3.99 \(\int \frac {\text {sech}^3(x)}{a+b \text {csch}(x)} \, dx\)

Optimal. Leaf size=95 \[ -\frac {a^2 b \log (a \sinh (x)+b)}{\left (a^2+b^2\right )^2}-\frac {\text {sech}^2(x) (b-a \sinh (x))}{2 \left (a^2+b^2\right )}-\frac {i a \log (-\sinh (x)+i)}{4 (a-i b)^2}+\frac {i a \log (\sinh (x)+i)}{4 (a+i b)^2} \]

[Out]

-1/4*I*a*ln(I-sinh(x))/(a-I*b)^2+1/4*I*a*ln(I+sinh(x))/(a+I*b)^2-a^2*b*ln(b+a*sinh(x))/(a^2+b^2)^2-1/2*sech(x)
^2*(b-a*sinh(x))/(a^2+b^2)

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 95, 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, 2837, 12, 823, 801} \[ -\frac {a^2 b \log (a \sinh (x)+b)}{\left (a^2+b^2\right )^2}-\frac {\text {sech}^2(x) (b-a \sinh (x))}{2 \left (a^2+b^2\right )}-\frac {i a \log (-\sinh (x)+i)}{4 (a-i b)^2}+\frac {i a \log (\sinh (x)+i)}{4 (a+i b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(a + b*Csch[x]),x]

[Out]

((-I/4)*a*Log[I - Sinh[x]])/(a - I*b)^2 + ((I/4)*a*Log[I + Sinh[x]])/(a + I*b)^2 - (a^2*b*Log[b + a*Sinh[x]])/
(a^2 + b^2)^2 - (Sech[x]^2*(b - a*Sinh[x]))/(2*(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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 2837

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 78, normalized size = 0.82 \[ \frac {-b \left (a^2+b^2\right ) \text {sech}^2(x)+a \left (a^2+b^2\right ) \tanh (x) \text {sech}(x)+2 a \left (\left (a^2-b^2\right ) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+a b (\log (\cosh (x))-\log (a \sinh (x)+b))\right )}{2 \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(a + b*Csch[x]),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.64, size = 675, normalized size = 7.11 \[ \frac {{\left (a^{3} + a b^{2}\right )} \cosh \relax (x)^{3} + {\left (a^{3} + a b^{2}\right )} \sinh \relax (x)^{3} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \relax (x)^{2} - {\left (2 \, a^{2} b + 2 \, b^{3} - 3 \, {\left (a^{3} + a b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{2} + {\left ({\left (a^{3} - a b^{2}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{3} - a b^{2}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{3} - a b^{2}\right )} \sinh \relax (x)^{4} + a^{3} - a b^{2} + 2 \, {\left (a^{3} - a b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{3} - a b^{2} + 3 \, {\left (a^{3} - a b^{2}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{2} + 4 \, {\left ({\left (a^{3} - a b^{2}\right )} \cosh \relax (x)^{3} + {\left (a^{3} - a b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) - {\left (a^{3} + a b^{2}\right )} \cosh \relax (x) - {\left (a^{2} b \cosh \relax (x)^{4} + 4 \, a^{2} b \cosh \relax (x) \sinh \relax (x)^{3} + a^{2} b \sinh \relax (x)^{4} + 2 \, a^{2} b \cosh \relax (x)^{2} + a^{2} b + 2 \, {\left (3 \, a^{2} b \cosh \relax (x)^{2} + a^{2} b\right )} \sinh \relax (x)^{2} + 4 \, {\left (a^{2} b \cosh \relax (x)^{3} + a^{2} b \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, {\left (a \sinh \relax (x) + b\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) + {\left (a^{2} b \cosh \relax (x)^{4} + 4 \, a^{2} b \cosh \relax (x) \sinh \relax (x)^{3} + a^{2} b \sinh \relax (x)^{4} + 2 \, a^{2} b \cosh \relax (x)^{2} + a^{2} b + 2 \, {\left (3 \, a^{2} b \cosh \relax (x)^{2} + a^{2} b\right )} \sinh \relax (x)^{2} + 4 \, {\left (a^{2} b \cosh \relax (x)^{3} + a^{2} b \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) - {\left (a^{3} + a b^{2} - 3 \, {\left (a^{3} + a b^{2}\right )} \cosh \relax (x)^{2} + 4 \, {\left (a^{2} b + b^{3}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{4} + a^{4} + 2 \, a^{2} b^{2} + b^{4} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x)^{2} + 4 \, {\left ({\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{3} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*csch(x)),x, algorithm="fricas")

[Out]

((a^3 + a*b^2)*cosh(x)^3 + (a^3 + a*b^2)*sinh(x)^3 - 2*(a^2*b + b^3)*cosh(x)^2 - (2*a^2*b + 2*b^3 - 3*(a^3 + a
*b^2)*cosh(x))*sinh(x)^2 + ((a^3 - a*b^2)*cosh(x)^4 + 4*(a^3 - a*b^2)*cosh(x)*sinh(x)^3 + (a^3 - a*b^2)*sinh(x
)^4 + a^3 - a*b^2 + 2*(a^3 - a*b^2)*cosh(x)^2 + 2*(a^3 - a*b^2 + 3*(a^3 - a*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^
3 - a*b^2)*cosh(x)^3 + (a^3 - a*b^2)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^3 + a*b^2)*cosh(x) - (a^
2*b*cosh(x)^4 + 4*a^2*b*cosh(x)*sinh(x)^3 + a^2*b*sinh(x)^4 + 2*a^2*b*cosh(x)^2 + a^2*b + 2*(3*a^2*b*cosh(x)^2
+ a^2*b)*sinh(x)^2 + 4*(a^2*b*cosh(x)^3 + a^2*b*cosh(x))*sinh(x))*log(2*(a*sinh(x) + b)/(cosh(x) - sinh(x)))
+ (a^2*b*cosh(x)^4 + 4*a^2*b*cosh(x)*sinh(x)^3 + a^2*b*sinh(x)^4 + 2*a^2*b*cosh(x)^2 + a^2*b + 2*(3*a^2*b*cosh
(x)^2 + a^2*b)*sinh(x)^2 + 4*(a^2*b*cosh(x)^3 + a^2*b*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) - (
a^3 + a*b^2 - 3*(a^3 + a*b^2)*cosh(x)^2 + 4*(a^2*b + b^3)*cosh(x))*sinh(x))/((a^4 + 2*a^2*b^2 + b^4)*cosh(x)^4
+ 4*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^3 + (a^4 + 2*a^2*b^2 + b^4)*sinh(x)^4 + a^4 + 2*a^2*b^2 + b^4 + 2
*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4 + 2*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^2
+ 4*((a^4 + 2*a^2*b^2 + b^4)*cosh(x)^3 + (a^4 + 2*a^2*b^2 + b^4)*cosh(x))*sinh(x))

________________________________________________________________________________________

giac [B]  time = 0.14, size = 218, normalized size = 2.29 \[ -\frac {a^{3} b \log \left ({\left | -a {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, b \right |}\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4}} + \frac {a^{2} b \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} {\left (a^{3} - a b^{2}\right )}}{4 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac {a^{2} b {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 2 \, a^{3} {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )} + 8 \, a^{2} b + 4 \, b^{3}}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*csch(x)),x, algorithm="giac")

[Out]

-a^3*b*log(abs(-a*(e^(-x) - e^x) + 2*b))/(a^5 + 2*a^3*b^2 + a*b^4) + 1/2*a^2*b*log((e^(-x) - e^x)^2 + 4)/(a^4
+ 2*a^2*b^2 + b^4) + 1/4*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))*(a^3 - a*b^2)/(a^4 + 2*a^2*b^2 + b^4) - 1/2
*(a^2*b*(e^(-x) - e^x)^2 + 2*a^3*(e^(-x) - e^x) + 2*a*b^2*(e^(-x) - e^x) + 8*a^2*b + 4*b^3)/((a^4 + 2*a^2*b^2
+ b^4)*((e^(-x) - e^x)^2 + 4))

________________________________________________________________________________________

maple [B]  time = 0.19, size = 275, normalized size = 2.89 \[ -\frac {a^{2} b \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 a \tanh \left (\frac {x}{2}\right )-b \right )}{\left (a^{2}+b^{2}\right )^{2}}-\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) a^{3}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) a \,b^{2}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) a^{2} b}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b^{3}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right ) a^{3}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tanh \left (\frac {x}{2}\right ) a \,b^{2}}{\left (a^{2}+b^{2}\right )^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\arctan \left (\tanh \left (\frac {x}{2}\right )\right ) a^{3}}{\left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tanh \left (\frac {x}{2}\right )\right ) a \,b^{2}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right ) a^{2} b}{\left (a^{2}+b^{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^3/(a+b*csch(x)),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.41, size = 161, normalized size = 1.69 \[ -\frac {a^{2} b \log \left (-2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} - a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a^{2} b \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a^{3} - a b^{2}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a e^{\left (-x\right )} - 2 \, b e^{\left (-2 \, x\right )} - a e^{\left (-3 \, x\right )}}{a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(a+b*csch(x)),x, algorithm="maxima")

[Out]

-a^2*b*log(-2*b*e^(-x) + a*e^(-2*x) - a)/(a^4 + 2*a^2*b^2 + b^4) + a^2*b*log(e^(-2*x) + 1)/(a^4 + 2*a^2*b^2 +
b^4) - (a^3 - a*b^2)*arctan(e^(-x))/(a^4 + 2*a^2*b^2 + b^4) + (a*e^(-x) - 2*b*e^(-2*x) - a*e^(-3*x))/(a^2 + b^
2 + 2*(a^2 + b^2)*e^(-2*x) + (a^2 + b^2)*e^(-4*x))

________________________________________________________________________________________

mupad [B]  time = 2.99, size = 256, normalized size = 2.69 \[ \frac {\frac {2\,b}{a^2+b^2}-\frac {2\,a\,{\mathrm {e}}^x}{a^2+b^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}-\frac {\frac {2\,\left (a^2\,b+b^3\right )}{{\left (a^2+b^2\right )}^2}-\frac {{\mathrm {e}}^x\,\left (a^3+a\,b^2\right )}{{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}+\frac {a\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}{2\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {a^2\,b\,\ln \left (a^6\,{\mathrm {e}}^{2\,x}-a^6-a^2\,b^4-14\,a^4\,b^2+a^2\,b^4\,{\mathrm {e}}^{2\,x}+14\,a^4\,b^2\,{\mathrm {e}}^{2\,x}+2\,a\,b^5\,{\mathrm {e}}^x+2\,a^5\,b\,{\mathrm {e}}^x+28\,a^3\,b^3\,{\mathrm {e}}^x\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {a\,\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^3*(a + b/sinh(x))),x)

[Out]

((2*b)/(a^2 + b^2) - (2*a*exp(x))/(a^2 + b^2))/(2*exp(2*x) + exp(4*x) + 1) - ((2*(a^2*b + b^3))/(a^2 + b^2)^2
- (exp(x)*(a*b^2 + a^3))/(a^2 + b^2)^2)/(exp(2*x) + 1) + (a*log(exp(x)*1i + 1)*1i)/(2*(a*b*2i - a^2 + b^2)) +
(a*log(exp(x) + 1i))/(2*(2*a*b - a^2*1i + b^2*1i)) - (a^2*b*log(a^6*exp(2*x) - a^6 - a^2*b^4 - 14*a^4*b^2 + a^
2*b^4*exp(2*x) + 14*a^4*b^2*exp(2*x) + 2*a*b^5*exp(x) + 2*a^5*b*exp(x) + 28*a^3*b^3*exp(x)))/(a^4 + b^4 + 2*a^
2*b^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{3}{\relax (x )}}{a + b \operatorname {csch}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**3/(a+b*csch(x)),x)

[Out]

Integral(sech(x)**3/(a + b*csch(x)), x)

________________________________________________________________________________________