3.100 \(\int \frac {\text {sech}^4(x)}{a+b \text {csch}(x)} \, dx\)

Optimal. Leaf size=104 \[ -\frac {\text {sech}^3(x) (b-a \sinh (x))}{3 \left (a^2+b^2\right )}-\frac {\text {sech}(x) \left (3 a^2 b-a \left (2 a^2-b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2}+\frac {2 a^3 b \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3872, 2866, 12, 2660, 618, 204} \[ \frac {2 a^3 b \tanh ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {\text {sech}^3(x) (b-a \sinh (x))}{3 \left (a^2+b^2\right )}-\frac {\text {sech}(x) \left (3 a^2 b-a \left (2 a^2-b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(a + b*Csch[x]),x]

[Out]

(2*a^3*b*ArcTanh[(a - b*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (Sech[x]^3*(b - a*Sinh[x]))/(3*(a^2 +
b^2)) - (Sech[x]*(3*a^2*b - a*(2*a^2 - b^2)*Sinh[x]))/(3*(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 204

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.63, size = 114, normalized size = 1.10 \[ -\frac {\left (a b^2-2 a^3\right ) \tanh (x)+b \left (a^2+b^2\right ) \text {sech}^3(x)-a \left (a^2+b^2\right ) \tanh (x) \text {sech}^2(x)+3 a^2 b \text {sech}(x)+\frac {6 a^3 b \tan ^{-1}\left (\frac {a-b \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}}{3 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^4/(a + b*Csch[x]),x]

[Out]

-1/3*((6*a^3*b*ArcTan[(a - b*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] + 3*a^2*b*Sech[x] + b*(a^2 + b^2)*
Sech[x]^3 + (-2*a^3 + a*b^2)*Tanh[x] - a*(a^2 + b^2)*Sech[x]^2*Tanh[x])/(a^2 + b^2)^2

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fricas [B]  time = 1.56, size = 1155, normalized size = 11.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*csch(x)),x, algorithm="fricas")

[Out]

-1/3*(6*(a^4*b + a^2*b^3)*cosh(x)^5 + 6*(a^4*b + a^2*b^3)*sinh(x)^5 + 4*a^5 + 2*a^3*b^2 - 2*a*b^4 - 6*(a^3*b^2
+ a*b^4)*cosh(x)^4 - 6*(a^3*b^2 + a*b^4 - 5*(a^4*b + a^2*b^3)*cosh(x))*sinh(x)^4 + 4*(5*a^4*b + 7*a^2*b^3 + 2
*b^5)*cosh(x)^3 + 4*(5*a^4*b + 7*a^2*b^3 + 2*b^5 + 15*(a^4*b + a^2*b^3)*cosh(x)^2 - 6*(a^3*b^2 + a*b^4)*cosh(x
))*sinh(x)^3 + 12*(a^5 + a^3*b^2)*cosh(x)^2 + 12*(a^5 + a^3*b^2 + 5*(a^4*b + a^2*b^3)*cosh(x)^3 - 3*(a^3*b^2 +
a*b^4)*cosh(x)^2 + (5*a^4*b + 7*a^2*b^3 + 2*b^5)*cosh(x))*sinh(x)^2 - 3*(a^3*b*cosh(x)^6 + 6*a^3*b*cosh(x)*si
nh(x)^5 + a^3*b*sinh(x)^6 + 3*a^3*b*cosh(x)^4 + 3*a^3*b*cosh(x)^2 + 3*(5*a^3*b*cosh(x)^2 + a^3*b)*sinh(x)^4 +
a^3*b + 4*(5*a^3*b*cosh(x)^3 + 3*a^3*b*cosh(x))*sinh(x)^3 + 3*(5*a^3*b*cosh(x)^4 + 6*a^3*b*cosh(x)^2 + a^3*b)*
sinh(x)^2 + 6*(a^3*b*cosh(x)^5 + 2*a^3*b*cosh(x)^3 + a^3*b*cosh(x))*sinh(x))*sqrt(a^2 + b^2)*log((a^2*cosh(x)^
2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) + a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(a^2 + b^2)*(a*cosh(x)
+ a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sinh(x) - a)) + 6*(a^4*b + a^2
*b^3)*cosh(x) + 6*(a^4*b + a^2*b^3 + 5*(a^4*b + a^2*b^3)*cosh(x)^4 - 4*(a^3*b^2 + a*b^4)*cosh(x)^3 + 2*(5*a^4*
b + 7*a^2*b^3 + 2*b^5)*cosh(x)^2 + 4*(a^5 + a^3*b^2)*cosh(x))*sinh(x))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*co
sh(x)^6 + 6*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)*sinh(x)^5 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sinh(x
)^6 + a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^4 + 3*(a^6 + 3*a^4*b^2
+ 3*a^2*b^4 + b^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^6 + 3*a^4*b^2 + 3*a^
2*b^4 + b^6)*cosh(x)^3 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x))*sinh(x)^3 + 3*(a^6 + 3*a^4*b^2 + 3*a^2
*b^4 + b^6)*cosh(x)^2 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^4
+ 6*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)
^5 + 2*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^3 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x))*sinh(x))

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giac [A]  time = 0.18, size = 174, normalized size = 1.67 \[ -\frac {a^{3} b \log \left (\frac {{\left | 2 \, a e^{x} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{x} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, a^{2} b e^{\left (5 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} + 10 \, a^{2} b e^{\left (3 \, x\right )} + 4 \, b^{3} e^{\left (3 \, x\right )} + 6 \, a^{3} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} + 2 \, a^{3} - a b^{2}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*csch(x)),x, algorithm="giac")

[Out]

-a^3*b*log(abs(2*a*e^x + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^x + 2*b + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 +
b^4)*sqrt(a^2 + b^2)) - 2/3*(3*a^2*b*e^(5*x) - 3*a*b^2*e^(4*x) + 10*a^2*b*e^(3*x) + 4*b^3*e^(3*x) + 6*a^3*e^(2
*x) + 3*a^2*b*e^x + 2*a^3 - a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(e^(2*x) + 1)^3)

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maple [A]  time = 0.18, size = 170, normalized size = 1.63 \[ -\frac {4 a^{3} b \arctanh \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{4}+4 a^{2} b^{2}+2 b^{4}\right ) \sqrt {a^{2}+b^{2}}}-\frac {2 \left (-a^{3} \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )+\left (2 a^{2} b +b^{3}\right ) \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )+\left (-\frac {2}{3} a^{3}+\frac {4}{3} a \,b^{2}\right ) \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )+2 a^{2} b \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-a^{3} \tanh \left (\frac {x}{2}\right )+\frac {4 a^{2} b}{3}+\frac {b^{3}}{3}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^4/(a+b*csch(x)),x)

[Out]

-4*a^3*b/(2*a^4+4*a^2*b^2+2*b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tanh(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))-2/(a^4+2*a
^2*b^2+b^4)*(-a^3*tanh(1/2*x)^5+(2*a^2*b+b^3)*tanh(1/2*x)^4+(-2/3*a^3+4/3*a*b^2)*tanh(1/2*x)^3+2*a^2*b*tanh(1/
2*x)^2-a^3*tanh(1/2*x)+4/3*a^2*b+1/3*b^3)/(tanh(1/2*x)^2+1)^3

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maxima [B]  time = 0.42, size = 226, normalized size = 2.17 \[ -\frac {a^{3} b \log \left (\frac {a e^{\left (-x\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-x\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, a^{2} b e^{\left (-x\right )} - 6 \, a^{3} e^{\left (-2 \, x\right )} + 3 \, a b^{2} e^{\left (-4 \, x\right )} + 3 \, a^{2} b e^{\left (-5 \, x\right )} - 2 \, a^{3} + a b^{2} + 2 \, {\left (5 \, a^{2} b + 2 \, b^{3}\right )} e^{\left (-3 \, x\right )}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-4 \, x\right )} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-6 \, x\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*csch(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.79, size = 269, normalized size = 2.59 \[ \frac {\frac {2\,a\,b^2}{{\left (a^2+b^2\right )}^2}-\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {4\,\left (a^3+a\,b^2\right )}{{\left (a^2+b^2\right )}^2}+\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b+b^3\right )}{3\,{\left (a^2+b^2\right )}^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\frac {8\,a}{3\,\left (a^2+b^2\right )}+\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2+b^2\right )}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {a^3\,b\,\ln \left (\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^2}-\frac {2\,a^2\,b\,\left (a-b\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{5/2}}\right )}{{\left (a^2+b^2\right )}^{5/2}}+\frac {a^3\,b\,\ln \left (\frac {2\,a^2\,b\,\left (a-b\,{\mathrm {e}}^x\right )}{{\left (a^2+b^2\right )}^{5/2}}+\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^2}\right )}{{\left (a^2+b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^4*(a + b/sinh(x))),x)

[Out]

((2*a*b^2)/(a^2 + b^2)^2 - (2*a^2*b*exp(x))/(a^2 + b^2)^2)/(exp(2*x) + 1) - ((4*(a*b^2 + a^3))/(a^2 + b^2)^2 +
(8*exp(x)*(a^2*b + b^3))/(3*(a^2 + b^2)^2))/(2*exp(2*x) + exp(4*x) + 1) + ((8*a)/(3*(a^2 + b^2)) + (8*b*exp(x
))/(3*(a^2 + b^2)))/(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1) - (a^3*b*log((2*a^2*b*exp(x))/(a^2 + b^2)^2 - (2*
a^2*b*(a - b*exp(x)))/(a^2 + b^2)^(5/2)))/(a^2 + b^2)^(5/2) + (a^3*b*log((2*a^2*b*(a - b*exp(x)))/(a^2 + b^2)^
(5/2) + (2*a^2*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 \[ \int \frac {\operatorname {sech}^{4}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**4/(a+b*csch(x)),x)

[Out]

Integral(sech(x)**4/(a + b*csch(x)), x)

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