3.2.51 \(\int \sqrt {-9+e^{2 t}} \, dt\) [151]

Optimal. Leaf size=30 \[ \sqrt {-9+e^{2 t}}-3 \tan ^{-1}\left (\frac {1}{3} \sqrt {-9+e^{2 t}}\right ) \]

[Out]

-3*arctan(1/3*(-9+exp(2*t))^(1/2))+(-9+exp(2*t))^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2320, 52, 65, 209} \begin {gather*} \sqrt {e^{2 t}-9}-3 \tan ^{-1}\left (\frac {1}{3} \sqrt {e^{2 t}-9}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-9 + E^(2*t)],t]

[Out]

Sqrt[-9 + E^(2*t)] - 3*ArcTan[Sqrt[-9 + E^(2*t)]/3]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

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

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {align*} \int \sqrt {-9+e^{2 t}} \, dt &=\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {-9+t}}{t} \, dt,t,e^{2 t}\right )\\ &=\sqrt {-9+e^{2 t}}-\frac {9}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-9+t} t} \, dt,t,e^{2 t}\right )\\ &=\sqrt {-9+e^{2 t}}-9 \text {Subst}\left (\int \frac {1}{9+t^2} \, dt,t,\sqrt {-9+e^{2 t}}\right )\\ &=\sqrt {-9+e^{2 t}}-3 \tan ^{-1}\left (\frac {1}{3} \sqrt {-9+e^{2 t}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 30, normalized size = 1.00 \begin {gather*} \sqrt {-9+e^{2 t}}-3 \tan ^{-1}\left (\frac {1}{3} \sqrt {-9+e^{2 t}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-9 + E^(2*t)],t]

[Out]

Sqrt[-9 + E^(2*t)] - 3*ArcTan[Sqrt[-9 + E^(2*t)]/3]

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 2.51, size = 32, normalized size = 1.07 \begin {gather*} \text {ConditionalExpression}\left [\sqrt {-9+E^{2 t}}-3 \text {ArcCos}\left [3 E^{-t}\right ],E^t>-3\text {\&\&}E^t<3\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Sqrt[E^(2*t) - 9],t]')

[Out]

ConditionalExpression[Sqrt[-9 + E ^ (2 t)] - 3 ArcCos[3 E ^ (-t)], E ^ t > -3 && E ^ t < 3]

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Maple [A]
time = 0.03, size = 23, normalized size = 0.77

method	result	size
derivativedivides	\(-3 \arctan \left (\frac {\sqrt {-9+{\mathrm e}^{2 t}}}{3}\right )+\sqrt {-9+{\mathrm e}^{2 t}}\)	\(23\)
default	\(-3 \arctan \left (\frac {\sqrt {-9+{\mathrm e}^{2 t}}}{3}\right )+\sqrt {-9+{\mathrm e}^{2 t}}\)	\(23\)
risch	\(-3 \arctan \left (\frac {\sqrt {-9+{\mathrm e}^{2 t}}}{3}\right )+\sqrt {-9+{\mathrm e}^{2 t}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9+exp(2*t))^(1/2),t,method=_RETURNVERBOSE)

[Out]

-3*arctan(1/3*(-9+exp(2*t))^(1/2))+(-9+exp(2*t))^(1/2)

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Maxima [A]
time = 0.34, size = 22, normalized size = 0.73 \begin {gather*} \sqrt {e^{\left (2 \, t\right )} - 9} - 3 \, \arctan \left (\frac {1}{3} \, \sqrt {e^{\left (2 \, t\right )} - 9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9+exp(2*t))^(1/2),t, algorithm="maxima")

[Out]

sqrt(e^(2*t) - 9) - 3*arctan(1/3*sqrt(e^(2*t) - 9))

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Fricas [A]
time = 0.33, size = 22, normalized size = 0.73 \begin {gather*} \sqrt {e^{\left (2 \, t\right )} - 9} - 3 \, \arctan \left (\frac {1}{3} \, \sqrt {e^{\left (2 \, t\right )} - 9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9+exp(2*t))^(1/2),t, algorithm="fricas")

[Out]

sqrt(e^(2*t) - 9) - 3*arctan(1/3*sqrt(e^(2*t) - 9))

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Sympy [A]
time = 0.62, size = 26, normalized size = 0.87 \begin {gather*} \begin {cases} \sqrt {e^{2 t} - 9} - 3 \operatorname {acos}{\left (3 e^{- t} \right )} & \text {for}\: e^{t} > -3 \wedge e^{t} < 3 \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9+exp(2*t))**(1/2),t)

[Out]

Piecewise((sqrt(exp(2*t) - 9) - 3*acos(3*exp(-t)), (exp(t) > -3) & (exp(t) < 3)))

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Giac [A]
time = 0.00, size = 28, normalized size = 0.93 \begin {gather*} \sqrt {\mathrm {e}^{2 t}-9}-3 \arctan \left (\frac {\sqrt {\mathrm {e}^{2 t}-9}}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9+exp(2*t))^(1/2),t)

[Out]

sqrt(e^(2*t) - 9) - 3*arctan(1/3*sqrt(e^(2*t) - 9))

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Mupad [B]
time = 0.23, size = 34, normalized size = 1.13 \begin {gather*} \left (\frac {3\,{\mathrm {e}}^{-t}\,\mathrm {asin}\left (3\,{\mathrm {e}}^{-t}\right )}{\sqrt {1-9\,{\mathrm {e}}^{-2\,t}}}+1\right )\,\sqrt {{\mathrm {e}}^{2\,t}-9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*t) - 9)^(1/2),t)

[Out]

((3*exp(-t)*asin(3*exp(-t)))/(1 - 9*exp(-2*t))^(1/2) + 1)*(exp(2*t) - 9)^(1/2)

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