Integrand size = 18, antiderivative size = 155 \[ \int e^{2 (a+b x)} \text {sech}^5(d+b x) \, dx=-\frac {4 e^{2 a+3 d+5 b x}}{b \left (1+e^{2 d+2 b x}\right )^4}-\frac {10 e^{2 a+d+3 b x}}{3 b \left (1+e^{2 d+2 b x}\right )^3}-\frac {5 e^{2 a-d+b x}}{2 b \left (1+e^{2 d+2 b x}\right )^2}+\frac {5 e^{2 a-d+b x}}{4 b \left (1+e^{2 d+2 b x}\right )}+\frac {5 e^{2 a-2 d} \arctan \left (e^{d+b x}\right )}{4 b} \] Output:
-4*exp(5*b*x+2*a+3*d)/b/(1+exp(2*b*x+2*d))^4-10/3*exp(3*b*x+2*a+d)/b/(1+ex p(2*b*x+2*d))^3-5/2*exp(b*x+2*a-d)/b/(1+exp(2*b*x+2*d))^2+5/4*exp(b*x+2*a- d)/b/(1+exp(2*b*x+2*d))+5/4*exp(2*a-2*d)*arctan(exp(b*x+d))/b
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.59 \[ \int e^{2 (a+b x)} \text {sech}^5(d+b x) \, dx=\frac {e^{2 a-2 d} \left (30 \left (-e^{d+b x}+\arctan \left (e^{d+b x}\right )\right )+5 e^{2 (d+b x)} \text {sech}(d+b x) (2+\tanh (d+b x))+2 e^{2 (d+b x)} \text {sech}^3(d+b x) (2+3 \tanh (d+b x))\right )}{24 b} \] Input:
Integrate[E^(2*(a + b*x))*Sech[d + b*x]^5,x]
Output:
(E^(2*a - 2*d)*(30*(-E^(d + b*x) + ArcTan[E^(d + b*x)]) + 5*E^(2*(d + b*x) )*Sech[d + b*x]*(2 + Tanh[d + b*x]) + 2*E^(2*(d + b*x))*Sech[d + b*x]^3*(2 + 3*Tanh[d + b*x])))/(24*b)
Time = 0.24 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.74, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2720, 27, 252, 252, 252, 215, 216}
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.
\(\displaystyle \int e^{2 (a+b x)} \text {sech}^5(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {32 e^{2 a+6 b x}}{\left (1+e^{2 b x}\right )^5}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {32 e^{2 a} \int \frac {e^{6 b x}}{\left (1+e^{2 b x}\right )^5}de^{b x}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {32 e^{2 a} \left (\frac {5}{8} \int \frac {e^{4 b x}}{\left (1+e^{2 b x}\right )^4}de^{b x}-\frac {e^{5 b x}}{8 \left (e^{2 b x}+1\right )^4}\right )}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {32 e^{2 a} \left (\frac {5}{8} \left (\frac {1}{2} \int \frac {e^{2 b x}}{\left (1+e^{2 b x}\right )^3}de^{b x}-\frac {e^{3 b x}}{6 \left (e^{2 b x}+1\right )^3}\right )-\frac {e^{5 b x}}{8 \left (e^{2 b x}+1\right )^4}\right )}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {32 e^{2 a} \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{\left (1+e^{2 b x}\right )^2}de^{b x}-\frac {e^{b x}}{4 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{3 b x}}{6 \left (e^{2 b x}+1\right )^3}\right )-\frac {e^{5 b x}}{8 \left (e^{2 b x}+1\right )^4}\right )}{b}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {32 e^{2 a} \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1+e^{2 b x}}de^{b x}+\frac {e^{b x}}{2 \left (e^{2 b x}+1\right )}\right )-\frac {e^{b x}}{4 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{3 b x}}{6 \left (e^{2 b x}+1\right )^3}\right )-\frac {e^{5 b x}}{8 \left (e^{2 b x}+1\right )^4}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {32 e^{2 a} \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \arctan \left (e^{b x}\right )+\frac {e^{b x}}{2 \left (e^{2 b x}+1\right )}\right )-\frac {e^{b x}}{4 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{3 b x}}{6 \left (e^{2 b x}+1\right )^3}\right )-\frac {e^{5 b x}}{8 \left (e^{2 b x}+1\right )^4}\right )}{b}\) |
Input:
Int[E^(2*(a + b*x))*Sech[d + b*x]^5,x]
Output:
(32*E^(2*a)*(-1/8*E^(5*b*x)/(1 + E^(2*b*x))^4 + (5*(-1/6*E^(3*b*x)/(1 + E^ (2*b*x))^3 + (-1/4*E^(b*x)/(1 + E^(2*b*x))^2 + (E^(b*x)/(2*(1 + E^(2*b*x)) ) + ArcTan[E^(b*x)]/2)/4)/2))/8))/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)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, 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 = 5.30 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95
method | result | size |
risch | \(-\frac {\left (-15 \,{\mathrm e}^{6 b x +6 a +6 d}+73 \,{\mathrm e}^{4 b x +6 a +4 d}+55 \,{\mathrm e}^{2 b x +6 a +2 d}+15 \,{\mathrm e}^{6 a}\right ) {\mathrm e}^{b x +4 a -d}}{12 \left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{4} b}+\frac {5 i \ln \left ({\mathrm e}^{b x +a}+i {\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{8 b}-\frac {5 i \ln \left ({\mathrm e}^{b x +a}-i {\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{8 b}\) | \(148\) |
Input:
int(exp(2*b*x+2*a)*sech(b*x+d)^5,x,method=_RETURNVERBOSE)
Output:
-1/12/(exp(2*b*x+2*a+2*d)+exp(2*a))^4/b*(-15*exp(6*b*x+6*a+6*d)+73*exp(4*b *x+6*a+4*d)+55*exp(2*b*x+6*a+2*d)+15*exp(6*a))*exp(b*x+4*a-d)+5/8*I*ln(exp (b*x+a)+I*exp(a-d))/b*exp(2*a-2*d)-5/8*I*ln(exp(b*x+a)-I*exp(a-d))/b*exp(2 *a-2*d)
Leaf count of result is larger than twice the leaf count of optimal. 1710 vs. \(2 (137) = 274\).
Time = 0.09 (sec) , antiderivative size = 1710, normalized size of antiderivative = 11.03 \[ \int e^{2 (a+b x)} \text {sech}^5(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)^5,x, algorithm="fricas")
Output:
1/12*(15*cosh(b*x + d)^7*cosh(-2*a + 2*d) + 15*(cosh(-2*a + 2*d) - sinh(-2 *a + 2*d))*sinh(b*x + d)^7 + 105*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b* x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^6 - 73*cosh(b*x + d)^5*cosh(-2*a + 2*d) + (315*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (315*cosh(b*x + d)^2 - 73)* sinh(-2*a + 2*d) - 73*cosh(-2*a + 2*d))*sinh(b*x + d)^5 + 5*(105*cosh(b*x + d)^3*cosh(-2*a + 2*d) - 73*cosh(b*x + d)*cosh(-2*a + 2*d) - (105*cosh(b* x + d)^3 - 73*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^4 - 55*cosh(b *x + d)^3*cosh(-2*a + 2*d) + 5*(105*cosh(b*x + d)^4*cosh(-2*a + 2*d) - 146 *cosh(b*x + d)^2*cosh(-2*a + 2*d) - (105*cosh(b*x + d)^4 - 146*cosh(b*x + d)^2 - 11)*sinh(-2*a + 2*d) - 11*cosh(-2*a + 2*d))*sinh(b*x + d)^3 + 5*(63 *cosh(b*x + d)^5*cosh(-2*a + 2*d) - 146*cosh(b*x + d)^3*cosh(-2*a + 2*d) - 33*cosh(b*x + d)*cosh(-2*a + 2*d) - (63*cosh(b*x + d)^5 - 146*cosh(b*x + d)^3 - 33*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^2 + 15*(cosh(b*x + d)^8*cosh(-2*a + 2*d) + (cosh(-2*a + 2*d) - sinh(-2*a + 2*d))*sinh(b*x + d)^8 + 8*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d) )*sinh(b*x + d)^7 + 4*cosh(b*x + d)^6*cosh(-2*a + 2*d) + 4*(7*cosh(b*x + d )^2*cosh(-2*a + 2*d) - (7*cosh(b*x + d)^2 + 1)*sinh(-2*a + 2*d) + cosh(-2* a + 2*d))*sinh(b*x + d)^6 + 8*(7*cosh(b*x + d)^3*cosh(-2*a + 2*d) + 3*cosh (b*x + d)*cosh(-2*a + 2*d) - (7*cosh(b*x + d)^3 + 3*cosh(b*x + d))*sinh(-2 *a + 2*d))*sinh(b*x + d)^5 + 6*cosh(b*x + d)^4*cosh(-2*a + 2*d) + 2*(35...
\[ \int e^{2 (a+b x)} \text {sech}^5(d+b x) \, dx=e^{2 a} \int e^{2 b x} \operatorname {sech}^{5}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)**5,x)
Output:
exp(2*a)*Integral(exp(2*b*x)*sech(b*x + d)**5, x)
Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.83 \[ \int e^{2 (a+b x)} \text {sech}^5(d+b x) \, dx=-\frac {5 \, \arctan \left (e^{\left (-b x - d\right )}\right ) e^{\left (2 \, a - 2 \, d\right )}}{4 \, b} + \frac {{\left (15 \, e^{\left (-b x - d\right )} - 73 \, e^{\left (-3 \, b x - 3 \, d\right )} - 55 \, e^{\left (-5 \, b x - 5 \, d\right )} - 15 \, e^{\left (-7 \, b x - 7 \, d\right )}\right )} e^{\left (2 \, a - 2 \, d\right )}}{12 \, b {\left (4 \, e^{\left (-2 \, b x - 2 \, d\right )} + 6 \, e^{\left (-4 \, b x - 4 \, d\right )} + 4 \, e^{\left (-6 \, b x - 6 \, d\right )} + e^{\left (-8 \, b x - 8 \, d\right )} + 1\right )}} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)^5,x, algorithm="maxima")
Output:
-5/4*arctan(e^(-b*x - d))*e^(2*a - 2*d)/b + 1/12*(15*e^(-b*x - d) - 73*e^( -3*b*x - 3*d) - 55*e^(-5*b*x - 5*d) - 15*e^(-7*b*x - 7*d))*e^(2*a - 2*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 = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.54 \[ \int e^{2 (a+b x)} \text {sech}^5(d+b x) \, dx=\frac {{\left (15 \, \arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (-2 \, d\right )} + \frac {{\left (15 \, e^{\left (7 \, b x + 7 \, d\right )} - 73 \, e^{\left (5 \, b x + 5 \, d\right )} - 55 \, e^{\left (3 \, b x + 3 \, d\right )} - 15 \, e^{\left (b x + d\right )}\right )} e^{\left (-2 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{4}}\right )} e^{\left (2 \, a\right )}}{12 \, b} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)^5,x, algorithm="giac")
Output:
1/12*(15*arctan(e^(b*x + d))*e^(-2*d) + (15*e^(7*b*x + 7*d) - 73*e^(5*b*x + 5*d) - 55*e^(3*b*x + 3*d) - 15*e^(b*x + d))*e^(-2*d)/(e^(2*b*x + 2*d) + 1)^4)*e^(2*a)/b
Time = 2.60 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.50 \[ \int e^{2 (a+b x)} \text {sech}^5(d+b x) \, dx=\frac {5\,{\mathrm {e}}^{2\,a-d+b\,x}}{4\,b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}+1\right )}-\frac {10\,{\mathrm {e}}^{2\,a+d+3\,b\,x}}{3\,b\,\left (3\,{\mathrm {e}}^{2\,d+2\,b\,x}+3\,{\mathrm {e}}^{4\,d+4\,b\,x}+{\mathrm {e}}^{6\,d+6\,b\,x}+1\right )}-\frac {5\,{\mathrm {e}}^{2\,a-d+b\,x}}{2\,b\,\left (2\,{\mathrm {e}}^{2\,d+2\,b\,x}+{\mathrm {e}}^{4\,d+4\,b\,x}+1\right )}-\frac {4\,{\mathrm {e}}^{2\,a+3\,d+5\,b\,x}}{b\,\left (4\,{\mathrm {e}}^{2\,d+2\,b\,x}+6\,{\mathrm {e}}^{4\,d+4\,b\,x}+4\,{\mathrm {e}}^{6\,d+6\,b\,x}+{\mathrm {e}}^{8\,d+8\,b\,x}+1\right )}+\frac {5\,\sqrt {{\mathrm {e}}^{4\,a-4\,d}}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}\,\sqrt {b^2}}{b\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )}{4\,\sqrt {b^2}} \] Input:
int(exp(2*a + 2*b*x)/cosh(d + b*x)^5,x)
Output:
(5*exp(2*a - d + b*x))/(4*b*(exp(2*d + 2*b*x) + 1)) - (10*exp(2*a + d + 3* b*x))/(3*b*(3*exp(2*d + 2*b*x) + 3*exp(4*d + 4*b*x) + exp(6*d + 6*b*x) + 1 )) - (5*exp(2*a - d + b*x))/(2*b*(2*exp(2*d + 2*b*x) + exp(4*d + 4*b*x) + 1)) - (4*exp(2*a + 3*d + 5*b*x))/(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)) + (5*exp(4*a - 4*d)^(1/2) *atan((exp(2*a)*exp(-d)*exp(b*x)*(b^2)^(1/2))/(b*(exp(4*a)*exp(-4*d))^(1/2 ))))/(4*(b^2)^(1/2))
Time = 0.27 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.31 \[ \int e^{2 (a+b x)} \text {sech}^5(d+b x) \, dx=\frac {e^{2 a} \left (15 e^{8 b x +8 d} \mathit {atan} \left (e^{b x +d}\right )+60 e^{6 b x +6 d} \mathit {atan} \left (e^{b x +d}\right )+90 e^{4 b x +4 d} \mathit {atan} \left (e^{b x +d}\right )+60 e^{2 b x +2 d} \mathit {atan} \left (e^{b x +d}\right )+15 \mathit {atan} \left (e^{b x +d}\right )+15 e^{7 b x +7 d}-73 e^{5 b x +5 d}-55 e^{3 b x +3 d}-15 e^{b x +d}\right )}{12 e^{2 d} b \left (e^{8 b x +8 d}+4 e^{6 b x +6 d}+6 e^{4 b x +4 d}+4 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(2*b*x+2*a)*sech(b*x+d)^5,x)
Output:
(e**(2*a)*(15*e**(8*b*x + 8*d)*atan(e**(b*x + d)) + 60*e**(6*b*x + 6*d)*at an(e**(b*x + d)) + 90*e**(4*b*x + 4*d)*atan(e**(b*x + d)) + 60*e**(2*b*x + 2*d)*atan(e**(b*x + d)) + 15*atan(e**(b*x + d)) + 15*e**(7*b*x + 7*d) - 7 3*e**(5*b*x + 5*d) - 55*e**(3*b*x + 3*d) - 15*e**(b*x + d)))/(12*e**(2*d)* b*(e**(8*b*x + 8*d) + 4*e**(6*b*x + 6*d) + 6*e**(4*b*x + 4*d) + 4*e**(2*b* x + 2*d) + 1))