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

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

[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)

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Rubi [A]  time = 0.270334, antiderivative size = 104, normalized size of antiderivative = 1., 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 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]

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 12

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

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 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 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])

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*}

Mathematica [A]  time = 0.61086, size = 114, normalized size = 1.1 \[ -\frac{\left (a b^2-2 a^3\right ) \tanh (x)+b \left (a^2+b^2\right ) \text{sech}^3(x)+\frac{6 a^3 b \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-a \left (a^2+b^2\right ) \tanh (x) \text{sech}^2(x)+3 a^2 b \text{sech}(x)}{3 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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])/(3*(a^2 + b^2)^2)

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Maple [A]  time = 0.04, size = 170, normalized size = 1.6 \begin{align*} -4\,{\frac{{a}^{3}b}{ \left ( 2\,{a}^{4}+4\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b-2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{-{a}^{3} \left ( \tanh \left ( x/2 \right ) \right ) ^{5}+ \left ( 2\,{a}^{2}b+{b}^{3} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{4}+ \left ( -2/3\,{a}^{3}+4/3\,a{b}^{2} \right ) \left ( \tanh \left ( x/2 \right ) \right ) ^{3}+2\,{a}^{2}b \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-{a}^{3}\tanh \left ( x/2 \right ) +4/3\,{a}^{2}b+1/3\,{b}^{3}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}} \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.80851, size = 2822, normalized size = 27.13 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{4}{\left (x \right )}}{a + b \operatorname{csch}{\left (x \right )}}\, dx \end{align*}

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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Giac [A]  time = 1.21967, size = 235, normalized size = 2.26 \begin{align*} -\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}} \end{align*}

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)