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

Optimal. Leaf size=111 \[ -\frac {2 a^3 b \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}+\frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\text {csch}(x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cosh (x)\right )}{3 \left (a^2-b^2\right )^2} \]

[Out]

-2*a^3*b*arctan((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a+b)^(5/2)/(a-b)^(5/2)-1/3*(3*a^2*b-a*(2*a^2+b^2)*cosh(x
))*csch(x)/(a^2-b^2)^2+1/3*(b-a*cosh(x))*csch(x)^3/(a^2-b^2)

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3872, 2866, 12, 2659, 205} \[ \frac {\text {csch}^3(x) (b-a \cosh (x))}{3 \left (a^2-b^2\right )}-\frac {\text {csch}(x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cosh (x)\right )}{3 \left (a^2-b^2\right )^2}-\frac {2 a^3 b \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(a + b*Sech[x]),x]

[Out]

(-2*a^3*b*ArcTan[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) - ((3*a^2*b - a*(2*a^2 +
b^2)*Cosh[x])*Csch[x])/(3*(a^2 - b^2)^2) + ((b - a*Cosh[x])*Csch[x]^3)/(3*(a^2 - b^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 205

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.60, size = 156, normalized size = 1.41 \[ \frac {\text {sech}(x) (a \cosh (x)+b) \left (\frac {48 a^3 b \tan ^{-1}\left (\frac {(b-a) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {2 b \tanh \left (\frac {x}{2}\right )}{(a-b)^2}+\frac {8 a \tanh \left (\frac {x}{2}\right )}{(a-b)^2}+\frac {2 (4 a+b) \coth \left (\frac {x}{2}\right )}{(a+b)^2}-\frac {\sinh (x) \text {csch}^4\left (\frac {x}{2}\right )}{2 (a+b)}+\frac {8 \sinh ^4\left (\frac {x}{2}\right ) \text {csch}^3(x)}{a-b}\right )}{24 (a+b \text {sech}(x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cosh[x])*Sech[x]*((48*a^3*b*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + (2*(4*a
+ b)*Coth[x/2])/(a + b)^2 + (8*Csch[x]^3*Sinh[x/2]^4)/(a - b) - (Csch[x/2]^4*Sinh[x])/(2*(a + b)) + (8*a*Tanh[
x/2])/(a - b)^2 - (2*b*Tanh[x/2])/(a - b)^2))/(24*(a + b*Sech[x]))

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 2340, normalized size = 21.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*sech(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)*s
inh(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)
*cosh(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)*sin
h(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)
), -2/3*(3*(a^4*b - a^2*b^3)*cosh(x)^5 + 3*(a^4*b - a^2*b^3)*sinh(x)^5 - 2*a^5 + a^3*b^2 + a*b^4 - 3*(a^3*b^2
- a*b^4)*cosh(x)^4 - 3*(a^3*b^2 - a*b^4 - 5*(a^4*b - a^2*b^3)*cosh(x))*sinh(x)^4 - 2*(5*a^4*b - 7*a^2*b^3 + 2*
b^5)*cosh(x)^3 - 2*(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 + 6*(a^5 - a^3*b^2)*cosh(x)^2 + 6*(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)*sinh(
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)*sin
h(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)*arctan(-(a*cosh(x) +
 a*sinh(x) + b)/sqrt(a^2 - b^2)) + 3*(a^4*b - a^2*b^3)*cosh(x) + 3*(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)*cosh(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)*co
sh(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))]

________________________________________________________________________________________

giac [A]  time = 0.14, size = 149, normalized size = 1.34 \[ -\frac {2 \, a^{3} b \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (3 \, a^{2} b e^{\left (5 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} - 10 \, a^{2} b e^{\left (3 \, x\right )} + 4 \, b^{3} e^{\left (3 \, x\right )} + 6 \, a^{3} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{x} - 2 \, a^{3} - a b^{2}\right )}}{3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3*b*arctan((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^(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)

________________________________________________________________________________________

maple [A]  time = 0.16, size = 154, normalized size = 1.39 \[ -\frac {a \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{24 \left (a -b \right )^{2}}+\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b}{24 \left (a -b \right )^{2}}+\frac {3 a \tanh \left (\frac {x}{2}\right )}{8 \left (a -b \right )^{2}}-\frac {\tanh \left (\frac {x}{2}\right ) b}{8 \left (a -b \right )^{2}}-\frac {2 a^{3} b \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{24 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{3}}+\frac {3 a}{8 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )}+\frac {b}{8 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/(a-b)^2*a*tanh(1/2*x)^3+1/24/(a-b)^2*tanh(1/2*x)^3*b+3/8/(a-b)^2*a*tanh(1/2*x)-1/8/(a-b)^2*tanh(1/2*x)*b
-2/(a-b)^2/(a+b)^2*a^3*b/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-1/24/(a+b)/tanh(1/2
*x)^3+3/8/(a+b)^2/tanh(1/2*x)*a+1/8/(a+b)^2/tanh(1/2*x)*b

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(a+b*sech(x)),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 details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 1.75, size = 295, normalized size = 2.66 \[ \frac {\frac {4\,\left (a\,b^2-a^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b-b^3\right )}{3\,{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {\frac {8\,a}{3\,\left (a^2-b^2\right )}-\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2-b^2\right )}}{3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1}+\frac {\frac {2\,a\,b^2}{{\left (a^2-b^2\right )}^2}-\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{2\,x}-1}+\frac {a^3\,b\,\ln \left (\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2-b^2\right )}^2}-\frac {2\,a^2\,b\,\left (a+b\,{\mathrm {e}}^x\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}}-\frac {a^3\,b\,\ln \left (\frac {2\,a^2\,b\,{\mathrm {e}}^x}{{\left (a^2-b^2\right )}^2}+\frac {2\,a^2\,b\,\left (a+b\,{\mathrm {e}}^x\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (b-a\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^4*(a + b/cosh(x))),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{4}{\relax (x )}}{a + b \operatorname {sech}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________