Integrand size = 20, antiderivative size = 103 \[ \int e^{c (a+b x)} \arctan (\text {sech}(a c+b c x)) \, dx=\frac {e^{a c+b c x} \arctan (\text {sech}(c (a+b x)))}{b c}+\frac {\left (1-\sqrt {2}\right ) \log \left (3-2 \sqrt {2}+e^{2 c (a+b x)}\right )}{2 b c}+\frac {\left (1+\sqrt {2}\right ) \log \left (3+2 \sqrt {2}+e^{2 c (a+b x)}\right )}{2 b c} \]
exp(b*c*x+a*c)*arctan(sech(c*(b*x+a)))/b/c+1/2*ln(3+exp(2*c*(b*x+a))-2*2^( 1/2))*(1-2^(1/2))/b/c+1/2*ln(3+exp(2*c*(b*x+a))+2*2^(1/2))*(1+2^(1/2))/b/c
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.41 \[ \int e^{c (a+b x)} \arctan (\text {sech}(a c+b c x)) \, dx=\frac {4 c (a+b x)+2 e^{c (a+b x)} \arctan \left (\frac {2 e^{c (a+b x)}}{1+e^{2 c (a+b x)}}\right )+\text {RootSum}\left [1+6 \text {$\#$1}^2+\text {$\#$1}^4\&,\frac {-a c-b c x+\log \left (e^{c (a+b x)}-\text {$\#$1}\right )-7 a c \text {$\#$1}^2-7 b c x \text {$\#$1}^2+7 \log \left (e^{c (a+b x)}-\text {$\#$1}\right ) \text {$\#$1}^2}{1+3 \text {$\#$1}^2}\&\right ]}{2 b c} \]
(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)
Time = 0.46 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {7281, 5730, 25, 2720, 27, 1576, 1141, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{c (a+b x)} \arctan (\text {sech}(a c+b c x)) \, dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle \frac {\int e^{a c+b x c} \arctan (\text {sech}(a c+b x c))d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 5730 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\text {sech}(a c+b c x))-\int -\frac {e^{a c+b x c} \text {sech}(a c+b x c) \tanh (a c+b x c)}{\text {sech}^2(a c+b x c)+1}d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {e^{a c+b x c} \text {sech}(a c+b x c) \tanh (a c+b x c)}{\text {sech}^2(a c+b x c)+1}d(a c+b x c)+e^{a c+b c x} \arctan (\text {sech}(a c+b c x))}{b c}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {2 e^{a c+b x c} \left (1-e^{2 a c+2 b x c}\right )}{1+6 e^{2 a c+2 b x c}+e^{4 a c+4 b x c}}de^{a c+b x c}+e^{a c+b c x} \arctan (\text {sech}(a c+b c x))}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\text {sech}(a c+b c x))-2 \int \frac {e^{a c+b x c} \left (1-e^{2 a c+2 b x c}\right )}{1+6 e^{2 a c+2 b x c}+e^{4 a c+4 b x c}}de^{a c+b x c}}{b c}\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\text {sech}(a c+b c x))-\int \frac {-a c-b x c+1}{1+7 e^{2 a c+2 b x c}}de^{2 a c+2 b x c}}{b c}\) |
\(\Big \downarrow \) 1141 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\text {sech}(a c+b c x))-\int \left (-\frac {2-\sqrt {2}}{2 \left (4-3 \sqrt {2}-\sqrt {2} e^{2 a c+2 b x c}\right )}-\frac {1+\sqrt {2}}{2 \left (3+2 \sqrt {2}+e^{2 a c+2 b x c}\right )}\right )de^{2 a c+2 b x c}}{b c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\text {sech}(a c+b c x))+\frac {1}{2} \left (1+\sqrt {2}\right ) \log \left (e^{2 a c+2 b c x}+3+2 \sqrt {2}\right )+\frac {1}{2} \left (1-\sqrt {2}\right ) \log \left (-\sqrt {2} e^{2 a c+2 b c x}+4-3 \sqrt {2}\right )}{b c}\) |
(E^(a*c + b*c*x)*ArcTan[Sech[a*c + b*c*x]] + ((1 + Sqrt[2])*Log[3 + 2*Sqrt [2] + E^(2*a*c + 2*b*c*x)])/2 + ((1 - Sqrt[2])*Log[4 - 3*Sqrt[2] - Sqrt[2] *E^(2*a*c + 2*b*c*x)])/2)/(b*c)
3.2.51.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[ (d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 1] || !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Simp[(a + b*ArcTan[u]) w, x] - Simp[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_.) /; Fre eQ[{c, d, m}, x]] && FalseQ[FunctionOfLinear[v*(a + b*ArcTan[u]), x]]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.97 (sec) , antiderivative size = 838, normalized size of antiderivative = 8.14
-1/2*I/b/c*exp(c*(b*x+a))*ln(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a)))-1/4/b/ c*Pi*csgn(I*exp(2*c*(b*x+a))+I+2*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/b/c*Pi*csgn(I*exp(2*c*(b*x+a))+I+2*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/b/c*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/b/c*Pi*c sgn(-I*exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-I)*csgn(I/(1+exp(2*c*(b*x+a))))*c sgn(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/b/c*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/b/c*Pi*cs gn(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/b/c*Pi*csgn(-I*exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-I)*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/b/c*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))-1/2/b/c*2^(1/2)*ln(exp(2*c*(b*x+a))+(2^(1/2)-1 )^2)+1/2/b/c*2^(1/2)*ln(exp(2*c*(b*x+a))+(1+2^(1/2))^2)-2*a/b+1/2*I/b/c*ex p(c*(b*x+a))*ln(exp(2*c*(b*x+a))+1-2*I*exp(c*(b*x+a)))+1/2/b/c*ln(exp(2*c* (b*x+a))+(2^(1/2)-1)^2)+1/2/b/c*ln(exp(2*c*(b*x+a))+(1+2^(1/2))^2)
Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (86) = 172\).
Time = 0.29 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.68 \[ \int e^{c (a+b x)} \arctan (\text {sech}(a c+b c x)) \, dx=\frac {2 \, {\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \arctan \left (\frac {2 \, {\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )}}{\cosh \left (b c x + a c\right )^{2} + 2 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )^{2} + 1}\right ) + \sqrt {2} \log \left (\frac {3 \, {\left (2 \, \sqrt {2} + 3\right )} \cosh \left (b c x + a c\right )^{2} - 4 \, {\left (3 \, \sqrt {2} + 4\right )} \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + 3 \, {\left (2 \, \sqrt {2} + 3\right )} \sinh \left (b c x + a c\right )^{2} + 2 \, \sqrt {2} + 3}{\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} + 3}\right ) + \log \left (\frac {2 \, {\left (\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} + 3\right )}}{\cosh \left (b c x + a c\right )^{2} - 2 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )^{2}}\right )}{2 \, b c} \]
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)*cosh (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)
\[ \int e^{c (a+b x)} \arctan (\text {sech}(a c+b c x)) \, dx=e^{a c} \int e^{b c x} \operatorname {atan}{\left (\operatorname {sech}{\left (a c + b c x \right )} \right )}\, dx \]
Time = 0.31 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.64 \[ \int e^{c (a+b x)} \arctan (\text {sech}(a c+b c x)) \, dx=\frac {\arctan \left (\operatorname {sech}\left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {3 \, \sqrt {2} \log \left (-\frac {2 \, \sqrt {2} - e^{\left (2 \, b c x + 2 \, a c\right )} - 3}{2 \, \sqrt {2} + e^{\left (2 \, b c x + 2 \, a c\right )} + 3}\right )}{8 \, b c} + \frac {\sqrt {2} \log \left (-\frac {2 \, \sqrt {2} - e^{\left (-2 \, b c x - 2 \, a c\right )} - 3}{2 \, \sqrt {2} + e^{\left (-2 \, b c x - 2 \, a c\right )} + 3}\right )}{8 \, b c} + \frac {\log \left (e^{\left (4 \, b c x + 4 \, a c\right )} + 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{2 \, b c} \]
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*sqrt(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)
Time = 0.29 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.50 \[ \int e^{c (a+b x)} \arctan (\text {sech}(a c+b c x)) \, dx=-\frac {{\left (\sqrt {2} e^{\left (-a c\right )} \log \left (-\frac {2 \, \sqrt {2} e^{\left (2 \, a c\right )} - e^{\left (2 \, b c x + 4 \, a c\right )} - 3 \, e^{\left (2 \, a c\right )}}{2 \, \sqrt {2} e^{\left (2 \, a c\right )} + e^{\left (2 \, b c x + 4 \, a c\right )} + 3 \, e^{\left (2 \, a c\right )}}\right ) - 2 \, \arctan \left (\frac {2}{e^{\left (b c x + a c\right )} + e^{\left (-b c x - a c\right )}}\right ) e^{\left (b c x\right )} - e^{\left (-a c\right )} \log \left (e^{\left (4 \, b c x + 4 \, a c\right )} + 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )\right )} e^{\left (a c\right )}}{2 \, b c} \]
-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)
Time = 0.96 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.31 \[ \int e^{c (a+b x)} \arctan (\text {sech}(a c+b c x)) \, dx=\frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\mathrm {atan}\left (\frac {1}{\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}+\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}}\right )}{b\,c}+\frac {\ln \left (8\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}-2\,\sqrt {2}-6\,\sqrt {2}\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}\right )\,\left (\sqrt {2}+1\right )}{2\,b\,c}-\frac {\ln \left (8\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}+2\,\sqrt {2}+6\,\sqrt {2}\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}\right )\,\left (\sqrt {2}-1\right )}{2\,b\,c} \]