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

Optimal. Leaf size=40 \[ -\frac {\left (a^2-b^2\right ) \log (a+b \tanh (x))}{b^3}+\frac {a \tanh (x)}{b^2}-\frac {\tanh ^2(x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 40, 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} \[ -\frac {\left (a^2-b^2\right ) \log (a+b \tanh (x))}{b^3}+\frac {a \tanh (x)}{b^2}-\frac {\tanh ^2(x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(a + b*Tanh[x]),x]

[Out]

-(((a^2 - b^2)*Log[a + b*Tanh[x]])/b^3) + (a*Tanh[x])/b^2 - Tanh[x]^2/(2*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}^4(x)}{a+b \tanh (x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1-\frac {x^2}{b^2}}{a+x} \, dx,x,b \tanh (x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a}{b^2}-\frac {x}{b^2}+\frac {-a^2+b^2}{b^2 (a+x)}\right ) \, dx,x,b \tanh (x)\right )}{b}\\ &=-\frac {\left (a^2-b^2\right ) \log (a+b \tanh (x))}{b^3}+\frac {a \tanh (x)}{b^2}-\frac {\tanh ^2(x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 49, normalized size = 1.22 \[ \frac {2 \left (a^2-b^2\right ) (\log (\cosh (x))-\log (a \cosh (x)+b \sinh (x)))+2 a b \tanh (x)+b^2 \text {sech}^2(x)}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.61, size = 430, normalized size = 10.75 \[ -\frac {2 \, {\left (a b - b^{2}\right )} \cosh \relax (x)^{2} + 4 \, {\left (a b - b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + 2 \, {\left (a b - b^{2}\right )} \sinh \relax (x)^{2} + 2 \, a b + {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{2} - b^{2}\right )} \sinh \relax (x)^{4} + 2 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} + a^{2} - b^{2}\right )} \sinh \relax (x)^{2} + a^{2} - b^{2} + 4 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{3} + {\left (a^{2} - b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{4} + 4 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x) \sinh \relax (x)^{3} + {\left (a^{2} - b^{2}\right )} \sinh \relax (x)^{4} + 2 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (3 \, {\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{2} + a^{2} - b^{2}\right )} \sinh \relax (x)^{2} + a^{2} - b^{2} + 4 \, {\left ({\left (a^{2} - b^{2}\right )} \cosh \relax (x)^{3} + {\left (a^{2} - b^{2}\right )} \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{b^{3} \cosh \relax (x)^{4} + 4 \, b^{3} \cosh \relax (x) \sinh \relax (x)^{3} + b^{3} \sinh \relax (x)^{4} + 2 \, b^{3} \cosh \relax (x)^{2} + b^{3} + 2 \, {\left (3 \, b^{3} \cosh \relax (x)^{2} + b^{3}\right )} \sinh \relax (x)^{2} + 4 \, {\left (b^{3} \cosh \relax (x)^{3} + b^{3} \cosh \relax (x)\right )} \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.14, size = 104, normalized size = 2.60 \[ -\frac {{\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a b^{3} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right )}{b^{3}} - \frac {2 \, {\left (a b + {\left (a b - b^{2}\right )} e^{\left (2 \, x\right )}\right )}}{b^{3} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 143, normalized size = 3.58 \[ -\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \tanh \left (\frac {x}{2}\right ) b +a \right ) a^{2}}{b^{3}}+\frac {\ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2 \tanh \left (\frac {x}{2}\right ) b +a \right )}{b}+\frac {2 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) a}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{b \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 a \tanh \left (\frac {x}{2}\right )}{b^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{3}}-\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 89, normalized size = 2.22 \[ \frac {2 \, {\left ({\left (a + b\right )} e^{\left (-2 \, x\right )} + a\right )}}{2 \, b^{2} e^{\left (-2 \, x\right )} + b^{2} e^{\left (-4 \, x\right )} + b^{2}} - \frac {{\left (a^{2} - b^{2}\right )} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{b^{3}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.27, size = 88, normalized size = 2.20 \[ \frac {\ln \left ({\mathrm {e}}^{2\,x}+1\right )\,\left (a+b\right )\,\left (a-b\right )}{b^3}-\frac {2\,\left (a-b\right )}{b^2\,\left ({\mathrm {e}}^{2\,x}+1\right )}-\frac {\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a+b\right )\,\left (a-b\right )}{b^3}-\frac {2}{b\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^4*(a + b*tanh(x))),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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