Integrand size = 20, antiderivative size = 285 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(d+b x) \, dx=-\frac {4 e^{\frac {5 (a-d)}{3}+\frac {14}{3} (d+b x)}}{b \left (1+e^{2 (d+b x)}\right )^4}-\frac {28 e^{\frac {5 (a-d)}{3}+\frac {8}{3} (d+b x)}}{9 b \left (1+e^{2 (d+b x)}\right )^3}-\frac {56 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)}}{27 b \left (1+e^{2 (d+b x)}\right )^2}+\frac {56 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)}}{81 b \left (1+e^{2 (d+b x)}\right )}-\frac {112 e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1-2 e^{\frac {2}{3} (d+b x)}}{\sqrt {3}}\right )}{81 \sqrt {3} b}+\frac {112 e^{\frac {5 (a-d)}{3}} \log \left (1+e^{\frac {2}{3} (d+b x)}\right )}{243 b}-\frac {56 e^{\frac {5 (a-d)}{3}} \log \left (1-e^{\frac {2}{3} (d+b x)}+e^{\frac {4}{3} (d+b x)}\right )}{243 b} \] Output:
-4*exp(5/3*a+3*d+14/3*b*x)/b/(1+exp(2*b*x+2*d))^4-28/9*exp(5/3*a+d+8/3*b*x )/b/(1+exp(2*b*x+2*d))^3-56/27*exp(5/3*a-d+2/3*b*x)/b/(1+exp(2*b*x+2*d))^2 +56/81*exp(5/3*a-d+2/3*b*x)/b/(1+exp(2*b*x+2*d))-112/243*3^(1/2)*exp(5/3*a -5/3*d)*arctan(1/3*(1-2*exp(2/3*b*x+2/3*d))*3^(1/2))/b+112/243*exp(5/3*a-5 /3*d)*ln(1+exp(2/3*b*x+2/3*d))/b-56/243*exp(5/3*a-5/3*d)*ln(1-exp(2/3*b*x+ 2/3*d)+exp(4/3*b*x+4/3*d))/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.30 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(d+b x) \, dx=\frac {e^{\frac {5}{3} (a+b x)} \left (-112 e^{d+b x} \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3},\frac {7}{3},-e^{2 (d+b x)}\right )+28 \text {sech}(d+b x) (5+3 \tanh (d+b x))+9 \text {sech}^3(d+b x) (5+9 \tanh (d+b x))\right )}{324 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*Sech[d + b*x]^5,x]
Output:
(E^((5*(a + b*x))/3)*(-112*E^(d + b*x)*Hypergeometric2F1[1, 4/3, 7/3, -E^( 2*(d + b*x))] + 28*Sech[d + b*x]*(5 + 3*Tanh[d + b*x]) + 9*Sech[d + b*x]^3 *(5 + 9*Tanh[d + b*x])))/(324*b)
Time = 0.51 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.55, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {2720, 27, 807, 817, 817, 817, 749, 750, 16, 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.
| \(\displaystyle \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(b x+d) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \int \frac {32 e^{\frac {5 a}{3}+\frac {19 b x}{3}}}{\left (1+e^{2 b x}\right )^5}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {96 e^{5 a/3} \int \frac {e^{\frac {19 b x}{3}}}{\left (1+e^{2 b x}\right )^5}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {48 e^{5 a/3} \int \frac {e^{3 b x}}{\left (1+e^{b x}\right )^5}de^{\frac {2 b x}{3}}}{b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \int \frac {e^{2 b x}}{\left (1+e^{b x}\right )^4}de^{\frac {2 b x}{3}}-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \int \frac {e^{b x}}{\left (1+e^{b x}\right )^3}de^{\frac {2 b x}{3}}-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \int \frac {1}{\left (1+e^{b x}\right )^2}de^{\frac {2 b x}{3}}-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \left (\frac {2}{3} \int \frac {1}{1+e^{b x}}de^{\frac {2 b x}{3}}+\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \int \left (2-e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}+\frac {1}{3} \int \frac {1}{1+e^{\frac {2 b x}{3}}}de^{\frac {2 b x}{3}}\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \int \left (2-e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {3}{2} \int 1de^{\frac {2 b x}{3}}-\frac {1}{2} \int \left (-1+2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}\right )+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {3}{2} \int 1de^{\frac {2 b x}{3}}+\frac {1}{2} \int \left (1-2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}\right )+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}-3 \int \frac {1}{-2-2 e^{\frac {2 b x}{3}}}d\left (-1+2 e^{\frac {2 b x}{3}}\right )\right )+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}+\sqrt {3} \arctan \left (\frac {2 e^{\frac {2 b x}{3}}-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {48 e^{5 a/3} \left (\frac {7}{12} \left (\frac {4}{9} \left (\frac {1}{6} \left (\frac {2}{3} \left (\frac {\arctan \left (\frac {2 e^{\frac {2 b x}{3}}-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )+\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {2 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )-\frac {e^{\frac {4 b x}{3}}}{9 \left (e^{b x}+1\right )^3}\right )-\frac {e^{\frac {7 b x}{3}}}{12 \left (e^{b x}+1\right )^4}\right )}{b}\) |
Input:
Int[E^((5*(a + b*x))/3)*Sech[d + b*x]^5,x]
Output:
(48*E^((5*a)/3)*(-1/12*E^((7*b*x)/3)/(1 + E^(b*x))^4 + (7*(-1/9*E^((4*b*x) /3)/(1 + E^(b*x))^3 + (4*(-1/6*E^((2*b*x)/3)/(1 + E^(b*x))^2 + (E^((2*b*x) /3)/(3*(1 + E^(b*x))) + (2*(ArcTan[(-1 + 2*E^((2*b*x)/3))/Sqrt[3]]/Sqrt[3] + Log[1 + E^((2*b*x)/3)]/3))/3)/6))/9))/12))/b
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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[(-(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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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[((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 complex when optimal does not.
Time = 5.36 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.79
| method | result | size |
| risch | \(\frac {4 \left (14 \,{\mathrm e}^{6 b x +6 d}-144 \,{\mathrm e}^{4 b x +4 d}-105 \,{\mathrm e}^{2 b x +2 d}-28\right ) {\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}}}{81 \left (1+{\mathrm e}^{2 b x +2 d}\right )^{4} b}-\frac {56 \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{243 b}+\frac {56 i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{243 b}-\frac {56 \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{243 b}-\frac {56 i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{243 b}+\frac {112 \,{\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \ln \left (1+{\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}\right )}{243 b}\) | \(224\) |
Input:
int(exp(5/3*b*x+5/3*a)*sech(b*x+d)^5,x,method=_RETURNVERBOSE)
Output:
4/81/(1+exp(2*b*x+2*d))^4/b*(14*exp(6*b*x+6*d)-144*exp(4*b*x+4*d)-105*exp( 2*b*x+2*d)-28)*exp(5/3*a-d+2/3*b*x)-56/243*ln(exp(2/3*b*x+2/3*d)-1/2+1/2*I *3^(1/2))/b*exp(5/3*a-5/3*d)+56/243*I*ln(exp(2/3*b*x+2/3*d)-1/2+1/2*I*3^(1 /2))/b*exp(5/3*a-5/3*d)*3^(1/2)-56/243*ln(exp(2/3*b*x+2/3*d)-1/2-1/2*I*3^( 1/2))/b*exp(5/3*a-5/3*d)-56/243*I*ln(exp(2/3*b*x+2/3*d)-1/2-1/2*I*3^(1/2)) /b*exp(5/3*a-5/3*d)*3^(1/2)+112/243*exp(5/3*a-5/3*d)*ln(1+exp(2/3*b*x+2/3* d))/b
Leaf count of result is larger than twice the leaf count of optimal. 11662 vs. \(2 (215) = 430\).
Time = 0.26 (sec) , antiderivative size = 11662, normalized size of antiderivative = 40.92 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^5,x, algorithm="fricas")
Output:
Too large to include
\[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(d+b x) \, dx=e^{\frac {5 a}{3}} \int e^{\frac {5 b x}{3}} \operatorname {sech}^{5}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)**5,x)
Output:
exp(5*a/3)*Integral(exp(5*b*x/3)*sech(b*x + d)**5, x)
Time = 0.13 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.81 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(d+b x) \, dx=\frac {56 \, {\left (2 \, \sqrt {3} \arctan \left (\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) - 2 \, \sqrt {3} \arctan \left (-\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) - \log \left (\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) - \log \left (-\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) + 2 \, \log \left (e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{243 \, b} + \frac {4 \, {\left (14 \, e^{\left (-\frac {4}{3} \, b x - \frac {4}{3} \, d\right )} - 144 \, e^{\left (-\frac {10}{3} \, b x - \frac {10}{3} \, d\right )} - 105 \, e^{\left (-\frac {16}{3} \, b x - \frac {16}{3} \, d\right )} - 28 \, e^{\left (-\frac {22}{3} \, b x - \frac {22}{3} \, d\right )}\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{81 \, b {\left (4 \, e^{\left (-2 \, b x - 2 \, d\right )} + 6 \, e^{\left (-4 \, b x - 4 \, d\right )} + 4 \, e^{\left (-6 \, b x - 6 \, d\right )} + e^{\left (-8 \, b x - 8 \, d\right )} + 1\right )}} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^5,x, algorithm="maxima")
Output:
56/243*(2*sqrt(3)*arctan(sqrt(3) + 2*e^(-1/3*b*x - 1/3*d)) - 2*sqrt(3)*arc tan(-sqrt(3) + 2*e^(-1/3*b*x - 1/3*d)) - log(sqrt(3)*e^(-1/3*b*x - 1/3*d) + e^(-2/3*b*x - 2/3*d) + 1) - log(-sqrt(3)*e^(-1/3*b*x - 1/3*d) + e^(-2/3* b*x - 2/3*d) + 1) + 2*log(e^(-2/3*b*x - 2/3*d) + 1))*e^(5/3*a - 5/3*d)/b + 4/81*(14*e^(-4/3*b*x - 4/3*d) - 144*e^(-10/3*b*x - 10/3*d) - 105*e^(-16/3 *b*x - 16/3*d) - 28*e^(-22/3*b*x - 22/3*d))*e^(5/3*a - 5/3*d)/(b*(4*e^(-2* b*x - 2*d) + 6*e^(-4*b*x - 4*d) + 4*e^(-6*b*x - 6*d) + e^(-8*b*x - 8*d) + 1))
Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.53 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(d+b x) \, dx=\frac {4 \, {\left (28 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {2}{3} \, b x\right )} - e^{\left (-\frac {2}{3} \, d\right )}\right )} e^{\left (\frac {2}{3} \, d\right )}\right ) e^{\left (-\frac {20}{3} \, d\right )} - 14 \, e^{\left (-\frac {20}{3} \, d\right )} \log \left (e^{\left (\frac {4}{3} \, b x\right )} - e^{\left (\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + e^{\left (-\frac {4}{3} \, d\right )}\right ) + 28 \, e^{\left (-\frac {20}{3} \, d\right )} \log \left (e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) + \frac {3 \, {\left (14 \, e^{\left (\frac {20}{3} \, b x + 6 \, d\right )} - 144 \, e^{\left (\frac {14}{3} \, b x + 4 \, d\right )} - 105 \, e^{\left (\frac {8}{3} \, b x + 2 \, d\right )} - 28 \, e^{\left (\frac {2}{3} \, b x\right )}\right )} e^{\left (-6 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{4}}\right )} e^{\left (\frac {5}{3} \, a + 5 \, d\right )}}{243 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^5,x, algorithm="giac")
Output:
4/243*(28*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(2/3*b*x) - e^(-2/3*d))*e^(2/3*d ))*e^(-20/3*d) - 14*e^(-20/3*d)*log(e^(4/3*b*x) - e^(2/3*b*x - 2/3*d) + e^ (-4/3*d)) + 28*e^(-20/3*d)*log(e^(2/3*b*x) + e^(-2/3*d)) + 3*(14*e^(20/3*b *x + 6*d) - 144*e^(14/3*b*x + 4*d) - 105*e^(8/3*b*x + 2*d) - 28*e^(2/3*b*x ))*e^(-6*d)/(e^(2*b*x + 2*d) + 1)^4)*e^(5/3*a + 5*d)/b
Time = 5.49 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.28 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(d+b x) \, dx=\frac {56\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}}{81\,b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}+1\right )}-\frac {28\,{\mathrm {e}}^{\frac {5\,a}{3}+d+\frac {8\,b\,x}{3}}}{9\,b\,\left (3\,{\mathrm {e}}^{2\,d+2\,b\,x}+3\,{\mathrm {e}}^{4\,d+4\,b\,x}+{\mathrm {e}}^{6\,d+6\,b\,x}+1\right )}-\frac {56\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}}{27\,b\,\left (2\,{\mathrm {e}}^{2\,d+2\,b\,x}+{\mathrm {e}}^{4\,d+4\,b\,x}+1\right )}+\frac {112\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (-\frac {224\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{243}-\frac {224\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{243}\right )}{243\,b}-\frac {4\,{\mathrm {e}}^{\frac {5\,a}{3}+3\,d+\frac {14\,b\,x}{3}}}{b\,\left (4\,{\mathrm {e}}^{2\,d+2\,b\,x}+6\,{\mathrm {e}}^{4\,d+4\,b\,x}+4\,{\mathrm {e}}^{6\,d+6\,b\,x}+{\mathrm {e}}^{8\,d+8\,b\,x}+1\right )}+\frac {112\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (-\frac {224\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{243}-\frac {224\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{243}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{243\,b}-\frac {112\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (\frac {224\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{243}-\frac {224\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{243}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{243\,b} \] Input:
int(exp((5*a)/3 + (5*b*x)/3)/cosh(d + b*x)^5,x)
Output:
(56*exp((5*a)/3 - d + (2*b*x)/3))/(81*b*(exp(2*d + 2*b*x) + 1)) - (28*exp( (5*a)/3 + d + (8*b*x)/3))/(9*b*(3*exp(2*d + 2*b*x) + 3*exp(4*d + 4*b*x) + exp(6*d + 6*b*x) + 1)) - (56*exp((5*a)/3 - d + (2*b*x)/3))/(27*b*(2*exp(2* d + 2*b*x) + exp(4*d + 4*b*x) + 1)) + (112*exp(5*a - 5*d)^(1/3)*log(- (224 *(exp(5*a)*exp(-5*d))^(1/3))/243 - (224*exp((5*a)/3)*exp((2*d)/3)*exp(-(5* d)/3)*exp((2*b*x)/3))/243))/(243*b) - (4*exp((5*a)/3 + 3*d + (14*b*x)/3))/ (b*(4*exp(2*d + 2*b*x) + 6*exp(4*d + 4*b*x) + 4*exp(6*d + 6*b*x) + exp(8*d + 8*b*x) + 1)) + (112*exp(5*a - 5*d)^(1/3)*log(- (224*((3^(1/2)*1i)/2 - 1 /2)*(exp(5*a)*exp(-5*d))^(1/3))/243 - (224*exp((5*a)/3)*exp((2*d)/3)*exp(- (5*d)/3)*exp((2*b*x)/3))/243)*((3^(1/2)*1i)/2 - 1/2))/(243*b) - (112*exp(5 *a - 5*d)^(1/3)*log((224*((3^(1/2)*1i)/2 + 1/2)*(exp(5*a)*exp(-5*d))^(1/3) )/243 - (224*exp((5*a)/3)*exp((2*d)/3)*exp(-(5*d)/3)*exp((2*b*x)/3))/243)* ((3^(1/2)*1i)/2 + 1/2))/(243*b)
\[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^5(d+b x) \, dx=\int e^{\frac {5 b x}{3}+\frac {5 a}{3}} \mathrm {sech}\left (b x +d \right )^{5}d x \] Input:
int(exp(5/3*b*x+5/3*a)*sech(b*x+d)^5,x)
Output:
int(e**((5*a + 5*b*x)/3)*sech(b*x + d)**5,x)