Integrand size = 20, antiderivative size = 200 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(d+b x) \, dx=-\frac {2 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{b \left (1+e^{2 (d+b x)}\right )}+\frac {10 e^{\frac {5 (a-d)}{3}} \arctan \left (e^{\frac {1}{3} (d+b x)}\right )}{3 b}-\frac {5 e^{\frac {5 (a-d)}{3}} \arctan \left (\sqrt {3}-2 e^{\frac {1}{3} (d+b x)}\right )}{3 b}+\frac {5 e^{\frac {5 (a-d)}{3}} \arctan \left (\sqrt {3}+2 e^{\frac {1}{3} (d+b x)}\right )}{3 b}-\frac {5 e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (\frac {\sqrt {3} e^{\frac {1}{3} (d+b x)}}{1+e^{\frac {2}{3} (d+b x)}}\right )}{\sqrt {3} b} \] Output:
-2*exp(5/3*b*x+5/3*a)/b/(1+exp(2*b*x+2*d))+10/3*exp(5/3*a-5/3*d)*arctan(ex p(1/3*b*x+1/3*d))/b+5/3*exp(5/3*a-5/3*d)*arctan(-3^(1/2)+2*exp(1/3*b*x+1/3 *d))/b+5/3*exp(5/3*a-5/3*d)*arctan(3^(1/2)+2*exp(1/3*b*x+1/3*d))/b-5/3*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 = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.22 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(d+b x) \, dx=\frac {12 e^{\frac {5}{3} (a+b x)+2 (d+b x)} \operatorname {Hypergeometric2F1}\left (\frac {11}{6},2,\frac {17}{6},-e^{2 (d+b x)}\right )}{11 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*Sech[d + b*x]^2,x]
Output:
(12*E^((5*(a + b*x))/3 + 2*(d + b*x))*Hypergeometric2F1[11/6, 2, 17/6, -E^ (2*(d + b*x))])/(11*b)
Time = 0.36 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.84, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {2720, 27, 817, 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.
| \(\displaystyle \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(b x+d) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \int \frac {4 e^{\frac {5 a}{3}+\frac {10 b x}{3}}}{\left (1+e^{2 b x}\right )^2}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {12 e^{5 a/3} \int \frac {e^{\frac {10 b x}{3}}}{\left (1+e^{2 b x}\right )^2}de^{\frac {b x}{3}}}{b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \int \frac {e^{\frac {4 b x}{3}}}{1+e^{2 b x}}de^{\frac {b x}{3}}-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
| \(\Big \downarrow \) 824 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \left (\frac {1}{3} \int \frac {1}{1+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1-\sqrt {3} e^{\frac {b x}{3}}}{2 \left (1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}+\frac {1}{3} \int -\frac {1+\sqrt {3} e^{\frac {b x}{3}}}{2 \left (1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}\right )}de^{\frac {b x}{3}}\right )-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \left (\frac {1}{3} \int \frac {1}{1+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1-\sqrt {3} e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1+\sqrt {3} e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \left (-\frac {1}{6} \int \frac {1-\sqrt {3} e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{6} \int \frac {1+\sqrt {3} e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-2 e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}+2 e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}+2 e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \left (\frac {1}{6} \left (-\int \frac {1}{-1-e^{\frac {2 b x}{3}}}d\left (-\sqrt {3}+2 e^{\frac {b x}{3}}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{6} \left (-\int \frac {1}{-1-e^{\frac {2 b x}{3}}}d\left (\sqrt {3}+2 e^{\frac {b x}{3}}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}+2 e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-2 e^{\frac {b x}{3}}}{1-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}-\arctan \left (\sqrt {3}-2 e^{\frac {b x}{3}}\right )\right )+\frac {1}{6} \left (\arctan \left (2 e^{\frac {b x}{3}}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}+2 e^{\frac {b x}{3}}}{1+\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}}de^{\frac {b x}{3}}\right )+\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )\right )-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {12 e^{5 a/3} \left (\frac {5}{6} \left (\frac {1}{3} \arctan \left (e^{\frac {b x}{3}}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (-\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )-\arctan \left (\sqrt {3}-2 e^{\frac {b x}{3}}\right )\right )+\frac {1}{6} \left (\arctan \left (2 e^{\frac {b x}{3}}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\sqrt {3} e^{\frac {b x}{3}}+e^{\frac {2 b x}{3}}+1\right )\right )\right )-\frac {e^{\frac {5 b x}{3}}}{6 \left (e^{2 b x}+1\right )}\right )}{b}\) |
Input:
Int[E^((5*(a + b*x))/3)*Sech[d + b*x]^2,x]
Output:
(12*E^((5*a)/3)*(-1/6*E^((5*b*x)/3)/(1 + E^(2*b*x)) + (5*(ArcTan[E^((b*x)/ 3)]/3 + (-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 + (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))/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[(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 = 0.54 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.70
| method | result | size |
| risch | \(-\frac {2 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}}}{b \left (1+{\mathrm e}^{2 b x +2 d}\right )}+\frac {5 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+i\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{3 b}-\frac {5 i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-i\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{3 b}+4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (20736 b^{4} \textit {\_Z}^{4}-3600 b^{2} \textit {\_Z}^{2}+625\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+\frac {1728 b^{3} \textit {\_R}^{3}}{125}-\frac {12 b \textit {\_R}}{5}\right )\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}\) | \(140\) |
Input:
int(exp(5/3*b*x+5/3*a)*sech(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
-2*exp(5/3*b*x+5/3*a)/b/(1+exp(2*b*x+2*d))+5/3*I*ln(exp(1/3*b*x+1/3*d)+I)/ b*exp(5/3*a-5/3*d)-5/3*I*ln(exp(1/3*b*x+1/3*d)-I)/b*exp(5/3*a-5/3*d)+4*sum (_R*ln(exp(1/3*b*x+1/3*d)+1728/125*b^3*_R^3-12/5*b*_R),_R=RootOf(20736*_Z^ 4*b^4-3600*_Z^2*b^2+625))*exp(5/3*a-5/3*d)
Leaf count of result is larger than twice the leaf count of optimal. 2654 vs. \(2 (154) = 308\).
Time = 0.12 (sec) , antiderivative size = 2654, normalized size of antiderivative = 13.27 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^2,x, algorithm="fricas")
Output:
-1/6*(12*cosh(1/3*b*x + 1/3*d)^5*cosh(-5/3*a + 5/3*d) + 12*(cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^5 - 12*cosh(1/3*b*x + 1/3*d)^5*sinh(-5/3*a + 5/3*d) + 60*(cosh(1/3*b*x + 1/3*d)*cosh(-5/3*a + 5 /3*d) - cosh(1/3*b*x + 1/3*d)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^ 4 + 120*(cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3 *d)^2*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^3 + 15*sqrt(1/3)*(b*cosh (1/3*b*x + 1/3*d)^6 + 6*b*cosh(1/3*b*x + 1/3*d)^5*sinh(1/3*b*x + 1/3*d) + 15*b*cosh(1/3*b*x + 1/3*d)^4*sinh(1/3*b*x + 1/3*d)^2 + 20*b*cosh(1/3*b*x + 1/3*d)^3*sinh(1/3*b*x + 1/3*d)^3 + 15*b*cosh(1/3*b*x + 1/3*d)^2*sinh(1/3* b*x + 1/3*d)^4 + 6*b*cosh(1/3*b*x + 1/3*d)*sinh(1/3*b*x + 1/3*d)^5 + b*sin h(1/3*b*x + 1/3*d)^6 + b)*sqrt((cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d ))/(b^2*cosh(-5/3*a + 5/3*d) + b^2*sinh(-5/3*a + 5/3*d)))*log(5*cosh(1/3*b *x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) + 5*(cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^2 + 15*sqrt(1/3)*(b*cosh(1/3*b*x + 1/3*d) + b*sinh(1/3*b*x + 1/3*d))*sqrt((cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3 *d))/(b^2*cosh(-5/3*a + 5/3*d) + b^2*sinh(-5/3*a + 5/3*d))) + 10*(cosh(1/3 *b*x + 1/3*d)*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)*sinh(-5/3*a + 5 /3*d))*sinh(1/3*b*x + 1/3*d) - 5*(cosh(1/3*b*x + 1/3*d)^2 + 1)*sinh(-5/3*a + 5/3*d) + 5*cosh(-5/3*a + 5/3*d)) - 15*sqrt(1/3)*(b*cosh(1/3*b*x + 1/3*d )^6 + 6*b*cosh(1/3*b*x + 1/3*d)^5*sinh(1/3*b*x + 1/3*d) + 15*b*cosh(1/3...
\[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(d+b x) \, dx=e^{\frac {5 a}{3}} \int e^{\frac {5 b x}{3}} \operatorname {sech}^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)**2,x)
Output:
exp(5*a/3)*Integral(exp(5*b*x/3)*sech(b*x + d)**2, x)
Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.78 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(d+b x) \, dx=-\frac {5 \, {\left (\sqrt {3} \log \left (\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) - \sqrt {3} \log \left (-\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) + 2 \, \arctan \left (\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) + 2 \, \arctan \left (-\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) + 4 \, \arctan \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right )\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{6 \, b} - \frac {2 \, e^{\left (-\frac {1}{3} \, b x + \frac {5}{3} \, a - 2 \, d\right )}}{b {\left (e^{\left (-2 \, b x - 2 \, d\right )} + 1\right )}} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^2,x, algorithm="maxima")
Output:
-5/6*(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 - 2* e^(-1/3*b*x + 5/3*a - 2*d)/(b*(e^(-2*b*x - 2*d) + 1))
Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.89 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(d+b x) \, dx=-\frac {{\left (5 \, \sqrt {3} e^{\left (-\frac {11}{3} \, d\right )} \log \left (\sqrt {3} e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 5 \, \sqrt {3} e^{\left (-\frac {11}{3} \, d\right )} \log \left (-\sqrt {3} e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 10 \, \arctan \left ({\left (\sqrt {3} e^{\left (-\frac {1}{3} \, d\right )} + 2 \, e^{\left (\frac {1}{3} \, b x\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (-\frac {11}{3} \, d\right )} - 10 \, \arctan \left (-{\left (\sqrt {3} e^{\left (-\frac {1}{3} \, d\right )} - 2 \, e^{\left (\frac {1}{3} \, b x\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (-\frac {11}{3} \, d\right )} - 20 \, \arctan \left (e^{\left (\frac {1}{3} \, b x + \frac {1}{3} \, d\right )}\right ) e^{\left (-\frac {11}{3} \, d\right )} + \frac {12 \, e^{\left (\frac {5}{3} \, b x - 2 \, d\right )}}{e^{\left (2 \, b x + 2 \, d\right )} + 1}\right )} e^{\left (\frac {5}{3} \, a + 2 \, d\right )}}{6 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d)^2,x, algorithm="giac")
Output:
-1/6*(5*sqrt(3)*e^(-11/3*d)*log(sqrt(3)*e^(1/3*b*x - 1/3*d) + e^(2/3*b*x) + e^(-2/3*d)) - 5*sqrt(3)*e^(-11/3*d)*log(-sqrt(3)*e^(1/3*b*x - 1/3*d) + e ^(2/3*b*x) + e^(-2/3*d)) - 10*arctan((sqrt(3)*e^(-1/3*d) + 2*e^(1/3*b*x))* e^(1/3*d))*e^(-11/3*d) - 10*arctan(-(sqrt(3)*e^(-1/3*d) - 2*e^(1/3*b*x))*e ^(1/3*d))*e^(-11/3*d) - 20*arctan(e^(1/3*b*x + 1/3*d))*e^(-11/3*d) + 12*e^ (5/3*b*x - 2*d)/(e^(2*b*x + 2*d) + 1))*e^(5/3*a + 2*d)/b
Time = 5.84 (sec) , antiderivative size = 452, normalized size of antiderivative = 2.26 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(d+b x) \, dx =\text {Too large to display} \] Input:
int(exp((5*a)/3 + (5*b*x)/3)/cosh(d + b*x)^2,x)
Output:
(5*(-exp(10*a - 10*d))^(1/6)*log((100*exp((10*a)/3)*exp(-(10*d)/3))/9 - (1 00*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*(-exp(10*a)*exp(-10*d) )^(1/6))/9))/(3*b) - (5*(-exp(10*a - 10*d))^(1/6)*log((100*exp((10*a)/3)*e xp(-(10*d)/3))/9 + (100*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*( -exp(10*a)*exp(-10*d))^(1/6))/9))/(3*b) - (2*exp((5*a)/3 + (5*b*x)/3))/(b* (exp(2*d + 2*b*x) + 1)) + (5*log((100*exp((10*a)/3)*exp(-(10*d)/3))/9 - (1 00*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))/9)*(-exp(10*a - 10*d))^(1/6)*((3^(1/2)*1i) /2 - 1/2))/(3*b) - (5*log((100*exp((10*a)/3)*exp(-(10*d)/3))/9 + (100*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))/9)*(-exp(10*a - 10*d))^(1/6)*((3^(1/2)*1i)/2 - 1/ 2))/(3*b) + (5*log((100*exp((10*a)/3)*exp(-(10*d)/3))/9 - (100*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)*e xp(-10*d))^(1/6))/9)*(-exp(10*a - 10*d))^(1/6)*((3^(1/2)*1i)/2 + 1/2))/(3* b) - (5*log((100*exp((10*a)/3)*exp(-(10*d)/3))/9 + (100*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))/9)*(-exp(10*a - 10*d))^(1/6)*((3^(1/2)*1i)/2 + 1/2))/(3*b)
\[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^2(d+b x) \, dx=\int e^{\frac {5 b x}{3}+\frac {5 a}{3}} \mathrm {sech}\left (b x +d \right )^{2}d x \] Input:
int(exp(5/3*b*x+5/3*a)*sech(b*x+d)^2,x)
Output:
int(e**((5*a + 5*b*x)/3)*sech(b*x + d)**2,x)