3.3.69 \(\int e^{a+b x} \text {sech}^2(a+b x) \, dx\) [269]

Optimal. Leaf size=40 \[ -\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}+\frac {2 \text {ArcTan}\left (e^{a+b x}\right )}{b} \]

[Out]

-2*exp(b*x+a)/b/(1+exp(2*b*x+2*a))+2*arctan(exp(b*x+a))/b

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Rubi [A]
time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2320, 12, 294, 209} \begin {gather*} \frac {2 \text {ArcTan}\left (e^{a+b x}\right )}{b}-\frac {2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sech[a + b*x]^2,x]

[Out]

(-2*E^(a + b*x))/(b*(1 + E^(2*a + 2*b*x))) + (2*ArcTan[E^(a + b*x)])/b

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 294

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 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 e^{a+b x} \text {sech}^2(a+b x) \, dx &=\frac {\text {Subst}\left (\int \frac {4 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {4 \text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}+\frac {2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}+\frac {2 \tan ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 36, normalized size = 0.90 \begin {gather*} \frac {2 \left (-\frac {e^{a+b x}}{1+e^{2 (a+b x)}}+\text {ArcTan}\left (e^{a+b x}\right )\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Sech[a + b*x]^2,x]

[Out]

(2*(-(E^(a + b*x)/(1 + E^(2*(a + b*x)))) + ArcTan[E^(a + b*x)]))/b

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Maple [A]
time = 0.63, size = 25, normalized size = 0.62

method	result	size
derivativedivides	\(\frac {-\frac {1}{\cosh \left (b x +a \right )}+2 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\)	\(25\)
default	\(\frac {-\frac {1}{\cosh \left (b x +a \right )}+2 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\)	\(25\)
risch	\(-\frac {2 \,{\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}+1\right )}+\frac {i \ln \left ({\mathrm e}^{b x +a}+i\right )}{b}-\frac {i \ln \left ({\mathrm e}^{b x +a}-i\right )}{b}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*sech(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/cosh(b*x+a)+2*arctan(exp(b*x+a)))

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Maxima [A]
time = 0.47, size = 37, normalized size = 0.92 \begin {gather*} \frac {2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} - \frac {2 \, e^{\left (b x + a\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^2,x, algorithm="maxima")

[Out]

2*arctan(e^(b*x + a))/b - 2*e^(b*x + a)/(b*(e^(2*b*x + 2*a) + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (37) = 74\).
time = 0.43, size = 105, normalized size = 2.62 \begin {gather*} \frac {2 \, {\left ({\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - \cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} + b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^2,x, algorithm="fricas")

[Out]

2*((cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*arctan(cosh(b*x + a) + sinh(b*x + a
)) - cosh(b*x + a) - sinh(b*x + a))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 +
b)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{a} \int e^{b x} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)**2,x)

[Out]

exp(a)*Integral(exp(b*x)*sech(a + b*x)**2, x)

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Giac [A]
time = 0.39, size = 35, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (\frac {e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1} - \arctan \left (e^{\left (b x + a\right )}\right )\right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(b*x+a)^2,x, algorithm="giac")

[Out]

-2*(e^(b*x + a)/(e^(2*b*x + 2*a) + 1) - arctan(e^(b*x + a)))/b

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Mupad [B]
time = 0.08, size = 48, normalized size = 1.20 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)/cosh(a + b*x)^2,x)

[Out]

(2*atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b))/(b^2)^(1/2) - (2*exp(a + b*x))/(b*(exp(2*a + 2*b*x) + 1))

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