3.60 \(\int \frac{\text{sech}^3(x)}{a+b \cosh (x)} \, dx\)
Optimal. Leaf size=87 \[ -\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^3 \sqrt{a-b} \sqrt{a+b}}+\frac{\left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^3}-\frac{b \tanh (x)}{a^2}+\frac{\tanh (x) \text{sech}(x)}{2 a} \]
[Out]
((a^2 + 2*b^2)*ArcTan[Sinh[x]])/(2*a^3) - (2*b^3*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a^3*Sqrt[a - b
]*Sqrt[a + b]) - (b*Tanh[x])/a^2 + (Sech[x]*Tanh[x])/(2*a)
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Rubi [A] time = 0.299632, antiderivative size = 87, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used =
{2802, 3055, 3001, 3770, 2659, 208} \[ -\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^3 \sqrt{a-b} \sqrt{a+b}}+\frac{\left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^3}-\frac{b \tanh (x)}{a^2}+\frac{\tanh (x) \text{sech}(x)}{2 a} \]
Antiderivative was successfully verified.
[In]
Int[Sech[x]^3/(a + b*Cosh[x]),x]
[Out]
((a^2 + 2*b^2)*ArcTan[Sinh[x]])/(2*a^3) - (2*b^3*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a^3*Sqrt[a - b
]*Sqrt[a + b]) - (b*Tanh[x])/a^2 + (Sech[x]*Tanh[x])/(2*a)
Rule 2802
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))
Rule 3001
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rule 2659
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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\text{sech}^3(x)}{a+b \cosh (x)} \, dx &=\frac{\text{sech}(x) \tanh (x)}{2 a}+\frac{\int \frac{\left (-2 b+a \cosh (x)+b \cosh ^2(x)\right ) \text{sech}^2(x)}{a+b \cosh (x)} \, dx}{2 a}\\ &=-\frac{b \tanh (x)}{a^2}+\frac{\text{sech}(x) \tanh (x)}{2 a}+\frac{\int \frac{\left (a^2+2 b^2+a b \cosh (x)\right ) \text{sech}(x)}{a+b \cosh (x)} \, dx}{2 a^2}\\ &=-\frac{b \tanh (x)}{a^2}+\frac{\text{sech}(x) \tanh (x)}{2 a}-\frac{b^3 \int \frac{1}{a+b \cosh (x)} \, dx}{a^3}+\frac{\left (a^2+2 b^2\right ) \int \text{sech}(x) \, dx}{2 a^3}\\ &=\frac{\left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^3}-\frac{b \tanh (x)}{a^2}+\frac{\text{sech}(x) \tanh (x)}{2 a}-\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a^3}\\ &=\frac{\left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^3}-\frac{2 b^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^3 \sqrt{a-b} \sqrt{a+b}}-\frac{b \tanh (x)}{a^2}+\frac{\text{sech}(x) \tanh (x)}{2 a}\\ \end{align*}
Mathematica [A] time = 0.198691, size = 82, normalized size = 0.94 \[ \frac{\frac{4 b^3 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+2 \left (a^2+2 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+a \tanh (x) (a \text{sech}(x)-2 b)}{2 a^3} \]
Antiderivative was successfully verified.
[In]
Integrate[Sech[x]^3/(a + b*Cosh[x]),x]
[Out]
(2*(a^2 + 2*b^2)*ArcTan[Tanh[x/2]] + (4*b^3*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + a
*(-2*b + a*Sech[x])*Tanh[x])/(2*a^3)
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Maple [A] time = 0.027, size = 146, normalized size = 1.7 \begin{align*} -2\,{\frac{{b}^{3}}{{a}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}b}{{a}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{1}{a}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-2\,{\frac{\tanh \left ( x/2 \right ) b}{{a}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{1}{a}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }+2\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ){b}^{2}}{{a}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sech(x)^3/(a+b*cosh(x)),x)
[Out]
-2*b^3/a^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-1/a/(tanh(1/2*x)^2+1)^2*tanh(1/2
*x)^3-2/a^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*b+1/a/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)-2/a^2/(tanh(1/2*x)^2+1)^2*
tanh(1/2*x)*b+1/a*arctan(tanh(1/2*x))+2/a^3*arctan(tanh(1/2*x))*b^2
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(x)^3/(a+b*cosh(x)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 3.53065, size = 3337, normalized size = 38.36 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(x)^3/(a+b*cosh(x)),x, algorithm="fricas")
[Out]
[(2*a^3*b - 2*a*b^3 + (a^4 - a^2*b^2)*cosh(x)^3 + (a^4 - a^2*b^2)*sinh(x)^3 + 2*(a^3*b - a*b^3)*cosh(x)^2 + (2
*a^3*b - 2*a*b^3 + 3*(a^4 - a^2*b^2)*cosh(x))*sinh(x)^2 + (b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(
x)^4 + 2*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 + b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 + b^3*cosh(x))*sinh(x))*
sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(
x) + 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) +
a)*sinh(x) + b)) + ((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^4 + 4*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)*sinh(x)^3 + (a^4 + a
^2*b^2 - 2*b^4)*sinh(x)^4 + a^4 + a^2*b^2 - 2*b^4 + 2*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2 + 2*(a^4 + a^2*b^2 - 2
*b^4 + 3*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2)*sinh(x)^2 + 4*((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^3 + (a^4 + a^2*b^2
- 2*b^4)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^4 - a^2*b^2)*cosh(x) - (a^4 - a^2*b^2 - 3*(a^4 - a^2
*b^2)*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x))*sinh(x))/(a^5 - a^3*b^2 + (a^5 - a^3*b^2)*cosh(x)^4 + 4*(a^5 - a^
3*b^2)*cosh(x)*sinh(x)^3 + (a^5 - a^3*b^2)*sinh(x)^4 + 2*(a^5 - a^3*b^2)*cosh(x)^2 + 2*(a^5 - a^3*b^2 + 3*(a^5
- a^3*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 - a^3*b^2)*cosh(x)^3 + (a^5 - a^3*b^2)*cosh(x))*sinh(x)), (2*a^3*b
- 2*a*b^3 + (a^4 - a^2*b^2)*cosh(x)^3 + (a^4 - a^2*b^2)*sinh(x)^3 + 2*(a^3*b - a*b^3)*cosh(x)^2 + (2*a^3*b - 2
*a*b^3 + 3*(a^4 - a^2*b^2)*cosh(x))*sinh(x)^2 + 2*(b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(x)^4 + 2
*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 + b^3)*sinh(x)^2 + 4*(b^3*cosh(x)^3 + b^3*cosh(x))*sinh(x))*sqrt(-a^
2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + ((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^
4 + 4*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)*sinh(x)^3 + (a^4 + a^2*b^2 - 2*b^4)*sinh(x)^4 + a^4 + a^2*b^2 - 2*b^4 +
2*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2 + 2*(a^4 + a^2*b^2 - 2*b^4 + 3*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2)*sinh(x)^
2 + 4*((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^3 + (a^4 + a^2*b^2 - 2*b^4)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x))
- (a^4 - a^2*b^2)*cosh(x) - (a^4 - a^2*b^2 - 3*(a^4 - a^2*b^2)*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x))*sinh(x)
)/(a^5 - a^3*b^2 + (a^5 - a^3*b^2)*cosh(x)^4 + 4*(a^5 - a^3*b^2)*cosh(x)*sinh(x)^3 + (a^5 - a^3*b^2)*sinh(x)^4
+ 2*(a^5 - a^3*b^2)*cosh(x)^2 + 2*(a^5 - a^3*b^2 + 3*(a^5 - a^3*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 - a^3*b^2
)*cosh(x)^3 + (a^5 - a^3*b^2)*cosh(x))*sinh(x))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{3}{\left (x \right )}}{a + b \cosh{\left (x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(x)**3/(a+b*cosh(x)),x)
[Out]
Integral(sech(x)**3/(a + b*cosh(x)), x)
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Giac [A] time = 1.19855, size = 120, normalized size = 1.38 \begin{align*} -\frac{2 \, b^{3} \arctan \left (\frac{b e^{x} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{\sqrt{-a^{2} + b^{2}} a^{3}} + \frac{{\left (a^{2} + 2 \, b^{2}\right )} \arctan \left (e^{x}\right )}{a^{3}} + \frac{a e^{\left (3 \, x\right )} + 2 \, b e^{\left (2 \, x\right )} - a e^{x} + 2 \, b}{a^{2}{\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sech(x)^3/(a+b*cosh(x)),x, algorithm="giac")
[Out]
-2*b^3*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*a^3) + (a^2 + 2*b^2)*arctan(e^x)/a^3 + (a*e^(3*x
) + 2*b*e^(2*x) - a*e^x + 2*b)/(a^2*(e^(2*x) + 1)^2)