Integrand size = 22, antiderivative size = 72 \[ \int e^{a+b x} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {2 e^{a+b x}}{b \left (1+e^{2 d+2 b x}\right )^2}-\frac {3 e^{a+b x}}{b \left (1+e^{2 d+2 b x}\right )}+\frac {e^{a-d} \arctan \left (e^{d+b x}\right )}{b} \] Output:
2*exp(b*x+a)/b/(1+exp(2*b*x+2*d))^2-3*exp(b*x+a)/b/(1+exp(2*b*x+2*d))+exp( a-d)*arctan(exp(b*x+d))/b
Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.76 \[ \int e^{a+b x} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {e^a \left (-\frac {e^{b x} \left (1+3 e^{2 (d+b x)}\right )}{\left (1+e^{2 (d+b x)}\right )^2}+e^{-d} \arctan \left (e^{d+b x}\right )\right )}{b} \] Input:
Integrate[E^(a + b*x)*Sech[d + b*x]^2*Tanh[d + b*x],x]
Output:
(E^a*(-((E^(b*x)*(1 + 3*E^(2*(d + b*x))))/(1 + E^(2*(d + b*x)))^2) + ArcTa n[E^(d + b*x)]/E^d))/b
Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2720, 27, 360, 27, 298, 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.
\(\displaystyle \int e^{a+b x} \tanh (b x+d) \text {sech}^2(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {4 e^{a+2 b x} \left (1-e^{2 b x}\right )}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 e^a \int \frac {e^{2 b x} \left (1-e^{2 b x}\right )}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 360 |
\(\displaystyle -\frac {4 e^a \left (-\frac {1}{4} \int -\frac {2 \left (1-2 e^{2 b x}\right )}{\left (1+e^{2 b x}\right )^2}de^{b x}-\frac {e^{b x}}{2 \left (e^{2 b x}+1\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 e^a \left (\frac {1}{2} \int \frac {1-2 e^{2 b x}}{\left (1+e^{2 b x}\right )^2}de^{b x}-\frac {e^{b x}}{2 \left (e^{2 b x}+1\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle -\frac {4 e^a \left (\frac {1}{2} \left (\frac {3 e^{b x}}{2 \left (e^{2 b x}+1\right )}-\frac {1}{2} \int \frac {1}{1+e^{2 b x}}de^{b x}\right )-\frac {e^{b x}}{2 \left (e^{2 b x}+1\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {4 e^a \left (\frac {1}{2} \left (\frac {3 e^{b x}}{2 \left (e^{2 b x}+1\right )}-\frac {1}{2} \arctan \left (e^{b x}\right )\right )-\frac {e^{b x}}{2 \left (e^{2 b x}+1\right )^2}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Sech[d + b*x]^2*Tanh[d + b*x],x]
Output:
(-4*E^a*(-1/2*E^(b*x)/(1 + E^(2*b*x))^2 + ((3*E^(b*x))/(2*(1 + E^(2*b*x))) - ArcTan[E^(b*x)]/2)/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[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 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[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 = 5.01 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.54
method | result | size |
risch | \(-\frac {\left (3 \,{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) {\mathrm e}^{b x +3 a}}{\left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}+\frac {i \ln \left ({\mathrm e}^{b x +a}+i {\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}-\frac {i \ln \left ({\mathrm e}^{b x +a}-i {\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}\) | \(111\) |
Input:
int(exp(b*x+a)*sech(b*x+d)^2*tanh(b*x+d),x,method=_RETURNVERBOSE)
Output:
-1/(exp(2*b*x+2*a+2*d)+exp(2*a))^2/b*(3*exp(2*b*x+2*a+2*d)+exp(2*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*e xp(a-d))/b*exp(a-d)
Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (66) = 132\).
Time = 0.10 (sec) , antiderivative size = 529, normalized size of antiderivative = 7.35 \[ \int e^{a+b x} \text {sech}^2(d+b x) \tanh (d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(b*x+a)*sech(b*x+d)^2*tanh(b*x+d),x, algorithm="fricas")
Output:
-(3*cosh(b*x + d)^3*cosh(-a + d) + 3*(cosh(-a + d) - sinh(-a + d))*sinh(b* x + d)^3 + 9*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sin h(b*x + d)^2 - (cosh(b*x + d)^4*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d ))*sinh(b*x + d)^4 + 4*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^3 + 2*cosh(b*x + d)^2*cosh(-a + d) + 2*(3*cosh(b*x + d)^2*cosh(-a + d) - (3*cosh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d))*s inh(b*x + d)^2 + 4*(cosh(b*x + d)^3*cosh(-a + d) + cosh(b*x + d)*cosh(-a + d) - (cosh(b*x + d)^3 + cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d) - (cos h(b*x + d)^4 + 2*cosh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d))*arctan( cosh(b*x + d) + sinh(b*x + d)) + cosh(b*x + d)*cosh(-a + d) + (9*cosh(b*x + d)^2*cosh(-a + d) - (9*cosh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d)) *sinh(b*x + d) - (3*cosh(b*x + d)^3 + cosh(b*x + d))*sinh(-a + 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)
\[ \int e^{a+b x} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=e^{a} \int e^{b x} \tanh {\left (b x + d \right )} \operatorname {sech}^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*sech(b*x+d)**2*tanh(b*x+d),x)
Output:
exp(a)*Integral(exp(b*x)*tanh(b*x + d)*sech(b*x + d)**2, x)
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.10 \[ \int e^{a+b x} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {\arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (a - d\right )}}{b} - \frac {3 \, e^{\left (3 \, b x + 5 \, a + 2 \, d\right )} + e^{\left (b x + 5 \, a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a + 4 \, d\right )} + 2 \, e^{\left (2 \, b x + 4 \, a + 2 \, d\right )} + e^{\left (4 \, a\right )}\right )}} \] Input:
integrate(exp(b*x+a)*sech(b*x+d)^2*tanh(b*x+d),x, algorithm="maxima")
Output:
arctan(e^(b*x + d))*e^(a - d)/b - (3*e^(3*b*x + 5*a + 2*d) + e^(b*x + 5*a) )/(b*(e^(4*b*x + 4*a + 4*d) + 2*e^(2*b*x + 4*a + 2*d) + e^(4*a)))
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int e^{a+b x} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {\arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (a - d\right )} - \frac {{\left (3 \, e^{\left (3 \, b x + a + 3 \, d\right )} + e^{\left (b x + a + d\right )}\right )} e^{\left (-d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{2}}}{b} \] Input:
integrate(exp(b*x+a)*sech(b*x+d)^2*tanh(b*x+d),x, algorithm="giac")
Output:
(arctan(e^(b*x + d))*e^(a - d) - (3*e^(3*b*x + a + 3*d) + e^(b*x + a + d)) *e^(-d)/(e^(2*b*x + 2*d) + 1)^2)/b
Timed out. \[ \int e^{a+b x} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x}\,\mathrm {tanh}\left (d+b\,x\right )}{{\mathrm {cosh}\left (d+b\,x\right )}^2} \,d x \] Input:
int((exp(a + b*x)*tanh(d + b*x))/cosh(d + b*x)^2,x)
Output:
int((exp(a + b*x)*tanh(d + b*x))/cosh(d + b*x)^2, x)
Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.49 \[ \int e^{a+b x} \text {sech}^2(d+b x) \tanh (d+b x) \, dx=\frac {e^{a} \left (e^{4 b x +4 d} \mathit {atan} \left (e^{b x +d}\right )+2 e^{2 b x +2 d} \mathit {atan} \left (e^{b x +d}\right )+\mathit {atan} \left (e^{b x +d}\right )-3 e^{3 b x +3 d}-e^{b x +d}\right )}{e^{d} b \left (e^{4 b x +4 d}+2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(b*x+a)*sech(b*x+d)^2*tanh(b*x+d),x)
Output:
(e**a*(e**(4*b*x + 4*d)*atan(e**(b*x + d)) + 2*e**(2*b*x + 2*d)*atan(e**(b *x + d)) + atan(e**(b*x + d)) - 3*e**(3*b*x + 3*d) - e**(b*x + d)))/(e**d* b*(e**(4*b*x + 4*d) + 2*e**(2*b*x + 2*d) + 1))