3.103 \(\int \frac {\text {sech}^6(x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=83 \[ \frac {\left (a^2-b^2\right )^2 \log (a+b \tanh (x))}{b^5}-\frac {a \left (a^2-2 b^2\right ) \tanh (x)}{b^4}+\frac {\left (a^2-2 b^2\right ) \tanh ^2(x)}{2 b^3}-\frac {a \tanh ^3(x)}{3 b^2}+\frac {\tanh ^4(x)}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3506, 697} \[ \frac {\left (a^2-2 b^2\right ) \tanh ^2(x)}{2 b^3}-\frac {a \left (a^2-2 b^2\right ) \tanh (x)}{b^4}+\frac {\left (a^2-b^2\right )^2 \log (a+b \tanh (x))}{b^5}-\frac {a \tanh ^3(x)}{3 b^2}+\frac {\tanh ^4(x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^6/(a + b*Tanh[x]),x]

[Out]

((a^2 - b^2)^2*Log[a + b*Tanh[x]])/b^5 - (a*(a^2 - 2*b^2)*Tanh[x])/b^4 + ((a^2 - 2*b^2)*Tanh[x]^2)/(2*b^3) - (
a*Tanh[x]^3)/(3*b^2) + Tanh[x]^4/(4*b)

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 3506

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

Rubi steps

\begin {align*} \int \frac {\text {sech}^6(x)}{a+b \tanh (x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-\frac {x^2}{b^2}\right )^2}{a+x} \, dx,x,b \tanh (x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {-a^3+2 a b^2}{b^4}-\frac {\left (-a^2+2 b^2\right ) x}{b^4}-\frac {a x^2}{b^4}+\frac {x^3}{b^4}+\frac {\left (-a^2+b^2\right )^2}{b^4 (a+x)}\right ) \, dx,x,b \tanh (x)\right )}{b}\\ &=\frac {\left (a^2-b^2\right )^2 \log (a+b \tanh (x))}{b^5}-\frac {a \left (a^2-2 b^2\right ) \tanh (x)}{b^4}+\frac {\left (a^2-2 b^2\right ) \tanh ^2(x)}{2 b^3}-\frac {a \tanh ^3(x)}{3 b^2}+\frac {\tanh ^4(x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 92, normalized size = 1.11 \[ \frac {-4 \left (a b \left (3 a^2-5 b^2\right ) \tanh (x)+3 \left (a^2-b^2\right )^2 (\log (\cosh (x))-\log (a \cosh (x)+b \sinh (x)))\right )+\text {sech}^2(x) \left (-6 a^2 b^2+4 a b^3 \tanh (x)+6 b^4\right )+3 b^4 \text {sech}^4(x)}{12 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^6/(a + b*Tanh[x]),x]

[Out]

(3*b^4*Sech[x]^4 + Sech[x]^2*(-6*a^2*b^2 + 6*b^4 + 4*a*b^3*Tanh[x]) - 4*(3*(a^2 - b^2)^2*(Log[Cosh[x]] - Log[a
*Cosh[x] + b*Sinh[x]]) + a*b*(3*a^2 - 5*b^2)*Tanh[x]))/(12*b^5)

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fricas [B]  time = 0.59, size = 1827, normalized size = 22.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^6/(a+b*tanh(x)),x, algorithm="fricas")

[Out]

1/3*(6*(a^3*b - a^2*b^2 - a*b^3 + b^4)*cosh(x)^6 + 36*(a^3*b - a^2*b^2 - a*b^3 + b^4)*cosh(x)*sinh(x)^5 + 6*(a
^3*b - a^2*b^2 - a*b^3 + b^4)*sinh(x)^6 + 6*(3*a^3*b - 2*a^2*b^2 - 5*a*b^3 + 4*b^4)*cosh(x)^4 + 6*(3*a^3*b - 2
*a^2*b^2 - 5*a*b^3 + 4*b^4 + 15*(a^3*b - a^2*b^2 - a*b^3 + b^4)*cosh(x)^2)*sinh(x)^4 + 6*a^3*b - 10*a*b^3 + 24
*(5*(a^3*b - a^2*b^2 - a*b^3 + b^4)*cosh(x)^3 + (3*a^3*b - 2*a^2*b^2 - 5*a*b^3 + 4*b^4)*cosh(x))*sinh(x)^3 + 2
*(9*a^3*b - 3*a^2*b^2 - 17*a*b^3 + 3*b^4)*cosh(x)^2 + 2*(45*(a^3*b - a^2*b^2 - a*b^3 + b^4)*cosh(x)^4 + 9*a^3*
b - 3*a^2*b^2 - 17*a*b^3 + 3*b^4 + 18*(3*a^3*b - 2*a^2*b^2 - 5*a*b^3 + 4*b^4)*cosh(x)^2)*sinh(x)^2 + 3*((a^4 -
2*a^2*b^2 + b^4)*cosh(x)^8 + 8*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^7 + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^8
+ 4*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^6 + 4*(a^4 - 2*a^2*b^2 + b^4 + 7*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x
)^6 + 8*(7*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^3 + 3*(a^4 - 2*a^2*b^2 + b^4)*cosh(x))*sinh(x)^5 + 6*(a^4 - 2*a^2*b
^2 + b^4)*cosh(x)^4 + 2*(35*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^4 + 3*a^4 - 6*a^2*b^2 + 3*b^4 + 30*(a^4 - 2*a^2*b^
2 + b^4)*cosh(x)^2)*sinh(x)^4 + a^4 - 2*a^2*b^2 + b^4 + 8*(7*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^5 + 10*(a^4 - 2*a
^2*b^2 + b^4)*cosh(x)^3 + 3*(a^4 - 2*a^2*b^2 + b^4)*cosh(x))*sinh(x)^3 + 4*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 +
4*(7*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^6 + 15*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^4 + a^4 - 2*a^2*b^2 + b^4 + 9*(a^
4 - 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^2 + 8*((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^7 + 3*(a^4 - 2*a^2*b^2 + b^4)*c
osh(x)^5 + 3*(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))*log(2*(a*cosh(x) +
b*sinh(x))/(cosh(x) - sinh(x))) - 3*((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^8 + 8*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sin
h(x)^7 + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^8 + 4*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^6 + 4*(a^4 - 2*a^2*b^2 + b^4 +
7*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^6 + 8*(7*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^3 + 3*(a^4 - 2*a^2*b^2 +
b^4)*cosh(x))*sinh(x)^5 + 6*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^4 + 2*(35*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^4 + 3*a
^4 - 6*a^2*b^2 + 3*b^4 + 30*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^4 + a^4 - 2*a^2*b^2 + b^4 + 8*(7*(a^4 -
2*a^2*b^2 + b^4)*cosh(x)^5 + 10*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^3 + 3*(a^4 - 2*a^2*b^2 + b^4)*cosh(x))*sinh(x
)^3 + 4*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 + 4*(7*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^6 + 15*(a^4 - 2*a^2*b^2 + b^4
)*cosh(x)^4 + a^4 - 2*a^2*b^2 + b^4 + 9*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2)*sinh(x)^2 + 8*((a^4 - 2*a^2*b^2 + b
^4)*cosh(x)^7 + 3*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^5 + 3*(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))*log(2*cosh(x)/(cosh(x) - sinh(x))) + 4*(9*(a^3*b - a^2*b^2 - a*b^3 + b^4)*cosh(x)^5 +
6*(3*a^3*b - 2*a^2*b^2 - 5*a*b^3 + 4*b^4)*cosh(x)^3 + (9*a^3*b - 3*a^2*b^2 - 17*a*b^3 + 3*b^4)*cosh(x))*sinh(x
))/(b^5*cosh(x)^8 + 8*b^5*cosh(x)*sinh(x)^7 + b^5*sinh(x)^8 + 4*b^5*cosh(x)^6 + 6*b^5*cosh(x)^4 + 4*b^5*cosh(x
)^2 + 4*(7*b^5*cosh(x)^2 + b^5)*sinh(x)^6 + 8*(7*b^5*cosh(x)^3 + 3*b^5*cosh(x))*sinh(x)^5 + b^5 + 2*(35*b^5*co
sh(x)^4 + 30*b^5*cosh(x)^2 + 3*b^5)*sinh(x)^4 + 8*(7*b^5*cosh(x)^5 + 10*b^5*cosh(x)^3 + 3*b^5*cosh(x))*sinh(x)
^3 + 4*(7*b^5*cosh(x)^6 + 15*b^5*cosh(x)^4 + 9*b^5*cosh(x)^2 + b^5)*sinh(x)^2 + 8*(b^5*cosh(x)^7 + 3*b^5*cosh(
x)^5 + 3*b^5*cosh(x)^3 + b^5*cosh(x))*sinh(x))

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giac [B]  time = 0.13, size = 316, normalized size = 3.81 \[ \frac {{\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a b^{5} + b^{6}} - \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right )}{b^{5}} + \frac {25 \, a^{4} e^{\left (8 \, x\right )} - 50 \, a^{2} b^{2} e^{\left (8 \, x\right )} + 25 \, b^{4} e^{\left (8 \, x\right )} + 100 \, a^{4} e^{\left (6 \, x\right )} + 24 \, a^{3} b e^{\left (6 \, x\right )} - 224 \, a^{2} b^{2} e^{\left (6 \, x\right )} - 24 \, a b^{3} e^{\left (6 \, x\right )} + 124 \, b^{4} e^{\left (6 \, x\right )} + 150 \, a^{4} e^{\left (4 \, x\right )} + 72 \, a^{3} b e^{\left (4 \, x\right )} - 348 \, a^{2} b^{2} e^{\left (4 \, x\right )} - 120 \, a b^{3} e^{\left (4 \, x\right )} + 246 \, b^{4} e^{\left (4 \, x\right )} + 100 \, a^{4} e^{\left (2 \, x\right )} + 72 \, a^{3} b e^{\left (2 \, x\right )} - 224 \, a^{2} b^{2} e^{\left (2 \, x\right )} - 136 \, a b^{3} e^{\left (2 \, x\right )} + 124 \, b^{4} e^{\left (2 \, x\right )} + 25 \, a^{4} + 24 \, a^{3} b - 50 \, a^{2} b^{2} - 40 \, a b^{3} + 25 \, b^{4}}{12 \, b^{5} {\left (e^{\left (2 \, x\right )} + 1\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^6/(a+b*tanh(x)),x, algorithm="giac")

[Out]

(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a*b^5 + b^6) - (a
^4 - 2*a^2*b^2 + b^4)*log(e^(2*x) + 1)/b^5 + 1/12*(25*a^4*e^(8*x) - 50*a^2*b^2*e^(8*x) + 25*b^4*e^(8*x) + 100*
a^4*e^(6*x) + 24*a^3*b*e^(6*x) - 224*a^2*b^2*e^(6*x) - 24*a*b^3*e^(6*x) + 124*b^4*e^(6*x) + 150*a^4*e^(4*x) +
72*a^3*b*e^(4*x) - 348*a^2*b^2*e^(4*x) - 120*a*b^3*e^(4*x) + 246*b^4*e^(4*x) + 100*a^4*e^(2*x) + 72*a^3*b*e^(2
*x) - 224*a^2*b^2*e^(2*x) - 136*a*b^3*e^(2*x) + 124*b^4*e^(2*x) + 25*a^4 + 24*a^3*b - 50*a^2*b^2 - 40*a*b^3 +
25*b^4)/(b^5*(e^(2*x) + 1)^4)

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maple [B]  time = 0.15, size = 438, normalized size = 5.28 \[ \frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \tanh \left (\frac {x}{2}\right ) b +a \right ) a^{4}}{b^{5}}-\frac {2 \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \tanh \left (\frac {x}{2}\right ) b +a \right ) a^{2}}{b^{3}}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \tanh \left (\frac {x}{2}\right ) b +a \right )}{b}-\frac {2 \left (\tanh ^{7}\left (\frac {x}{2}\right )\right ) a^{3}}{b^{4} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {4 \left (\tanh ^{7}\left (\frac {x}{2}\right )\right ) a}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {2 \left (\tanh ^{6}\left (\frac {x}{2}\right )\right ) a^{2}}{b^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {4 \left (\tanh ^{6}\left (\frac {x}{2}\right )\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {6 \left (\tanh ^{5}\left (\frac {x}{2}\right )\right ) a^{3}}{b^{4} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {28 \left (\tanh ^{5}\left (\frac {x}{2}\right )\right ) a}{3 b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {4 \left (\tanh ^{4}\left (\frac {x}{2}\right )\right ) a^{2}}{b^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {4 \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {6 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) a^{3}}{b^{4} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {28 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) a}{3 b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) a^{2}}{b^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {4 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {2 \tanh \left (\frac {x}{2}\right ) a^{3}}{b^{4} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}+\frac {4 \tanh \left (\frac {x}{2}\right ) a}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{4}}-\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right ) a^{4}}{b^{5}}+\frac {2 \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{3}}-\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^6/(a+b*tanh(x)),x)

[Out]

1/b^5*ln(a*tanh(1/2*x)^2+2*tanh(1/2*x)*b+a)*a^4-2/b^3*ln(a*tanh(1/2*x)^2+2*tanh(1/2*x)*b+a)*a^2+1/b*ln(a*tanh(
1/2*x)^2+2*tanh(1/2*x)*b+a)-2/b^4/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^7*a^3+4/b^2/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^
7*a+2/b^3/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^6*a^2-4/b/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^6-6/b^4/(tanh(1/2*x)^2+1)^
4*tanh(1/2*x)^5*a^3+28/3/b^2/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^5*a+4/b^3/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^4*a^2-4
/b/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^4-6/b^4/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^3*a^3+28/3/b^2/(tanh(1/2*x)^2+1)^4*
tanh(1/2*x)^3*a+2/b^3/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^2*a^2-4/b/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^2-2/b^4/(tanh(
1/2*x)^2+1)^4*tanh(1/2*x)*a^3+4/b^2/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)*a-1/b^5*ln(tanh(1/2*x)^2+1)*a^4+2/b^3*ln(t
anh(1/2*x)^2+1)*a^2-1/b*ln(tanh(1/2*x)^2+1)

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maxima [B]  time = 0.42, size = 204, normalized size = 2.46 \[ -\frac {2 \, {\left (3 \, a^{3} - 5 \, a b^{2} + {\left (9 \, a^{3} + 3 \, a^{2} b - 17 \, a b^{2} - 3 \, b^{3}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (3 \, a^{3} + 2 \, a^{2} b - 5 \, a b^{2} - 4 \, b^{3}\right )} e^{\left (-4 \, x\right )} + 3 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} e^{\left (-6 \, x\right )}\right )}}{3 \, {\left (4 \, b^{4} e^{\left (-2 \, x\right )} + 6 \, b^{4} e^{\left (-4 \, x\right )} + 4 \, b^{4} e^{\left (-6 \, x\right )} + b^{4} e^{\left (-8 \, x\right )} + b^{4}\right )}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{b^{5}} - \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^6/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

-2/3*(3*a^3 - 5*a*b^2 + (9*a^3 + 3*a^2*b - 17*a*b^2 - 3*b^3)*e^(-2*x) + 3*(3*a^3 + 2*a^2*b - 5*a*b^2 - 4*b^3)*
e^(-4*x) + 3*(a^3 + a^2*b - a*b^2 - b^3)*e^(-6*x))/(4*b^4*e^(-2*x) + 6*b^4*e^(-4*x) + 4*b^4*e^(-6*x) + b^4*e^(
-8*x) + b^4) + (a^4 - 2*a^2*b^2 + b^4)*log(-(a - b)*e^(-2*x) - a - b)/b^5 - (a^4 - 2*a^2*b^2 + b^4)*log(e^(-2*
x) + 1)/b^5

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mupad [B]  time = 1.29, size = 169, normalized size = 2.04 \[ \frac {4}{b\,\left (4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1\right )}+\frac {2\,{\left (a-b\right )}^2}{b^3\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )}+\frac {8\,\left (a-3\,b\right )}{3\,b^2\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}+\frac {2\,\left (a+b\right )\,{\left (a-b\right )}^2}{b^4\,\left ({\mathrm {e}}^{2\,x}+1\right )}+\frac {\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )\,{\left (a+b\right )}^2\,{\left (a-b\right )}^2}{b^5}-\frac {\ln \left ({\mathrm {e}}^{2\,x}+1\right )\,{\left (a+b\right )}^2\,{\left (a-b\right )}^2}{b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^6*(a + b*tanh(x))),x)

[Out]

4/(b*(4*exp(2*x) + 6*exp(4*x) + 4*exp(6*x) + exp(8*x) + 1)) + (2*(a - b)^2)/(b^3*(2*exp(2*x) + exp(4*x) + 1))
+ (8*(a - 3*b))/(3*b^2*(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1)) + (2*(a + b)*(a - b)^2)/(b^4*(exp(2*x) + 1))
+ (log(a - b + a*exp(2*x) + b*exp(2*x))*(a + b)^2*(a - b)^2)/b^5 - (log(exp(2*x) + 1)*(a + b)^2*(a - b)^2)/b^5

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{6}{\relax (x )}}{a + b \tanh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**6/(a+b*tanh(x)),x)

[Out]

Integral(sech(x)**6/(a + b*tanh(x)), x)

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