Integrand size = 16, antiderivative size = 95 \[ \int e^{a+b x} \text {sech}^4(a+b x) \, dx=-\frac {8 e^{3 a+3 b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )^2}+\frac {e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}+\frac {\arctan \left (e^{a+b x}\right )}{b} \] Output:
-8/3*exp(3*b*x+3*a)/b/(1+exp(2*b*x+2*a))^3-2*exp(b*x+a)/b/(1+exp(2*b*x+2*a ))^2+exp(b*x+a)/b/(1+exp(2*b*x+2*a))+arctan(exp(b*x+a))/b
Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int e^{a+b x} \text {sech}^4(a+b x) \, dx=\frac {e^{a+b x} \left (-3-8 e^{2 (a+b x)}+3 e^{4 (a+b x)}\right )}{3 b \left (1+e^{2 (a+b x)}\right )^3}+\frac {\arctan \left (e^{a+b x}\right )}{b} \] Input:
Integrate[E^(a + b*x)*Sech[a + b*x]^4,x]
Output:
(E^(a + b*x)*(-3 - 8*E^(2*(a + b*x)) + 3*E^(4*(a + b*x))))/(3*b*(1 + E^(2* (a + b*x)))^3) + ArcTan[E^(a + b*x)]/b
Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12, 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.
\(\displaystyle \int e^{a+b x} \text {sech}^4(a+b x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {16 e^{4 a+4 b x}}{\left (1+e^{2 a+2 b x}\right )^4}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {16 \int \frac {e^{4 a+4 b x}}{\left (1+e^{2 a+2 b x}\right )^4}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {16 \left (\frac {1}{2} \int \frac {e^{2 a+2 b x}}{\left (1+e^{2 a+2 b x}\right )^3}de^{a+b x}-\frac {e^{3 a+3 b x}}{6 \left (e^{2 a+2 b x}+1\right )^3}\right )}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {16 \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{\left (1+e^{2 a+2 b x}\right )^2}de^{a+b x}-\frac {e^{a+b x}}{4 \left (e^{2 a+2 b x}+1\right )^2}\right )-\frac {e^{3 a+3 b x}}{6 \left (e^{2 a+2 b x}+1\right )^3}\right )}{b}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {16 \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1+e^{2 a+2 b x}}de^{a+b x}+\frac {e^{a+b x}}{2 \left (e^{2 a+2 b x}+1\right )}\right )-\frac {e^{a+b x}}{4 \left (e^{2 a+2 b x}+1\right )^2}\right )-\frac {e^{3 a+3 b x}}{6 \left (e^{2 a+2 b x}+1\right )^3}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {16 \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \arctan \left (e^{a+b x}\right )+\frac {e^{a+b x}}{2 \left (e^{2 a+2 b x}+1\right )}\right )-\frac {e^{a+b x}}{4 \left (e^{2 a+2 b x}+1\right )^2}\right )-\frac {e^{3 a+3 b x}}{6 \left (e^{2 a+2 b x}+1\right )^3}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Sech[a + b*x]^4,x]
Output:
(16*(-1/6*E^(3*a + 3*b*x)/(1 + E^(2*a + 2*b*x))^3 + (-1/4*E^(a + b*x)/(1 + E^(2*a + 2*b*x))^2 + (E^(a + b*x)/(2*(1 + E^(2*a + 2*b*x))) + ArcTan[E^(a + 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]]]
Time = 4.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.39
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 \cosh \left (b x +a \right )^{3}}+\frac {\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}+\arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(37\) |
default | \(\frac {-\frac {1}{3 \cosh \left (b x +a \right )^{3}}+\frac {\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}+\arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(37\) |
risch | \(\frac {{\mathrm e}^{b x +a} \left (3 \,{\mathrm e}^{4 b x +4 a}-8 \,{\mathrm e}^{2 b x +2 a}-3\right )}{3 b \left (1+{\mathrm e}^{2 b x +2 a}\right )^{3}}+\frac {i \ln \left ({\mathrm e}^{b x +a}+i\right )}{2 b}-\frac {i \ln \left ({\mathrm e}^{b x +a}-i\right )}{2 b}\) | \(82\) |
Input:
int(exp(b*x+a)*sech(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
1/b*(-1/3/cosh(b*x+a)^3+1/2*sech(b*x+a)*tanh(b*x+a)+arctan(exp(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (86) = 172\).
Time = 0.09 (sec) , antiderivative size = 513, normalized size of antiderivative = 5.40 \[ \int e^{a+b x} \text {sech}^4(a+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(b*x+a)*sech(b*x+a)^4,x, algorithm="fricas")
Output:
1/3*(3*cosh(b*x + a)^5 + 15*cosh(b*x + a)*sinh(b*x + a)^4 + 3*sinh(b*x + a )^5 + 2*(15*cosh(b*x + a)^2 - 4)*sinh(b*x + a)^3 - 8*cosh(b*x + a)^3 + 6*( 5*cosh(b*x + a)^3 - 4*cosh(b*x + a))*sinh(b*x + a)^2 + 3*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^4 + 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 + 3*cosh( b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 + 6*cosh(b*x + a)^2 + 1)* sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 + 6*(cosh(b*x + a)^5 + 2*cosh(b*x + a) ^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a )) + 3*(5*cosh(b*x + a)^4 - 8*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - 3*cosh( b*x + a))/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh( b*x + a)^6 + 3*b*cosh(b*x + a)^4 + 3*(5*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^4 + 4*(5*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^3 + 3*b*c osh(b*x + a)^2 + 3*(5*b*cosh(b*x + a)^4 + 6*b*cosh(b*x + a)^2 + b)*sinh(b* x + a)^2 + 6*(b*cosh(b*x + a)^5 + 2*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*s inh(b*x + a) + b)
\[ \int e^{a+b x} \text {sech}^4(a+b x) \, dx=e^{a} \int e^{b x} \operatorname {sech}^{4}{\left (a + b x \right )}\, dx \] Input:
integrate(exp(b*x+a)*sech(b*x+a)**4,x)
Output:
exp(a)*Integral(exp(b*x)*sech(a + b*x)**4, x)
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.87 \[ \int e^{a+b x} \text {sech}^4(a+b x) \, dx=\frac {\arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac {3 \, e^{\left (5 \, b x + 5 \, a\right )} - 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a\right )} + 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \] Input:
integrate(exp(b*x+a)*sech(b*x+a)^4,x, algorithm="maxima")
Output:
arctan(e^(b*x + a))/b + 1/3*(3*e^(5*b*x + 5*a) - 8*e^(3*b*x + 3*a) - 3*e^( b*x + a))/(b*(e^(6*b*x + 6*a) + 3*e^(4*b*x + 4*a) + 3*e^(2*b*x + 2*a) + 1) )
Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.63 \[ \int e^{a+b x} \text {sech}^4(a+b x) \, dx=\frac {\frac {3 \, e^{\left (5 \, b x + 5 \, a\right )} - 8 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{3}} + 3 \, \arctan \left (e^{\left (b x + a\right )}\right )}{3 \, b} \] Input:
integrate(exp(b*x+a)*sech(b*x+a)^4,x, algorithm="giac")
Output:
1/3*((3*e^(5*b*x + 5*a) - 8*e^(3*b*x + 3*a) - 3*e^(b*x + a))/(e^(2*b*x + 2 *a) + 1)^3 + 3*arctan(e^(b*x + a)))/b
Time = 2.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.37 \[ \int e^{a+b x} \text {sech}^4(a+b x) \, dx=\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}-\frac {8\,{\mathrm {e}}^{3\,a+3\,b\,x}}{3\,b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1\right )}+\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \] Input:
int(exp(a + b*x)/cosh(a + b*x)^4,x)
Output:
atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b)/(b^2)^(1/2) - (2*exp(a + b*x))/(b*(2 *exp(2*a + 2*b*x) + exp(4*a + 4*b*x) + 1)) - (8*exp(3*a + 3*b*x))/(3*b*(3* exp(2*a + 2*b*x) + 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) + 1)) + exp(a + b *x)/(b*(exp(2*a + 2*b*x) + 1))
Time = 0.24 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.55 \[ \int e^{a+b x} \text {sech}^4(a+b x) \, dx=\frac {3 e^{6 b x +6 a} \mathit {atan} \left (e^{b x +a}\right )+9 e^{4 b x +4 a} \mathit {atan} \left (e^{b x +a}\right )+9 e^{2 b x +2 a} \mathit {atan} \left (e^{b x +a}\right )+3 \mathit {atan} \left (e^{b x +a}\right )+3 e^{5 b x +5 a}-8 e^{3 b x +3 a}-3 e^{b x +a}}{3 b \left (e^{6 b x +6 a}+3 e^{4 b x +4 a}+3 e^{2 b x +2 a}+1\right )} \] Input:
int(exp(b*x+a)*sech(b*x+a)^4,x)
Output:
(3*e**(6*a + 6*b*x)*atan(e**(a + b*x)) + 9*e**(4*a + 4*b*x)*atan(e**(a + b *x)) + 9*e**(2*a + 2*b*x)*atan(e**(a + b*x)) + 3*atan(e**(a + b*x)) + 3*e* *(5*a + 5*b*x) - 8*e**(3*a + 3*b*x) - 3*e**(a + b*x))/(3*b*(e**(6*a + 6*b* x) + 3*e**(4*a + 4*b*x) + 3*e**(2*a + 2*b*x) + 1))