Integrand size = 16, antiderivative size = 103 \[ \int e^{a+b x} \text {sech}^4(d+b x) \, dx=-\frac {8 e^{a+2 d+3 b x}}{3 b \left (1+e^{2 d+2 b x}\right )^3}-\frac {2 e^{a+b x}}{b \left (1+e^{2 d+2 b x}\right )^2}+\frac {e^{a+b x}}{b \left (1+e^{2 d+2 b x}\right )}+\frac {e^{a-d} \arctan \left (e^{d+b x}\right )}{b} \] Output:
-8/3*exp(3*b*x+a+2*d)/b/(1+exp(2*b*x+2*d))^3-2*exp(b*x+a)/b/(1+exp(2*b*x+2 *d))^2+exp(b*x+a)/b/(1+exp(2*b*x+2*d))+exp(a-d)*arctan(exp(b*x+d))/b
Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.69 \[ \int e^{a+b x} \text {sech}^4(d+b x) \, dx=\frac {e^{a+b x} \left (-3-8 e^{2 (d+b x)}+3 e^{4 (d+b x)}\right )}{3 b \left (1+e^{2 (d+b x)}\right )^3}+\frac {e^{a-d} \arctan \left (e^{d+b x}\right )}{b} \] Input:
Integrate[E^(a + b*x)*Sech[d + b*x]^4,x]
Output:
(E^(a + b*x)*(-3 - 8*E^(2*(d + b*x)) + 3*E^(4*(d + b*x))))/(3*b*(1 + E^(2* (d + b*x)))^3) + (E^(a - d)*ArcTan[E^(d + b*x)])/b
Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2720, 27, 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.
\(\displaystyle \int e^{a+b x} \text {sech}^4(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {16 e^{a+4 b x}}{\left (1+e^{2 b x}\right )^4}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {16 e^a \int \frac {e^{4 b x}}{\left (1+e^{2 b x}\right )^4}de^{b x}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{2} \int \frac {e^{2 b x}}{\left (1+e^{2 b x}\right )^3}de^{b x}-\frac {e^{3 b x}}{6 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{\left (1+e^{2 b x}\right )^2}de^{b x}-\frac {e^{b x}}{4 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{3 b x}}{6 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1+e^{2 b x}}de^{b x}+\frac {e^{b x}}{2 \left (e^{2 b x}+1\right )}\right )-\frac {e^{b x}}{4 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{3 b x}}{6 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \arctan \left (e^{b x}\right )+\frac {e^{b x}}{2 \left (e^{2 b x}+1\right )}\right )-\frac {e^{b x}}{4 \left (e^{2 b x}+1\right )^2}\right )-\frac {e^{3 b x}}{6 \left (e^{2 b x}+1\right )^3}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Sech[d + b*x]^4,x]
Output:
(16*E^a*(-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))/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 = 2.33 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.23
method | result | size |
risch | \(-\frac {\left (-3 \,{\mathrm e}^{4 b x +4 a +4 d}+8 \,{\mathrm e}^{2 b x +4 a +2 d}+3 \,{\mathrm e}^{4 a}\right ) {\mathrm e}^{b x +3 a}}{3 \left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{3} b}+\frac {i \ln \left ({\mathrm e}^{b x +a}+i {\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}-\frac {i \ln \left ({\mathrm e}^{b x +a}-i {\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}\) | \(127\) |
Input:
int(exp(b*x+a)*sech(b*x+d)^4,x,method=_RETURNVERBOSE)
Output:
-1/3/(exp(2*b*x+2*a+2*d)+exp(2*a))^3/b*(-3*exp(4*b*x+4*a+4*d)+8*exp(2*b*x+ 4*a+2*d)+3*exp(4*a))*exp(b*x+3*a)+1/2*I*ln(exp(b*x+a)+I*exp(a-d))/b*exp(a- d)-1/2*I*ln(exp(b*x+a)-I*exp(a-d))/b*exp(a-d)
Leaf count of result is larger than twice the leaf count of optimal. 996 vs. \(2 (93) = 186\).
Time = 0.10 (sec) , antiderivative size = 996, normalized size of antiderivative = 9.67 \[ \int e^{a+b x} \text {sech}^4(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(b*x+a)*sech(b*x+d)^4,x, algorithm="fricas")
Output:
1/3*(3*cosh(b*x + d)^5*cosh(-a + d) + 3*(cosh(-a + d) - sinh(-a + d))*sinh (b*x + d)^5 + 15*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d)) *sinh(b*x + d)^4 - 8*cosh(b*x + d)^3*cosh(-a + d) + 2*(15*cosh(b*x + d)^2* cosh(-a + d) - (15*cosh(b*x + d)^2 - 4)*sinh(-a + d) - 4*cosh(-a + d))*sin h(b*x + d)^3 + 6*(5*cosh(b*x + d)^3*cosh(-a + d) - 4*cosh(b*x + d)*cosh(-a + d) - (5*cosh(b*x + d)^3 - 4*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d)^ 2 + 3*(cosh(b*x + d)^6*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b *x + d)^6 + 6*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*si nh(b*x + d)^5 + 3*cosh(b*x + d)^4*cosh(-a + d) + 3*(5*cosh(b*x + d)^2*cosh (-a + d) - (5*cosh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d))*sinh(b*x + d)^4 + 4*(5*cosh(b*x + d)^3*cosh(-a + d) + 3*cosh(b*x + d)*cosh(-a + d) - (5*cosh(b*x + d)^3 + 3*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d)^3 + 3*c osh(b*x + d)^2*cosh(-a + d) + 3*(5*cosh(b*x + d)^4*cosh(-a + d) + 6*cosh(b *x + d)^2*cosh(-a + d) - (5*cosh(b*x + d)^4 + 6*cosh(b*x + d)^2 + 1)*sinh( -a + d) + cosh(-a + d))*sinh(b*x + d)^2 + 6*(cosh(b*x + d)^5*cosh(-a + d) + 2*cosh(b*x + d)^3*cosh(-a + d) + cosh(b*x + d)*cosh(-a + d) - (cosh(b*x + d)^5 + 2*cosh(b*x + d)^3 + cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^6 + 3*cosh(b*x + d)^4 + 3*cosh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d))*arctan(cosh(b*x + d) + sinh(b*x + d)) - 3*cosh(b*x + d)*c osh(-a + d) + 3*(5*cosh(b*x + d)^4*cosh(-a + d) - 8*cosh(b*x + d)^2*cos...
\[ \int e^{a+b x} \text {sech}^4(d+b x) \, dx=e^{a} \int e^{b x} \operatorname {sech}^{4}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*sech(b*x+d)**4,x)
Output:
exp(a)*Integral(exp(b*x)*sech(b*x + d)**4, x)
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.06 \[ \int e^{a+b x} \text {sech}^4(d+b x) \, dx=\frac {\arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (a - d\right )}}{b} + \frac {3 \, e^{\left (5 \, b x + 7 \, a + 4 \, d\right )} - 8 \, e^{\left (3 \, b x + 7 \, a + 2 \, d\right )} - 3 \, e^{\left (b x + 7 \, a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a + 6 \, d\right )} + 3 \, e^{\left (4 \, b x + 6 \, a + 4 \, d\right )} + 3 \, e^{\left (2 \, b x + 6 \, a + 2 \, d\right )} + e^{\left (6 \, a\right )}\right )}} \] Input:
integrate(exp(b*x+a)*sech(b*x+d)^4,x, algorithm="maxima")
Output:
arctan(e^(b*x + d))*e^(a - d)/b + 1/3*(3*e^(5*b*x + 7*a + 4*d) - 8*e^(3*b* x + 7*a + 2*d) - 3*e^(b*x + 7*a))/(b*(e^(6*b*x + 6*a + 6*d) + 3*e^(4*b*x + 6*a + 4*d) + 3*e^(2*b*x + 6*a + 2*d) + e^(6*a)))
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.73 \[ \int e^{a+b x} \text {sech}^4(d+b x) \, dx=\frac {1}{3} \, {\left (\frac {3 \, \arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (-5 \, d\right )}}{b} + \frac {{\left (3 \, e^{\left (5 \, b x + 4 \, d\right )} - 8 \, e^{\left (3 \, b x + 2 \, d\right )} - 3 \, e^{\left (b x\right )}\right )} e^{\left (-4 \, d\right )}}{b {\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{3}}\right )} e^{\left (a + 4 \, d\right )} \] Input:
integrate(exp(b*x+a)*sech(b*x+d)^4,x, algorithm="giac")
Output:
1/3*(3*arctan(e^(b*x + d))*e^(-5*d)/b + (3*e^(5*b*x + 4*d) - 8*e^(3*b*x + 2*d) - 3*e^(b*x))*e^(-4*d)/(b*(e^(2*b*x + 2*d) + 1)^3))*e^(a + 4*d)
Timed out. \[ \int e^{a+b x} \text {sech}^4(d+b x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x}}{{\mathrm {cosh}\left (d+b\,x\right )}^4} \,d x \] Input:
int(exp(a + b*x)/cosh(d + b*x)^4,x)
Output:
int(exp(a + b*x)/cosh(d + b*x)^4, x)
Time = 0.21 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.50 \[ \int e^{a+b x} \text {sech}^4(d+b x) \, dx=\frac {e^{a} \left (3 e^{6 b x +6 d} \mathit {atan} \left (e^{b x +d}\right )+9 e^{4 b x +4 d} \mathit {atan} \left (e^{b x +d}\right )+9 e^{2 b x +2 d} \mathit {atan} \left (e^{b x +d}\right )+3 \mathit {atan} \left (e^{b x +d}\right )+3 e^{5 b x +5 d}-8 e^{3 b x +3 d}-3 e^{b x +d}\right )}{3 e^{d} b \left (e^{6 b x +6 d}+3 e^{4 b x +4 d}+3 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(b*x+a)*sech(b*x+d)^4,x)
Output:
(e**a*(3*e**(6*b*x + 6*d)*atan(e**(b*x + d)) + 9*e**(4*b*x + 4*d)*atan(e** (b*x + d)) + 9*e**(2*b*x + 2*d)*atan(e**(b*x + d)) + 3*atan(e**(b*x + d)) + 3*e**(5*b*x + 5*d) - 8*e**(3*b*x + 3*d) - 3*e**(b*x + d)))/(3*e**d*b*(e* *(6*b*x + 6*d) + 3*e**(4*b*x + 4*d) + 3*e**(2*b*x + 2*d) + 1))