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

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Rubi [A]  time = 0.06617, antiderivative size = 40, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.157174, 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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Maple [B]  time = 0.039, size = 143, normalized size = 3.6 \begin{align*} -{\frac{{a}^{2}}{{b}^{3}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) b+a \right ) }+{\frac{1}{b}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) b+a \right ) }+2\,{\frac{a \left ( \tanh \left ( x/2 \right ) \right ) ^{3}}{{b}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}}{b \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{a\tanh \left ( x/2 \right ) }{{b}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{1}{b}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) } \end{align*}

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 = 1.6785, size = 120, normalized size = 3. \begin{align*} \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}} \end{align*}

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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Fricas [B]  time = 2.43367, size = 1088, normalized size = 27.2 \begin{align*} -\frac{2 \,{\left (a b - b^{2}\right )} \cosh \left (x\right )^{2} + 4 \,{\left (a b - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + 2 \,{\left (a b - b^{2}\right )} \sinh \left (x\right )^{2} + 2 \, a b +{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{4} + 4 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} +{\left (a^{2} - b^{2}\right )} \sinh \left (x\right )^{4} + 2 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (3 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + a^{2} - b^{2}\right )} \sinh \left (x\right )^{2} + a^{2} - b^{2} + 4 \,{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{3} +{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \,{\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) -{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{4} + 4 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} +{\left (a^{2} - b^{2}\right )} \sinh \left (x\right )^{4} + 2 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (3 \,{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + a^{2} - b^{2}\right )} \sinh \left (x\right )^{2} + a^{2} - b^{2} + 4 \,{\left ({\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{3} +{\left (a^{2} - b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{b^{3} \cosh \left (x\right )^{4} + 4 \, b^{3} \cosh \left (x\right ) \sinh \left (x\right )^{3} + b^{3} \sinh \left (x\right )^{4} + 2 \, b^{3} \cosh \left (x\right )^{2} + b^{3} + 2 \,{\left (3 \, b^{3} \cosh \left (x\right )^{2} + b^{3}\right )} \sinh \left (x\right )^{2} + 4 \,{\left (b^{3} \cosh \left (x\right )^{3} + b^{3} \cosh \left (x\right )\right )} \sinh \left (x\right )} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{4}{\left (x \right )}}{a + b \tanh{\left (x \right )}}\, dx \end{align*}

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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Giac [B]  time = 1.20692, size = 140, normalized size = 3.5 \begin{align*} -\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}} \end{align*}

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)