Integrand size = 24, antiderivative size = 69 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^2(d+b x) \, dx=\frac {4 e^{a+b x}}{b \left (1-e^{4 d+4 b x}\right )}-\frac {2 e^{a-d} \arctan \left (e^{d+b x}\right )}{b}-\frac {2 e^{a-d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
4*exp(b*x+a)/b/(1-exp(4*b*x+4*d))-2*exp(a-d)*arctan(exp(b*x+d))/b-2*exp(a- d)*arctanh(exp(b*x+d))/b
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^2(d+b x) \, dx=\frac {2 e^{a-d} \left (-\frac {2 e^{d+b x}}{-1+e^{4 (d+b x)}}-\arctan \left (e^{d+b x}\right )-\text {arctanh}\left (e^{d+b x}\right )\right )}{b} \] Input:
Integrate[E^(a + b*x)*Csch[d + b*x]^2*Sech[d + b*x]^2,x]
Output:
(2*E^(a - d)*((-2*E^(d + b*x))/(-1 + E^(4*(d + b*x))) - ArcTan[E^(d + b*x) ] - ArcTanh[E^(d + b*x)]))/b
Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2720, 27, 817, 756, 216, 219}
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.
\(\displaystyle \int e^{a+b x} \text {csch}^2(b x+d) \text {sech}^2(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {16 e^{a+4 b x}}{\left (1-e^{4 b x}\right )^2}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {16 e^a \int \frac {e^{4 b x}}{\left (1-e^{4 b x}\right )^2}de^{b x}}{b}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {16 e^a \left (\frac {e^{b x}}{4 \left (1-e^{4 b x}\right )}-\frac {1}{4} \int \frac {1}{1-e^{4 b x}}de^{b x}\right )}{b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{4} \left (-\frac {1}{2} \int \frac {1}{1-e^{2 b x}}de^{b x}-\frac {1}{2} \int \frac {1}{1+e^{2 b x}}de^{b x}\right )+\frac {e^{b x}}{4 \left (1-e^{4 b x}\right )}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{4} \left (-\frac {1}{2} \int \frac {1}{1-e^{2 b x}}de^{b x}-\frac {1}{2} \arctan \left (e^{b x}\right )\right )+\frac {e^{b x}}{4 \left (1-e^{4 b x}\right )}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{4} \left (-\frac {1}{2} \arctan \left (e^{b x}\right )-\frac {1}{2} \text {arctanh}\left (e^{b x}\right )\right )+\frac {e^{b x}}{4 \left (1-e^{4 b x}\right )}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Csch[d + b*x]^2*Sech[d + b*x]^2,x]
Output:
(16*E^a*(E^(b*x)/(4*(1 - E^(4*b*x))) + (-1/2*ArcTan[E^(b*x)] - ArcTanh[E^( b*x)]/2)/4))/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)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, 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 = 64.27 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.38
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{b x +5 a}}{\left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) \left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) b}+\frac {\ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{b}-\frac {\ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{b}+\frac {i \ln \left ({\mathrm e}^{b x +a}-i {\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{b}-\frac {i \ln \left ({\mathrm e}^{b x +a}+i {\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{b}\) | \(164\) |
Input:
int(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
4/(exp(2*b*x+2*a+2*d)+exp(2*a))/(-exp(2*b*x+2*a+2*d)+exp(2*a))/b*exp(b*x+5 *a)+ln(exp(b*x+a)-exp(a-d))/b*exp(a-d)-ln(exp(b*x+a)+exp(a-d))/b*exp(a-d)+ I*ln(exp(b*x+a)-I*exp(a-d))/b*exp(a-d)-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. 740 vs. \(2 (61) = 122\).
Time = 0.10 (sec) , antiderivative size = 740, normalized size of antiderivative = 10.72 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^2(d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^2,x, algorithm="fricas")
Output:
-(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 + 6*(cosh(b*x + d)^2*cosh(-a + d) - cosh(b*x + d)^2*sinh(-a + d))*sinh(b*x + d)^2 + 4*(cosh(b*x + d)^3*cosh(-a + d) - cosh(b*x + d)^3*si nh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^4 - 1)*sinh(-a + d) - cosh(-a + d))*arctan(cosh(b*x + d) + sinh(b*x + d)) + 4*cosh(b*x + d)*cosh(-a + d) + (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 + 6*(cosh(b*x + d)^2*cosh(-a + d) - cosh(b*x + d)^2*sinh(-a + d) )*sinh(b*x + d)^2 + 4*(cosh(b*x + d)^3*cosh(-a + d) - cosh(b*x + d)^3*sinh (-a + d))*sinh(b*x + d) - (cosh(b*x + d)^4 - 1)*sinh(-a + d) - cosh(-a + d ))*log(cosh(b*x + d) + sinh(b*x + d) + 1) - (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 + 6*(cosh(b*x + d)^2*c osh(-a + d) - cosh(b*x + d)^2*sinh(-a + d))*sinh(b*x + d)^2 + 4*(cosh(b*x + d)^3*cosh(-a + d) - cosh(b*x + d)^3*sinh(-a + d))*sinh(b*x + d) - (cosh( b*x + d)^4 - 1)*sinh(-a + d) - cosh(-a + d))*log(cosh(b*x + d) + sinh(b*x + d) - 1) + 4*(cosh(-a + d) - sinh(-a + d))*sinh(b*x + d) - 4*cosh(b*x + d )*sinh(-a + d))/(b*cosh(b*x + d)^4 + 4*b*cosh(b*x + d)^3*sinh(b*x + d) + 6 *b*cosh(b*x + d)^2*sinh(b*x + d)^2 + 4*b*cosh(b*x + d)*sinh(b*x + d)^3 ...
\[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^2(d+b x) \, dx=e^{a} \int e^{b x} \operatorname {csch}^{2}{\left (b x + d \right )} \operatorname {sech}^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*csch(b*x+d)**2*sech(b*x+d)**2,x)
Output:
exp(a)*Integral(exp(b*x)*csch(b*x + d)**2*sech(b*x + d)**2, x)
Time = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.42 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^2(d+b x) \, dx=-\frac {2 \, \arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (a - d\right )}}{b} - \frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right )}{b} + \frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} - e^{a}\right )}{b} - \frac {4 \, e^{\left (b x + 5 \, a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a + 4 \, d\right )} - e^{\left (4 \, a\right )}\right )}} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^2,x, algorithm="maxima")
Output:
-2*arctan(e^(b*x + d))*e^(a - d)/b - e^(a - d)*log(e^(b*x + a + d) + e^a)/ b + e^(a - d)*log(e^(b*x + a + d) - e^a)/b - 4*e^(b*x + 5*a)/(b*(e^(4*b*x + 4*a + 4*d) - e^(4*a)))
Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^2(d+b x) \, dx=-{\left (\frac {2 \, \arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (-5 \, d\right )}}{b} + \frac {e^{\left (-5 \, d\right )} \log \left (e^{\left (b x + d\right )} + 1\right )}{b} - \frac {e^{\left (-5 \, d\right )} \log \left ({\left | e^{\left (b x + d\right )} - 1 \right |}\right )}{b} + \frac {4 \, e^{\left (b x - 4 \, d\right )}}{b {\left (e^{\left (4 \, b x + 4 \, d\right )} - 1\right )}}\right )} e^{\left (a + 4 \, d\right )} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^2,x, algorithm="giac")
Output:
-(2*arctan(e^(b*x + d))*e^(-5*d)/b + e^(-5*d)*log(e^(b*x + d) + 1)/b - e^( -5*d)*log(abs(e^(b*x + d) - 1))/b + 4*e^(b*x - 4*d)/(b*(e^(4*b*x + 4*d) - 1)))*e^(a + 4*d)
Time = 3.91 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.81 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^2(d+b x) \, dx=\frac {4\,{\mathrm {e}}^{a+b\,x}}{b-b\,{\mathrm {e}}^{4\,d+4\,b\,x}}-\frac {{\left ({\mathrm {e}}^{4\,a-4\,d}\right )}^{1/4}\,\ln \left (-4\,{\left ({\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}\right )}^{5/4}-4\,{\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-4\,d}\,{\mathrm {e}}^{b\,x}\right )}{b}+\frac {{\left ({\mathrm {e}}^{4\,a-4\,d}\right )}^{1/4}\,\ln \left (4\,{\left ({\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}\right )}^{5/4}-4\,{\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-4\,d}\,{\mathrm {e}}^{b\,x}\right )}{b}-\frac {{\left ({\mathrm {e}}^{4\,a-4\,d}\right )}^{1/4}\,\left (\ln \left (-4\,{\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-4\,d}\,{\mathrm {e}}^{b\,x}-{\left ({\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}\right )}^{5/4}\,4{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left (-4\,{\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-4\,d}\,{\mathrm {e}}^{b\,x}+{\left ({\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}\right )}^{5/4}\,4{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{b} \] Input:
int(exp(a + b*x)/(cosh(d + b*x)^2*sinh(d + b*x)^2),x)
Output:
(4*exp(a + b*x))/(b - b*exp(4*d + 4*b*x)) - (exp(4*a - 4*d)^(1/4)*log(- 4* (exp(4*a)*exp(-4*d))^(5/4) - 4*exp(5*a)*exp(-4*d)*exp(b*x)))/b + (exp(4*a - 4*d)^(1/4)*log(4*(exp(4*a)*exp(-4*d))^(5/4) - 4*exp(5*a)*exp(-4*d)*exp(b *x)))/b - (exp(4*a - 4*d)^(1/4)*(log(- (exp(4*a)*exp(-4*d))^(5/4)*4i - 4*e xp(5*a)*exp(-4*d)*exp(b*x))*1i - log((exp(4*a)*exp(-4*d))^(5/4)*4i - 4*exp (5*a)*exp(-4*d)*exp(b*x))*1i))/b
Time = 0.26 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.90 \[ \int e^{a+b x} \text {csch}^2(d+b x) \text {sech}^2(d+b x) \, dx=\frac {e^{a} \left (-2 e^{4 b x +4 d} \mathit {atan} \left (e^{b x +d}\right )+2 \mathit {atan} \left (e^{b x +d}\right )+e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )-e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )-4 e^{b x +d}-\mathrm {log}\left (e^{b x +d}-1\right )+\mathrm {log}\left (e^{b x +d}+1\right )\right )}{e^{d} b \left (e^{4 b x +4 d}-1\right )} \] Input:
int(exp(b*x+a)*csch(b*x+d)^2*sech(b*x+d)^2,x)
Output:
(e**a*( - 2*e**(4*b*x + 4*d)*atan(e**(b*x + d)) + 2*atan(e**(b*x + d)) + e **(4*b*x + 4*d)*log(e**(b*x + d) - 1) - e**(4*b*x + 4*d)*log(e**(b*x + d) + 1) - 4*e**(b*x + d) - log(e**(b*x + d) - 1) + log(e**(b*x + d) + 1)))/(e **d*b*(e**(4*b*x + 4*d) - 1))