Integrand size = 24, antiderivative size = 84 \[ \int e^{2 (a+b x)} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=-\frac {2 e^{2 a-2 d}}{b \left (1+e^{2 d+2 b x}\right )^2}+\frac {6 e^{2 a-2 d}}{b \left (1+e^{2 d+2 b x}\right )}+\frac {2 e^{2 a-2 d} \log \left (1+e^{2 d+2 b x}\right )}{b} \] Output:
-2*exp(2*a-2*d)/b/(1+exp(2*b*x+2*d))^2+6*exp(2*a-2*d)/b/(1+exp(2*b*x+2*d)) +2*exp(2*a-2*d)*ln(1+exp(2*b*x+2*d))/b
Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64 \[ \int e^{2 (a+b x)} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {2 e^{2 a-2 d} \left (\frac {2+3 e^{2 (d+b x)}}{\left (1+e^{2 (d+b x)}\right )^2}+\log \left (1+e^{2 (d+b x)}\right )\right )}{b} \] Input:
Integrate[E^(2*(a + b*x))*Sech[d + b*x]^2*Tanh[d + b*x],x]
Output:
(2*E^(2*a - 2*d)*((2 + 3*E^(2*(d + b*x)))/(1 + E^(2*(d + b*x)))^2 + Log[1 + E^(2*(d + b*x))]))/b
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.52, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2720, 27, 354, 86, 2009}
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.
\(\displaystyle \int e^{2 (a+b x)} \tanh (b x+d) \text {sech}^2(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {4 e^{2 a+3 b x} \left (1-e^{2 b x}\right )}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 e^{2 a} \int \frac {e^{3 b x} \left (1-e^{2 b x}\right )}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {2 e^{2 a} \int \frac {e^{2 b x} \left (1-e^{2 b x}\right )}{\left (1+e^{2 b x}\right )^3}de^{2 b x}}{b}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {2 e^{2 a} \int \left (\frac {3}{\left (1+e^{2 b x}\right )^2}-\frac {2}{\left (1+e^{2 b x}\right )^3}+\frac {1}{-1-e^{2 b x}}\right )de^{2 b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 e^{2 a} \left (-\frac {3}{e^{2 b x}+1}+\frac {1}{\left (e^{2 b x}+1\right )^2}-\log \left (e^{2 b x}+1\right )\right )}{b}\) |
Input:
Int[E^(2*(a + b*x))*Sech[d + b*x]^2*Tanh[d + b*x],x]
Output:
(-2*E^(2*a)*((1 + E^(2*b*x))^(-2) - 3/(1 + E^(2*b*x)) - Log[1 + E^(2*b*x)] ))/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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
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 = 5.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20
method | result | size |
risch | \(-\frac {4 \,{\mathrm e}^{2 a -2 d} a}{b}+\frac {2 \left (3 \,{\mathrm e}^{2 b x +2 a +2 d}+2 \,{\mathrm e}^{2 a}\right ) {\mathrm e}^{4 a -2 d}}{\left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}+\frac {2 \ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{2 a -2 d}}{b}\) | \(101\) |
Input:
int(exp(2*b*x+2*a)*sech(b*x+d)^2*tanh(b*x+d),x,method=_RETURNVERBOSE)
Output:
-4/b*exp(2*a-2*d)*a+2/(exp(2*b*x+2*a+2*d)+exp(2*a))^2/b*(3*exp(2*b*x+2*a+2 *d)+2*exp(2*a))*exp(4*a-2*d)+2*ln(exp(2*b*x+2*a)+exp(2*a-2*d))/b*exp(2*a-2 *d)
Leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (78) = 156\).
Time = 0.08 (sec) , antiderivative size = 520, normalized size of antiderivative = 6.19 \[ \int e^{2 (a+b x)} \text {sech}^2(d+b x) \tanh (d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)^2*tanh(b*x+d),x, algorithm="fricas")
Output:
2*(3*cosh(b*x + d)^2*cosh(-2*a + 2*d) + 3*(cosh(-2*a + 2*d) - sinh(-2*a + 2*d))*sinh(b*x + d)^2 + (cosh(b*x + d)^4*cosh(-2*a + 2*d) + (cosh(-2*a + 2 *d) - sinh(-2*a + 2*d))*sinh(b*x + d)^4 + 4*(cosh(b*x + d)*cosh(-2*a + 2*d ) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^3 + 2*cosh(b*x + d)^2*co sh(-2*a + 2*d) + 2*(3*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (3*cosh(b*x + d)^ 2 + 1)*sinh(-2*a + 2*d) + cosh(-2*a + 2*d))*sinh(b*x + d)^2 + 4*(cosh(b*x + d)^3*cosh(-2*a + 2*d) + cosh(b*x + d)*cosh(-2*a + 2*d) - (cosh(b*x + d)^ 3 + cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d) - (cosh(b*x + d)^4 + 2* cosh(b*x + d)^2 + 1)*sinh(-2*a + 2*d) + cosh(-2*a + 2*d))*log(2*cosh(b*x + d)/(cosh(b*x + d) - sinh(b*x + d))) + 6*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d) - (3*cosh(b*x + d)^2 + 2)*s inh(-2*a + 2*d) + 2*cosh(-2*a + 2*d))/(b*cosh(b*x + d)^4 + 4*b*cosh(b*x + d)*sinh(b*x + d)^3 + b*sinh(b*x + d)^4 + 2*b*cosh(b*x + d)^2 + 2*(3*b*cosh (b*x + d)^2 + b)*sinh(b*x + d)^2 + 4*(b*cosh(b*x + d)^3 + b*cosh(b*x + d)) *sinh(b*x + d) + b)
\[ \int e^{2 (a+b x)} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=e^{2 a} \int e^{2 b x} \tanh {\left (b x + d \right )} \operatorname {sech}^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)**2*tanh(b*x+d),x)
Output:
exp(2*a)*Integral(exp(2*b*x)*tanh(b*x + d)*sech(b*x + d)**2, x)
Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.18 \[ \int e^{2 (a+b x)} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=4 \, x e^{\left (2 \, a - 2 \, d\right )} + \frac {4 \, d e^{\left (2 \, a - 2 \, d\right )}}{b} + \frac {2 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-2 \, b x - 2 \, d\right )} + 1\right )}{b} - \frac {2 \, {\left (e^{\left (-2 \, b x - 2 \, d\right )} + 2\right )} e^{\left (2 \, a - 2 \, d\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, d\right )} + e^{\left (-4 \, b x - 4 \, d\right )} + 1\right )}} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)^2*tanh(b*x+d),x, algorithm="maxima")
Output:
4*x*e^(2*a - 2*d) + 4*d*e^(2*a - 2*d)/b + 2*e^(2*a - 2*d)*log(e^(-2*b*x - 2*d) + 1)/b - 2*(e^(-2*b*x - 2*d) + 2)*e^(2*a - 2*d)/(b*(2*e^(-2*b*x - 2*d ) + e^(-4*b*x - 4*d) + 1))
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.80 \[ \int e^{2 (a+b x)} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {2 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right ) - \frac {{\left (3 \, e^{\left (4 \, b x + 2 \, a + 4 \, d\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (-2 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{2}}}{b} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)^2*tanh(b*x+d),x, algorithm="giac")
Output:
(2*e^(2*a - 2*d)*log(e^(2*b*x + 2*d) + 1) - (3*e^(4*b*x + 2*a + 4*d) - e^( 2*a))*e^(-2*d)/(e^(2*b*x + 2*d) + 1)^2)/b
Time = 2.69 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int e^{2 (a+b x)} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {2\,{\mathrm {e}}^{2\,a-2\,d}\,\ln \left ({\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,b\,x}+1\right )}{b}-\frac {2\,{\mathrm {e}}^{2\,a-2\,d}}{b\,\left (2\,{\mathrm {e}}^{2\,d+2\,b\,x}+{\mathrm {e}}^{4\,d+4\,b\,x}+1\right )}+\frac {6\,{\mathrm {e}}^{2\,a-2\,d}}{b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}+1\right )} \] Input:
int((exp(2*a + 2*b*x)*tanh(d + b*x))/cosh(d + b*x)^2,x)
Output:
(2*exp(2*a - 2*d)*log(exp(2*d)*exp(2*b*x) + 1))/b - (2*exp(2*a - 2*d))/(b* (2*exp(2*d + 2*b*x) + exp(4*d + 4*b*x) + 1)) + (6*exp(2*a - 2*d))/(b*(exp( 2*d + 2*b*x) + 1))
Time = 0.26 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44 \[ \int e^{2 (a+b x)} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {e^{2 a} \left (2 e^{4 b x +4 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-3 e^{4 b x +4 d}+4 e^{2 b x +2 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )+2 \,\mathrm {log}\left (e^{2 b x +2 d}+1\right )+1\right )}{e^{2 d} b \left (e^{4 b x +4 d}+2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(2*b*x+2*a)*sech(b*x+d)^2*tanh(b*x+d),x)
Output:
(e**(2*a)*(2*e**(4*b*x + 4*d)*log(e**(2*b*x + 2*d) + 1) - 3*e**(4*b*x + 4* d) + 4*e**(2*b*x + 2*d)*log(e**(2*b*x + 2*d) + 1) + 2*log(e**(2*b*x + 2*d) + 1) + 1))/(e**(2*d)*b*(e**(4*b*x + 4*d) + 2*e**(2*b*x + 2*d) + 1))