2} \, dx,x,x^2\right )-4 \int \frac {e^{x^2}}{(-1+x)^2} \, dx-4 \int \frac {e^{x^2}}{-1+x} \, dx+4 \int \frac {e^{x^2}}{(1+x)^2} \, dx-4 \int \frac {e^{x^2}}{1+x} \, dx+\frac {45}{8} \int \frac {1}{-1+x^2} \, dx+10 \int e^{x^2} \, dx+10 \int \frac {e^{x^2}}{-1+x} \, dx-16 \int \frac {e^{x^2}}{x} \, dx-18 \int \frac {e^{x^2}}{1+x} \, dx+20 \int e^{x^2} \, dx-30 \int e^{x^2} \, dx+30 \int \frac {e^{x^2}}{1+x} \, dx-75 \int \frac {1}{-1+x^2} \, dx-\operatorname {Subst}\left (\int \frac {e^{2 x}}{x^2} \, dx,x,x^2\right )\\ &=\frac {2 e^{x^2}}{(1-x)^2}+\frac {e^{x^2}}{1-x}+\frac {16}{x^2}+\frac {8 e^{x^2}}{x^2}+\frac {e^{2 x^2}}{x^2}-\frac {75}{x}-\frac {10 e^{x^2}}{x}+\frac {45 x}{8}+\frac {2 e^{x^2}}{(1+x)^2}+\frac {11 e^{x^2}}{1+x}+\frac {16}{\left (1-x^2\right )^2}+\frac {e^{2 x^2}}{\left (1-x^2\right )^2}+\frac {10}{x \left (1-x^2\right )^2}-\frac {23 x}{2 \left (1-x^2\right )^2}+\frac {9 x^3}{4 \left (1-x^2\right )^2}-\frac {3 x^5}{4 \left (1-x^2\right )^2}+\frac {16}{1-x^2}+\frac {3 e^{2 x^2}}{1-x^2}+\frac {25}{x \left (1-x^2\right )}-\frac {609 x}{8 \left (1-x^2\right )}+\frac {15 x^3}{8 \left (1-x^2\right )}+2 \text {Ei}\left (2 x^2\right )-2 e^2 \text {Ei}\left (-2 \left (1-x^2\right )\right )+2 \int \frac {e^{x^2}}{-1+x} \, dx-2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{(-1+x)^2} \, dx,x,x^2\right )-2 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,x^2\right )-4 \int \frac {e^{x^2}}{-1+x} \, dx-4 \int \frac {e^{x^2}}{1+x} \, dx+6 \operatorname {Subst}\left (\int \frac {e^{2 x}}{-1+x} \, dx,x,x^2\right )-8 \int \frac {e^{x^2}}{-1+x} \, dx-8 \int \frac {e^{x^2}}{1+x} \, dx+10 \int \frac {e^{x^2}}{-1+x} \, dx-18 \int \frac {e^{x^2}}{1+x} \, dx+30 \int \frac {e^{x^2}}{1+x} \, dx\\ &=\frac {2 e^{x^2}}{(1-x)^2}+\frac {e^{x^2}}{1-x}+\frac {16}{x^2}+\frac {8 e^{x^2}}{x^2}+\frac {e^{2 x^2}}{x^2}-\frac {75}{x}-\frac {10 e^{x^2}}{x}+\frac {45 x}{8}+\frac {2 e^{x^2}}{(1+x)^2}+\frac {11 e^{x^2}}{1+x}+\frac {16}{\left (1-x^2\right )^2}+\frac {e^{2 x^2}}{\left (1-x^2\right )^2}+\frac {10}{x \left (1-x^2\right )^2}-\frac {23 x}{2 \left (1-x^2\right )^2}+\frac {9 x^3}{4 \left (1-x^2\right )^2}-\frac {3 x^5}{4 \left (1-x^2\right )^2}+\frac {16}{1-x^2}+\frac {e^{2 x^2}}{1-x^2}+\frac {25}{x \left (1-x^2\right )}-\frac {609 x}{8 \left (1-x^2\right )}+\frac {15 x^3}{8 \left (1-x^2\right )}+4 e^2 \text {Ei}\left (-2 \left (1-x^2\right )\right )+2 \int \frac {e^{x^2}}{-1+x} \, dx-4 \int \frac {e^{x^2}}{-1+x} \, dx-4 \int \frac {e^{x^2}}{1+x} \, dx-4 \operatorname {Subst}\left (\int \frac {e^{2 x}}{-1+x} \, dx,x,x^2\right )-8 \int \frac {e^{x^2}}{-1+x} \, dx-8 \int \frac {e^{x^2}}{1+x} \, dx+10 \int \frac {e^{x^2}}{-1+x} \, dx-18 \int \frac {e^{x^2}}{1+x} \, dx+30 \int \frac {e^{x^2}}{1+x} \, dx\\ &=\frac {2 e^{x^2}}{(1-x)^2}+\frac {e^{x^2}}{1-x}+\frac {16}{x^2}+\frac {8 e^{x^2}}{x^2}+\frac {e^{2 x^2}}{x^2}-\frac {75}{x}-\frac {10 e^{x^2}}{x}+\frac {45 x}{8}+\frac {2 e^{x^2}}{(1+x)^2}+\frac {11 e^{x^2}}{1+x}+\frac {16}{\left (1-x^2\right )^2}+\frac {e^{2 x^2}}{\left (1-x^2\right )^2}+\frac {10}{x \left (1-x^2\right )^2}-\frac {23 x}{2 \left (1-x^2\right )^2}+\frac {9 x^3}{4 \left (1-x^2\right )^2}-\frac {3 x^5}{4 \left (1-x^2\right )^2}+\frac {16}{1-x^2}+\frac {e^{2 x^2}}{1-x^2}+\frac {25}{x \left (1-x^2\right )}-\frac {609 x}{8 \left (1-x^2\right )}+\frac {15 x^3}{8 \left (1-x^2\right )}+2 \int \frac {e^{x^2}}{-1+x} \, dx-4 \int \frac {e^{x^2}}{-1+x} \, dx-4 \int \frac {e^{x^2}}{1+x} \, dx-8 \int \frac {e^{x^2}}{-1+x} \, dx-8 \int \frac {e^{x^2}}{1+x} \, dx+10 \int \frac {e^{x^2}}{-1+x} \, dx-18 \int \frac {e^{x^2}}{1+x} \, dx+30 \int \frac {e^{x^2}}{1+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 55, normalized size = 2.20 \begin {gather*} \frac {16+e^{2 x^2}-40 x+43 x^3-6 x^5+3 x^7+2 e^{x^2} \left (4-5 x+5 x^3\right )}{x^2 \left (-1+x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 - 40*x - 96*x^2 + 157*x^3 - 111*x^5 - 9*x^7 + 3*x^9 + E^(2*x^2)*(2 - 10*x^2 + 4*x^4) + E^x^2*(16
- 10*x - 64*x^2 + 60*x^3 + 16*x^4 - 70*x^5 + 20*x^7))/(-x^3 + 3*x^5 - 3*x^7 + x^9),x]

[Out]

(16 + E^(2*x^2) - 40*x + 43*x^3 - 6*x^5 + 3*x^7 + 2*E^x^2*(4 - 5*x + 5*x^3))/(x^2*(-1 + x^2)^2)

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fricas [B]  time = 0.90, size = 57, normalized size = 2.28 \begin {gather*} \frac {3 \, x^{7} - 6 \, x^{5} + 43 \, x^{3} + 2 \, {\left (5 \, x^{3} - 5 \, x + 4\right )} e^{\left (x^{2}\right )} - 40 \, x + e^{\left (2 \, x^{2}\right )} + 16}{x^{6} - 2 \, x^{4} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-10*x^2+2)*exp(x^2)^2+(20*x^7-70*x^5+16*x^4+60*x^3-64*x^2-10*x+16)*exp(x^2)+3*x^9-9*x^7-111*x
^5+157*x^3-96*x^2-40*x+32)/(x^9-3*x^7+3*x^5-x^3),x, algorithm="fricas")

[Out]

(3*x^7 - 6*x^5 + 43*x^3 + 2*(5*x^3 - 5*x + 4)*e^(x^2) - 40*x + e^(2*x^2) + 16)/(x^6 - 2*x^4 + x^2)

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giac [B]  time = 0.41, size = 63, normalized size = 2.52 \begin {gather*} \frac {3 \, x^{7} - 6 \, x^{5} + 10 \, x^{3} e^{\left (x^{2}\right )} + 43 \, x^{3} - 10 \, x e^{\left (x^{2}\right )} - 40 \, x + e^{\left (2 \, x^{2}\right )} + 8 \, e^{\left (x^{2}\right )} + 16}{x^{6} - 2 \, x^{4} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-10*x^2+2)*exp(x^2)^2+(20*x^7-70*x^5+16*x^4+60*x^3-64*x^2-10*x+16)*exp(x^2)+3*x^9-9*x^7-111*x
^5+157*x^3-96*x^2-40*x+32)/(x^9-3*x^7+3*x^5-x^3),x, algorithm="giac")

[Out]

(3*x^7 - 6*x^5 + 10*x^3*e^(x^2) + 43*x^3 - 10*x*e^(x^2) - 40*x + e^(2*x^2) + 8*e^(x^2) + 16)/(x^6 - 2*x^4 + x^
2)

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maple [B]  time = 0.48, size = 60, normalized size = 2.40

method	result	size
norman	\(\frac {16+{\mathrm e}^{2 x^{2}}-40 x +43 x^{3}-6 x^{5}+3 x^{7}+10 x^{3} {\mathrm e}^{x^{2}}-10 \,{\mathrm e}^{x^{2}} x +8 \,{\mathrm e}^{x^{2}}}{x^{2} \left (x^{2}-1\right )^{2}}\)	\(60\)
risch	\(3 x +\frac {40 x^{3}-40 x +16}{x^{2} \left (x^{4}-2 x^{2}+1\right )}+\frac {{\mathrm e}^{2 x^{2}}}{x^{2} \left (x^{2}-1\right )^{2}}+\frac {2 \left (5 x^{3}-5 x +4\right ) {\mathrm e}^{x^{2}}}{x^{2} \left (x^{2}-1\right )^{2}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^4-10*x^2+2)*exp(x^2)^2+(20*x^7-70*x^5+16*x^4+60*x^3-64*x^2-10*x+16)*exp(x^2)+3*x^9-9*x^7-111*x^5+157
*x^3-96*x^2-40*x+32)/(x^9-3*x^7+3*x^5-x^3),x,method=_RETURNVERBOSE)

[Out]

(16+exp(x^2)^2-40*x+43*x^3-6*x^5+3*x^7+10*x^3*exp(x^2)-10*exp(x^2)*x+8*exp(x^2))/x^2/(x^2-1)^2

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maxima [B]  time = 0.55, size = 205, normalized size = 8.20 \begin {gather*} 3 \, x - \frac {5 \, {\left (15 \, x^{4} - 25 \, x^{2} + 8\right )}}{x^{5} - 2 \, x^{3} + x} + \frac {8 \, {\left (6 \, x^{4} - 9 \, x^{2} + 2\right )}}{x^{6} - 2 \, x^{4} + x^{2}} - \frac {3 \, {\left (9 \, x^{3} - 7 \, x\right )}}{8 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} + \frac {9 \, {\left (5 \, x^{3} - 3 \, x\right )}}{8 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} + \frac {157 \, {\left (3 \, x^{3} - 5 \, x\right )}}{8 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} + \frac {111 \, {\left (x^{3} + x\right )}}{8 \, {\left (x^{4} - 2 \, x^{2} + 1\right )}} - \frac {24 \, {\left (2 \, x^{2} - 3\right )}}{x^{4} - 2 \, x^{2} + 1} + \frac {2 \, {\left (5 \, x^{3} - 5 \, x + 4\right )} e^{\left (x^{2}\right )} + e^{\left (2 \, x^{2}\right )}}{x^{6} - 2 \, x^{4} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-10*x^2+2)*exp(x^2)^2+(20*x^7-70*x^5+16*x^4+60*x^3-64*x^2-10*x+16)*exp(x^2)+3*x^9-9*x^7-111*x
^5+157*x^3-96*x^2-40*x+32)/(x^9-3*x^7+3*x^5-x^3),x, algorithm="maxima")

[Out]

3*x - 5*(15*x^4 - 25*x^2 + 8)/(x^5 - 2*x^3 + x) + 8*(6*x^4 - 9*x^2 + 2)/(x^6 - 2*x^4 + x^2) - 3/8*(9*x^3 - 7*x
)/(x^4 - 2*x^2 + 1) + 9/8*(5*x^3 - 3*x)/(x^4 - 2*x^2 + 1) + 157/8*(3*x^3 - 5*x)/(x^4 - 2*x^2 + 1) + 111/8*(x^3
+ x)/(x^4 - 2*x^2 + 1) - 24*(2*x^2 - 3)/(x^4 - 2*x^2 + 1) + (2*(5*x^3 - 5*x + 4)*e^(x^2) + e^(2*x^2))/(x^6 -
2*x^4 + x^2)

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mupad [B]  time = 1.25, size = 52, normalized size = 2.08 \begin {gather*} 3\,x+\frac {8\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+x^3\,\left (10\,{\mathrm {e}}^{x^2}+40\right )-x\,\left (10\,{\mathrm {e}}^{x^2}+40\right )+16}{x^2\,{\left (x^2-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x^2)*(4*x^4 - 10*x^2 + 2) - 40*x - 96*x^2 + 157*x^3 - 111*x^5 - 9*x^7 + 3*x^9 + exp(x^2)*(60*x^3 -
64*x^2 - 10*x + 16*x^4 - 70*x^5 + 20*x^7 + 16) + 32)/(x^3 - 3*x^5 + 3*x^7 - x^9),x)

[Out]

3*x + (8*exp(x^2) + exp(2*x^2) + x^3*(10*exp(x^2) + 40) - x*(10*exp(x^2) + 40) + 16)/(x^2*(x^2 - 1)^2)

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sympy [B]  time = 0.24, size = 105, normalized size = 4.20 \begin {gather*} 3 x + \frac {\left (x^{6} - 2 x^{4} + x^{2}\right ) e^{2 x^{2}} + \left (10 x^{9} - 30 x^{7} + 8 x^{6} + 30 x^{5} - 16 x^{4} - 10 x^{3} + 8 x^{2}\right ) e^{x^{2}}}{x^{12} - 4 x^{10} + 6 x^{8} - 4 x^{6} + x^{4}} + \frac {40 x^{3} - 40 x + 16}{x^{6} - 2 x^{4} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**4-10*x**2+2)*exp(x**2)**2+(20*x**7-70*x**5+16*x**4+60*x**3-64*x**2-10*x+16)*exp(x**2)+3*x**9-
9*x**7-111*x**5+157*x**3-96*x**2-40*x+32)/(x**9-3*x**7+3*x**5-x**3),x)

[Out]

3*x + ((x**6 - 2*x**4 + x**2)*exp(2*x**2) + (10*x**9 - 30*x**7 + 8*x**6 + 30*x**5 - 16*x**4 - 10*x**3 + 8*x**2
)*exp(x**2))/(x**12 - 4*x**10 + 6*x**8 - 4*x**6 + x**4) + (40*x**3 - 40*x + 16)/(x**6 - 2*x**4 + x**2)

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