3.205 \(\int \frac {\text {sech}^2(x)}{(a+b \sinh (x))^2} \, dx\)

Optimal. Leaf size=93 \[ -\frac {6 a b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b \text {sech}(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}(x) \left (\left (a^2-2 b^2\right ) \sinh (x)+3 a b\right )}{\left (a^2+b^2\right )^2} \]

[Out]

-6*a*b^2*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2)-b*sech(x)/(a^2+b^2)/(a+b*sinh(x))+sech(x)*
(3*a*b+(a^2-2*b^2)*sinh(x))/(a^2+b^2)^2

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Rubi [A]  time = 0.17, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = {2694, 2866, 12, 2660, 618, 206} \[ -\frac {6 a b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b \text {sech}(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}(x) \left (\left (a^2-2 b^2\right ) \sinh (x)+3 a b\right )}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(a + b*Sinh[x])^2,x]

[Out]

(-6*a*b^2*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (b*Sech[x])/((a^2 + b^2)*(a + b*Sinh
[x])) + (Sech[x]*(3*a*b + (a^2 - 2*b^2)*Sinh[x]))/(a^2 + b^2)^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 2694

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m +
1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + p + 2)*Sin[e + f*x]), x], x] /; F
reeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
 + (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
 + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\text {sech}^2(x)}{(a+b \sinh (x))^2} \, dx &=-\frac {b \text {sech}(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-\frac {\int \frac {\text {sech}^2(x) (-a+2 b \sinh (x))}{a+b \sinh (x)} \, dx}{a^2+b^2}\\ &=-\frac {b \text {sech}(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}(x) \left (3 a b+\left (a^2-2 b^2\right ) \sinh (x)\right )}{\left (a^2+b^2\right )^2}+\frac {\int \frac {3 a b^2}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac {b \text {sech}(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}(x) \left (3 a b+\left (a^2-2 b^2\right ) \sinh (x)\right )}{\left (a^2+b^2\right )^2}+\frac {\left (3 a b^2\right ) \int \frac {1}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac {b \text {sech}(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}(x) \left (3 a b+\left (a^2-2 b^2\right ) \sinh (x)\right )}{\left (a^2+b^2\right )^2}+\frac {\left (6 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{\left (a^2+b^2\right )^2}\\ &=-\frac {b \text {sech}(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}(x) \left (3 a b+\left (a^2-2 b^2\right ) \sinh (x)\right )}{\left (a^2+b^2\right )^2}-\frac {\left (12 a b^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{\left (a^2+b^2\right )^2}\\ &=-\frac {6 a b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b \text {sech}(x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\text {sech}(x) \left (3 a b+\left (a^2-2 b^2\right ) \sinh (x)\right )}{\left (a^2+b^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 94, normalized size = 1.01 \[ \frac {\frac {6 a b^2 \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+a^2 \tanh (x)-\frac {b^3 \cosh (x)}{a+b \sinh (x)}+2 a b \text {sech}(x)-b^2 \tanh (x)}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(a + b*Sinh[x])^2,x]

[Out]

((6*a*b^2*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] + 2*a*b*Sech[x] - (b^3*Cosh[x])/(a + b*
Sinh[x]) + a^2*Tanh[x] - b^2*Tanh[x])/(a^2 + b^2)^2

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fricas [B]  time = 1.94, size = 802, normalized size = 8.62 \[ -\frac {2 \, a^{4} b - 2 \, a^{2} b^{3} - 4 \, b^{5} + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x)^{3} + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} \sinh \relax (x)^{3} + 6 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cosh \relax (x)^{2} + 6 \, {\left (a^{4} b + a^{2} b^{3} + 3 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{2} + 3 \, {\left (a b^{3} \cosh \relax (x)^{4} + a b^{3} \sinh \relax (x)^{4} + 2 \, a^{2} b^{2} \cosh \relax (x)^{3} + 2 \, a^{2} b^{2} \cosh \relax (x) - a b^{3} + 2 \, {\left (2 \, a b^{3} \cosh \relax (x) + a^{2} b^{2}\right )} \sinh \relax (x)^{3} + 6 \, {\left (a b^{3} \cosh \relax (x)^{2} + a^{2} b^{2} \cosh \relax (x)\right )} \sinh \relax (x)^{2} + 2 \, {\left (2 \, a b^{3} \cosh \relax (x)^{3} + 3 \, a^{2} b^{2} \cosh \relax (x)^{2} + a^{2} b^{2}\right )} \sinh \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) - b}\right ) - 2 \, {\left (2 \, a^{5} + a^{3} b^{2} - a b^{4}\right )} \cosh \relax (x) - 2 \, {\left (2 \, a^{5} + a^{3} b^{2} - a b^{4} - 9 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cosh \relax (x)^{2} - 6 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7} - {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cosh \relax (x)^{4} - {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \sinh \relax (x)^{4} - 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cosh \relax (x)^{3} - 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6} + 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{3} - 6 \, {\left ({\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cosh \relax (x)^{2} + {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{2} - 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cosh \relax (x) - 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6} + 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cosh \relax (x)^{3} + 3 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^4*b - 2*a^2*b^3 - 4*b^5 + 6*(a^3*b^2 + a*b^4)*cosh(x)^3 + 6*(a^3*b^2 + a*b^4)*sinh(x)^3 + 6*(a^4*b + a^2
*b^3)*cosh(x)^2 + 6*(a^4*b + a^2*b^3 + 3*(a^3*b^2 + a*b^4)*cosh(x))*sinh(x)^2 + 3*(a*b^3*cosh(x)^4 + a*b^3*sin
h(x)^4 + 2*a^2*b^2*cosh(x)^3 + 2*a^2*b^2*cosh(x) - a*b^3 + 2*(2*a*b^3*cosh(x) + a^2*b^2)*sinh(x)^3 + 6*(a*b^3*
cosh(x)^2 + a^2*b^2*cosh(x))*sinh(x)^2 + 2*(2*a*b^3*cosh(x)^3 + 3*a^2*b^2*cosh(x)^2 + a^2*b^2)*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)*sin
h(x) - b)) - 2*(2*a^5 + a^3*b^2 - a*b^4)*cosh(x) - 2*(2*a^5 + a^3*b^2 - a*b^4 - 9*(a^3*b^2 + a*b^4)*cosh(x)^2
- 6*(a^4*b + a^2*b^3)*cosh(x))*sinh(x))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5
+ b^7)*cosh(x)^4 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*sinh(x)^4 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*c
osh(x)^3 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x))*sinh(x)^3
 - 6*((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^2 + (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x))*sinh(x)
^2 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x) - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 2*(a^6*b + 3*a
^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^3 + 3*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^2)*sinh(x))

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giac [A]  time = 0.17, size = 167, normalized size = 1.80 \[ \frac {3 \, a b^{2} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, a b^{2} e^{\left (3 \, x\right )} + 3 \, a^{2} b e^{\left (2 \, x\right )} - 2 \, a^{3} e^{x} + a b^{2} e^{x} + a^{2} b - 2 \, b^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b e^{\left (4 \, x\right )} + 2 \, a e^{\left (3 \, x\right )} + 2 \, a e^{x} - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a*b^2*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 +
 b^4)*sqrt(a^2 + b^2)) + 2*(3*a*b^2*e^(3*x) + 3*a^2*b*e^(2*x) - 2*a^3*e^x + a*b^2*e^x + a^2*b - 2*b^3)/((a^4 +
 2*a^2*b^2 + b^4)*(b*e^(4*x) + 2*a*e^(3*x) + 2*a*e^x - b))

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maple [A]  time = 0.07, size = 138, normalized size = 1.48 \[ -\frac {2 b^{2} \left (\frac {-\frac {b^{2} \tanh \left (\frac {x}{2}\right )}{a}-b}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a}-\frac {3 a \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{2}}-\frac {2 \left (\left (-a^{2}+b^{2}\right ) \tanh \left (\frac {x}{2}\right )-2 a b \right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(a+b*sinh(x))^2,x)

[Out]

-2*b^2/(a^2+b^2)^2*((-b^2/a*tanh(1/2*x)-b)/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)-3*a/(a^2+b^2)^(1/2)*arctanh(1/2
*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2)))-2/(a^4+2*a^2*b^2+b^4)*((-a^2+b^2)*tanh(1/2*x)-2*a*b)/(tanh(1/2*x)^2+1
)

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maxima [B]  time = 0.42, size = 215, normalized size = 2.31 \[ \frac {3 \, a b^{2} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, a^{2} b e^{\left (-2 \, x\right )} - 3 \, a b^{2} e^{\left (-3 \, x\right )} + a^{2} b - 2 \, b^{3} + {\left (2 \, a^{3} - a b^{2}\right )} e^{\left (-x\right )}\right )}}{a^{4} b + 2 \, a^{2} b^{3} + b^{5} + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} e^{\left (-x\right )} + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} e^{\left (-3 \, x\right )} - {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-4 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a*b^2*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a
^2 + b^2)) + 2*(3*a^2*b*e^(-2*x) - 3*a*b^2*e^(-3*x) + a^2*b - 2*b^3 + (2*a^3 - a*b^2)*e^(-x))/(a^4*b + 2*a^2*b
^3 + b^5 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*e^(-x) + 2*(a^5 + 2*a^3*b^2 + a*b^4)*e^(-3*x) - (a^4*b + 2*a^2*b^3 + b^
5)*e^(-4*x))

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mupad [B]  time = 0.97, size = 302, normalized size = 3.25 \[ \frac {\frac {6\,a^4\,b^4\,{\mathrm {e}}^{2\,x}}{\left (a^3+a\,b^2\right )\,\left (a^3\,b^3+a\,b^5\right )}-\frac {2\,\left (2\,a^2\,b^6-a^4\,b^4\right )}{\left (a^3+a\,b^2\right )\,\left (a^3\,b^3+a\,b^5\right )}+\frac {6\,a^3\,b^5\,{\mathrm {e}}^{3\,x}}{\left (a^3+a\,b^2\right )\,\left (a^3\,b^3+a\,b^5\right )}+\frac {2\,a\,{\mathrm {e}}^x\,\left (a^2\,b^6-2\,a^4\,b^4\right )}{b\,\left (a^3+a\,b^2\right )\,\left (a^3\,b^3+a\,b^5\right )}}{2\,a\,{\mathrm {e}}^x-b+2\,a\,{\mathrm {e}}^{3\,x}+b\,{\mathrm {e}}^{4\,x}}-\frac {3\,a\,b^2\,\ln \left (-\frac {6\,a\,b\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^2}-\frac {6\,a\,b\,\left (b-a\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{5/2}}\right )}{{\left (a^2+b^2\right )}^{5/2}}+\frac {3\,a\,b^2\,\ln \left (\frac {6\,a\,b\,\left (b-a\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{5/2}}-\frac {6\,a\,b\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^2}\right )}{{\left (a^2+b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(a + b*sinh(x))^2),x)

[Out]

((6*a^4*b^4*exp(2*x))/((a*b^2 + a^3)*(a*b^5 + a^3*b^3)) - (2*(2*a^2*b^6 - a^4*b^4))/((a*b^2 + a^3)*(a*b^5 + a^
3*b^3)) + (6*a^3*b^5*exp(3*x))/((a*b^2 + a^3)*(a*b^5 + a^3*b^3)) + (2*a*exp(x)*(a^2*b^6 - 2*a^4*b^4))/(b*(a*b^
2 + a^3)*(a*b^5 + a^3*b^3)))/(2*a*exp(x) - b + 2*a*exp(3*x) + b*exp(4*x)) - (3*a*b^2*log(- (6*a*b*exp(x))/(a^2
 + b^2)^2 - (6*a*b*(b - a*exp(x)))/(a^2 + b^2)^(5/2)))/(a^2 + b^2)^(5/2) + (3*a*b^2*log((6*a*b*(b - a*exp(x)))
/(a^2 + b^2)^(5/2) - (6*a*b*exp(x))/(a^2 + b^2)^2))/(a^2 + b^2)^(5/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{2}{\relax (x )}}{\left (a + b \sinh {\relax (x )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(a+b*sinh(x))**2,x)

[Out]

Integral(sech(x)**2/(a + b*sinh(x))**2, x)

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