3.110 \(\int \frac {\text {sech}^5(x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=102 \[ -\frac {a \left (2 a^2-3 b^2\right ) \tan ^{-1}(\sinh (x))}{2 b^4}+\frac {\left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {\cosh (x) (a \tanh (x)+b)}{\sqrt {a^2-b^2}}\right )}{b^4}-\frac {\left (a^2-b^2\right ) \text {sech}(x)}{b^3}+\frac {a \tanh (x) \text {sech}(x)}{2 b^2}+\frac {\text {sech}^3(x)}{3 b} \]

[Out]

-1/2*a*(2*a^2-3*b^2)*arctan(sinh(x))/b^4+(a^2-b^2)^(3/2)*arctan(cosh(x)*(b+a*tanh(x))/(a^2-b^2)^(1/2))/b^4-(a^
2-b^2)*sech(x)/b^3+1/3*sech(x)^3/b+1/2*a*sech(x)*tanh(x)/b^2

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Rubi [A]  time = 0.17, antiderivative size = 109, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3510, 3486, 3768, 3770, 3509, 206} \[ -\frac {\left (a^2-b^2\right ) \text {sech}(x)}{b^3}-\frac {a \left (a^2-b^2\right ) \tan ^{-1}(\sinh (x))}{b^4}+\frac {\left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {\cosh (x) (a \tanh (x)+b)}{\sqrt {a^2-b^2}}\right )}{b^4}+\frac {a \tan ^{-1}(\sinh (x))}{2 b^2}+\frac {a \tanh (x) \text {sech}(x)}{2 b^2}+\frac {\text {sech}^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^5/(a + b*Tanh[x]),x]

[Out]

(a*ArcTan[Sinh[x]])/(2*b^2) - (a*(a^2 - b^2)*ArcTan[Sinh[x]])/b^4 + ((a^2 - b^2)^(3/2)*ArcTan[(Cosh[x]*(b + a*
Tanh[x]))/Sqrt[a^2 - b^2]])/b^4 - ((a^2 - b^2)*Sech[x])/b^3 + Sech[x]^3/(3*b) + (a*Sech[x]*Tanh[x])/(2*b^2)

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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3509

Int[sec[(e_.) + (f_.)*(x_)]/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[f^(-1), Subst[Int[1/(a^
2 + b^2 - x^2), x], x, (b - a*Tan[e + f*x])/Sec[e + f*x]], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0]

Rule 3510

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[d^2/b^2, I
nt[(d*Sec[e + f*x])^(m - 2)*(a - b*Tan[e + f*x]), x], x] + Dist[(d^2*(a^2 + b^2))/b^2, Int[(d*Sec[e + f*x])^(m
- 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 116, normalized size = 1.14 \[ \frac {-6 \left (a \left (2 a^2-3 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+2 \sqrt {a-b} \sqrt {a+b} \left (b^2-a^2\right ) \tan ^{-1}\left (\frac {a \tanh \left (\frac {x}{2}\right )+b}{\sqrt {a-b} \sqrt {a+b}}\right )\right )+3 b \text {sech}(x) \left (-2 a^2+a b \tanh (x)+2 b^2\right )+2 b^3 \text {sech}^3(x)}{6 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^5/(a + b*Tanh[x]),x]

[Out]

(-6*(a*(2*a^2 - 3*b^2)*ArcTan[Tanh[x/2]] + 2*Sqrt[a - b]*Sqrt[a + b]*(-a^2 + b^2)*ArcTan[(b + a*Tanh[x/2])/(Sq
rt[a - b]*Sqrt[a + b])]) + 2*b^3*Sech[x]^3 + 3*b*Sech[x]*(-2*a^2 + 2*b^2 + a*b*Tanh[x]))/(6*b^4)

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fricas [B]  time = 0.76, size = 2043, normalized size = 20.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^5/(a+b*tanh(x)),x, algorithm="fricas")

[Out]

[-1/3*(3*(2*a^2*b - a*b^2 - 2*b^3)*cosh(x)^5 + 15*(2*a^2*b - a*b^2 - 2*b^3)*cosh(x)*sinh(x)^4 + 3*(2*a^2*b - a
*b^2 - 2*b^3)*sinh(x)^5 + 4*(3*a^2*b - 5*b^3)*cosh(x)^3 + 2*(6*a^2*b - 10*b^3 + 15*(2*a^2*b - a*b^2 - 2*b^3)*c
osh(x)^2)*sinh(x)^3 + 6*(5*(2*a^2*b - a*b^2 - 2*b^3)*cosh(x)^3 + 2*(3*a^2*b - 5*b^3)*cosh(x))*sinh(x)^2 + 3*((
a^2 - b^2)*cosh(x)^6 + 6*(a^2 - b^2)*cosh(x)*sinh(x)^5 + (a^2 - b^2)*sinh(x)^6 + 3*(a^2 - b^2)*cosh(x)^4 + 3*(
5*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^4 + 4*(5*(a^2 - b^2)*cosh(x)^3 + 3*(a^2 - b^2)*cosh(x))*sinh(x)^3
+ 3*(a^2 - b^2)*cosh(x)^2 + 3*(5*(a^2 - b^2)*cosh(x)^4 + 6*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2
- b^2 + 6*((a^2 - b^2)*cosh(x)^5 + 2*(a^2 - b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*sinh(x))*sqrt(-a^2 + b^2)*l
og(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - 2*sqrt(-a^2 + b^2)*(cosh(x) + sinh(x))
- a + b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a - b)) + 3*((2*a^3 - 3*a*b^2)*
cosh(x)^6 + 6*(2*a^3 - 3*a*b^2)*cosh(x)*sinh(x)^5 + (2*a^3 - 3*a*b^2)*sinh(x)^6 + 3*(2*a^3 - 3*a*b^2)*cosh(x)^
4 + 3*(2*a^3 - 3*a*b^2 + 5*(2*a^3 - 3*a*b^2)*cosh(x)^2)*sinh(x)^4 + 4*(5*(2*a^3 - 3*a*b^2)*cosh(x)^3 + 3*(2*a^
3 - 3*a*b^2)*cosh(x))*sinh(x)^3 + 2*a^3 - 3*a*b^2 + 3*(2*a^3 - 3*a*b^2)*cosh(x)^2 + 3*(5*(2*a^3 - 3*a*b^2)*cos
h(x)^4 + 2*a^3 - 3*a*b^2 + 6*(2*a^3 - 3*a*b^2)*cosh(x)^2)*sinh(x)^2 + 6*((2*a^3 - 3*a*b^2)*cosh(x)^5 + 2*(2*a^
3 - 3*a*b^2)*cosh(x)^3 + (2*a^3 - 3*a*b^2)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) + 3*(2*a^2*b + a*b^2 -
2*b^3)*cosh(x) + 3*(5*(2*a^2*b - a*b^2 - 2*b^3)*cosh(x)^4 + 2*a^2*b + a*b^2 - 2*b^3 + 4*(3*a^2*b - 5*b^3)*cosh
(x)^2)*sinh(x))/(b^4*cosh(x)^6 + 6*b^4*cosh(x)*sinh(x)^5 + b^4*sinh(x)^6 + 3*b^4*cosh(x)^4 + 3*b^4*cosh(x)^2 +
3*(5*b^4*cosh(x)^2 + b^4)*sinh(x)^4 + b^4 + 4*(5*b^4*cosh(x)^3 + 3*b^4*cosh(x))*sinh(x)^3 + 3*(5*b^4*cosh(x)^
4 + 6*b^4*cosh(x)^2 + b^4)*sinh(x)^2 + 6*(b^4*cosh(x)^5 + 2*b^4*cosh(x)^3 + b^4*cosh(x))*sinh(x)), -1/3*(3*(2*
a^2*b - a*b^2 - 2*b^3)*cosh(x)^5 + 15*(2*a^2*b - a*b^2 - 2*b^3)*cosh(x)*sinh(x)^4 + 3*(2*a^2*b - a*b^2 - 2*b^3
)*sinh(x)^5 + 4*(3*a^2*b - 5*b^3)*cosh(x)^3 + 2*(6*a^2*b - 10*b^3 + 15*(2*a^2*b - a*b^2 - 2*b^3)*cosh(x)^2)*si
nh(x)^3 + 6*(5*(2*a^2*b - a*b^2 - 2*b^3)*cosh(x)^3 + 2*(3*a^2*b - 5*b^3)*cosh(x))*sinh(x)^2 + 6*((a^2 - b^2)*c
osh(x)^6 + 6*(a^2 - b^2)*cosh(x)*sinh(x)^5 + (a^2 - b^2)*sinh(x)^6 + 3*(a^2 - b^2)*cosh(x)^4 + 3*(5*(a^2 - b^2
)*cosh(x)^2 + a^2 - b^2)*sinh(x)^4 + 4*(5*(a^2 - b^2)*cosh(x)^3 + 3*(a^2 - b^2)*cosh(x))*sinh(x)^3 + 3*(a^2 -
b^2)*cosh(x)^2 + 3*(5*(a^2 - b^2)*cosh(x)^4 + 6*(a^2 - b^2)*cosh(x)^2 + a^2 - b^2)*sinh(x)^2 + a^2 - b^2 + 6*(
(a^2 - b^2)*cosh(x)^5 + 2*(a^2 - b^2)*cosh(x)^3 + (a^2 - b^2)*cosh(x))*sinh(x))*sqrt(a^2 - b^2)*arctan(sqrt(a^
2 - b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) + 3*((2*a^3 - 3*a*b^2)*cosh(x)^6 + 6*(2*a^3 - 3*a*b^2)*cosh(x)*s
inh(x)^5 + (2*a^3 - 3*a*b^2)*sinh(x)^6 + 3*(2*a^3 - 3*a*b^2)*cosh(x)^4 + 3*(2*a^3 - 3*a*b^2 + 5*(2*a^3 - 3*a*b
^2)*cosh(x)^2)*sinh(x)^4 + 4*(5*(2*a^3 - 3*a*b^2)*cosh(x)^3 + 3*(2*a^3 - 3*a*b^2)*cosh(x))*sinh(x)^3 + 2*a^3 -
3*a*b^2 + 3*(2*a^3 - 3*a*b^2)*cosh(x)^2 + 3*(5*(2*a^3 - 3*a*b^2)*cosh(x)^4 + 2*a^3 - 3*a*b^2 + 6*(2*a^3 - 3*a
*b^2)*cosh(x)^2)*sinh(x)^2 + 6*((2*a^3 - 3*a*b^2)*cosh(x)^5 + 2*(2*a^3 - 3*a*b^2)*cosh(x)^3 + (2*a^3 - 3*a*b^2
)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) + 3*(2*a^2*b + a*b^2 - 2*b^3)*cosh(x) + 3*(5*(2*a^2*b - a*b^2 -
2*b^3)*cosh(x)^4 + 2*a^2*b + a*b^2 - 2*b^3 + 4*(3*a^2*b - 5*b^3)*cosh(x)^2)*sinh(x))/(b^4*cosh(x)^6 + 6*b^4*co
sh(x)*sinh(x)^5 + b^4*sinh(x)^6 + 3*b^4*cosh(x)^4 + 3*b^4*cosh(x)^2 + 3*(5*b^4*cosh(x)^2 + b^4)*sinh(x)^4 + b^
4 + 4*(5*b^4*cosh(x)^3 + 3*b^4*cosh(x))*sinh(x)^3 + 3*(5*b^4*cosh(x)^4 + 6*b^4*cosh(x)^2 + b^4)*sinh(x)^2 + 6*
(b^4*cosh(x)^5 + 2*b^4*cosh(x)^3 + b^4*cosh(x))*sinh(x))]

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giac [A]  time = 0.15, size = 152, normalized size = 1.49 \[ -\frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \arctan \left (e^{x}\right )}{b^{4}} + \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \arctan \left (\frac {a e^{x} + b e^{x}}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {6 \, a^{2} e^{\left (5 \, x\right )} - 3 \, a b e^{\left (5 \, x\right )} - 6 \, b^{2} e^{\left (5 \, x\right )} + 12 \, a^{2} e^{\left (3 \, x\right )} - 20 \, b^{2} e^{\left (3 \, x\right )} + 6 \, a^{2} e^{x} + 3 \, a b e^{x} - 6 \, b^{2} e^{x}}{3 \, b^{3} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^5/(a+b*tanh(x)),x, algorithm="giac")

[Out]

-(2*a^3 - 3*a*b^2)*arctan(e^x)/b^4 + 2*(a^4 - 2*a^2*b^2 + b^4)*arctan((a*e^x + b*e^x)/sqrt(a^2 - b^2))/(sqrt(a
^2 - b^2)*b^4) - 1/3*(6*a^2*e^(5*x) - 3*a*b*e^(5*x) - 6*b^2*e^(5*x) + 12*a^2*e^(3*x) - 20*b^2*e^(3*x) + 6*a^2*
e^x + 3*a*b*e^x - 6*b^2*e^x)/(b^3*(e^(2*x) + 1)^3)

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maple [B]  time = 0.16, size = 316, normalized size = 3.10 \[ \frac {2 \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{4}}{b^{4} \sqrt {a^{2}-b^{2}}}-\frac {4 \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{b^{2} \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\frac {2 a \tanh \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}-\frac {a \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 \left (\tanh ^{4}\left (\frac {x}{2}\right )\right ) a^{2}}{b^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {4 \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {4 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) a^{2}}{b^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {4 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {\tanh \left (\frac {x}{2}\right ) a}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 a^{2}}{b^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {8}{3 b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) a^{3}}{b^{4}}+\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) a}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^5/(a+b*tanh(x)),x)

[Out]

2/b^4/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tanh(1/2*x)+2*b)/(a^2-b^2)^(1/2))*a^4-4/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(
2*a*tanh(1/2*x)+2*b)/(a^2-b^2)^(1/2))*a^2+2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tanh(1/2*x)+2*b)/(a^2-b^2)^(1/2))-
1/b^2/(tanh(1/2*x)^2+1)^3*a*tanh(1/2*x)^5-2/b^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^4*a^2+4/b/(tanh(1/2*x)^2+1)^3*
tanh(1/2*x)^4-4/b^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^2*a^2+4/b/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^2+1/b^2/(tanh(1/
2*x)^2+1)^3*tanh(1/2*x)*a-2/b^3/(tanh(1/2*x)^2+1)^3*a^2+8/3/b/(tanh(1/2*x)^2+1)^3-2/b^4*arctan(tanh(1/2*x))*a^
3+3/b^2*arctan(tanh(1/2*x))*a

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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(sech(x)^5/(a+b*tanh(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?

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mupad [B]  time = 5.22, size = 265, normalized size = 2.60 \[ \frac {\ln \left (\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}+a^3\,{\mathrm {e}}^x-b^3\,{\mathrm {e}}^x-a\,b^2\,{\mathrm {e}}^x+a^2\,b\,{\mathrm {e}}^x\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{b^4}-\frac {8\,{\mathrm {e}}^x}{3\,b\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}-\frac {\ln \left (\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-a^3\,{\mathrm {e}}^x+b^3\,{\mathrm {e}}^x+a\,b^2\,{\mathrm {e}}^x-a^2\,b\,{\mathrm {e}}^x\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{b^4}-\frac {2\,{\mathrm {e}}^x\,\left (3\,a-4\,b\right )}{3\,b^2\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )}+\frac {{\mathrm {e}}^x\,\left (-2\,a^2+a\,b+2\,b^2\right )}{b^3\,\left ({\mathrm {e}}^{2\,x}+1\right )}+\frac {a\,\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,\left (2\,a^2-3\,b^2\right )\,1{}\mathrm {i}}{2\,b^4}-\frac {a\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,\left (2\,a^2-3\,b^2\right )\,1{}\mathrm {i}}{2\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^5*(a + b*tanh(x))),x)

[Out]

(log((-(a + b)^3*(a - b)^3)^(1/2) + a^3*exp(x) - b^3*exp(x) - a*b^2*exp(x) + a^2*b*exp(x))*(-(a + b)^3*(a - b)
^3)^(1/2))/b^4 - (8*exp(x))/(3*b*(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1)) - (log((-(a + b)^3*(a - b)^3)^(1/2)
- a^3*exp(x) + b^3*exp(x) + a*b^2*exp(x) - a^2*b*exp(x))*(-(a + b)^3*(a - b)^3)^(1/2))/b^4 - (2*exp(x)*(3*a -
4*b))/(3*b^2*(2*exp(2*x) + exp(4*x) + 1)) + (exp(x)*(a*b - 2*a^2 + 2*b^2))/(b^3*(exp(2*x) + 1)) + (a*log(exp(
x) - 1i)*(2*a^2 - 3*b^2)*1i)/(2*b^4) - (a*log(exp(x) + 1i)*(2*a^2 - 3*b^2)*1i)/(2*b^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{5}{\relax (x )}}{a + b \tanh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**5/(a+b*tanh(x)),x)

[Out]

Integral(sech(x)**5/(a + b*tanh(x)), x)

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