3.70.30 \(\int \frac {-80 x+e^{4 x^2} (-40 x-160 x^3-160 x^5)}{1+2 x^2+x^4} \, dx\)

Optimal. Leaf size=23 \[ \frac {20 \left (2+e^{4 x^2}\right ) x}{-\frac {1}{x}-x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.56, antiderivative size = 35, normalized size of antiderivative = 1.52, number of steps used = 11, number of rules used = 8, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {28, 6742, 261, 6715, 2199, 2194, 2177, 2178} \begin {gather*} -20 e^{4 x^2}+\frac {20 e^{4 x^2}}{x^2+1}+\frac {40}{x^2+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-80*x + E^(4*x^2)*(-40*x - 160*x^3 - 160*x^5))/(1 + 2*x^2 + x^4),x]

[Out]

-20*E^(4*x^2) + 40/(1 + x^2) + (20*E^(4*x^2))/(1 + x^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 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 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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, 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} &=\int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{\left (1+x^2\right )^2} \, dx\\ &=\int \left (-\frac {80 x}{\left (1+x^2\right )^2}-\frac {40 e^{4 x^2} x \left (1+2 x^2\right )^2}{\left (1+x^2\right )^2}\right ) \, dx\\ &=-\left (40 \int \frac {e^{4 x^2} x \left (1+2 x^2\right )^2}{\left (1+x^2\right )^2} \, dx\right )-80 \int \frac {x}{\left (1+x^2\right )^2} \, dx\\ &=\frac {40}{1+x^2}-20 \operatorname {Subst}\left (\int \frac {e^{4 x} (1+2 x)^2}{(1+x)^2} \, dx,x,x^2\right )\\ &=\frac {40}{1+x^2}-20 \operatorname {Subst}\left (\int \left (4 e^{4 x}+\frac {e^{4 x}}{(1+x)^2}-\frac {4 e^{4 x}}{1+x}\right ) \, dx,x,x^2\right )\\ &=\frac {40}{1+x^2}-20 \operatorname {Subst}\left (\int \frac {e^{4 x}}{(1+x)^2} \, dx,x,x^2\right )-80 \operatorname {Subst}\left (\int e^{4 x} \, dx,x,x^2\right )+80 \operatorname {Subst}\left (\int \frac {e^{4 x}}{1+x} \, dx,x,x^2\right )\\ &=-20 e^{4 x^2}+\frac {40}{1+x^2}+\frac {20 e^{4 x^2}}{1+x^2}+\frac {80 \text {Ei}\left (4 \left (1+x^2\right )\right )}{e^4}-80 \operatorname {Subst}\left (\int \frac {e^{4 x}}{1+x} \, dx,x,x^2\right )\\ &=-20 e^{4 x^2}+\frac {40}{1+x^2}+\frac {20 e^{4 x^2}}{1+x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 22, normalized size = 0.96 \begin {gather*} -\frac {20 \left (-2+e^{4 x^2} x^2\right )}{1+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80*x + E^(4*x^2)*(-40*x - 160*x^3 - 160*x^5))/(1 + 2*x^2 + x^4),x]

[Out]

(-20*(-2 + E^(4*x^2)*x^2))/(1 + x^2)

________________________________________________________________________________________

fricas [A]  time = 0.73, size = 21, normalized size = 0.91 \begin {gather*} -\frac {20 \, {\left (x^{2} e^{\left (4 \, x^{2}\right )} - 2\right )}}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-160*x^5-160*x^3-40*x)*exp(x^2)^4-80*x)/(x^4+2*x^2+1),x, algorithm="fricas")

[Out]

-20*(x^2*e^(4*x^2) - 2)/(x^2 + 1)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 21, normalized size = 0.91 \begin {gather*} -\frac {20 \, {\left (x^{2} e^{\left (4 \, x^{2}\right )} - 2\right )}}{x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-160*x^5-160*x^3-40*x)*exp(x^2)^4-80*x)/(x^4+2*x^2+1),x, algorithm="giac")

[Out]

-20*(x^2*e^(4*x^2) - 2)/(x^2 + 1)

________________________________________________________________________________________

maple [A]  time = 0.09, size = 22, normalized size = 0.96

method	result	size
norman	\(\frac {-20 x^{2} {\mathrm e}^{4 x^{2}}+40}{x^{2}+1}\)	\(22\)
risch	\(\frac {40}{x^{2}+1}-\frac {20 x^{2} {\mathrm e}^{4 x^{2}}}{x^{2}+1}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-160*x^5-160*x^3-40*x)*exp(x^2)^4-80*x)/(x^4+2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

(-20*x^2*exp(x^2)^4+40)/(x^2+1)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {20 \, x^{2} e^{\left (4 \, x^{2}\right )}}{x^{2} + 1} + \frac {20 \, e^{\left (-4\right )} E_{2}\left (-4 \, x^{2} - 4\right )}{x^{2} + 1} + \frac {40}{x^{2} + 1} + 40 \, \int \frac {x e^{\left (4 \, x^{2}\right )}}{x^{4} + 2 \, x^{2} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-160*x^5-160*x^3-40*x)*exp(x^2)^4-80*x)/(x^4+2*x^2+1),x, algorithm="maxima")

[Out]

-20*x^2*e^(4*x^2)/(x^2 + 1) + 20*e^(-4)*exp_integral_e(2, -4*x^2 - 4)/(x^2 + 1) + 40/(x^2 + 1) + 40*integrate(
x*e^(4*x^2)/(x^4 + 2*x^2 + 1), x)

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 20, normalized size = 0.87 \begin {gather*} -\frac {20\,x^2\,\left ({\mathrm {e}}^{4\,x^2}+2\right )}{x^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(80*x + exp(4*x^2)*(40*x + 160*x^3 + 160*x^5))/(2*x^2 + x^4 + 1),x)

[Out]

-(20*x^2*(exp(4*x^2) + 2))/(x^2 + 1)

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 24, normalized size = 1.04 \begin {gather*} - \frac {20 x^{2} e^{4 x^{2}}}{x^{2} + 1} + \frac {80}{2 x^{2} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-160*x**5-160*x**3-40*x)*exp(x**2)**4-80*x)/(x**4+2*x**2+1),x)

[Out]

-20*x**2*exp(4*x**2)/(x**2 + 1) + 80/(2*x**2 + 2)

________________________________________________________________________________________