3.2.63 \(\int \frac {e^{t^2} t}{1+t^2} \, dt\) [163]

Optimal. Leaf size=13 \[ \frac {\text {Ei}\left (1+t^2\right )}{2 e} \]

[Out]

1/2*Ei(t^2+1)/exp(1)

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Rubi [A]
time = 0.05, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6847, 2209} \begin {gather*} \frac {\text {Ei}\left (t^2+1\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^t^2*t)/(1 + t^2),t]

[Out]

ExpIntegralEi[1 + t^2]/(2*E)

Rule 2209

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

Rule 6847

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]

Rubi steps

\begin {align*} \int \frac {e^{t^2} t}{1+t^2} \, dt &=\frac {1}{2} \text {Subst}\left (\int \frac {e^t}{1+t} \, dt,t,t^2\right )\\ &=\frac {\text {Ei}\left (1+t^2\right )}{2 e}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 13, normalized size = 1.00 \begin {gather*} \frac {\text {Ei}\left (1+t^2\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^t^2*t)/(1 + t^2),t]

[Out]

ExpIntegralEi[1 + t^2]/(2*E)

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[t*(E^t^2/(t^2 + 1)),t]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [A]
time = 0.06, size = 14, normalized size = 1.08

method	result	size
derivativedivides	\(-\frac {{\mathrm e}^{-1} \expIntegral \left (1, -t^{2}-1\right )}{2}\)	\(14\)
default	\(-\frac {{\mathrm e}^{-1} \expIntegral \left (1, -t^{2}-1\right )}{2}\)	\(14\)
risch	\(-\frac {{\mathrm e}^{-1} \expIntegral \left (1, -t^{2}-1\right )}{2}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(t^2)*t/(t^2+1),t,method=_RETURNVERBOSE)

[Out]

-1/2*exp(-1)*Ei(1,-t^2-1)

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Maxima [A]
time = 0.36, size = 13, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \, e^{\left (-1\right )} E_{1}\left (-t^{2} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(t^2)*t/(t^2+1),t, algorithm="maxima")

[Out]

-1/2*e^(-1)*exp_integral_e(1, -t^2 - 1)

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Fricas [A]
time = 0.33, size = 10, normalized size = 0.77 \begin {gather*} \frac {1}{2} \, {\rm Ei}\left (t^{2} + 1\right ) e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(t^2)*t/(t^2+1),t, algorithm="fricas")

[Out]

1/2*Ei(t^2 + 1)*e^(-1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {t e^{t^{2}}}{t^{2} + 1}\, dt \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(t**2)*t/(t**2+1),t)

[Out]

Integral(t*exp(t**2)/(t**2 + 1), t)

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Giac [A]
time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} \frac {\frac {1}{2} \mathrm {Ei}\left (t^{2}+1\right )}{\mathrm {e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(t^2)*t/(t^2+1),t)

[Out]

1/2*Ei(t^2 + 1)*e^(-1)

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Mupad [B]
time = 0.12, size = 10, normalized size = 0.77 \begin {gather*} \frac {{\mathrm {e}}^{-1}\,\mathrm {ei}\left (t^2+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((t*exp(t^2))/(t^2 + 1),t)

[Out]

(exp(-1)*ei(t^2 + 1))/2

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