Integrand size = 20, antiderivative size = 286 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(d+b x) \, dx=-\frac {8 e^{\frac {5 (a-d)}{3}+\frac {11}{3} (d+b x)}}{3 b \left (1+e^{2 (d+b x)}\right )^3}-\frac {22 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{9 b \left (1+e^{2 (d+b x)}\right )^2}+\frac {55 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{27 b \left (1+e^{2 (d+b x)}\right )}+\frac {55 e^{\frac {5 (a-d)}{3}} \arctan \left (e^{\frac {1}{3} (d+b x)}\right )}{81 b}-\frac {55 e^{\frac {5 (a-d)}{3}} \arctan \left (\sqrt {3}-2 e^{\frac {1}{3} (d+b x)}\right )}{162 b}+\frac {55 e^{\frac {5 (a-d)}{3}} \arctan \left (\sqrt {3}+2 e^{\frac {1}{3} (d+b x)}\right )}{162 b}-\frac {55 e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (\frac {\sqrt {3} e^{\frac {1}{3} (d+b x)}}{1+e^{\frac {2}{3} (d+b x)}}\right )}{54 \sqrt {3} b} \] Output:
-8/3*exp(5/3*a+2*d+11/3*b*x)/b/(1+exp(2*b*x+2*d))^3-22/9*exp(5/3*b*x+5/3*a )/b/(1+exp(2*b*x+2*d))^2+55/27*exp(5/3*b*x+5/3*a)/b/(1+exp(2*b*x+2*d))+55/ 81*exp(5/3*a-5/3*d)*arctan(exp(1/3*b*x+1/3*d))/b+55/162*exp(5/3*a-5/3*d)*a rctan(-3^(1/2)+2*exp(1/3*b*x+1/3*d))/b+55/162*exp(5/3*a-5/3*d)*arctan(3^(1 /2)+2*exp(1/3*b*x+1/3*d))/b-55/162*3^(1/2)*exp(5/3*a-5/3*d)*arctanh(3^(1/2 )*exp(1/3*b*x+1/3*d)/(1+exp(2/3*b*x+2/3*d)))/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.24 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(d+b x) \, dx=\frac {e^{\frac {5}{3} (a+b x)} \left (4 e^{2 (d+b x)} \operatorname {Hypergeometric2F1}\left (\frac {11}{6},2,\frac {17}{6},-e^{2 (d+b x)}\right )+\text {sech}^2(d+b x) (5+6 \tanh (d+b x))\right )}{18 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*Sech[d + b*x]^4,x]
Output:
(E^((5*(a + b*x))/3)*(4*E^(2*(d + b*x))*Hypergeometric2F1[11/6, 2, 17/6, - E^(2*(d + b*x))] + Sech[d + b*x]^2*(5 + 6*Tanh[d + b*x])))/(18*b)
Time = 0.40 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.78, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {2720, 27, 817, 817, 819, 824, 27, 216, 1142, 25, 1083, 217, 1103}
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.
| \(\displaystyle \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(b x+d) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \int \frac {16 e^{\frac {5 a}{3}+\frac {16 b x}{3}}}{\left (1+e^{2 b x}\right )^4}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {48 e^{5 a/3} \int \frac {e^{\frac {16 b x}{3}}}{\left (1+e^{2 b x}\right )^4}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \int \frac {e^{\frac {10 b x}{3}}}{\left (1+e^{2 b x}\right )^3}de^{\frac {b x}{3}}-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \int \frac {e^{\frac {4 b x}{3}}}{\left (1+e^{2 b x}\right )^2}de^{\frac {b x}{3}}-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 819 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \int \frac {e^{\frac {4 b x}{3}}}{1+e^{2 b x}}de^{\frac {b x}{3}}+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 824 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{3} \int \frac {1}{1+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1-\sqrt {3} e^{\frac {b x}{3}}}{2 \left (1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1+\sqrt {3} e^{\frac {b x}{3}}}{2 \left (1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{3} \int \frac {1}{1+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1-\sqrt {3} e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1+\sqrt {3} e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \left (-\frac {1}{6} \int \frac {1-\sqrt {3} e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1+\sqrt {3} e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}+2 e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}+2 e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (-\int \frac {1}{-1-e^{\frac {2 b x}{3}}}d\left (-\sqrt {3}+2 e^{\frac {b x}{3}}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (-\int \frac {1}{-1-e^{\frac {2 b x}{3}}}d\left (\sqrt {3}+2 e^{\frac {b x}{3}}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}+2 e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\arctan \left (\sqrt {3}-2 e^{\frac {b x}{3}}\right )\right )+\frac {1}{6} \left (\arctan \left (2 e^{\frac {b x}{3}}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}+2 e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {11}{18} \left (\frac {5}{12} \left (\frac {1}{6} \left (\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )-\arctan \left (\sqrt {3}-2 e^{\frac {b x}{3}}\right )\right )+\frac {1}{6} \left (\arctan \left (2 e^{\frac {b x}{3}}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )\right )\right )+\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )-\frac {e^{\frac {5 b x}{3}}}{12 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{\frac {11 b x}{3}}}{18 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
Input:
Int[E^((5*(a + b*x))/3)*Sech[d + b*x]^4,x]
Output:
(48*E^((5*a)/3)*(-1/18*E^((11*b*x)/3)/(1 + E^(2*b*x))^3 + (11*(-1/12*E^((5 *b*x)/3)/(1 + E^(2*b*x))^2 + (5*(E^((5*b*x)/3)/(6*(1 + E^(2*b*x))) + (ArcT an[E^((b*x)/3)]/3 + (-ArcTan[Sqrt[3] - 2*E^((b*x)/3)] + (Sqrt[3]*Log[1 - S qrt[3]*E^((b*x)/3) + E^((2*b*x)/3)])/2)/6 + (ArcTan[Sqrt[3] + 2*E^((b*x)/3 )] - (Sqrt[3]*Log[1 + Sqrt[3]*E^((b*x)/3) + E^((2*b*x)/3)])/2)/6)/6))/12)) /18))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
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] - Simp[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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator [Rt[a/b, n]], s = Denominator[Rt[a/b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k - 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] ; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m)) Int[1/(r^2 + s^2*x^2), x] + 2*(r^(m + 1)/(a*n*s^m)) Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGt Q[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, 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]]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.40 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.57
| method | result | size |
| risch | \(\frac {\left (55 \,{\mathrm e}^{4 b x +4 d}-28 \,{\mathrm e}^{2 b x +2 d}-11\right ) {\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}}}{27 \left (1+{\mathrm e}^{2 b x +2 d}\right )^{3} b}+16 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (45137758519296 b^{4} \textit {\_Z}^{4}-20323353600 b^{2} \textit {\_Z}^{2}+9150625\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {17414258688 b^{3} \textit {\_R}^{3}}{166375}-\frac {2592 b \textit {\_R}}{55}\right )\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}+\frac {55 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+i\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{162 b}-\frac {55 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-i\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{162 b}\) | \(164\) |
Input:
int(exp(5/3*b*x+5/3*a)*sech(b*x+d)^4,x,method=_RETURNVERBOSE)
Output:
1/27/(1+exp(2*b*x+2*d))^3/b*(55*exp(4*b*x+4*d)-28*exp(2*b*x+2*d)-11)*exp(5 /3*b*x+5/3*a)+16*sum(_R*ln(exp(1/3*b*x+1/3*d)+17414258688/166375*b^3*_R^3- 2592/55*b*_R),_R=RootOf(45137758519296*_Z^4*b^4-20323353600*_Z^2*b^2+91506 25))*exp(5/3*a-5/3*d)+55/162*I*ln(exp(1/3*b*x+1/3*d)+I)/b*exp(5/3*a-5/3*d) -55/162*I*ln(exp(1/3*b*x+1/3*d)-I)/b*exp(5/3*a-5/3*d)
Leaf count of result is larger than twice the leaf count of optimal. 9595 vs. \(2 (211) = 422\).
Time = 0.20 (sec) , antiderivative size = 9595, normalized size of antiderivative = 33.55 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^4,x, algorithm="fricas")
Output:
Too large to include
\[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(d+b x) \, dx=e^{\frac {5 a}{3}} \int e^{\frac {5 b x}{3}} \operatorname {sech}^{4}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)**4,x)
Output:
exp(5*a/3)*Integral(exp(5*b*x/3)*sech(b*x + d)**4, x)
Time = 0.14 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.72 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(d+b x) \, dx=-\frac {55 \, {\left (\sqrt {3} \log \left (\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) - \sqrt {3} \log \left (-\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) + 2 \, \arctan \left (\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) + 2 \, \arctan \left (-\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) + 4 \, \arctan \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right )\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{324 \, b} + \frac {{\left (55 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} - 28 \, e^{\left (-\frac {7}{3} \, b x - \frac {7}{3} \, d\right )} - 11 \, e^{\left (-\frac {13}{3} \, b x - \frac {13}{3} \, d\right )}\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{27 \, b {\left (3 \, e^{\left (-2 \, b x - 2 \, d\right )} + 3 \, e^{\left (-4 \, b x - 4 \, d\right )} + e^{\left (-6 \, b x - 6 \, d\right )} + 1\right )}} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^4,x, algorithm="maxima")
Output:
-55/324*(sqrt(3)*log(sqrt(3)*e^(-1/3*b*x - 1/3*d) + e^(-2/3*b*x - 2/3*d) + 1) - sqrt(3)*log(-sqrt(3)*e^(-1/3*b*x - 1/3*d) + e^(-2/3*b*x - 2/3*d) + 1 ) + 2*arctan(sqrt(3) + 2*e^(-1/3*b*x - 1/3*d)) + 2*arctan(-sqrt(3) + 2*e^( -1/3*b*x - 1/3*d)) + 4*arctan(e^(-1/3*b*x - 1/3*d)))*e^(5/3*a - 5/3*d)/b + 1/27*(55*e^(-1/3*b*x - 1/3*d) - 28*e^(-7/3*b*x - 7/3*d) - 11*e^(-13/3*b*x - 13/3*d))*e^(5/3*a - 5/3*d)/(b*(3*e^(-2*b*x - 2*d) + 3*e^(-4*b*x - 4*d) + e^(-6*b*x - 6*d) + 1))
Time = 0.12 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.71 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(d+b x) \, dx=-\frac {{\left (55 \, \sqrt {3} e^{\left (-\frac {17}{3} \, d\right )} \log \left (\sqrt {3} e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 55 \, \sqrt {3} e^{\left (-\frac {17}{3} \, d\right )} \log \left (-\sqrt {3} e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 110 \, \arctan \left ({\left (\sqrt {3} e^{\left (-\frac {1}{3} \, d\right )} + 2 \, e^{\left (\frac {1}{3} \, b x\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (-\frac {17}{3} \, d\right )} - 110 \, \arctan \left (-{\left (\sqrt {3} e^{\left (-\frac {1}{3} \, d\right )} - 2 \, e^{\left (\frac {1}{3} \, b x\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (-\frac {17}{3} \, d\right )} - 220 \, \arctan \left (e^{\left (\frac {1}{3} \, b x + \frac {1}{3} \, d\right )}\right ) e^{\left (-\frac {17}{3} \, d\right )} + \frac {12 \, {\left (11 \, e^{\left (\frac {5}{3} \, b x\right )} - 55 \, e^{\left (\frac {17}{3} \, b x + 4 \, d\right )} + 28 \, e^{\left (\frac {11}{3} \, b x + 2 \, d\right )}\right )} e^{\left (-4 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{3}}\right )} e^{\left (\frac {5}{3} \, a + 4 \, d\right )}}{324 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^4,x, algorithm="giac")
Output:
-1/324*(55*sqrt(3)*e^(-17/3*d)*log(sqrt(3)*e^(1/3*b*x - 1/3*d) + e^(2/3*b* x) + e^(-2/3*d)) - 55*sqrt(3)*e^(-17/3*d)*log(-sqrt(3)*e^(1/3*b*x - 1/3*d) + e^(2/3*b*x) + e^(-2/3*d)) - 110*arctan((sqrt(3)*e^(-1/3*d) + 2*e^(1/3*b *x))*e^(1/3*d))*e^(-17/3*d) - 110*arctan(-(sqrt(3)*e^(-1/3*d) - 2*e^(1/3*b *x))*e^(1/3*d))*e^(-17/3*d) - 220*arctan(e^(1/3*b*x + 1/3*d))*e^(-17/3*d) + 12*(11*e^(5/3*b*x) - 55*e^(17/3*b*x + 4*d) + 28*e^(11/3*b*x + 2*d))*e^(- 4*d)/(e^(2*b*x + 2*d) + 1)^3)*e^(5/3*a + 4*d)/b
Time = 5.88 (sec) , antiderivative size = 542, normalized size of antiderivative = 1.90 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(d+b x) \, dx=\text {Too large to display} \] Input:
int(exp((5*a)/3 + (5*b*x)/3)/cosh(d + b*x)^4,x)
Output:
(55*(-exp(10*a - 10*d))^(1/6)*log((3025*exp((10*a)/3)*exp(-(10*d)/3))/6561 - (3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*(-exp(10*a)*exp( -10*d))^(1/6))/6561))/(162*b) - (55*(-exp(10*a - 10*d))^(1/6)*log((3025*ex p((10*a)/3)*exp(-(10*d)/3))/6561 + (3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/ 3)*exp((b*x)/3)*(-exp(10*a)*exp(-10*d))^(1/6))/6561))/(162*b) + (55*exp((5 *a)/3 + (5*b*x)/3))/(27*b*(exp(2*d + 2*b*x) + 1)) - (8*exp((5*a)/3 + 2*d + (11*b*x)/3))/(3*b*(3*exp(2*d + 2*b*x) + 3*exp(4*d + 4*b*x) + exp(6*d + 6* b*x) + 1)) - (22*exp((5*a)/3 + (5*b*x)/3))/(9*b*(2*exp(2*d + 2*b*x) + exp( 4*d + 4*b*x) + 1)) + (55*log((3025*exp((10*a)/3)*exp(-(10*d)/3))/6561 - (3 025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2)*1i)/2 - 1/2 )*(-exp(10*a)*exp(-10*d))^(1/6))/6561)*(-exp(10*a - 10*d))^(1/6)*((3^(1/2) *1i)/2 - 1/2))/(162*b) - (55*log((3025*exp((10*a)/3)*exp(-(10*d)/3))/6561 + (3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2)*1i)/2 - 1/2)*(-exp(10*a)*exp(-10*d))^(1/6))/6561)*(-exp(10*a - 10*d))^(1/6)*((3^( 1/2)*1i)/2 - 1/2))/(162*b) + (55*log((3025*exp((10*a)/3)*exp(-(10*d)/3))/6 561 - (3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2)*1i) /2 + 1/2)*(-exp(10*a)*exp(-10*d))^(1/6))/6561)*(-exp(10*a - 10*d))^(1/6)*( (3^(1/2)*1i)/2 + 1/2))/(162*b) - (55*log((3025*exp((10*a)/3)*exp(-(10*d)/3 ))/6561 + (3025*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2) *1i)/2 + 1/2)*(-exp(10*a)*exp(-10*d))^(1/6))/6561)*(-exp(10*a - 10*d))^...
\[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^4(d+b x) \, dx=\int e^{\frac {5 b x}{3}+\frac {5 a}{3}} \mathrm {sech}\left (b x +d \right )^{4}d x \] Input:
int(exp(5/3*b*x+5/3*a)*sech(b*x+d)^4,x)
Output:
int(e**((5*a + 5*b*x)/3)*sech(b*x + d)**4,x)