3.107 \(\int \frac{\cosh ^2(x)}{a+b \tanh (x)} \, dx\)

Optimal. Leaf size=91 \[ \frac{b^3 \log (a+b \tanh (x))}{\left (a^2-b^2\right )^2}-\frac{\cosh ^2(x) (b-a \tanh (x))}{2 \left (a^2-b^2\right )}-\frac{(a+2 b) \log (1-\tanh (x))}{4 (a+b)^2}+\frac{(a-2 b) \log (\tanh (x)+1)}{4 (a-b)^2} \]

[Out]

-((a + 2*b)*Log[1 - Tanh[x]])/(4*(a + b)^2) + ((a - 2*b)*Log[1 + Tanh[x]])/(4*(a - b)^2) + (b^3*Log[a + b*Tanh
[x]])/(a^2 - b^2)^2 - (Cosh[x]^2*(b - a*Tanh[x]))/(2*(a^2 - b^2))

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Rubi [A]  time = 0.1455, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3506, 741, 801} \[ \frac{b^3 \log (a+b \tanh (x))}{\left (a^2-b^2\right )^2}-\frac{\cosh ^2(x) (b-a \tanh (x))}{2 \left (a^2-b^2\right )}-\frac{(a+2 b) \log (1-\tanh (x))}{4 (a+b)^2}+\frac{(a-2 b) \log (\tanh (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(a + b*Tanh[x]),x]

[Out]

-((a + 2*b)*Log[1 - Tanh[x]])/(4*(a + b)^2) + ((a - 2*b)*Log[1 + Tanh[x]])/(4*(a - b)^2) + (b^3*Log[a + b*Tanh
[x]])/(a^2 - b^2)^2 - (Cosh[x]^2*(b - a*Tanh[x]))/(2*(a^2 - b^2))

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 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, 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]

Rubi steps

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

Mathematica [A]  time = 0.112653, size = 75, normalized size = 0.82 \[ \frac{a \left (a^2-b^2\right ) \sinh (2 x)+\left (b^3-a^2 b\right ) \cosh (2 x)+2 a^3 x-6 a b^2 x+4 b^3 \log (a \cosh (x)+b \sinh (x))}{4 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^2/(a + b*Tanh[x]),x]

[Out]

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

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Maple [B]  time = 0.046, size = 175, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,a-2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+2\,{\frac{1}{ \left ( 4\,a-4\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) }}+{\frac{a}{2\, \left ( a-b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{b}{ \left ( a-b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{b}^{3}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) b+a \right ) }+{\frac{1}{2\,b+2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+2\,{\frac{1}{ \left ( 4\,a+4\,b \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) }}-{\frac{a}{2\, \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{b}{ \left ( a+b \right ) ^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(a+b*tanh(x)),x)

[Out]

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

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Maxima [A]  time = 1.22641, size = 116, normalized size = 1.27 \begin{align*} \frac{b^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a + 2 \, b\right )} x}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{e^{\left (2 \, x\right )}}{8 \,{\left (a + b\right )}} - \frac{e^{\left (-2 \, x\right )}}{8 \,{\left (a - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.30382, size = 818, normalized size = 8.99 \begin{align*} \frac{{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{4} + 4 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} +{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \left (x\right )^{4} + 4 \,{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x \cosh \left (x\right )^{2} - a^{3} - a^{2} b + a b^{2} + b^{3} + 2 \,{\left (3 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x\right )} \sinh \left (x\right )^{2} + 8 \,{\left (b^{3} \cosh \left (x\right )^{2} + 2 \, b^{3} \cosh \left (x\right ) \sinh \left (x\right ) + b^{3} \sinh \left (x\right )^{2}\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 ) + 4 \,{\left ({\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{3} + 2 \,{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right )}{8 \,{\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2/(a+b*tanh(x)),x, algorithm="fricas")

[Out]

1/8*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 - a^2*b - a*b^2 + b^3)*cosh(x)*sinh(x)^3 + (a^3 - a^2*b -
a*b^2 + b^3)*sinh(x)^4 + 4*(a^3 - 3*a*b^2 - 2*b^3)*x*cosh(x)^2 - a^3 - a^2*b + a*b^2 + b^3 + 2*(3*(a^3 - a^2*b
- a*b^2 + b^3)*cosh(x)^2 + 2*(a^3 - 3*a*b^2 - 2*b^3)*x)*sinh(x)^2 + 8*(b^3*cosh(x)^2 + 2*b^3*cosh(x)*sinh(x)
+ b^3*sinh(x)^2)*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x))) + 4*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^3
+ 2*(a^3 - 3*a*b^2 - 2*b^3)*x*cosh(x))*sinh(x))/((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4 - 2*a^2*b^2 + b^4
)*cosh(x)*sinh(x) + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh ^{2}{\left (x \right )}}{a + b \tanh{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**2/(a+b*tanh(x)),x)

[Out]

Integral(cosh(x)**2/(a + b*tanh(x)), x)

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Giac [A]  time = 1.21046, size = 150, normalized size = 1.65 \begin{align*} \frac{b^{3} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a - 2 \, b\right )} x}{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac{{\left (2 \, a e^{\left (2 \, x\right )} - 4 \, b e^{\left (2 \, x\right )} + a - b\right )} e^{\left (-2 \, x\right )}}{8 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{e^{\left (2 \, x\right )}}{8 \,{\left (a + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2/(a+b*tanh(x)),x, algorithm="giac")

[Out]

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