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

Optimal. Leaf size=84 \[ \frac{a^2 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^2}-\frac{\cosh ^2(x) (b-a \tanh (x))}{2 \left (a^2-b^2\right )}+\frac{a \log (1-\tanh (x))}{4 (a+b)^2}-\frac{a \log (\tanh (x)+1)}{4 (a-b)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.161791, antiderivative size = 84, 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 = {3516, 1647, 801} \[ \frac{a^2 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^2}-\frac{\cosh ^2(x) (b-a \tanh (x))}{2 \left (a^2-b^2\right )}+\frac{a \log (1-\tanh (x))}{4 (a+b)^2}-\frac{a \log (\tanh (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Tanh[x]),x]

[Out]

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

Rule 3516

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

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

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{\sinh ^2(x)}{a+b \tanh (x)} \, dx &=b \operatorname{Subst}\left (\int \frac{x^2}{(a+x) \left (-b^2+x^2\right )^2} \, dx,x,b \tanh (x)\right )\\ &=-\frac{\cosh ^2(x) (b-a \tanh (x))}{2 \left (a^2-b^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{\frac{a^2 b^2}{a^2-b^2}-\frac{a b^2 x}{a^2-b^2}}{(a+x) \left (-b^2+x^2\right )} \, dx,x,b \tanh (x)\right )}{2 b}\\ &=-\frac{\cosh ^2(x) (b-a \tanh (x))}{2 \left (a^2-b^2\right )}+\frac{\operatorname{Subst}\left (\int \left (-\frac{a b}{2 (a+b)^2 (b-x)}+\frac{2 a^2 b^2}{(a-b)^2 (a+b)^2 (a+x)}-\frac{a b}{2 (a-b)^2 (b+x)}\right ) \, dx,x,b \tanh (x)\right )}{2 b}\\ &=\frac{a \log (1-\tanh (x))}{4 (a+b)^2}-\frac{a \log (1+\tanh (x))}{4 (a-b)^2}+\frac{a^2 b \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 = 0.221769, size = 73, normalized size = 0.87 \[ \frac{\left (b^3-a^2 b\right ) \cosh (2 x)+a \left (-2 x \left (a^2+b^2\right )+\left (a^2-b^2\right ) \sinh (2 x)+4 a b \log (a \cosh (x)+b \sinh (x))\right )}{4 (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 145, normalized size = 1.7 \begin{align*} -4\,{\frac{1}{ \left ( 8\,a-8\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{2}}}+8\,{\frac{1}{ \left ( 16\,a-16\,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{{a}^{2}b}{ \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 ) }+4\,{\frac{1}{ \left ( 8\,a+8\,b \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{2}}}+8\,{\frac{1}{ \left ( 16\,a+16\,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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/(8*a-8*b)/(tanh(1/2*x)+1)^2+8/(16*a-16*b)/(tanh(1/2*x)+1)-1/2*a/(a-b)^2*ln(tanh(1/2*x)+1)+a^2*b/(a-b)^2/(a+
b)^2*ln(a*tanh(1/2*x)^2+2*tanh(1/2*x)*b+a)+4/(8*a+8*b)/(tanh(1/2*x)-1)^2+8/(16*a+16*b)/(tanh(1/2*x)-1)+1/2*a/(
a+b)^2*ln(tanh(1/2*x)-1)

________________________________________________________________________________________

Maxima [A]  time = 1.16775, size = 112, normalized size = 1.33 \begin{align*} \frac{a^{2} b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{a 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(sinh(x)^2/(a+b*tanh(x)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.63341, size = 826, normalized size = 9.83 \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} + 2 \, a^{2} b + a b^{2}\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} + 2 \, a^{2} b + a b^{2}\right )} x\right )} \sinh \left (x\right )^{2} + 8 \,{\left (a^{2} b \cosh \left (x\right )^{2} + 2 \, a^{2} b \cosh \left (x\right ) \sinh \left (x\right ) + a^{2} b \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} + 2 \, a^{2} b + a b^{2}\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(sinh(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 + 2*a^2*b + a*b^2)*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 + 2*a^2*b + a*b^2)*x)*sinh(x)^2 + 8*(a^2*b*cosh(x)^2 + 2*a^2*b*cosh(x)*sinh
(x) + a^2*b*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)*cos
h(x)^3 - 2*(a^3 + 2*a^2*b + a*b^2)*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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.21264, size = 136, normalized size = 1.62 \begin{align*} \frac{a^{2} b \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{a x}{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (2 \, a 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(sinh(x)^2/(a+b*tanh(x)),x, algorithm="giac")

[Out]

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