\(\int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx\) [92]

Optimal result

Integrand size = 12, antiderivative size = 109 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx=\frac {x}{a^2}-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))} \]

[Out]

x/a^2-2*b*(2*a^2-b^2)*arctan((a-b)^(1/2)*tanh(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^2/(a-b)^(3/2)/(a+b)^(3/2)/d+b^2*ta
nh(d*x+c)/a/(a^2-b^2)/d/(a+b*sech(d*x+c))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3870, 4004, 3916, 2738, 214} \[ \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx=-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}+\frac {x}{a^2} \]

[In]

Int[(a + b*Sech[c + d*x])^(-2),x]

[Out]

x/a^2 - (2*b*(2*a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tanh[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(3/2)*(a + b)^(3/
2)*d) + (b^2*Tanh[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sech[c + d*x]))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3870

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[b^2*Cot[c + d*x]*((a + b*Csc[c + d*x])^(n +
1)/(a*d*(n + 1)*(a^2 - b^2))), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}-\frac {\int \frac {-a^2+b^2+a b \text {sech}(c+d x)}{a+b \text {sech}(c+d x)} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {x}{a^2}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}-\frac {\left (b \left (2 a^2-b^2\right )\right ) \int \frac {\text {sech}(c+d x)}{a+b \text {sech}(c+d x)} \, dx}{a^2 \left (a^2-b^2\right )} \\ & = \frac {x}{a^2}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{1+\frac {a \cosh (c+d x)}{b}} \, dx}{a^2 \left (a^2-b^2\right )} \\ & = \frac {x}{a^2}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))}+\frac {\left (2 i \left (2 a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a^2 \left (a^2-b^2\right ) d} \\ & = \frac {x}{a^2}-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.86 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx=\frac {a \left (\left (a^2-b^2\right )^{3/2} (c+d x)+\left (4 a^2 b-2 b^3\right ) \arctan \left (\frac {(-a+b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )\right ) \cosh (c+d x)+b \left (\left (a^2-b^2\right )^{3/2} (c+d x)+\left (4 a^2 b-2 b^3\right ) \arctan \left (\frac {(-a+b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+a b \sqrt {a^2-b^2} \sinh (c+d x)\right )}{a^2 (a-b) (a+b) \sqrt {a^2-b^2} d (b+a \cosh (c+d x))} \]

[In]

Integrate[(a + b*Sech[c + d*x])^(-2),x]

[Out]

(a*((a^2 - b^2)^(3/2)*(c + d*x) + (4*a^2*b - 2*b^3)*ArcTan[((-a + b)*Tanh[(c + d*x)/2])/Sqrt[a^2 - b^2]])*Cosh
[c + d*x] + b*((a^2 - b^2)^(3/2)*(c + d*x) + (4*a^2*b - 2*b^3)*ArcTan[((-a + b)*Tanh[(c + d*x)/2])/Sqrt[a^2 -
b^2]] + a*b*Sqrt[a^2 - b^2]*Sinh[c + d*x]))/(a^2*(a - b)*(a + b)*Sqrt[a^2 - b^2]*d*(b + a*Cosh[c + d*x]))

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.50

method	result	size
derivativedivides	\(\frac {-\frac {2 b \left (-\frac {a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}+\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}}{d}\)	\(164\)
default	\(\frac {-\frac {2 b \left (-\frac {a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}+\frac {\ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}}{d}\)	\(164\)
risch	\(\frac {x}{a^{2}}-\frac {2 b^{2} \left ({\mathrm e}^{d x +c} b +a \right )}{d \,a^{2} \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 d x +2 c} a +2 \,{\mathrm e}^{d x +c} b +a \right )}-\frac {2 b \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {2 b \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}\)	\(368\)

[In]

int(1/(a+b*sech(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/a^2*b*(-a*b/(a^2-b^2)*tanh(1/2*d*x+1/2*c)/(tanh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b+a+b)+(2*a^2
-b^2)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))+1/a^2*ln(1+tanh(1
/2*d*x+1/2*c))-1/a^2*ln(tanh(1/2*d*x+1/2*c)-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 544 vs. \(2 (100) = 200\).

Time = 0.29 (sec) , antiderivative size = 1207, normalized size of antiderivative = 11.07 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a+b*sech(d*x+c))^2,x, algorithm="fricas")

[Out]

[-(2*a^3*b^2 - 2*a*b^4 - (a^5 - 2*a^3*b^2 + a*b^4)*d*x*cosh(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*x*sinh(d*
x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*x + (2*a^3*b - a*b^3 + (2*a^3*b - a*b^3)*cosh(d*x + c)^2 + (2*a^3*b - a
*b^3)*sinh(d*x + c)^2 + 2*(2*a^2*b^2 - b^4)*cosh(d*x + c) + 2*(2*a^2*b^2 - b^4 + (2*a^3*b - a*b^3)*cosh(d*x +
c))*sinh(d*x + c))*sqrt(-a^2 + b^2)*log((a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) - a^2
+ 2*b^2 + 2*(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(-a^2 + b^2)*(a*cosh(d*x + c) + a*sinh(d*x + c) +
b))/(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + 2*b*cosh(d*x + c) + 2*(a*cosh(d*x + c) + b)*sinh(d*x + c) + a))
+ 2*(a^2*b^3 - b^5 - (a^4*b - 2*a^2*b^3 + b^5)*d*x)*cosh(d*x + c) + 2*(a^2*b^3 - b^5 - (a^5 - 2*a^3*b^2 + a*b^
4)*d*x*cosh(d*x + c) - (a^4*b - 2*a^2*b^3 + b^5)*d*x)*sinh(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x +
c)^2 + (a^7 - 2*a^5*b^2 + a^3*b^4)*d*sinh(d*x + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cosh(d*x + c) + (a^7
- 2*a^5*b^2 + a^3*b^4)*d + 2*((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)*
sinh(d*x + c)), -(2*a^3*b^2 - 2*a*b^4 - (a^5 - 2*a^3*b^2 + a*b^4)*d*x*cosh(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b
^4)*d*x*sinh(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*x - 2*(2*a^3*b - a*b^3 + (2*a^3*b - a*b^3)*cosh(d*x + c)
^2 + (2*a^3*b - a*b^3)*sinh(d*x + c)^2 + 2*(2*a^2*b^2 - b^4)*cosh(d*x + c) + 2*(2*a^2*b^2 - b^4 + (2*a^3*b - a
*b^3)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cosh(d*x + c) + a*sinh(d*x + c) + b)/sqrt(a^2 -
b^2)) + 2*(a^2*b^3 - b^5 - (a^4*b - 2*a^2*b^3 + b^5)*d*x)*cosh(d*x + c) + 2*(a^2*b^3 - b^5 - (a^5 - 2*a^3*b^2
+ a*b^4)*d*x*cosh(d*x + c) - (a^4*b - 2*a^2*b^3 + b^5)*d*x)*sinh(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos
h(d*x + c)^2 + (a^7 - 2*a^5*b^2 + a^3*b^4)*d*sinh(d*x + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cosh(d*x + c)
+ (a^7 - 2*a^5*b^2 + a^3*b^4)*d + 2*((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b
^5)*d)*sinh(d*x + c))]

Sympy [F]

\[ \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx=\int \frac {1}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

integrate(1/(a+b*sech(d*x+c))**2,x)

[Out]

Integral((a + b*sech(c + d*x))**(-2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(a+b*sech(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.23 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (2 \, a^{2} b - b^{3}\right )} \arctan \left (\frac {a e^{\left (d x + c\right )} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (b^{3} e^{\left (d x + c\right )} + a b^{2}\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (d x + c\right )} + a\right )}} - \frac {d x + c}{a^{2}}}{d} \]

[In]

integrate(1/(a+b*sech(d*x+c))^2,x, algorithm="giac")

[Out]

-(2*(2*a^2*b - b^3)*arctan((a*e^(d*x + c) + b)/sqrt(a^2 - b^2))/((a^4 - a^2*b^2)*sqrt(a^2 - b^2)) + 2*(b^3*e^(
d*x + c) + a*b^2)/((a^4 - a^2*b^2)*(a*e^(2*d*x + 2*c) + 2*b*e^(d*x + c) + a)) - (d*x + c)/a^2)/d

Mupad [B] (verification not implemented)

Time = 2.55 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.72 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^2} \, dx=\frac {\frac {2\,b^2}{d\,\left (a\,b^2-a^3\right )}+\frac {2\,b^3\,{\mathrm {e}}^{c+d\,x}}{a\,d\,\left (a\,b^2-a^3\right )}}{a+2\,b\,{\mathrm {e}}^{c+d\,x}+a\,{\mathrm {e}}^{2\,c+2\,d\,x}}+\frac {x}{a^2}+\frac {b\,\ln \left (\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (2\,a^2\,b-b^3\right )}{a^3\,\left (a^2-b^2\right )}-\frac {2\,b\,\left (2\,a^2-b^2\right )\,\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}{a^3\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}-\frac {b\,\ln \left (\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (2\,a^2\,b-b^3\right )}{a^3\,\left (a^2-b^2\right )}+\frac {2\,b\,\left (2\,a^2-b^2\right )\,\left (a+b\,{\mathrm {e}}^{c+d\,x}\right )}{a^3\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}} \]

[In]

int(1/(a + b/cosh(c + d*x))^2,x)

[Out]

((2*b^2)/(d*(a*b^2 - a^3)) + (2*b^3*exp(c + d*x))/(a*d*(a*b^2 - a^3)))/(a + 2*b*exp(c + d*x) + a*exp(2*c + 2*d
*x)) + x/a^2 + (b*log((2*exp(c + d*x)*(2*a^2*b - b^3))/(a^3*(a^2 - b^2)) - (2*b*(2*a^2 - b^2)*(a + b*exp(c + d
*x)))/(a^3*(a + b)^(3/2)*(b - a)^(3/2)))*(2*a^2 - b^2))/(a^2*d*(a + b)^(3/2)*(b - a)^(3/2)) - (b*log((2*exp(c
+ d*x)*(2*a^2*b - b^3))/(a^3*(a^2 - b^2)) + (2*b*(2*a^2 - b^2)*(a + b*exp(c + d*x)))/(a^3*(a + b)^(3/2)*(b - a
)^(3/2)))*(2*a^2 - b^2))/(a^2*d*(a + b)^(3/2)*(b - a)^(3/2))