Integrand size = 24, antiderivative size = 102 \[ \int e^{2 (a+b x)} \text {sech}(d+b x) \tanh ^2(d+b x) \, dx=\frac {2 e^{2 a-d+b x}}{b}-\frac {2 e^{2 a-d+b x}}{b \left (1+e^{2 d+2 b x}\right )^2}+\frac {5 e^{2 a-d+b x}}{b \left (1+e^{2 d+2 b x}\right )}-\frac {5 e^{2 a-2 d} \arctan \left (e^{d+b x}\right )}{b} \] Output:
2*exp(b*x+2*a-d)/b-2*exp(b*x+2*a-d)/b/(1+exp(2*b*x+2*d))^2+5*exp(b*x+2*a-d )/b/(1+exp(2*b*x+2*d))-5*exp(2*a-2*d)*arctan(exp(b*x+d))/b
Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.68 \[ \int e^{2 (a+b x)} \text {sech}(d+b x) \tanh ^2(d+b x) \, dx=\frac {e^{2 a-2 d} \left (\frac {e^{d+b x} \left (5+9 e^{2 (d+b x)}+2 e^{4 (d+b x)}\right )}{\left (1+e^{2 (d+b x)}\right )^2}-5 \arctan \left (e^{d+b x}\right )\right )}{b} \] Input:
Integrate[E^(2*(a + b*x))*Sech[d + b*x]*Tanh[d + b*x]^2,x]
Output:
(E^(2*a - 2*d)*((E^(d + b*x)*(5 + 9*E^(2*(d + b*x)) + 2*E^(4*(d + b*x))))/ (1 + E^(2*(d + b*x)))^2 - 5*ArcTan[E^(d + b*x)]))/b
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2720, 27, 366, 27, 360, 25, 299, 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.
\(\displaystyle \int e^{2 (a+b x)} \tanh ^2(b x+d) \text {sech}(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {2 e^{2 a+2 b x} \left (1-e^{2 b x}\right )^2}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 e^{2 a} \int \frac {e^{2 b x} \left (1-e^{2 b x}\right )^2}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 366 |
\(\displaystyle \frac {2 e^{2 a} \left (\frac {e^{3 b x}}{\left (e^{2 b x}+1\right )^2}-\frac {1}{4} \int \frac {4 e^{2 b x} \left (2-e^{2 b x}\right )}{\left (1+e^{2 b x}\right )^2}de^{b x}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 e^{2 a} \left (\frac {e^{3 b x}}{\left (e^{2 b x}+1\right )^2}-\int \frac {e^{2 b x} \left (2-e^{2 b x}\right )}{\left (1+e^{2 b x}\right )^2}de^{b x}\right )}{b}\) |
\(\Big \downarrow \) 360 |
\(\displaystyle \frac {2 e^{2 a} \left (\frac {1}{2} \int -\frac {3-2 e^{2 b x}}{1+e^{2 b x}}de^{b x}+\frac {3 e^{b x}}{2 \left (e^{2 b x}+1\right )}+\frac {e^{3 b x}}{\left (e^{2 b x}+1\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 e^{2 a} \left (-\frac {1}{2} \int \frac {3-2 e^{2 b x}}{1+e^{2 b x}}de^{b x}+\frac {3 e^{b x}}{2 \left (e^{2 b x}+1\right )}+\frac {e^{3 b x}}{\left (e^{2 b x}+1\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {2 e^{2 a} \left (\frac {1}{2} \left (2 e^{b x}-5 \int \frac {1}{1+e^{2 b x}}de^{b x}\right )+\frac {3 e^{b x}}{2 \left (e^{2 b x}+1\right )}+\frac {e^{3 b x}}{\left (e^{2 b x}+1\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2 e^{2 a} \left (\frac {1}{2} \left (2 e^{b x}-5 \arctan \left (e^{b x}\right )\right )+\frac {3 e^{b x}}{2 \left (e^{2 b x}+1\right )}+\frac {e^{3 b x}}{\left (e^{2 b x}+1\right )^2}\right )}{b}\) |
Input:
Int[E^(2*(a + b*x))*Sech[d + b*x]*Tanh[d + b*x]^2,x]
Output:
(2*E^(2*a)*(E^(3*b*x)/(1 + E^(2*b*x))^2 + (3*E^(b*x))/(2*(1 + E^(2*b*x))) + (2*E^(b*x) - 5*ArcTan[E^(b*x)])/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)^(-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)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
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 = 1.88 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{b x +2 a -d}}{b}+\frac {\left (5 \,{\mathrm e}^{2 b x +2 a +2 d}+3 \,{\mathrm e}^{2 a}\right ) {\mathrm e}^{b x +4 a -d}}{\left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}+\frac {5 i \ln \left ({\mathrm e}^{b x +a}-i {\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{2 b}-\frac {5 i \ln \left ({\mathrm e}^{b x +a}+i {\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{2 b}\) | \(135\) |
Input:
int(exp(2*b*x+2*a)*sech(b*x+d)*tanh(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
2*exp(b*x+2*a-d)/b+1/(exp(2*b*x+2*a+2*d)+exp(2*a))^2/b*(5*exp(2*b*x+2*a+2* d)+3*exp(2*a))*exp(b*x+4*a-d)+5/2*I*ln(exp(b*x+a)-I*exp(a-d))/b*exp(2*a-2* d)-5/2*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. 770 vs. \(2 (95) = 190\).
Time = 0.10 (sec) , antiderivative size = 770, normalized size of antiderivative = 7.55 \[ \int e^{2 (a+b x)} \text {sech}(d+b x) \tanh ^2(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)*tanh(b*x+d)^2,x, algorithm="fricas")
Output:
(2*cosh(b*x + d)^5*cosh(-2*a + 2*d) + 2*(cosh(-2*a + 2*d) - sinh(-2*a + 2* d))*sinh(b*x + d)^5 + 10*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*s inh(-2*a + 2*d))*sinh(b*x + d)^4 + 9*cosh(b*x + d)^3*cosh(-2*a + 2*d) + (2 0*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (20*cosh(b*x + d)^2 + 9)*sinh(-2*a + 2*d) + 9*cosh(-2*a + 2*d))*sinh(b*x + d)^3 + (20*cosh(b*x + d)^3*cosh(-2*a + 2*d) + 27*cosh(b*x + d)*cosh(-2*a + 2*d) - (20*cosh(b*x + d)^3 + 27*cos h(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^2 - 5*(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*(cos h(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*cosh(-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))*arctan(cosh(b*x + d) + sinh(b*x + d)) + 5*cosh(b*x + d)*cosh(-2* a + 2*d) + (10*cosh(b*x + d)^4*cosh(-2*a + 2*d) + 27*cosh(b*x + d)^2*cosh( -2*a + 2*d) - (10*cosh(b*x + d)^4 + 27*cosh(b*x + d)^2 + 5)*sinh(-2*a + 2* d) + 5*cosh(-2*a + 2*d))*sinh(b*x + d) - (2*cosh(b*x + d)^5 + 9*cosh(b*x + d)^3 + 5*cosh(b*x + d))*sinh(-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*(...
\[ \int e^{2 (a+b x)} \text {sech}(d+b x) \tanh ^2(d+b x) \, dx=e^{2 a} \int e^{2 b x} \tanh ^{2}{\left (b x + d \right )} \operatorname {sech}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)*tanh(b*x+d)**2,x)
Output:
exp(2*a)*Integral(exp(2*b*x)*tanh(b*x + d)**2*sech(b*x + d), x)
Time = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.90 \[ \int e^{2 (a+b x)} \text {sech}(d+b x) \tanh ^2(d+b x) \, dx=\frac {5 \, \arctan \left (e^{\left (-b x - d\right )}\right ) e^{\left (2 \, a - 2 \, d\right )}}{b} + \frac {{\left (9 \, e^{\left (-2 \, b x - 2 \, d\right )} + 5 \, e^{\left (-4 \, b x - 4 \, d\right )} + 2\right )} e^{\left (2 \, a - 2 \, d\right )}}{b {\left (e^{\left (-b x - d\right )} + 2 \, e^{\left (-3 \, b x - 3 \, d\right )} + e^{\left (-5 \, b x - 5 \, d\right )}\right )}} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)*tanh(b*x+d)^2,x, algorithm="maxima")
Output:
5*arctan(e^(-b*x - d))*e^(2*a - 2*d)/b + (9*e^(-2*b*x - 2*d) + 5*e^(-4*b*x - 4*d) + 2)*e^(2*a - 2*d)/(b*(e^(-b*x - d) + 2*e^(-3*b*x - 3*d) + e^(-5*b *x - 5*d)))
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int e^{2 (a+b x)} \text {sech}(d+b x) \tanh ^2(d+b x) \, dx=-\frac {5 \, \arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (2 \, a - 2 \, d\right )} - \frac {{\left (5 \, e^{\left (3 \, b x + 2 \, a + 3 \, d\right )} + 3 \, e^{\left (b x + 2 \, a + d\right )}\right )} e^{\left (-2 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{2}} - 2 \, e^{\left (b x + 2 \, a - d\right )}}{b} \] Input:
integrate(exp(2*b*x+2*a)*sech(b*x+d)*tanh(b*x+d)^2,x, algorithm="giac")
Output:
-(5*arctan(e^(b*x + d))*e^(2*a - 2*d) - (5*e^(3*b*x + 2*a + 3*d) + 3*e^(b* x + 2*a + d))*e^(-2*d)/(e^(2*b*x + 2*d) + 1)^2 - 2*e^(b*x + 2*a - d))/b
Time = 2.68 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.33 \[ \int e^{2 (a+b x)} \text {sech}(d+b x) \tanh ^2(d+b x) \, dx=\frac {2\,{\mathrm {e}}^{2\,a-d+b\,x}}{b}+\frac {5\,{\mathrm {e}}^{2\,a-d+b\,x}}{b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}+1\right )}-\frac {2\,{\mathrm {e}}^{2\,a-d+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,d+2\,b\,x}+{\mathrm {e}}^{4\,d+4\,b\,x}+1\right )}-\frac {5\,\sqrt {{\mathrm {e}}^{4\,a-4\,d}}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}\,\sqrt {b^2}}{b\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )}{\sqrt {b^2}} \] Input:
int((exp(2*a + 2*b*x)*tanh(d + b*x)^2)/cosh(d + b*x),x)
Output:
(2*exp(2*a - d + b*x))/b + (5*exp(2*a - d + b*x))/(b*(exp(2*d + 2*b*x) + 1 )) - (2*exp(2*a - d + b*x))/(b*(2*exp(2*d + 2*b*x) + exp(4*d + 4*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))))/(b^2)^(1/2)
Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.24 \[ \int e^{2 (a+b x)} \text {sech}(d+b x) \tanh ^2(d+b x) \, dx=\frac {e^{2 a} \left (-5 e^{4 b x +4 d} \mathit {atan} \left (e^{b x +d}\right )-10 e^{2 b x +2 d} \mathit {atan} \left (e^{b x +d}\right )-5 \mathit {atan} \left (e^{b x +d}\right )+2 e^{5 b x +5 d}+9 e^{3 b x +3 d}+5 e^{b x +d}\right )}{e^{2 d} b \left (e^{4 b x +4 d}+2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(2*b*x+2*a)*sech(b*x+d)*tanh(b*x+d)^2,x)
Output:
(e**(2*a)*( - 5*e**(4*b*x + 4*d)*atan(e**(b*x + d)) - 10*e**(2*b*x + 2*d)* atan(e**(b*x + d)) - 5*atan(e**(b*x + d)) + 2*e**(5*b*x + 5*d) + 9*e**(3*b *x + 3*d) + 5*e**(b*x + d)))/(e**(2*d)*b*(e**(4*b*x + 4*d) + 2*e**(2*b*x + 2*d) + 1))