3.77 \(\int \frac{e^{x^2} (1+4 x^2+x^3+5 x^4+2 x^6)}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=24 \[ e^{x^2} x+\frac{e^{x^2}}{2 \left (x^2+1\right )} \]

[Out]

E^x^2*x + E^x^2/(2*(1 + x^2))

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Rubi [A]  time = 0.35703, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6742, 2204, 2212, 6715, 2177, 2178} \[ e^{x^2} x+\frac{e^{x^2}}{2 \left (x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[(E^x^2*(1 + 4*x^2 + x^3 + 5*x^4 + 2*x^6))/(1 + x^2)^2,x]

[Out]

E^x^2*x + E^x^2/(2*(1 + x^2))

Rule 6742

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

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 2212

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

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 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

Rubi steps

\begin{align*} \int \frac{e^{x^2} \left (1+4 x^2+x^3+5 x^4+2 x^6\right )}{\left (1+x^2\right )^2} \, dx &=\int \left (e^{x^2}+2 e^{x^2} x^2-\frac{e^{x^2} x}{\left (1+x^2\right )^2}+\frac{e^{x^2} x}{1+x^2}\right ) \, dx\\ &=2 \int e^{x^2} x^2 \, dx+\int e^{x^2} \, dx-\int \frac{e^{x^2} x}{\left (1+x^2\right )^2} \, dx+\int \frac{e^{x^2} x}{1+x^2} \, dx\\ &=e^{x^2} x+\frac{1}{2} \sqrt{\pi } \text{erfi}(x)-\frac{1}{2} \operatorname{Subst}\left (\int \frac{e^x}{(1+x)^2} \, dx,x,x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{e^x}{1+x} \, dx,x,x^2\right )-\int e^{x^2} \, dx\\ &=e^{x^2} x+\frac{e^{x^2}}{2 \left (1+x^2\right )}+\frac{\text{Ei}\left (1+x^2\right )}{2 e}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{e^x}{1+x} \, dx,x,x^2\right )\\ &=e^{x^2} x+\frac{e^{x^2}}{2 \left (1+x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.068602, size = 20, normalized size = 0.83 \[ \frac{1}{2} e^{x^2} \left (\frac{1}{x^2+1}+2 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^x^2*(1 + 4*x^2 + x^3 + 5*x^4 + 2*x^6))/(1 + x^2)^2,x]

[Out]

(E^x^2*(2*x + (1 + x^2)^(-1)))/2

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Maple [A]  time = 0.005, size = 24, normalized size = 1. \begin{align*}{\frac{ \left ( 2\,{x}^{3}+2\,x+1 \right ){{\rm e}^{{x}^{2}}}}{2\,{x}^{2}+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2)*(2*x^6+5*x^4+x^3+4*x^2+1)/(x^2+1)^2,x)

[Out]

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

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Maxima [A]  time = 1.54735, size = 31, normalized size = 1.29 \begin{align*} \frac{{\left (2 \, x^{3} + 2 \, x + 1\right )} e^{\left (x^{2}\right )}}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*(2*x^6+5*x^4+x^3+4*x^2+1)/(x^2+1)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.17282, size = 55, normalized size = 2.29 \begin{align*} \frac{{\left (2 \, x^{3} + 2 \, x + 1\right )} e^{\left (x^{2}\right )}}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*(2*x^6+5*x^4+x^3+4*x^2+1)/(x^2+1)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.108063, size = 20, normalized size = 0.83 \begin{align*} \frac{\left (2 x^{3} + 2 x + 1\right ) e^{x^{2}}}{2 x^{2} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x**2)*(2*x**6+5*x**4+x**3+4*x**2+1)/(x**2+1)**2,x)

[Out]

(2*x**3 + 2*x + 1)*exp(x**2)/(2*x**2 + 2)

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Giac [A]  time = 1.06218, size = 41, normalized size = 1.71 \begin{align*} \frac{2 \, x^{3} e^{\left (x^{2}\right )} + 2 \, x e^{\left (x^{2}\right )} + e^{\left (x^{2}\right )}}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*(2*x^6+5*x^4+x^3+4*x^2+1)/(x^2+1)^2,x, algorithm="giac")

[Out]

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