3.151 \(\int e^{c (a+b x)} \tan ^{-1}(\text{sech}(a c+b c x)) \, dx\)

Optimal. Leaf size=103 \[ \frac{\left (1-\sqrt{2}\right ) \log \left (e^{2 c (a+b x)}+3-2 \sqrt{2}\right )}{2 b c}+\frac{\left (1+\sqrt{2}\right ) \log \left (e^{2 c (a+b x)}+3+2 \sqrt{2}\right )}{2 b c}+\frac{e^{a c+b c x} \tan ^{-1}(\text{sech}(c (a+b x)))}{b c} \]

[Out]

(E^(a*c + b*c*x)*ArcTan[Sech[c*(a + b*x)]])/(b*c) + ((1 - Sqrt[2])*Log[3 - 2*Sqrt[2] + E^(2*c*(a + b*x))])/(2*
b*c) + ((1 + Sqrt[2])*Log[3 + 2*Sqrt[2] + E^(2*c*(a + b*x))])/(2*b*c)

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Rubi [A]  time = 0.145389, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {2194, 5207, 2282, 12, 1247, 632, 31} \[ \frac{\left (1-\sqrt{2}\right ) \log \left (e^{2 c (a+b x)}+3-2 \sqrt{2}\right )}{2 b c}+\frac{\left (1+\sqrt{2}\right ) \log \left (e^{2 c (a+b x)}+3+2 \sqrt{2}\right )}{2 b c}+\frac{e^{a c+b c x} \tan ^{-1}(\text{sech}(c (a+b x)))}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*ArcTan[Sech[a*c + b*c*x]],x]

[Out]

(E^(a*c + b*c*x)*ArcTan[Sech[c*(a + b*x)]])/(b*c) + ((1 - Sqrt[2])*Log[3 - 2*Sqrt[2] + E^(2*c*(a + b*x))])/(2*
b*c) + ((1 + Sqrt[2])*Log[3 + 2*Sqrt[2] + E^(2*c*(a + b*x))])/(2*b*c)

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 5207

Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcTan[u], w, x] - Dist
[b, Int[SimplifyIntegrand[(w*D[u, x])/(1 + u^2), x], x], x] /; InverseFunctionFreeQ[w, x]] /; FreeQ[{a, b}, x]
&& InverseFunctionFreeQ[u, x] &&  !MatchQ[v, ((c_.) + (d_.)*x)^(m_.) /; FreeQ[{c, d, m}, x]] && FalseQ[Functi
onOfLinear[v*(a + b*ArcTan[u]), x]]

Rule 2282

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]]]

Rule 12

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

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int e^{c (a+b x)} \tan ^{-1}(\text{sech}(a c+b c x)) \, dx &=\frac{\operatorname{Subst}\left (\int e^x \tan ^{-1}(\text{sech}(x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int \frac{e^x \text{sech}(x) \tanh (x)}{1+\text{sech}^2(x)} \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int \frac{2 x \left (-1+x^2\right )}{1+6 x^2+x^4} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{2 \operatorname{Subst}\left (\int \frac{x \left (-1+x^2\right )}{1+6 x^2+x^4} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int \frac{-1+x}{1+6 x+x^2} \, dx,x,e^{2 a c+2 b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{\left (1-\sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{3-2 \sqrt{2}+x} \, dx,x,e^{2 a c+2 b c x}\right )}{2 b c}+\frac{\left (1+\sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{3+2 \sqrt{2}+x} \, dx,x,e^{2 a c+2 b c x}\right )}{2 b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\text{sech}(c (a+b x)))}{b c}+\frac{\left (1-\sqrt{2}\right ) \log \left (3-2 \sqrt{2}+e^{2 a c+2 b c x}\right )}{2 b c}+\frac{\left (1+\sqrt{2}\right ) \log \left (3+2 \sqrt{2}+e^{2 a c+2 b c x}\right )}{2 b c}\\ \end{align*}

Mathematica [C]  time = 0.147943, size = 145, normalized size = 1.41 \[ \frac{\text{RootSum}\left [\text{$\#$1}^4+6 \text{$\#$1}^2+1\& ,\frac{7 \text{$\#$1}^2 \log \left (e^{c (a+b x)}-\text{$\#$1}\right )-7 \text{$\#$1}^2 a c-7 \text{$\#$1}^2 b c x+\log \left (e^{c (a+b x)}-\text{$\#$1}\right )-a c-b c x}{3 \text{$\#$1}^2+1}\& \right ]+4 c (a+b x)+2 e^{c (a+b x)} \tan ^{-1}\left (\frac{2 e^{c (a+b x)}}{e^{2 c (a+b x)}+1}\right )}{2 b c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*c*(a + b*x) + 2*E^(c*(a + b*x))*ArcTan[(2*E^(c*(a + b*x)))/(1 + E^(2*c*(a + b*x)))] + RootSum[1 + 6*#1^2 +
#1^4 & , (-(a*c) - b*c*x + Log[E^(c*(a + b*x)) - #1] - 7*a*c*#1^2 - 7*b*c*x*#1^2 + 7*Log[E^(c*(a + b*x)) - #1]
*#1^2)/(1 + 3*#1^2) & ])/(2*b*c)

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Maple [C]  time = 0.609, size = 842, normalized size = 8.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*arctan(sech(b*c*x+a*c)),x)

[Out]

-1/2*I/c/b*exp(c*(b*x+a))*ln(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a)))-1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))+1+2*I*
exp(c*(b*x+a)))/(1+exp(2*c*(b*x+a))))^3*exp(c*(b*x+a))+1/4/c/b*Pi*csgn(I/(1+exp(2*c*(b*x+a))))*csgn(I*(exp(2*c
*(b*x+a))+1+2*I*exp(c*(b*x+a)))/(1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))+1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))+1+2
*I*exp(c*(b*x+a))))*csgn(I*(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a)))/(1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))-1/4/
c/b*Pi*csgn(I*(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a))))*csgn(I/(1+exp(2*c*(b*x+a))))*csgn(I*(exp(2*c*(b*x+a))+1
+2*I*exp(c*(b*x+a)))/(1+exp(2*c*(b*x+a))))*exp(c*(b*x+a))+1/4/c/b*Pi*csgn(I*(-exp(2*c*(b*x+a))-1+2*I*exp(c*(b*
x+a))))*csgn(I/(1+exp(2*c*(b*x+a))))*csgn(I*(-exp(2*c*(b*x+a))-1+2*I*exp(c*(b*x+a)))/(1+exp(2*c*(b*x+a))))*exp
(c*(b*x+a))+1/4/c/b*Pi*csgn(I*(-exp(2*c*(b*x+a))-1+2*I*exp(c*(b*x+a))))*csgn(I*(-exp(2*c*(b*x+a))-1+2*I*exp(c*
(b*x+a)))/(1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))-1/4/c/b*Pi*csgn(I/(1+exp(2*c*(b*x+a))))*csgn(I*(-exp(2*c*(b*x
+a))-1+2*I*exp(c*(b*x+a)))/(1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))-1/4/c/b*Pi*csgn(I*(-exp(2*c*(b*x+a))-1+2*I*e
xp(c*(b*x+a)))/(1+exp(2*c*(b*x+a))))^3*exp(c*(b*x+a))-1/2/c/b*ln(exp(2*c*(b*x+a))+(2^(1/2)-1)^2)*2^(1/2)+1/2/c
/b*ln(exp(2*c*(b*x+a))+(1+2^(1/2))^2)*2^(1/2)-2*a/b+1/2/c/b*ln(exp(2*c*(b*x+a))+(2^(1/2)-1)^2)+1/2/c/b*ln(exp(
2*c*(b*x+a))+(1+2^(1/2))^2)+1/2*I/c/b*exp(c*(b*x+a))*ln(exp(2*c*(b*x+a))+1-2*I*exp(c*(b*x+a)))

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Maxima [A]  time = 1.57521, size = 228, normalized size = 2.21 \begin{align*} \frac{\arctan \left (\operatorname{sech}\left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac{3 \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (2 \, b c x + 2 \, a c\right )} - 3}{2 \, \sqrt{2} + e^{\left (2 \, b c x + 2 \, a c\right )} + 3}\right )}{8 \, b c} + \frac{\sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (-2 \, b c x - 2 \, a c\right )} - 3}{2 \, \sqrt{2} + e^{\left (-2 \, b c x - 2 \, a c\right )} + 3}\right )}{8 \, b c} + \frac{\log \left (e^{\left (4 \, b c x + 4 \, a c\right )} + 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{2 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(sech(b*c*x+a*c)),x, algorithm="maxima")

[Out]

arctan(sech(b*c*x + a*c))*e^((b*x + a)*c)/(b*c) - 3/8*sqrt(2)*log(-(2*sqrt(2) - e^(2*b*c*x + 2*a*c) - 3)/(2*sq
rt(2) + e^(2*b*c*x + 2*a*c) + 3))/(b*c) + 1/8*sqrt(2)*log(-(2*sqrt(2) - e^(-2*b*c*x - 2*a*c) - 3)/(2*sqrt(2) +
e^(-2*b*c*x - 2*a*c) + 3))/(b*c) + 1/2*log(e^(4*b*c*x + 4*a*c) + 6*e^(2*b*c*x + 2*a*c) + 1)/(b*c)

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Fricas [B]  time = 2.06773, size = 741, normalized size = 7.19 \begin{align*} \frac{2 \,{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \arctan \left (\frac{2 \,{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )}}{\cosh \left (b c x + a c\right )^{2} + 2 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )^{2} + 1}\right ) + \sqrt{2} \log \left (\frac{3 \,{\left (2 \, \sqrt{2} + 3\right )} \cosh \left (b c x + a c\right )^{2} - 4 \,{\left (3 \, \sqrt{2} + 4\right )} \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + 3 \,{\left (2 \, \sqrt{2} + 3\right )} \sinh \left (b c x + a c\right )^{2} + 2 \, \sqrt{2} + 3}{\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} + 3}\right ) + \log \left (\frac{2 \,{\left (\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} + 3\right )}}{\cosh \left (b c x + a c\right )^{2} - 2 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )^{2}}\right )}{2 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(sech(b*c*x+a*c)),x, algorithm="fricas")

[Out]

1/2*(2*(cosh(b*c*x + a*c) + sinh(b*c*x + a*c))*arctan(2*(cosh(b*c*x + a*c) + sinh(b*c*x + a*c))/(cosh(b*c*x +
a*c)^2 + 2*cosh(b*c*x + a*c)*sinh(b*c*x + a*c) + sinh(b*c*x + a*c)^2 + 1)) + sqrt(2)*log((3*(2*sqrt(2) + 3)*co
sh(b*c*x + a*c)^2 - 4*(3*sqrt(2) + 4)*cosh(b*c*x + a*c)*sinh(b*c*x + a*c) + 3*(2*sqrt(2) + 3)*sinh(b*c*x + a*c
)^2 + 2*sqrt(2) + 3)/(cosh(b*c*x + a*c)^2 + sinh(b*c*x + a*c)^2 + 3)) + log(2*(cosh(b*c*x + a*c)^2 + sinh(b*c*
x + a*c)^2 + 3)/(cosh(b*c*x + a*c)^2 - 2*cosh(b*c*x + a*c)*sinh(b*c*x + a*c) + sinh(b*c*x + a*c)^2)))/(b*c)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*atan(sech(b*c*x+a*c)),x)

[Out]

Timed out

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Giac [A]  time = 1.13004, size = 208, normalized size = 2.02 \begin{align*} -\frac{{\left (\sqrt{2} e^{\left (-a c\right )} \log \left (-\frac{2 \, \sqrt{2} e^{\left (2 \, a c\right )} - e^{\left (2 \, b c x + 4 \, a c\right )} - 3 \, e^{\left (2 \, a c\right )}}{2 \, \sqrt{2} e^{\left (2 \, a c\right )} + e^{\left (2 \, b c x + 4 \, a c\right )} + 3 \, e^{\left (2 \, a c\right )}}\right ) - 2 \, \arctan \left (\frac{2}{e^{\left (b c x + a c\right )} + e^{\left (-b c x - a c\right )}}\right ) e^{\left (b c x\right )} - e^{\left (-a c\right )} \log \left (e^{\left (4 \, b c x + 4 \, a c\right )} + 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )\right )} e^{\left (a c\right )}}{2 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(sech(b*c*x+a*c)),x, algorithm="giac")

[Out]

-1/2*(sqrt(2)*e^(-a*c)*log(-(2*sqrt(2)*e^(2*a*c) - e^(2*b*c*x + 4*a*c) - 3*e^(2*a*c))/(2*sqrt(2)*e^(2*a*c) + e
^(2*b*c*x + 4*a*c) + 3*e^(2*a*c))) - 2*arctan(2/(e^(b*c*x + a*c) + e^(-b*c*x - a*c)))*e^(b*c*x) - e^(-a*c)*log
(e^(4*b*c*x + 4*a*c) + 6*e^(2*b*c*x + 2*a*c) + 1))*e^(a*c)/(b*c)