Integrand size = 20, antiderivative size = 103 \[ \int e^{c (a+b x)} \arctan (\cosh (a c+b c x)) \, dx=\frac {e^{a c+b c x} \arctan (\cosh (c (a+b x)))}{b c}-\frac {\left (1-\sqrt {2}\right ) \log \left (3-2 \sqrt {2}+e^{2 c (a+b x)}\right )}{2 b c}-\frac {\left (1+\sqrt {2}\right ) \log \left (3+2 \sqrt {2}+e^{2 c (a+b x)}\right )}{2 b c} \]
exp(b*c*x+a*c)*arctan(cosh(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 = 146, normalized size of antiderivative = 1.42 \[ \int e^{c (a+b x)} \arctan (\cosh (a c+b c x)) \, dx=\frac {-4 c (a+b x)+2 e^{c (a+b x)} \arctan \left (\frac {1}{2} e^{-c (a+b x)} \left (1+e^{2 c (a+b x)}\right )\right )+\text {RootSum}\left [1+6 \text {$\#$1}^2+\text {$\#$1}^4\&,\frac {a c+b c x-\log \left (e^{c (a+b x)}-\text {$\#$1}\right )+7 a c \text {$\#$1}^2+7 b c x \text {$\#$1}^2-7 \log \left (e^{c (a+b x)}-\text {$\#$1}\right ) \text {$\#$1}^2}{1+3 \text {$\#$1}^2}\&\right ]}{2 b c} \]
(-4*c*(a + b*x) + 2*E^(c*(a + b*x))*ArcTan[(1 + E^(2*c*(a + b*x)))/(2*E^(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 = 0.49 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {7281, 5730, 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.
\(\displaystyle \int e^{c (a+b x)} \arctan (\cosh (a c+b c x)) \, dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle \frac {\int e^{a c+b x c} \arctan (\cosh (a c+b x c))d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 5730 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\cosh (a c+b c x))-\int \frac {e^{a c+b x c} \sinh (a c+b x c)}{\cosh ^2(a c+b x c)+1}d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\cosh (a c+b c x))-\int -\frac {2 e^{a c+b x c} \left (1-e^{2 a c+2 b x c}\right )}{1+6 e^{2 a c+2 b x c}+e^{4 a c+4 b x c}}de^{a c+b x c}}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {e^{a c+b x c} \left (1-e^{2 a c+2 b x c}\right )}{1+6 e^{2 a c+2 b x c}+e^{4 a c+4 b x c}}de^{a c+b x c}+e^{a c+b c x} \arctan (\cosh (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle \frac {\int \frac {-a c-b x c+1}{1+7 e^{2 a c+2 b x c}}de^{2 a c+2 b x c}+e^{a c+b c x} \arctan (\cosh (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 1141 |
\(\displaystyle \frac {\int \left (-\frac {2-\sqrt {2}}{2 \left (4-3 \sqrt {2}-\sqrt {2} e^{2 a c+2 b x c}\right )}-\frac {1+\sqrt {2}}{2 \left (3+2 \sqrt {2}+e^{2 a c+2 b x c}\right )}\right )de^{2 a c+2 b x c}+e^{a c+b c x} \arctan (\cosh (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\cosh (a c+b c x))-\frac {1}{2} \left (1+\sqrt {2}\right ) \log \left (e^{2 a c+2 b c x}+3+2 \sqrt {2}\right )-\frac {1}{2} \left (1-\sqrt {2}\right ) \log \left (-\sqrt {2} e^{2 a c+2 b c x}+4-3 \sqrt {2}\right )}{b c}\) |
(E^(a*c + b*c*x)*ArcTan[Cosh[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)
3.2.48.3.1 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 = 5.59 (sec) , antiderivative size = 1371, normalized size of antiderivative = 13.31
-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))+2*exp(c*(b*x+a))-I)*csgn(I*exp(-c*(b*x+a))*( exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a))))^2*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I *exp(-c*(b*x+a)))*csgn(-I*exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-I)*csgn(I*exp( -c*(b*x+a))*(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a))))*exp(c*(b*x+a))+1/4/b/ c*Pi*csgn(I*exp(-c*(b*x+a))*(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a))))^3*exp (c*(b*x+a))-1/4/b/c*Pi*csgn(I*exp(-c*(b*x+a)))*csgn(I*exp(-c*(b*x+a))*(exp (2*c*(b*x+a))+1+2*I*exp(c*(b*x+a))))^2*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I*ex p(-c*(b*x+a))*(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a))))*csgn(exp(-c*(b*x+a) )*(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a))))^2*exp(c*(b*x+a))+1/4/b/c*Pi*csg n(I*exp(-c*(b*x+a))*(-exp(2*c*(b*x+a))-1+2*I*exp(c*(b*x+a))))^3*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(-c* (b*x+a))*(-exp(2*c*(b*x+a))-1+2*I*exp(c*(b*x+a))))^2*exp(c*(b*x+a))+1/4/b/ c*Pi*csgn(I*exp(-c*(b*x+a)))*csgn(I*exp(-c*(b*x+a))*(-exp(2*c*(b*x+a))-1+2 *I*exp(c*(b*x+a))))^2*exp(c*(b*x+a))+1/4/b/c*Pi*csgn(I*exp(-c*(b*x+a)))*cs gn(I*exp(2*c*(b*x+a))+I+2*exp(c*(b*x+a)))*csgn(I*exp(-c*(b*x+a))*(-exp(2*c *(b*x+a))-1+2*I*exp(c*(b*x+a))))*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I*exp(-c*( b*x+a))*(-exp(2*c*(b*x+a))-1+2*I*exp(c*(b*x+a))))*csgn(exp(-c*(b*x+a))*(-e xp(2*c*(b*x+a))-1+2*I*exp(c*(b*x+a))))^2*exp(c*(b*x+a))+1/4/b/c*Pi*csgn(ex p(-c*(b*x+a))*(exp(2*c*(b*x+a))+1+2*I*exp(c*(b*x+a))))^3*exp(c*(b*x+a))...
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (86) = 172\).
Time = 0.27 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.15 \[ \int e^{c (a+b x)} \arctan (\cosh (a c+b c x)) \, dx=\frac {2 \, {\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \arctan \left (\cosh \left (b c x + a c\right )\right ) + \sqrt {2} \log \left (-\frac {3 \, {\left (2 \, \sqrt {2} - 3\right )} \cosh \left (b c x + a c\right )^{2} - 4 \, {\left (3 \, \sqrt {2} - 4\right )} \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + 3 \, {\left (2 \, \sqrt {2} - 3\right )} \sinh \left (b c x + a c\right )^{2} + 2 \, \sqrt {2} - 3}{\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} + 3}\right ) - \log \left (\frac {2 \, {\left (\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} + 3\right )}}{\cosh \left (b c x + a c\right )^{2} - 2 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )^{2}}\right )}{2 \, b c} \]
1/2*(2*(cosh(b*c*x + a*c) + sinh(b*c*x + a*c))*arctan(cosh(b*c*x + a*c)) + sqrt(2)*log(-(3*(2*sqrt(2) - 3)*cosh(b*c*x + a*c)^2 - 4*(3*sqrt(2) - 4)*c osh(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)
\[ \int e^{c (a+b x)} \arctan (\cosh (a c+b c x)) \, dx=e^{a c} \int e^{b c x} \operatorname {atan}{\left (\cosh {\left (a c + b c x \right )} \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.27 \[ \int e^{c (a+b x)} \arctan (\cosh (a c+b c x)) \, dx=\frac {\arctan \left (\cosh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {\sqrt {2} \log \left (-\frac {2 \, \sqrt {2} - e^{\left (-2 \, b c x - 2 \, a c\right )} - 3}{2 \, \sqrt {2} + e^{\left (-2 \, b c x - 2 \, a c\right )} + 3}\right )}{2 \, b c} - \frac {2 \, {\left (b c x + a c\right )}}{b c} - \frac {\log \left (6 \, e^{\left (-2 \, b c x - 2 \, a c\right )} + e^{\left (-4 \, b c x - 4 \, a c\right )} + 1\right )}{2 \, b c} \]
arctan(cosh(b*c*x + a*c))*e^((b*x + a)*c)/(b*c) - 1/2*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) - 2*(b*c*x + a*c)/(b*c) - 1/2*log(6*e^(-2*b*c*x - 2*a*c) + e^(-4*b*c*x - 4*a*c) + 1)/(b*c)
Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.50 \[ \int e^{c (a+b x)} \arctan (\cosh (a c+b c x)) \, dx=\frac {{\left (\sqrt {2} e^{\left (-a c\right )} \log \left (-\frac {2 \, \sqrt {2} e^{\left (2 \, a c\right )} - e^{\left (2 \, b c x + 4 \, a c\right )} - 3 \, e^{\left (2 \, a c\right )}}{2 \, \sqrt {2} e^{\left (2 \, a c\right )} + e^{\left (2 \, b c x + 4 \, a c\right )} + 3 \, e^{\left (2 \, a c\right )}}\right ) + 2 \, \arctan \left (\frac {1}{2} \, e^{\left (b c x + a c\right )} + \frac {1}{2} \, e^{\left (-b c x - a c\right )}\right ) e^{\left (b c x\right )} - e^{\left (-a c\right )} \log \left (e^{\left (4 \, b c x + 4 \, a c\right )} + 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )\right )} e^{\left (a c\right )}}{2 \, b c} \]
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(1/2*e^(b*c*x + a*c) + 1/2*e^(-b*c*x - a*c))*e^(b*c*x) - e^(-a*c)*lo g(e^(4*b*c*x + 4*a*c) + 6*e^(2*b*c*x + 2*a*c) + 1))*e^(a*c)/(b*c)
Time = 0.34 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.29 \[ \int e^{c (a+b x)} \arctan (\cosh (a c+b c x)) \, dx=\frac {\ln \left (-8\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}-2\,\sqrt {2}-6\,\sqrt {2}\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}\right )\,\left (\sqrt {2}-1\right )}{2\,b\,c}-\frac {\ln \left (2\,\sqrt {2}-8\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}+6\,\sqrt {2}\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}\right )\,\left (\sqrt {2}+1\right )}{2\,b\,c}+\frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}+\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}\right )}{b\,c} \]